Induction Heating
Induction Heating
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by the/an author.
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             c   Ali Kamil Makki Al-Shaikhli
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                            LOUGHBOROUGH
                       UNIVERSITY OF TECHNOLOGY
                                LIBRARY
         AUTHOR/FILING TITLE                                                 1
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               This book was bound by
               Badminton Press
               18 Half Croft, Syston, Leicester, LE? 8LD
               Telephone: Leicester 10533) 602918.
          A NOVEL METHOD FOR THE DESIGN OF
by
A DOCTORAL THESIS
Supervisor:    Dr L Hobson
               Department of Electronic and Electrical Engineering
Synopsis
means that it can be used on desk top computers and even to some
lamination packs and the current distribution within the work coil
conductor itself.
                             iii
ACKNOWLEDGEMENTS
I wish to thank the Head and Staff of the Electronic and Electrical
the French and German references and Mrs Ann Hammond for the typing
of this thesis.
CONTENTS
                                                                   page no.
   Title Page
                                                                      i
   Synopsis
                                                                      ii
   Acknowledgements
                                                                      iii
   Contents
                                                                     iv
   List of Princip~l Symbols                                         VJ
   1.    INTRODUCTION
                                                                         1
References 238
R                         Reluctance (A/Wb)
     m
N                         Nwnber of turns
I Current (A)
V Voltage (V)
f Frequency (Hz)
11       ~
             llollr       Absolute permeability (H/m)
11                        Relative permeability
     r
z                         Impedance ((l)
R Resistance (n)
X Reactance W>
r Radius (m)
9. J,ength (m)
0        ~~
          or
                          Penetration depth (m)
p Resistivity ((lm)
VA Volt-ampere (VA)
p. F.                     Power factor
                             vii
7 Efficiency (%)
T Temperature (OC)
Suffix
r Return
c Coil
t Total
w Workpiece
g Air gap
CHAPTER 1
INTRODUCTION
Induction heating was first used in industry to nelt metals but now
with the billet and induces eddy currents within it.   The eddy currents
                                                           2
in turn produce heat as a result of the Joule heating or I R losses.
current.
and the ease of control and the ability to repeat the working cycle
make induction heating ideal for a fully automated process. When hot
metal is not required the heater can be switched off thus eliminating
                                              PYROMETER
                                                                CTION
                                                              COIL.
                                                                        RETRACTABLE
                                                                           STOP.
                                                                          REJECTS CHUTE
        TABLE
UNSCRAMBLER.
                                                                                               -.........
                                                                                          HOT PIECES
                                                                                          10 FORMI~
                     BASE UNIT/                                                           PROCESS.
                    !MAY HOUSE TRANSFORMERS
                     AND CAPACITORS.!
           I
           I          '
               ''
                                                          A.C. Supply
within certain parts of the metal forming industry [1.1]. Other areas
[1.4] and vaughan and Williamson [1.5, 1.6). The method assumes a
uniform magnetic field strength along the length of the workpiece and
billet has been carried out. The method is very simple and can
CHAPTER 2
WORK COILS
particular problems.
coil design.
large air gap between the coil and the workpiece is required so
radiated heat losses from the billet to the coil can be reduced.
The magnetic flux produced by the work coil, ~t' has alternative parallel
WORKPIECE
AIR GAP
DD DD DD COIL
magnetic reluctances R
                       mw' Rme and Rmg   The return flux <fJ has a reluctance
                                                            r
of R
    mr
      .   The effect of <fJ on the coil exterior is usually
                           c
                .·.
ignored   [2.l,2.2] because it is numerically far less than <fJ • The
                                                                r
effect of <fJ on the coil interior is represented by R        The
             c                                        me
assembly total reluctance equals R   in series with the parallel
          +                       mr
connection of R , R    and R      The position of R   relative to R
               mw   mg      me                          me            mr
does not .alter the performance of the magnetic circuit seriously
because, the coil reluctance is generally far higher than the external
law:
               N I =             ~t Rmt
                c c
                                                           R   R
                                              R   (R        mg mw
                                                        +
                                               me    mr   R   + I{
                                                           mg     mw
                           =     ~t   [               R    R           J   (2 .l)
                                          R   + R   +  m9: mw
                                           me    mr   R    + R
                                                       mg     mw
law:
               V       = jwNc~t                                            (2.2)
                   c
from
                       V
                           c
               z       I                                                   (2.3)
                           c
       X                   N2             w
                   =                                                       (2. 4)
           r                c         R
                                       mr
                           N2          w
       X           =                                                       (2. 5)
        g                   c         R
                                       mg
                           'N2          w
       zw          =       J c        R                                    ( 2. 6)
                                       mw
                                       w
       zc          =       'N2
                           J c        --
                                      R                                    ( 2. 7)
                                      me
                                                               8
                                                           R
                                                               mr
                                                                    _j
                                                                                                                     ..
                                              <j>c                                   <j>g                 <j>w
                    mmf             R                                        R                             R
                                         me                                      mg                            mw
"
                                     z
                                         c
                                     I
                     I
            I
                c
                           R
                               c
                                               X
                                                   c
                                                       '                                                                         I
r---' T T T
                                                                                                1     R
                                                                                                          w
                                                                                                ..,
    V
                                                           • X
        c                                                  . r                                                       f-Z w
                                                                                                 .
                                                                                                 •    X
                                                                                                          w
                                                                                                 •
                                                                                                                 I
•
                                                                             X
                                                                                 g
                                                                         T           T
                                                                                                                                 I
                         Fig. 2.3             The equivalent electrical circuit             '                                    I
                                                                       ')
                                             t
                                c
         R
          mg
                    =         2 2                                                    (2. 9}
                        11 7r(r.-r}
                          0   c w
                                 t
                                         c
        R
         mw
                    =                                                                (2.10}
                                 t
        R
                                     c
            me                                                                       (2.11}
                        11 rrk
                         o
                                     or
                                 r c c
                                                 (1-j}
to give:
                                     r
        X
                                         c
            r       ~                                                               (2.12}
                        2 (0. 45+k} ' '
                         (r2 - r 2 }
        X       =
                           c     w
         g          ~            t
                                         c
                                                                                    (2.13}
                                                        2
                                                   11 r
        zw      = R + ·x = KR                        r w
                   w  J w                                        (Q   + jP}         (2.14}
                                                    i
                                                        w
                                                         ..•
                                                   k or
        zc      = R   + jX = KR                     r c c
                    c     c                                           (1 + j}       (2.15}
                                                        1-
                                                             c
where                                    2
             KR =       WIJ 1TN
                             0           c                                          (2.16}
                                                                                --·----.
                                                       10
and the efficiency and power factor. As the coil number of turns
                                       ·x [·x
                                       J rs J gs
                                                   +Z]
                                                     ws
     zs       =       z           +-=C-_,2.::...__  _;.:.,=--                                 (2.17)
                          CS
     z       is of a complex value;
         s
     Z       = R      + jX                                                                    (2.18)
         s        s                s
                                            2
    iz s I    =                   +    l(
                                            s                                                 (2.19)
         1
                          WS
              =                                                                               (2.20)
                      R
                          s
                          R
                              §
     P.F. =                                                                                   (2. 21)
                      TZ'.T
                        s
The volt-ampere, VA, of the coil can be calculated from the power input
to the coil P :
             c
                                               ll
    VA = - -                                                                         (2.22)
                      P.F.
    V= Z I                                                                           (2.23)
    c   c c
                                                           2
    multiply by the voltage and divide by N ,
                                           c
                                   zc
                      V I                                                            (2,24)
                          c c      7
                                   . c
                                             take the square root
    v             =           r'{vA) z                                               (2.25)
         CS                              S
Since the number of turns is now derived, the impedance of the coil is;
            2
     z = Z N                                                                         (2. 27)
          s c
               V
     I        =-
                c                                                                    (2.28)
         c            z
                                        0.92
    f.J   ~ --=-   k           =                                                                                      (2.29)
                       r
                                        lkc
                                          N    t
                                    =
                                              c c
where                      k                                                                                          (2.30)
                               c              t
                                                  c
are shown in Figs. 2.4 and 2.5 and the difference between them is
                                                      2
                                    W1J11r N
                                      • c c
                   X =                                                                                                (2. 31)
                   r                 0.9 k
                                            e
in Fig. 2.5. The methods of Baker and Vaughan and Williamson were
reference [2.5] .
                                   l3
                         zc
                          I
        I
            c   /R
                     c             \
                              ~T
                                                 1      R
                                                             I
V
    c                                            l• X
                                                                 r zw
                                                  ) w
                                                         I
                                 X
                                   g
                                   ••
                FA
                     + FC                                              (2.32)
              FB + H
field strength.
where [z] is the square matrix comprising all of the coil and billet
self and mutual impedances, [r] is the column matrix of unknown coil
and billet segment currents and [v] is the column matrix of driving
300 unknowns for the IBM 3033 computer [2.9) which has a memory
when dealing with curved surfaces and when the penetration depth is
the skin depth 6[2.9) from the surface and using larger subdivisions
Kolbe and Reiss [2.6] employed the mutually coupled circuit method
in order to determine:
this has been determined one can then calculate the temperature
distribution at the time t = 6t, and with it the new material's
the time t = 26t and the corresponding materials values are then
computer, even though the thermal conductivity of the material was ignored.
Dudley and Burke [2.7] used the mutually coupled circuit method to
see Appendix 3. The second case is that of a sheet coil and billet;
required the single turn and the billet to be divided into a number
This method has been used in the solution of other problems and
nodal parameters.
techniques, both, are not easily applied by any one other than
must (2.9]
the coil
previously.
Nemoto and Tabuchi [2.15] used the finite element method to analyse
other hand, the accuracy derived from these equations will be minimal.
this change partially into account. Moreover, this method can not
the program.
the flux and the power in an induction heating system during the heating
                                      4
By omitting terms from (6xl               onwards the simple expression for
d2$
        is obtained by adding equations (2.33a and b)
   2
dx
              1
                        [$(x + Ax} - 2$(x) + $(x- Axl]                             (2.34)
                   2
            (Ax)
(2.35a}
                                                                                   (2. 35b}
Equations (2.35a and b) are known respectively as the forward
i.e.
       dx - _!_
       d<P- 2llx            [<P(x + llx) - <P(x- llxl]                                         (2.35c)
the x-y plane into sets of equal rectangles of sides llx and lly,
                              2
                    =       (u>
                              2
                p            ax       i    .
                                          ,J
(2.36)
                                  1
                        =                      [ <Pi+l ,]. - 2 <P·l.,J. + <P._l  ·]            (2.37)
                                                                           l.   ,]
                                               24
i,i+l
                                                      P(iAX jl\y)
                                      i-l,j                   i+i ,j
i,j-1
                                      1                                                 (2.38)
Similarly,                        =
                                      (t.y)2
                            i,j
aay4>   at p
               •
               1s
= (2. 39)
                     ~i
Reichert           (2.19] developed a finite difference computer program to
determine the field and the temperature distribution in an inductively
for the inexperienced user. Also, the computer used was not
Finally, the program's running time was too long. This might not
work load.
l. Over the heating cycle time span, the change in the physical
facility is essential.
                   "
3.     The problem of non-uniform power density induced along
2. 3 Conclusion
For many years, very simple models, based on the electrical equivalent
Almost all of the existing work coil design methods are based on
density along the load. These methods, too, either assume a uniform
CHAP'l'ER 3
INITIAL INVESTrGATlJ:ONS
The aim is to gain experience in the use of work coil design methods,
effects, the reactance of the external flux path and the nonuniformity
of the workpiece.
these methods.
workpiece.
PRIME 400 system and employed to assess the effect of the external
were obtained from reference [3.2]. This file was inserted into
and the expressions for the components of the circuit are those
Fig. 2. 5.
( START )
            in terms of K/N 2
                            c
                ,....,. . . . . I
I
        Write
( STOP
The three computer programs were employed to assess the effects of:
Coil resistivity                                         -8
                                          = 2   X    10       rlm
The values used for each of the following factors are as follows unless
The magnetic load was assumed to be EN3 mild steel with a surface
                           0                                           0
 temperature of 600 C and a centre temperature of 20 C.                          This corresponds
The results obtained from the three computer programs are shown
in Tables (3.1-3.7).
Table 3.1 shows that when dealing with a nonmagnetic load the three
circuits give values which are all within 10% of each other for
factor is more than 15%. The long coil, Fig. 2.5, tends to produce
better results than the short coil, Fig. 2.4, when compared with
        EQUIV       SHORT         LONG       MAX %     EQUIV          SHORT       LONG       MAX%        EQUIV        SHORT    LONG       MAX %
                                             DIFF                                            DIFF                                         DIFF
N         151             156      145         7.6         10           11            10       10             93       65            62     49
 c
I        1709         1636        1678         4.5     8401           8189        8280        2.6            674      1589       674       136
    c
p. F. .266 .28 .27 5.3 .054 .056 .055 3.7 .675 .286 .674 136
P.F.
    7           38.3
                o. 26                    0.31
                                              29            32.1
                                                            19.2
                                                                                      43.8
                                                                                      .256
                                                                                                             34. 2
                                                                                                         o. 302
                                                                                                                              28.1
18
N                    10                       12                 20                     10                     11               11
 c
                                                                                                II
I               8832                     7014               25.9                      8876                   7171             23.8
    c
7 44.8 35 28 49 39 25.6
                            k
                                r       = 1.0        k
                                                         r    =   1.5                 Max %               k
                                                                                                              r    = 1.0 I           k
                                                                                                                                         r    = 1.5              Max %
                                                                                      Diff            '                      I                                   Diff
                                                                                                      i
N                                        64                    66                      3.13           i             62                         62                 0
 c
le                                  1607                     1549                      3.75                        674                        674                 0
        k
            e
                =·   0.'8   k
                                    e
                                        = 1.2      %Diff              k    = 0. 8
                                                                          e.            k
                                                                                            e
                                                                                                =   1.2       %Diff          k. =
                                                                                                                               e.:
                                                                                                                                       o. 8 ke = 1.2.                 %Diff
    P.F.
        l      0.274    0.2S4            7.9        .OS6   .OSl          9.8    .67S
                                                                                        96.9
                                                                                        .675
                                                                                                  0
                                                                                                  0
I
                             Table 3. 7        The effect of k
                                                                    c
The magnetic load results obtained from the three circuits have
the snort and long coil circuits are equal, but a different number
cases·, but different in the latter case of the short coil circuit.
                 '
            The effect of the empirical factor k , which allows for the spacing
                                                           r
and negligible when adopting the long coil circuit. However, the
3G
to 20%, while in the current can reach 26% and the efficiency 32%.
estimating the current for magnetic materials when they will vary
The effect of the factor k, which is for the return flux, on the
circuit of Fig. 2.3 "EQUIV" is shown on Table 3.6. When this constant
was varied between the maximum and minimum values (0.1 and -0.22),
the results did not fluctuate by more than 10%, apart from when
case of the nonmagnetic load was within 10% on the number of turns
The above work showed that the three circuits may give different answers.
by these circuits, and here again, this influence depends on the case
                                   37
circuit and the right values of the empirical factors suitable for
a given configuration.
of its imperfections.
                                   r
                                        c
        X       K                                                                                (3.1)
            r   R
                      2 (0. 45 + k)
                               2            r2)
                          (r            -
                               c             w
        X     = KR                                                                               ( 3. 2)
            g                      R.
                                        c
                      0.92
                                              oc        r
                                                            c
                                                                (1   +   j)                      (3. 3)
                          ,.'k                     R.
                                                        c
                             c
                               2
where               W)l TTN                                                                      (3.4)
                      0        c
                                              39
where E(r) and H*(r)are the r.m.s. values of the electric field
       E(r)      =-    p
                           r
                               J{r)                                  ( 3. 6)
                                                                ·lO
                                 R
                                 ··m,.
      t                r
$c $g $n $2 $1
mmf           R
               me
                                             R
                                              mg                      ~-mn                  Rm2   ~ ml
zc
                                                                \
                                                   X
                   I                                   c                       I
                       c                                                           w
                                                                       X
          V                                                                r
          c
                                                                                       X
                                                                                       g
sr ( 3. 8)
           I=NI=HR.                                                                     (3. 9)
            w c c 0 w
                                        2
                          H H*R-                                                        (3.10)
                            0       0   w
(3.12)
(3.13)
(3.14)
(3.15)
solution is essential.
(3.16)
(3.17)
                                                                                       (3.18)
                                              43
(3.19)
where 61, 62 and 63 are the skin depths in the 1st, 2nd and 3rd
( 3. 20)
          J (r)
           n
                   =? [c 2un      n- 2
                                         .r'(/2j; ) + c
                                          o        un        2n- 1
                                                                     .K'(fi); ) ]
                                                                      o      un
                                                                                      (3. 21)
the junctions between the surface of the billet and the air gap.
It is known that the magnetic and electric field strengths are both
(C /H) to (C /H), and therefore to solve for H(r)/H and J(r)/H too.
  1 O·      n o                                    o          o
By substitution into equation (:LU), the impedances can be found.                         The
          r = r                                                                       (3.22)
                  w
and       H (r)       H                                                               ( 3. 23)
                          0
                                    r                    r
i .. e.   H = c2.Io(ffj
           0
                                    6;)   + c3.Ko (/2]   6;)                          (3.24)
                                          44
(3.25)
and (3.26)
(3.27)
i.e. ( 3. 28)
and
        p       12                    r
            1
                      c r· <mJlil
                     "l"o      ....!:.,
            01
(3.29)
Defining,
                            r.
        X ..
            1,J        12jrl                                       (3.30)
and (3.31)
then
(3.32)
_c_2 = A.Io(x2,l).I~(xl,l)
                                                                   (3.33)
H                                11
    0
                                45
(3.34}
where
                                                                     (3.35}
At this point all the electric circuit components were defined.
The computer program "EQUIV" was improved to cater for this derivation.
circuit was that shown in Fig. 3.3 and the calculation was as stated
of these resistivities.
(3.36)
        u       =    ___AF~--+ CF                                             (3.37)
            X        BF + H
                           X
be provided.
                        START     )
/              Read the Input Data
                                                        /
    Divide the billet into number of cylinders
                   Set TTE=O
                         I
                         ?              NO
                       TTE= 1                 I   TTE   1
YES
( STOP
    ~
                                                              %   Diff between
           f
0                               1          2      5      8        1 and 8
                    inders
p
      1P.F.
                                40       51.7   36.5
38
~"''"""
                                                              %   Diff between
                                1          2      5      8        1 and 8
           N                    93         84     84     67          39
               c
           I                   674        663    663    833          19
               c
The above tables showed the differences between a uniform load and,
67% and the power factor difference is up to 38%· Hence the correct
previous chapter, as this would require much time and effort. The
as both the uniformity of the material and the simple geometry of the
was written for any user, as one simply needs to run the program and answer
The program was based, mainly, on the computer program "BIEEDY"   [3. 4   1·
This piece of research has been selected because it is well documented
and explained.
equations:
                                  as
             ~E.dl\ =
             w
                           s
                             J ( -at) ds                                     (3.38)
( 3. 40)
becomes:
             VxH = J                                                         (3.41)
                       c
As:          E   pJ                                                          (3.42)
and          B = IJH                                                         (3.43)
the resultant equation together with equation (3.43) into equation (3.38)
so as to yield:
            H           ~    H       ~   0                                                               (3.45)
                y                X
and         H       ~   H                                                                                (3.46)
                             z
                     oH      aH         3H     oH       oH    3H
                    (- -z- - - Jy)_. + ( :x
      VxH                               - - - - -z)] +
                                                     . (__x - ---2!.) k                                  (3. 47)
                            ay           az                 az        ax         ax    ay
As the change of the magnetic field occurs only within the depth of
(3. 48)
                               oH .
            VxH         ~    - a;{J                                                                      (3.49)
            CJlp aH d£. ~
                 ax
                                             s
                                                 f   ;t   (JlH)ds                                        (3.50)
            J       ~-
                        aH
                                                                                                         (3.52)
                        ax
                                                                           3
The average volume power density (W/m) at any point                                         [3.3) when
sinusoidal magnetic field is in operation is:
            p                        pJ J*                                                               (3.53)
                V
                                   53
A program "slab" has been written to predict the magnetic field and
changes rapidly the mesh nodes should be close together, i.e. near the
more gradually the nodes need to be spaced further apart, i.e. towards
the centre of the load.
The surface and line integrals of equation (3.51) are taken over the
of uniform material. Following Gibson [3.4] who considered the more gen-
eral case of non-uniform load, the finite difference equation for a uniform
load will be
                                                           54
2:.2 SLX
i= 1 2 3 4 5 6 7 8 MX
                                                       .            ~·       -p           r,               -
                                    2         3                4                                               ~
j=l           1
      2
              10                 11        12                  13   fl4       fl5         fL6         7            8
      3
          "19                       20     21               22      03        04          b5         6             7
                                               .
                                                                                                                       .!_ SLy
                                                                                                                       2
      1
              28                    29     30                  31   b2        ~3          ~4         5             6
          I
      5
          p7                        38     39                  40   ~1        ~2          ~3                       5
                                                                                                    1'14
      7
              55                    56        57               58   159       ~0      ~1            r>2            3
      8
              64                    65     66                  67   tsB      169      no            r,l            2
      9
          173                       74        75               76       77       78   179           ~0             1
 MY 182                             83     84                  85   ~6       187      ~8            ~9             0
                           I                                   I
                           I                i ,j+l             I
                       i                                       I
 i                         I
                           I
                                                               I
                                                               I
- Y2                                                       I
i-
                           I
             I ----                                    __ TI _____
        ------
                           I
                                            f.--                           -- -
                .          I                                   I
- Y2                       I                                   I
                           I                                   I
        i-l,j              I                i,j                I           ~+l,j
                           I
                                                               I
flYl                       I                                   I
-2-                        I
                                                               I
 t-
flY 1
-2-
                           I
        -----,.----- ----- -+- ----
                           I
                           I
                                                               I
                                                               I
                                                                            ---
                           I                                   I
 ~                                                             I
                           I                i,j-1
                           I                                   I
        -    fl;l
                           I   t. xl                 tJ.'x2    I    flx2
                                                                    -2-
                           I
                                 2                     2
                                                              ·~·
                           I                                   I
                           I                                   I
                                            H      -   H
                                       + [< i,j-1      i,j) (llxl +
                                                llyl              2
(3.54)
                             aH
       as y =       0       -=    0
                             ay
                    1
       as   X   = - SLX      H= H
                    2             0
       as y = !_SLY         H = H
                    2             0
Equation (3.54) was applied to each node of the grid and the set of
this the magnetic field strength at each node of the quarter section
density within each mesh, the power generated at each node and the
total power induced in the load which is four times the total power
        Jx =
                                                                             (3.55)
 similarly,     JY =
                                                                             (3.57)
 and      Jy +
                                                                            (3. 58)
                           Jx + [ Jx +      ~     dy   J
Average Jx    = AvJx =                                                      ( 3. 59)
                                   2
                           H2 - Hl + Hll - HlO
                       =
                                      2dx                                  (3. 60)
                           Jy +[Jy + ~ dx)
Average J·y = AvJy
                                  2                                        (3. 61)
                           HlO - Hl + Hll - H2
                       =
                                  2dy                                      (3.62)
                               3
The volume power density p (W/m ) was calculated by the expression (3.63).
                                               58
Jx
                                                                 Jy +   ~d
                                                                        8X X
                 dy           Jy
                         19                        Jx
                         I
                                             Jx + - - dy
                                                    ay      ~1
                         I                                 I
                                                X
      1
 =-
  8         p 4dxdxdydy (Av Jy Av Jy* + Av Jx Av Jx* )                         (3.64)
      1
 =z-         p(Av Jy Av Jy* + Av Jx Av Jx*]                                    ( 3. 65)
             1                           3
      p =   2 p J J*               W/m                                         (3.66)
                                  5~
subroutine SETUP.
the power generated at each node and the total power induced
The user has the choice of printing only the values of the outputs
( START
                                ,,
                                                                    <
l'       1-   - - - - -      - --- -         -   -   -   -   - -
I_             Calculate and print the total power in
               the slab, power.at each nOde, power
               density within each mesh and the magnetic
               field strength at each node
( STOP
                       Subroutine GAUSS
                     solves the matrix by
!._.__
                   the Gaussian elimination                    '-
method
are required.
                                      62
'~'
0·8
0·6
0·4
POWER (W)
140
120
100
80
60
40
20
       OL---~1----~2----3~~~4~~;---~6----~7--L-~8------+
                                                                 y ( X10- mJ
      Fig. 3.8    Power distribution down the centre line of the cross-sectional face
                                63
has not achieved the market share that perhaps one might expect
work coil design are required which can be adapted to solve problems
turns. The distribution of the power density induced on the load can
CHAPTER 4
an equation has been derived for the surface magnetic field distribution
ensuing chapter.
from the surface of a semi-infinite metal slab. Given that the slab
and carrying current -I. Both the conductor and the image will produce
           7 7 7 /stab 7 7 ~I 7 7 7 7 7 ~7 7
                                                 I                                    0rp
                                                 I
                                         Qymage
(a l
                                    X....
                                             '   ......                   H      Hr
                                                          ''
                                                             ' ......
                                                            2Hz --...:
                                                                      _...,...-; p
                                                                  /
                                                              /
                                                          /
                                             /
                                                     /
                                                                  H           'Hr
                                    ......                                                  •
                          (bl
( 4. l)
HE = 2H z (4.2)
but H = H cos a                                                            ( 4. 3)
     z
                         I                   h
         = 2
     ~                2               >1,2             i                   (4.4)
                211;{        + 'z'i              + z
          I          h
     Hp = 1T-                                                              ( 4. 5)
                 h2 + z 2
(4.6)
( 4. 7)
the current penetration depth into the slab and the conductor is small
compared with the height h and the thickness of the slab, the case
                                              67
near a metal plate [4.2]. The solution of this problem (4.3, 4.41
                                                                                (4.8)
                     2         2         2
                 h       - a       + z
                                                                                (4.9)
                               2
                         + z
                         2
     h           - a                                                            (4.10)
         e
from a flat plane and the above derivation may be applied to a single
turn coil surrounding a cylinder, provided that the distance between the
in the cylinder.
current I amperes.
                            Gfl
           1          2            ~               N
                                     ,...
   I
   I
          $I
                     <P           <P .... 0        <P
   I                               I
   .
   .... z1~
   I
   I
          Ze.
                     •'
                                   I
                                   I
                                   I
                                   I
   I
   1-           Za                .,
                                   I
                                   I
                          ZN
   I
   I
   I
sum of the current per unit length induced by each individual conductor.
(4.11)
Where APl' AP 2 , Ap)' .•• , ~N are the current per unit length induced
then
                                                                 (4.12)
                                                   69
zl = lzll
        ' = lzl + dl
        z2
                                                                       (4.13)
z3 = lzl + 2dl
        z        =   lz         + (N-l)dl
            cl            1
and ( 4. 14)
                          I h        X=N-1
                                e                       1
                           1T          I                               (4.15)
                                      X=O
        JP =          /2      HP    exp J. ( wt + 4
                                                  ")                   (4.16)
                      0
                     fi                       1T                 1T
i.e. JP=~ HP {cos(wt +                        4>   + jsin(wt +
                                                                 4)}   (4.17)
( 4.18)
                                                                       (4.19)
                                 70
           0
          = -J                                                        (4.20)
            ffp
at that point.
and Maslen [4.5] for the case' of large cylinder. The agreement was very
good and the difference did not exceed 2%. Callaghan and Maslen .•
for a large cylindrical load and high frequency nearly equals that
culate and plot the current per unit length and the surface power
                   I
                             Read: a, h, t, f,
                                     p'   u
                                               r'
                                                     I
                                                         c'
                                                              p
                                                                  w   J
                                     Set N = 2
                                     z - 0.0
                                     p = 0.0
                             p       = !J.z* (PD)
                                 z                        z
                             p = p + p
                                                    z
                                           ?
                   YES
                                     z+ IJ.z<t
                                               NO
                                         ?
                        NO
L----..j   N N+l                      p ;: p
YES
/ Write N _/
STOP
(PD) p (4.22)
coil. When the billet is inside the coil, it·s starting edge is at
The temperatures of the rest of the load ranges between these two
assumed by dividing the load into (X) equal divisions and taking
                      pf .-p.~
        ~p                                                                                ( 4. 23)
                       X - l
(4. 25)
Since the current per unit length induced at any point in the load
below
        (PD)       =                                                      (4.26)
               n
the resistivity and the induced power density, the latter being
will gradually increase until they reach a maximum. This case was
is shown below.
                        (          STAR'!'
                                                                I
                 Read LM, a., h ,                        R..,
      I          Fi'   f\i'
                           1   1
                                    p2~' llri'Ici'Ni'
                                   i=l, LM
                                                         1
z- 0
                                        I
                 Incremented the index by
                                        l
                 Calculate A    (PD)
                            z '      z
                                        ?
           YES
                               z + R,i          ~   o.
                                            NO
                                        ?
           NO
                         ,..        i   ~   LM
YES
                                        ?
            NO
                                    i   ~   LM
YES
( STOP
Figs. 4.3 and 4.4 show that increasing the number of turns will increase
the current per unit length induced in the load and increases the
of 15 turns. The above result agrees with the fact that the current
which will result in a very uniform power density induced in the load.
the air gap. Decreasing the air gap increases the current per unit
decreases the uniformity as can be seen from the upper curves on these
graphs.
Although the distributions of the current per unit length and the power
the current per unit length and the power density induced in the load;
this can be read off Figs. 4.7 and 4.8.   The curves represent 3 very
                             78
similar coils differing only in the size of the conductor. The air gap
the induced power, the thinner the conductor the more the non-
uniformity of the induced power, and the less the efficiency. This
Figs. 4.10 and 4.11, while the current per unit length induced from
Fig. 4.11 shows the axial change of resistivity. The power density
45
'10
35
  a
';g     30
..;
.c'
<-1
        25
 ""<:
 Q)
.-<
                                                                              N=45
<-1
·.-!
 <:
"
'"'
 Q)
        20
0.
<-1                                                                            30
<:
Q)
'"'     15
'"'
"
u
'0
Q)
u
"
'0
<:                                                                            15
H
        10
         Fig. 4.3   Current per unit length along the load due to lOOOA
                    flowing through coils of different number of turns, tl,
                    at h = 40mm and the conductor radius is a = lomm.
                                                      80
4.5
4.0
3.5
             3.0
N
      8
    ......
    :3:
    ..\<:
             2.5
     0
    "''
     ;>,
    ....+J
     Ul
     .:Q)    2.0
    'Cl
     ":.
     Q)
     0
                                                                                        N=45
     0.
    'Cl
             1.5
     Q)
     ".:"
    'Cl
    H
             1.0
                                 '
                                                                                        . 30
              5
                                                                                         15
40_
35
             30
 ~
 ,;
 8
  ,;   -
 .r:
  +J
       -     25
                                                                                           h=35
   0>
   <:
 ....'"
                                                                                           55
 ....+J
  <:                                                                                       75
  ::1
             20
  ...
  '0."
 +J
  <:
  ......'"
  ::1        15
.,u
.,u'<:"
  ::1
H
             10
             0
                  0      .1      .2     .3      :'1    .5       .6       .7   .8    .9   1.0
                                        Distance along the load, z,     (m)
                  Fig. 4.5    current per unit length along the load due to lOOOA
                              flowing through 40 turns coils at different heights, h,
                              and the conductor radius is a   = lOmm.
                                                  82
3.5
             2.5
     ~
N
      E
    '.>:
     ;3:
     0'      2.0
    "'>.'
    ....+J
     00
     <:
    .gj
    ...Ql    1.5
     ~
    0
    a.                                                                                  h=35
"'u Ql
"'"<:
H
             1.0
                                                                                        55
                                                                                        75
0.5
40_
35
a
-....                                                                                  h~37.5
~ 30 a~7.5
.,;'                                                                            h~4o
                                                                                a~lo
   '
...,
.<::
 "'c:
....QJ
         25
...,
....
 c:
 "
 H
 QJ
 0.      20
...,
 c:
 QJ
 H
 H
 "u
'tl
 QJ
         15
 u
 "c:
'tl
H
10
               Fig. 4.7   Current per unit length along the load due to lOOOA
                          flowing through 40 turns coils of different conductor
                          radii, a, at air gap of 30mm.
                                            84
3,5_
3.0
 ~
                                                                                   h=37.5
"'~      2.5                                                                       a=7.5
 :;::
 .!<
                                                                            h=40
 Q
                                                                            a=lO
 p.
   .
  ;:.,
 .._,
         2.0
 ....
  Ul
 "QJ
 '0
 ":<
  QJ
  0
  0.     1.5
 '0
  QJ
  u
  "'
 '0
 H
  "
         1.0
0.5
"''"
 <.J
"'"'<::
H           10
                 Fig. 4.9   Current per unit length along the load due to lOOOA
                            flowing through 40 turns coil of lOmm conductor
                            radius at h = 40mm.
                                                                     -H
                                                p          lLxlO
                                                                  -0
                                                          lOxlO
             6
                                                                -8
                                                          8xl
             5                                                   -8
"''-8                                                     6xl0
  3:
  -"
  0
  0.
   :>.       4                                                   -8
  ....+Jrn                                                4xl0
   "'"
  '0
    H
    ""
    0
    0.       3
  '0                                                             -8
    Q)                                                    2xl0
    u
   "
  '0
   H
    "
             2
             0.,1~~~~~~~~~~~~~~~~~~~~~~~~~~~
                 0     .1     .2      .3            :'1           .5      .6   .7             .8   .9   1.0
                                           Distance along the load, z,          (m)
5.0.
5.5.
5.0
'1.5
N
     s          4.0
    '-
    :3:
    "'
    Cl          3.5
    ';.,"
    ....'-'Ul
                3,0
    'd"
      QJ
     ...
      QJ                                                                                          10. 4xl0
      :>        2.5
      0
      0.
    'd
      QJ
      u                                  -8                                                       7.3xl0
    'd
      "         2.0        =   7. 3xl0
    H "
                                                                                                  4.8xl0
                1.5
                       - - 2.7xl0 -8
                1.0
05
An equation has been derived for the distribution of the induced current
the coil and, also, it does take the conductor diameter and the spacing
and the agreement was very good. This proves that the former is based
This theory was used to investigate the effect of the load non-
the power density induced in the load. The computer program neither
The application of this theory showed that the uniformity of the power
coil pitch and the air gap. The magnitude of the power density depends
largely on the number of turns, the air gap and, of course, the physical
The next logical step in the quest for a flexible coil design method,
necessary to divide the work into two parts. The first is to test the
where the air gap and the spacings between the turns were large in
above the aluminium slab; and the second was the examination of the
a single one.
The experimental rig of Fig. 5.1 was employed to study the influence
different parts of the system. This has been done by shielding the
41 SV                                                                          INDUCTIVELY
                                   ~
60A                                ~
                                   ~                                           HEATED
SOHz                                                                    r-
                                   ~                                           IVORKPIECE
                                   1
                                                                        '---
                                                                    .
                                04
The conductor was placed over the centre of the load. If this was
the load and not the point directly below the conductor. This
The function of the wooden frame, Fig. 5.1, was to hold stable the
Four conductors were used to carry the current above the slab. The
Those conductors were chosen because the induction heating work coil
about lOmm.
                                        lmrn
                              -
                                         •
                                   ~
                                    '          Bmrn
           1 Omrn
                                        _,
                              V                                               (1)
                                                                                    24 mm.
water pa th    /    (8)
                                                                    T
                                                                   4mrn
                          [/                                        -~        (2)
sol id copper /                                                    4mrn
                    (7)
                                                                    - 1-      (3)
                                                                   4mrn
                          ~Smm                  I
                                                T     3rnm   1l
                          j       (6)          (5)           (4)          I
                          1----                 lOmm _ _ . {
induction heating_
this slab was greater than 4 times the skin depth. This load
there was no worry about the change in permeability at this early stage
of the practical work; and secondly the thickness of this slab was
more than 4 times the skin depth, hence the current beyond this depth
approximately
                                         X
                                         0
                  J       = J       e
                      X         S
If x =       4o    then J
                                X
                                        0.018 J . i.e. J
                                               S           X
                                                               is small.
The surface current density induced on the load was measured by the
frame containing 2 brass pins was the type of probe suitable for
use on the surface of the aluminium slab. The distance between the
pins was 30.5mm and each one was connected to an insulated constantan
thin wire (O.lmm diameter), which was kept close to the surface. The
two wires were twisted together in the centre and the other ends were
Care was taken while constructing this probe as the constantan wire
should occur in any part other than the two pins. A coaxial wire
The current density on the surface of the conductor was also measured to
of the current density induced on the load. The probe was of 30mm
were laid along the conductor length and their ends were cleaned and
On the rectangular conductor; nine probes were used and they were
criteria: firstly, (J) and (p) must not change over the probe length;
the skin depth. The first condition was satisfied as the load was
uniform and the current's direction was known and the second too was
satisfied by using probes of O.lmm diameter wire when the skin depth
             V
                                                                       (5.1)
            p~
when there was aluminium slab beneath it and when there was no load.
The results are shown in Figs. 5.6 and 5.7. It can be seen that the
the thickness of the conductor was much smaller than the penetration
0.0
 -::+----- - - .
0.0
1.o1--r-
This was due to the proximity effect on the conductor itself. That
the surface current density induced on the slab; hence the surface
                h
         I          e                                                 (4.9)
     A
         1T             2
                + z
lu3
                       h
                           e                                                  (5.2)
               h2 +    a           2
                               z
                 e    cosa
where a and   B are   functions of h , see Figs. 5.8 and 5.9, and
                                    e
a is the angle between the line joining the conductor and point P
 •
 0
      Rec tan gular conductor
      Circular
                ""'""'"'        ... ··~··}
      Circular conductor of 15mm diameter
      Circular conductor of 28 mm diameter
                                                   Above aluminium
                                                        slab                0
                                                                                    the effective hei·ght !he)
                                                                                Rectangular conductor
                                                                                Circular conductor of 6mm diameter
                                                                                                                   }
50
Equation (4.9)
0 Measurements
  a
.....
 :;::
                  " ' '\
Cl
0.
    -     30           .\
                            \
..,-
....."'                         \
 {))
 "
 Q)
                                    \
'0
                                         \
:."
Q)
0
          20
                                             '\
"'
'0
                                                  \
Q)
"                                                     '\
"
'0
"
H
          10
                                    40                      80            120
                                                                                      ---
                                                                                      0          ---
                                                                                                 160       260
                                                       Distance along the load, z,        (mm)
90
80 Equation (4. 9)
                    \
N                       \
     fl
    '
    :;:
            60              \
                                \
    Q
    0.
        -                           \
     >.
    +J
        -   so
                                        \
    .....                                   \
     ""'
     Q)
    '0
                                                \
     '"'
     Q)
                                                    \
    :.      40
    0                                                   \
    "'
    '0
    Q)
                                                            \
    u
'0  "                                                           \
    "
    H       30                                                      \
20
10
                                                    40                                        120
                                                                                                        --- -----160             200
180
160
                                                                                                                        Equation (4.9)
      \
120       \
              \
                  \
                   \
lOO                I
                       \
                           \
                            \
so                          \
                                \
                                \
60                                  \
                                    I
                                        \
                                            \
40                                              \
                                                    \
                                                        \
                                                            \
20                                                              \
                                                                    ''
                                                                         '   ......   ...,
      Fig. 5.12                         Surface power density distribution along the alumini tun
                                        slab due to lOOOA flowing through circular conductor
                                                                              of 6mm diameter a.t h               c:   42mm
            500
                  \
                      \
                                                                                                         Equation (4.9)
                       \
                        I                                                                                Equation (5.2) with
                        I                                                                                a = 0. 2 and B = o. 85
                        1
    ,3:E                                                                                             0
N
                                                                                                         Measurements
            4           1
                        1
    Q
    0..
                            I
                            I
     >,
    .j..)
    .....
     Ul
                            I
     c:
     <))
    'tJ
     '~"    3
     0
     0.
    'tJ
     <))
                                I
     u                          I
    H
     "c:
    'tJ
                                I
                                I
                                I
                                  \
                                    I
                                    I
                                    I
                                      I
                                      I
                                        I
            lO                          I
                                            \
                                                \
            50                                      \
                                                        \
                                                            ''
                                                            40
                                                                  .....
                                                                          --    80             120
                                                                          Distance ulong the load, z,    (mm)
                                                                                                                160           200
50
45
                                                                                                   Equation (4.9)
40
                                                                                                   Equation (5.2) with
                                                                                                   a = 0. 9 and S = 1. 2
                                                                                               0
                                                                                                   Me as uremen ts
35
30
     "   '\
              \
25
                   \
                       \
                           \
                               \
20                                 \
                                       \
                                           \
15                                             \
                                                   \
                                                       \
                                                           \
10                                                             '\
                                                                    "\
                                                                         ~
                                                                                                           0
                                                                                                                 --- ---
                           40                                  80                        120              160              200
                                                       Distance along the load, z, (mm)
80
N
      8
    ':;;                                                                                                      Equation     (4. 9)
               70
     0                                                                                                        Equation (5. 2) with
    "'
     >,
                                                                                                              a = 0.8 and (3 = 1.2
    ....'"'                                                                                       0
                                                                                                              Measurements
     "'
      ~
     <lJ
    '0        60
     ...
     <lJ
     ~
     0
     "'
    '0
    <lJ
    u
    "
    '0
    ~
    H
              so
                    '\
                         \
                             \
              40                 \
                                     \
                                         \
                                             \
              30                                 \
                                                     \
                                                         \
              20
                                                             \
                                                                 \
                                                                     \
                                                                         \
                                                                             \.. ·o
              10                                                               '\
                                                                                  "
                                                 40                             80                    120
                                                                                                                      --    160      200
                                                                                Distance alonq the load, z,                  (mm)
N              140
       s                                                                                            Equation (4.9)
    '3:
                                                                                                    Equation ( 5. 2) with
     0                                                                                              a = 0.65 and B = l.l8
    "'><'                                                                                       0
                                                                                                    Measurements
    ........
      Ul
               120
     "
    '0
      (])
     ..,
     (])
     "
     0
'0   "'
    (])
               lOO
    u
'0  "'c              \
    H                    \
                             \
               8
                                 \
                                     \
                                         \
                                             \
               6                                 \
                                                     \
                                                         \
                                                             \
               40
20
300
Equation (4. 9)
  s
-..._                                                                           Equation (5.2) with
:?:
                                                                                Cl = 0.28 and     s
                                                                                               = 0.94
Q
             250                                                          0
                                                                                Measurements
"'
  >,
....+-'<Jl
  "
 OJ
'd
 H
 OJ
  "'
  0          200
 "'
'd
 OJ
                       0
  u
'd
  "                    I
 H
  "
                       I
             150       I
                       I
                       I
                           I
                           I
             lOO
                           \
                               \
                               \
                                   \
                                   \
                                       \
              5                            \
                                               \
                                                   \
                                                       "
                                           40                 80              120                lGO      200
                                                           Distance along the load, z,     (mm)
      ;>,
         .
    ....'Ul"'
      ~
      (J)
    'd
      H
      (J)
      ~
      0
      0.
    'd
      (J)
      ()
      ~
    'd
      ~
    H
                1
                                                                                             ---
                                 40            80               120              160          200
                                         Distance along the load, z,      (mm)
70
                                                                                          Equation (4.9)
60
                                                                                       Equation. (5.2) with
                                                                                      a = 0. 85 and S = l. 2
0 Measurements
so
40 " '\
               \
                   \
                       \
30                         \
                                \
                                    \
                                        \
20                                          \
                                                \
                                                    \
                                                        \
10                                                          ~
                                                                ~
                               40
                                                                                      -------
                                                            80           120                   160         200
                                            Distance along the load, z,            (mm)
Equation (4.9)
lOO
 :>,-
'"'
.....
 Ul           \
"Q)
                  \
"' so
><                    \
~                         \
0
0.
                              \
"'
Q)
{)
~       60                        \
"                                     \
                                      \
                                          \
        40
                                              \
                                                  \
                                                      \
                                                          \
        20
                                                              ""
                                                                   " """'   -....
200
\ Equation (4.9)
180
        \                                                                                                 Equation (5.2) with
160         I
             I
             I
140--
              \
               \
120
               \
                I
                 I
lOO
                 I
                 I
 80
                   \
                     \
                       \
 60
                         \
                           \
                             \
 40                            \
                                   \
                                       \
                                           \
 20                                            \.
                                                    '\.
                                                          " .......
                                 40                                   80                         120                       160        200
                                                               Distance along the load,                           z'   (mm)
45
40 0 Measurements
             35
C•l
      G
  ':;:
      Cl
      p.     30
      ;>,
  ,.,"-'
      Ul
  '0  ""'    25
      "':<
      Q)
      0
   0.
  '0
             20
   u
   ::l
      "'
  '0
  H
      "
             15
lO
           140
                                                                       Equation• (5.2) with
                                                                         = 0.5 and    s
                                                                                      = 1.2
e                                                                      "
'-         120                                                     0   Measurements
~
Cl
"'
 ><
+J         lOO
·.-<
 Ul
"OJ
'0
 'OJ;."'    so
 0
 0.
'0
 OJ
 u
"'
'0
           60
"
H
40
20
                                                               0
                                                                        0
                                      1 'l\
                                      '   '
180
                                                                  Equation (5.2) with
                                                                  a= 0.35 and  S = 1.07
                                                              0   Measurements
160
140
120
lOO
so
60
40
20
-lOO
360
280
"'  ~
           240
....""UJ
-1-l
                       0
 "
 Q)
'r1
 k         200
 ~
  0
  ""
'r1
  Q)
  u
  "'
 'Cl       160
  H"
120
so
.40
The difference between the measurements and the values of the power
distance z:
                        h
       cos a                                                                   ( 5. 3)
                  .{2   + z
                              2
                                                                               ( 5. 4)
                                        i :::-:
                                        2                                       -3        2
      8 = -(2.642 X 10-l) + (7.736 X l0- )h               -   (1.539   X   10        )h
                                                     e                                    e
(5.5)
slab, the same constants can be applied. This will be shown later on
per unit length and can be applied within a range of lOO% to 10%
of the maximum induced power density, the error within this band
      h       =   ~2     a
                             2
          e
both conductors, see Figs. 5.6 and 5.7, does not affect the applicability
The experimental rig and the circuit diagram were identical to those
V (z) ( 5. 6)
The above equation was then used to calculate the voltage from each
different points along the load. If the load was a slab then the
from equations (5.1) and (4.21). The equation suitable for the
power densities along the load have now been calculated and the results
is shown below.
             / R e a d : Np , Z.,
                               ~
                                  vi' i=l, NP, Il,
                       r , N, d, £, p' F, 11 r' Rol, NSC
                        2                                                                     /
              Call the subroutine                    E~2ACF    to calculate A , ... ,A
                                                                             1         5
                                                        I2
                              Calculate RA =            - '    SLO, CON
                                                        Il
Set Z = zero
                          r
                               NO
                                        ~
                                        "Z             NP
                                                                   YES
                                                                              I
                  2       2         3    4
 V= [A +A Z+A Z +A z +A Z +A Z ]                              V= ~SLO*Z) + coN]                       ~
      1 2    3    3    4    5
                                                                    *RA
             *RA
                          I                                                   I
                                Store Z and V in arrays
                                                 7
                                                              NO
                                         z >- £                                   z   z + 0. 25
YES
                                                 7
                      1                                              2
                                        NSC = 7
             I                                                                          I
Calculate p01-1er density
for slab (PD)
                                                                    rCalculate     power density
                                                                         for cylinder (PD)
                                                                                                      I
                      1                                                                 I
             [                   Store Z, PD in arrays
              /                     Plot V against z
                                    Plot PD against Z                                             /
                                             STOP
                                                                         11
                 The flow chart of the program                                ~1'-SC-FIT"
Then. five similar conductors were connected in parallel above the slab,
at identical air gaps to those which were applied for the single
conductor. The coil pitch (d) between these conductors was 60mm,
so that the proximity effect between them would be small. A total current
on the load was then measured. The current per conductor was 600A
which was less than that for the single conductor. This is because
it was not necessary to draw maximum current of 50001\ from the trans-
Although many experiments have been done; only one graph is given,
measured induced surface power density and that calculated from the
was small. In this case the current was distributed uniforrnlJ through
method can be applied even when the conductors are in close proximity
to each other, and when the current flows nonuniform/J~·~. This Fig.,
also shows the nonuniformi:ty of the power density along the load.
200
180
       160
 ~
 "'
 !:140
 ~
 D120
      -
 a..
  >-
  ~100
   11)
   c:
  QJ
  "0
  ... eo
  QJ
  ~
  0..60
  "0
  QJ
  u
  ~40
  -
  c:
20
40
36
  ;:;-
   E       32
  --
  3:
...0   ~
           28
  ..-
  X
       ~
 D         24
 0..
 .....>.   20
  Vl
  c
  QJ
 D
  '-
  QJ
           16
  )::
  0
 0..
 "Cl       12
  QJ
 u                                  0
 ::J
"Cl
 c
._,        8
           0
                0                        80
                                         100   120 140           160   180      200
                         Distance along the load ,z, (mm)
aluminium slab. The air gap between the conductors and the load was
varied between 89-23mm. The coil pitch was between 62-llnun, that is
a total current of 3000A was used in those involving more than one
conductor.
did not exceed 8%. Only one graph is shown, Fig. 5.27,
because all the results prove one thing, that is the validity of
for an air gap of 36mm and coil pitch of llmm. The results, Figs. 5.28
and 5.29, are represented in graphs which match the pattern recorded
conductors ..
from single conductor was studied. It was found that the equation of
l.O
o.oj
The other
conductors are
in this side
l.O
these factors.
slab due to more than one conductor have shown that the superposition
being applied and nor does the positioning of the conductor alter
factors; among them are the accuracy of the instruments, and the
power density implies that the calculation of the voltage and the current
induced on the load will be lower than this margin of error as the
correct, when the coil pitch is large the induced power on the load
will increase when one decreases the air gap, and when the coil pitch is
the surrounding metal work from being heated by the stray flux is well
outside of the coil. This prevents the flux from spreading away from
the outside of the coil and linking with surrounding metallic objects.
exist in the system, such as off-the-bar forging and vessel heating. Tl1is
load and to   e:·:<"~mine   the   ~-:;upcrpo~;ition met:-lud   on a wurl:pi(•Cc' ,)[ t.hi:-_; ;:i11d.
-------------------------------------,
the Curie point. The main problem associated with these applications
formula (5.7)
               FA
             - =  -
             FB + H
                      + FC                                         (5. 7)
It is very accurate at low magnetic field and around the knee of the
magnetic field apart from the empirical formula of Thorn ton    [s.lo J
                                L.L
this formula lies in the necessity to know the efficiency and the power
on the conductor above such a load, so that the effect of the conductor
distribution was first measured when there was no load beneath the
conductor, then the same measurements were repeated with the mild steel
slab under the conductor.   The results are shown in Figs. 5.30 u.nd 5.31
                                                                 0.~.        -·1'-----------,..-+-
'
___,o_.-"oCl------ • l.O
circular conductor did not exceed 4% when the height h was 37mm,and
less than this for a greater height. For the rectangular conductor,
this difference was within 8% only for a 22rnm air gap and less than
this, when the air gap was larger. As the change on the conductor
current distribution due to the mild steel slab was small, the influence
insignificant.
The accurate calculation of the surface power density induced on the mild
magnetic field strength [5.4, 5.7]. This formula has been used in
formula for an EN3 mild steel can be found in reference [5.8]. This
             2.08                      -6
                   + (1.257   X   10        )                                (5.8)
          1380 + H
With the aid of a magnetic field strength probe, H-probe [5.11] the
a (lOO x 3)mm perspex sheet of 30mm width. The two ends of the
                 NAw!l H                                                          ( 5. 9)
                        0
w 2nf (rad/s)
The following method was taken to measure the power density on the
                    H =                (A/m)                                      (5.10)
2.   The corresponding permeability was calculated by substituting
                                                                      -8
3.   The resistivity of the mild steel was taken as 16.78 x 10             Qm
                 -r;;t;;
                 -     1ff\1 \1
                           o r
                                                                      (5.11)
J ~ (5.12)
                 ~ pJ2o
        PD                                                            (5.13)
                       2
the surface of the load.          However, it was noticed that the maximum
                                                     l 3')
           420
N
     s
    '-
    3:
    0
      .    240
    p.
    :>,
    +'
    ·.-!
     <Jl
    c:Q)
    'tJ
    k
    Q)
     ~     180
    0
     0.
    'tJ
     Q)
     u
     ~
    'tJ
    c:
    H
60
                Fig. 5.32   Surface power density distribution along the       mild steel
                            slab due to lOOOA flowing through circular conductor of
                                    Gmm diameter at different heights (h)
            .l(>O
320
2SO
N
     8
    '3       240
    "'><-    200
    "'
    ·rl
     <Jl
    <::
    "'
    '1j
    ...
    :."      160
    0
     0.
    '1j
    "u
    "<::
    '1j
    H
             120
so
40
                    Fig. 5.33   Surface power density distribution along the mild steel
                                slab due to lOOOA flowing through circular conductor of
                                         l5mm diameter at different heights (h)
                                                      l·i l
320
               280J h=37mm
N
      a
    '~
    Q
    "'         240
    ..,;..,'
    ..,
     Ul
     ""'
    '0
     H         200
     "'0
     ~
     0.
    '0
     "'
     lJ
     "
    '0         160
    H
     "
                     Fig. 5.34   Surface power density distribution along the mild steel
                                 slab due to lOOOA flowing through circular conductor of
                                            2Bmm diameter at different heights (h")
                                                  .,,. '
160
140
120
 El
"3:
~
          lOO
Cl
0.
>,
'"'
.....
 Ul
 <:       80
 QJ
'0
 :.'0"'
 QJ
 0.
          60
'0
 QJ
 u
 "
'0
H
 "
          40
20
               Fig. 5.35   Surface power density distribution along the mild steel
                           slab due to lOOOA flowing through rectangular conductor
                                          at 4lmm air gap.
                                     i·i.'
power density induced on the surface of the load, i.e. beneath the
                  K
          PD =                                                       (5.14)
                  h
                      e
The factors which affect the power induced on a nonmagnetic load have
1. The induced power density decreases with the increase of both z and h.
conductor.
method on the mild steel slab. This has been done by using 5 circular
The voltages across the J-probe and the H-probe induced by a single
for the PRI~E 400 system. These values were then substituted into the
of conductors.
                                             144
          450
400
          350
"'~                             •
 ~
  Cl
  p..
          300
   >,-
  .j.J
 .....
   Ul
  "
  QJ
 'd
          250
  '"
  QJ
  "
  0
  0.
 'd
  QJ
  u       200
  "
 'd
 ....."
 !3
.....s    150
 "'"
::;:
lOO
50
                        20          40        60        80      lOO
                               The effective height, h , (mm)
                                                      e
                                                                           /
               /           Read N,    X. ,       V. , Il, I2,
                                       ~          ~
NT, D, L
                  /              Write A , ... , AB
                                        1                                  /
                      Calculate       RA~         r 2 ;r ,
                                          1
                       Slope of last part of curve                 =   SLO
                   The consta.·nt of the straight line                 ~     CON
                               Distance X = zero
                                           I
             NO
                               ~                             YES
                  /                  Write X, VX
                                                                           /
                      NO
0 YES
( STOP
method are shown in Figs. 5.37 and 5.38 for the circular conductors.
The agreement between the two values are very good and the
discrepancy did not exceed 6% for air gaps between 80mm and 22mm.
with very good accuracy, Fig. 5.41. The difference between the
good agreement between the measured and predicted values for the
workpiece.
                                        j   ·1 "i
700
          •
600
500
                                    •
400
300
                                                            0
200 0
lOO
40 80 160 200
      Fig. 5.37   Power density distribution along the mild steel slab
                  due to 5 circular conductors at air gap of 77mm, coil pitch
                  of 60mm, and a current of 600A/conductor.
             1200
                                              0
N
     s
    "3:      1000               0     0
    Q'
    p.                                               "
    ..,;.,
    ....Ul    800
                                                          0
     <:
     (])
    'tJ
     ";.
     (])
     0
     p.      600
    'tJ
     (])
     lJ
     ~
    'tJ
     <:
    H
             400
200
                                     40              80                               160              200
                                             Distance along the load, z,   (mm)
                    Fig. 5.38       Power density distribution along the mild steel slab due to
                                    5 circular conductors at air gap of l9mm, coil pitch of 60mm
                                    and a current of 600A/conductor
" a
           1400
                                   "
                                        •
           1200
                                             0
N
     8
    "-
    3:
    Cl     1000
    0.
     :>.
    -1-'
    ·rl
     Ul
     "
     Q)
    '0
            800
     ><
     Q)
     "
     0
     "'
    '0
     Q)
            600
     u
     :J
    '0
    H
     "
            400
200
                  Fig. 5.41       Power density distribution along the mild steel slab
                                  due to 5 rectangular conductors at air gap of 4lmm,
                                  coil pitch of 3lmm and a current of 600A/conductor.
lamination packs are installed around the coil. The main reason
for using them is to channel the flux in the region outside the
which will heat these objects and, also, might result in sparking.
flux guides due to the reduction on the stray losses and the
magnetizing current.
of conductors.
The experiments were carried out on the same aluminium slab employed
hl
                                  l                    conductor
                                                         '-
           1
           h                                            g
slab
The pack was then covered with a fibre glass tape to maintain a
edges.
and the conductor was held constant at 40mm. The current in the
The pack was above the centre of the aluminium slab, where the
to equation (5.2). The differences between the two values did not
exceed 5%; this proves that the practical equation (5.2) can be
used even when there are laminations above the conductor. The
        35                        1. 39                      1.16
        20                        1.46                       1.19
        ll                        1. 51                      1. 25
          5                       1. 58                      1.3
160
0
~    120
~
Q
~
~
~
·M   loo
w
~
w
~
~
w
~
0
~
     so
~
w
u
~
~
~
H
     60
40
20
           Fig. 5.43   Surface power density distribution along the aluminium slab
                       due to lOOOA flowing through circular conductor           of
                       2Smm at h = 54mm with lamination pack at hl = 35mm
        lSO
s       140
'-
"'
0
p.
>·      120
+'
·rl
 Ul
0:
"
'd
 ....
"0:.    lOO
"'
'd
"u
"
'd
H
 0:      so
60
40
20
              Fig. 5.44   Surface power density distribution along the aluminium slab
                          due to lOOOA flowing through circular conductor of
                          28mm diameter at h = 54mm with lamination pack at hl      20mm
         200
         180
                                                                      Equation (5.2) with
                                                                      "= 0.55 and 8  = 1.51
                                                              0       Measurements
         160
  e
......
;::_                              .,,
 Cl
 0.
         140
 ...>o
·4
 Ul
  ~
  "
  u
         lOO
  "
 'd
 H
  "
          so
60
40
20
               Fig. 5.45   Surface:! power density distribution along the aluminium slab
                           due to lOOOA flowing through circular conductor of
                           28mm ut h = 54mm 'Nith lamination pack at hl = llrnm
                                                    1 ~.i 7
          225
                                                              _____      Equation (5.2) with
                                                                        a = 0. 55 and B = l. 58
          200                                                       0   Measurements
"''-G
  e
  0
  "'..,
   >.
          175
  .....
   tl)
   c
   Q)
  '0
          150
   H
   Q)
   "
   0
   0,
  '0
   Q)
   u      125
   "c
  '0
  H
lOO
75
50
25
            Fig. 5.46   Surface power density distribution along the aluminium slab
                        due to lOOOA flowing through circular conductor of
                        28mm diameter at h    =   54mm with lamination pack at hl       5mm
                                               l.SB
200
0 Measurements
160
s
'-
3:
Q'
        140
p,
  '
...."
+J
"'<::
 QJ     120
'0
 1-<
 QJ
 ~
 0
 o,
'0      lOO
 ru
 lJ
 "<::
'0
H
so
60
40
         20
                                           •
              Fig. 5.47   Surface power density distribution along the aluminium slab
                          due to lOOOA flowing through rectangular conductor at g = 40mrn
                                     with lamination pack at hl = 35rnm
                                          l~;'_}
200
                                                   0     Heasurements
160
140
120
loo
80
60
40
20
      Fig.   5.48   Surface power density distribution laong the aluminium slab
                    due to lOOOA flowing through rect<mgular. conductor at
                    g = 40mm with lamination pack at hl = 20mm
                                                        Equation (5.2) with
                                                        a = 0.35 and S = 1.25
0 Measurements
125
lOO '
75
50
25
40 80 160 200
      Fig. 5.49   Surface power density distribution along the aluminium slab
                  due to lOOOA flowing through rectangular conductor
                  at g = 40mm with lamination pack at hl = ll mm
                                                    161
250
       225
                                                                  Equation (5.2) with
                                                                      =    0. 35 and B   =   l. 31
       200
                                                                  "
                                                              0   Measurements
 s
":;:
Q
       175
"'><
.jJ
·.-<
 Ul
"<ll
'd
       150
.
k
:.
0
0.
'd
<ll
       125
u
"
'd
"
H
lOO
75
so
25
            Fig. 5.50   Surface power density distribution rtlong the aluminium slab
                        due to lOOOA flowin9 through rectangular conductor
                        at q   =   tJOmm with lamination pack ut hl   ==   Smm
              hl (mm)      The load surface power   The load surface power
                          density induced from      density induced from
                           circular conductor       rectangular conductor
                               (W/m 2 )                   (W/m2)
35 170 200
20 189 213
ll 203 233
5 223 256
Table 5.2 shows the values of the power density induced on the load
the conductor itself in order to find out whether the change in the
for hl = 5mm. The readings are normalized with respect to the reading
at probe 5, see Fig. 5.3, and the results are shown in Fig. 5.52.
 l.O
                                                                 +·
1.0
Hence the change in the power induced on the load was due mainly
which are not exactly beneath them, as the power density induced
One lamination pack was placed above one edge of the load and the
limits of these two cases. These curves show the power density
induced on two parts of the load, near the two edges, one is covered
with one pack of laminations and the other edge is left exposed.
__________________________....
                                                       ll''
             lUO
160
     "
     QJ
    '0
             lOO
     "
     QJ
     3
     0
     0.
    '0        so
     QJ
     ()
     "
    '0
    H
     "
              60
                  Fig. 5.53   Surface power density distribution along the aluminium slab
                              due to lOOOA flowing through circular conductor
                              of 28mm at h   =   54mm with lamination pack at hl      =    35mm above one
                                                       edge of the slab
             '-
                                              l ()(;
225
175
       150
rl
"'c:
aJ
d
><
aJ
:.     125
0
"'
       lOO
75
        so
                                   0
        25
                                          •
                            40            80
                                                                     .      1 0
                                                                                •
                                                                                            200
                                       Distance along the load, z,       (mm)
             Fig. 5.54   Surface power density distribution along the aluminium slab
                         due to lOOOA flowing through circular conductor of
                         28mm diameter at h    =   54mm with lamination pack at hl   5 mm
                                         above one edge of the slab
             200
180
"'......8
    3:
    Cl
       .     140
   "':;.,.
   .._,
   .....     120
    {))
    "
    Q)
   '0
    ...
    Q)
    :>
    0
    "'
   '0
    Q)
             lOO
    u
    "
   '0
   H
    "
             80
60
40
20
                   Fig. 5.55   Surface power density distribution along the aluminium slab
                               due to lOOOA flowing through rectangular conductor at
                               g = 40mm with lamination pack at hl = 35mm above
                                          one edge of the slab
                                                        lbU
225
  e
           17!0
'i!:
Q
"''
 :>,
....'"'   150
 "'<:
'd"
                        ,
 "";.
 0
          125
 0.
'd
 '"
 u
H
 "<:
'd
          lOO
75
50
25
                  Fig. 5.56   Surface power density distribution along the aluminium slab
                              due to lOOOA flowing through rectangular conductor at
                              g   =    40mm with lamination pack at hl   =   5mm above
                                                 one edge of the slab
The power density induced on the first part is higher than that in
the second, at given points near the conductor. While the opposite
with the maximum value. This point should be taken into account
distribution.
Table 5.3 shows the power density induced on the load at those two
parts directly beneath the conductor. From this table it can be seen
that the power density induced on the other end varies only slightly
This phenomenon was investigated over all parts of the load in the
width. The lamination pack was situated above the 5th part as
The readings are not shown as the changes are small and cannot, therefore,
be shown in graph form. The shape of the curves would have been
conductor, which was above the slab and below the lamination pack
~~~~~~~~~~-----lamination pack
conductor
                   I            I          I          I         1
                                                                    50mm
                   I            I          I          I         I
           part
                   I            r          I          I
                                                                I                 slab
                                I          I                    I
                   I
            ( 1)   I     ( 2)   I   ( 3)   I    (4)   I   (5)   I
                                                                     (6)
                                                      I         I
                                '          '
Fig. 5.57 The lamination pack above part (5) of the load
(5)
( 4)
( 3)
                                                                           ( 3)
                                                                           ( 4)
(5)
1200
            1100       ..
                            0
1000 (J
N
     s
    '-
    :;:     900
Cl
    "'>.    800
    ....
    '-'
    if)
     c:Q)
    '0      700
    ...
    Q)
    :.0
    0.
            600
    '0
    Q)
    ()
    "'c:
    '0
    H       500
400
300
200
lOO
Although the numerical values and percentages are for the particular
be drawn.
must be taken into account when using this method with ferromagnetic
it's effect on the workpiece nor the proximity effect on the conductors
experimental work on the mild steel slab. Finally, the work with
along the load. This method has not yet been practically proved
along t!l"' lead. Part of the workpiece in this case will be hot and
the other part cold. This problem cannot be tackled in the laboratory
due to the high cost of building a heater capable of producing the high
of forming al umini urn have caused a very wide growth in it~· s use
front face; that part which enters the die first. In order to
operation:-
die opening,
billet is not only forced into the die but also moves along
to the press container, the rear end of the billet becomes progressively
cooler as the extrusion proceeds rather than the front end. This,
of course, is the exact reversal of the direct extrusion and care must
the theory because of the capital cost and the limitations of the
have been carried out to find the effect of single turn and multiturn
cylinder of 50.8mm radius and 340mm length had it's surface machined
in Chapter 5.
The experimental rig is shown in Fig. 6.1. The single turn and
multiturn coils have been constructed from a water cooled copper tube
of 6mm diameter. Three single turn coils of 70.8, 80.8 and lOO.Smm
The single turn coil was around the middle of the cylinder which
distances along the cylinder were measured by the current density probes.
Fig. 6.1   The experimental•rig
The probes were made from thin constantan wires wrapped around the
The results of the single turn coils are shown in Figs. 6.2-6.4.
These graphs show the measured surface power density together with
that the practical equation (5.2) which was developed for aluminium
and the measured surface current density into the equations given
in Appendix l.
The results of the single conductor show that the practical equation
slab. The discrepancy between the measured power density and that
calculated by the practical equation (5. 2) did not exceed 10% ~or
the range from maximum to 10% of the maximum power density. This
 60
                    \
                        \
  40
                            \
                                \          ~
                                            \.
                                    '\.
  20
                                          "'     ...........
                                                                                 ~
20 40 60 80 lOO
Fig. 6.2   Surface power density distribution along the aluminium cylinder
           due to 400A flowing through single turn coil at air gap of 20mm
                                                                                  Hl2
         140
                                                                                             Equation (4. 9)
C)
0...
     -   lOO
               \
                   \
....>.                 \
                           \
 VI
 c:      80                    \
 QJ
C)                                 \
                                       \
 '-
 QJ                                        \
 3:
 0                                             \
0...     60                                        \
"'Cl                                                   \
 QJ
 u
                                                           \
 :::>                                                          \
"'Cl
 c       40                                                        \
~
         20
                                                                           "" '
                                                                                  ''
                                                   20                             40        60           80            100
                                                                           Distance along the    load, z , (m m)
64
56 Equation (4.9)
0
       -
0...
       -
    >.40
    Vl
    c
 CJ
"'Cl
     r...
             32
     CJ
     3
     0
     c..
"'Cl
 CJ
     u       24
     ::::J
    "'Cl
- c
16
                                                                        -
                            20            40                 60              0     10
                                 Distance along the load, z, (mm l
                  Fig.6.4    Power density distribution along the aluminium
                             cylinder due to 400A flowing in single conductor
                            .at air gap of           SO mm
                                L, ,. ,
of the coil was around the middle of the cylinder and this was the
on the load is shown in Fig. 6.5 together with the predicted values
which were derived with the aid of the computer program "W-SC-FIT".
5.
4.
·-
 .r::;       4.
3
    -""
         ~
0
0..
....>.
    VI
    c
    Q)
    0
    '-
    Q)       2.
    :;.:
    0
    0..
    "0       2.
    Q)
    u
    :::>
    "0
-c
1.
                                                                                             0
             0
                   0   10     20      30   40     50   60        70     80   90   100   110 120
                                   Distance along the            load, z, (mm).
a constant surface power density with no heat loss from the surface
to be:
                                                                     2
                                      2                         -8     1 J     [13 n (r/R))
         Tr          (PD)~2T+        r        l
                                              - - 2   I     e
                                                                     n     o
                                                                                              }   (6.1)
                            A       2R2       4
                                                      n=l
                 =
                       At
         1
                            2      (dimensionless)                                                (6. 2)
                      yCR
t time (s)
      y
                                          3
                 = density (kg/m )
J 0 (x) is Bessel's function of the first kind and zero order and the
equation (6.1).                 Baker [6.2) showed that the the transient time
is equivalent to      T   = 0.25,          at this point the summation term will be very
small and as such can be ignored. Once the transient time has
                          R
      Ts - Tc        PD 2A                                                      (6.6)
rewritten as
                                           2
                                       r
      Tr    = Tc   + (Ts - Tc)
                                       R2                                       (6. 7)
As was stated earlier, the above equations are for a cylinder with power
                                 R
        Tr - Tc =       (PD)     2A                                                       (6. 8)
(6. 9)
(6.11)
less than the total input power [6.4) and equation (6.8) will be:
                                                PD
        Tr - Tc                                                                           (6.12)
                                                p
                                                    n
                                -8          4             4
        PR= 5.67       X   10        E(Ts       - Tb )                                    (6.13)
where PR                                        2
                 the radiated power density (W/m )
                                                                   2 4
         £   :   the emissivity coefficient (W/m                    K ) and
The billet does not enter the die .directly after leaving the heater,
                                                                         2
                      R                             ro              -a       T       Jo(ran/R)
             =~
                                                                         n
              2 0f
                                                    ~
      Tr                  r 'f(r ')dr' +            L           e
                                                                                          2
                R                                  n=l                                J     (a )
                                                                                          o n
                                    o
                                        JR        r'f(r'p (a r'/R)dr')
                                                                o        n
                                                                                                       (6 .14)
where the a 's are the positive roots of J ' (aR) = J (aR)                                         0
                n                                                            o                1
                                              2
                                          r
      f(r)          Tc + (Ts - Tc)                                                                     (6.15)
                                              2
                                         R
[6. 2] as:                                                                       2
                                                                     -AS t
                                                          ro
                                                                        n
                                        {.!:.                       e R2
      Tc (t)         Tc + (Ts-Tc)                 + 4     I                          }                 ( 6. 16)
                                                                S~Jo(Sn)
                                          2
                                                         n=l
                                                                                 2
                                                                     -AS t
                                                          ro
                                                                        n
                                        {.!:.      l            e            R2
      Ts ( t)        Tc + (Ts-Tc)                 +-      .'i                            }             (6.1•7)
                                                                         s~
                                          2         4
                                                         ri=l
The quantities inside the parentheses are shown in Fig. 6.6.
expression;
                                                                mrz
                                                         cos
                                                                 9.
        T    -1 !9. f (z') dz' + 2
                 £     0                    £
                                        t                nnz•
                                    I
                                    0
                                            f(z 1 )cos    9.     dz'                 (6.18)
The first term is the steady-state temperature and the second term
is the transient. The transient die away with time constants T',
where
                             1                                                       (6.19)
            T'       :::::
                             e
 soaking period.
                                 l_ <)   .J
the two ends will drop to 90% of the initial difference for a billet
was 300mm [6.4]. Hence, a billet leaving the heater with axial
and radial temperature differences, will enter the die with uniform
power density.
                          l.O
                         o. 8                JTs(t) - TcJ
                                               Ts - Tc
0.6
0.1
                                                            JTc(t) - TcJ
                                                              Ts - Tc
                          o. 2
1.0
                         0.8
...,
       0
E-<              0       0.6
           E-<
...,
             ""
           E-<
                         0.4
E-<""
                         0.2
                               0
                                       0.5      l.O     1.5       2.0    2.5     3.0
                                                        t/t'
(6.20)
the length of the cylinder at the moment when it leaves the heater
i.e. at the start of the soaking period. The output of this part
with the exception that different coil pitches inst<~ad of one value for
the pitch apply and the plotting of the temperature distribution. The
Calculate RA ;
Set z zero
NO YES
               2    2    3   4
V= [ A +A Z+A z +A z +A z +ASZ]                                              V       [(SLO*Z) + CON)
      1  2   3    3    4
                                                                                         *RA
             *RA
                                                                            NO
                                                                                         z       z + 0. 25
                                       Plot V against               z
                                       Plot PD against z
                              Plot the temperature distribution
( STOP )
made from 6mm diameter water cooled copper tubing which meant
aluminium cylindrical billet of radius 50.8mm and the air gap was
of the billet was then decided upon and details of the work coil
pitch are given in Table 6.1. The eleven turn work coil was con-
temperature taper and the results are shown in Fig. 6.8 correlation
3.0
N
      s
    '-                            0
    ~
    Cl        2.5
    0.
    .,....,
     Ul
     <:
     "'
    'd
              2.0
     "'"
     ~
     0
     0.
    'd
    u"'
     ~
    'd
     <:
    H         1.5
1.0
0. 5
                0
                     0           40             80            120                160       200
                                Distance along the load, z,   (mm)
Pitch no. 1 2 3 4 5 6 7 8 9 10
mm 7 8 11 23 23 23 25 23 25 32
been built. A 30 turn, 160mm inner diameter coil with uniform pitch
               11
The program         W-FC-TEMP" was used to design a coil sui table to produce
              '
           -i
                                                                                                              2.50
~   .00   -,I
          -                                                                                                   2.00
l .50     ---,
                                                                                                              1 .50
          ~
l .00
                                                                                                              1 .. 00
          '
    .sa
                                                                                                              .50
    .ea   -
                                                                                                         I    .00       CO
                                                                                                             .1
                  ra,dius                                                                     length
                                                                                .7
.9
1. 1 1. 1. •
                            Fig. 6.9   Temperature distribution due to the coil mentioned in Table 6.1
                     ~ 'J l   ·'
\
''
I
           lOO
"' El
  "c
            80
  "'"'
   :>.
  +-'
  ·.-I
   (})
   "
   <lJ      60
  "'....
   <lJ
   :>
   0
   0.
  'tl
   <lJ      40
   u
   "
  'tl
  H
   "
            20
                      •
shown in Table 6.2 with a coil current of 400A.             The practical
Pitch no.! 1 2 3 4 5 6 7 8 9 lO 11
mm 10.5 10.5 10.5 10.5 21 10.5 32.5 10.5 32.5 10.5 32.5
 Table 6.2    The inter turn spacings of the coil which produces
               the surface power density distribution shown in
                              Fig. 6.12
The correlation between the calculated and measured values were very
good and the discrepancy did not exceed 10%. The predicted
curve is not very linear, this is dne to the fact that the
spacing between the turns can be varied by 10.5mm, which is the pitch
                                                                    11
coil design technique;        a nontu1iform coil design program          Superposition 11
5.0
                              0
                 4.5
                                      0
                 4.0
                                             0
N
      s
    ...__
    :;:          3.5
    -"'
                                                    0
    0
    "'           3.0
                                                          0
    ....,:;.,
    ·.-<
     Ul
     <::
     Q)                                                          0
    '0          2.5
     ...
     Q)
    "0..
    0
    '0          20
    Q)
    {)
    "<::
    '0
    H
                1,5                                                                0
1.0 0
@.5
                  Fig. 6.12       Power density distribution along the aluminium cylinder due
                                  to 12 turns versatile coil at air gap of 30mm,
                                  nonuniform coil pitch and a current of 400A.
                                                                                                               2.50
.:.90 :
      :::::. r),   ---.:
                                                                                                               2.00
  '    _,V             I
                    ~
                                                                                                                   .50
  • 00                 I
                                                                                                               1 . 00
                                                                                                               .50
  .00
             • i                                                                                               .00
                                                                                                             •1
.3
                                                                                          .5
                           radius
                                                                                       length
                                                                                .7
.9
1 •1 1 •1
                                    Fig. 6.13   Temperature distribution due to the coil mentioned· in Table 6.2
curve to fit the readings of the single conductor, the practical
workpiece. The magnetic field strength induced from all turns are
a f unction of H 2 ; t h at 1s
                            .
PD ( 6. 21)
suggested coil pitches and the single turn equation parameters were
6.6 Conclusion
CHAPTER 7
7.1 Conclusion
The variation of the permeability has been taken into account in the
magnetic workpiece.
Magnetic field strength probe has been constructed and used to measure
inside a nonuniform coil have been predicted and plotted by the use
of the computer.
of different parameters.
of the temperature difference between the cold bar and the partly
showed that this technique can taKe the nonuniformity of the load's
for this application, since this method does not assume the coil
recommended.
 In the present economic climate abrupt changes in the price and
is recommended.
Appendix 1
by Induction
magnetic field HR along it's length (the symbol HR is used for the surface
necessary to take a ring of unit length, radius r (r<R) and thickness dr.
<j>r (Al.l)
     dE           -~                                                                            (Al. 2)
                         at
                   a {                 r
i.e. dE
                   at
                              )l   f       27TXH dx }
                                   0
             ~
                 -21T~    t
                          0
                                    aH
                                   x - dx
                                    at
                                                                                                (Al. 3)
                  2nrp
     R                                                                                          (Al. 4)
         r         dr
and the current is:
     I            ~
                        J dr                                     (1\l.S)
         r
                        dE
also I            ~                                              (ill. G)
         r               R
                            r
                       -21TJ.I
                                     0
                                         tx      aH dx
                                                 at
                                                                 (1\l. 7)
                                     211rp
                                      dr
                                           arr
     J
              -J.I
                  pr            fr       X-
                                           at
                                                 d
                                                     X           (Al.    8)
                            0
                       dH
     J                                                           (1\l.   9)
                       dr
     r dH ~ ~                    Jr       x aH dx                (Al.lO)
       dr   p
                                 0
                                            at
jwH (Al.ll)
     r dr ::;
             dH             H.
                            p
                                     Jr    x jwH dx              (1\l. 12)
                                     0
      r dH = junt                  Jr   X       H dx                               (Al.l3)
        dr    p
                                   0
limits yields:
               2
              d H       dH
      r                +- = jWIJ Hr                                                (Al.l4)
                   2    dr   p
              dr
let   k
                       12                                                          (Al.l5)
                       0
         w~
i.e. k2 = -                                                                        (Al. 16)
                       p
                       1 dH                 2
                   + --- j k                    H = 0                              (Al.l7)
                     r dr
                               =   00                                              (Al.l9)
As the magnetic field strength in the centre of the cylinder
to
(Al. 20)
                             H
                              R
i.e.       c   =
                                                                     (Al.23)
                   ber kR + j bei kR
                          kr + j bci kr]
       H           H Lber                                            (Al.24)
        r           R
                      ber kR + j bei kR
                       ber -
                              12 r              12
                                      + j bei l f r
                               0
i.e. H = HR
      r
                       (                                             (Al.25)
                       ber
                              12 R
                              -       + j bei
                                                12
                                                -R
                                  0              0
i.e. J
                    12
                   =--
                                          ber   - 12 6
                                                       r + j be1.
                                                               ''./2 r
                                                                  6      (Al. 28)
           r               6   HR                    2            2
                                          ber       - R + j bei - R
                                                     6            6
                                   r
     ...
     "'r
                   = 21f)J     J       C(ber k x + j bei k x)x dx        (Al. 29)
                               0
               r
               0
                          x ber x dx            =    x bei'x             (Al. 30)
               r
               0
                          x bei x dx            = -x ber'x               (Al. 31)
to yield:
     q,            =
                       2 TI)JCr (bei'kr - j ber'kr)
                                                                         (Al. 32)
           r               k
Substitute C from equation (A1.23) into equation (A1.32) to produce:
                 2n)lH r
                          R      bei'kr - j ber'kr
                               [ ber kR + j bei kR
                                                          J   (Al. 33)
      <j>r            k
                                 bei'kR - j ber'kR
                                                              (1\l. 34)
                               [ ber kR + j bei kR
(1\1.35)
where
        p
                  2       ber kR bei'kR - ber'kR bei kR
                                                              (1\l. 36)
                 kR                   2       .2
                                  ber kR + be1 kR
and
        p                                                     (Al. 38)
             r
      lO
      0·9
      0·8
      07
      0·6
....~ 0·5
 "'>
()I
      04
'd
 ~     0-3
 "' 0·2
       0·1
                                                4               6        7   8
             0      1          2         3
                        workpiece diameter to current depth ratio, 2r /ISw
                                                                             w
                        2                                                        (Al. 40)
                    J       2nrp dr
The total power per unit length of the cylinder can be calculated
integrating
                        JR2nlcl 2k 2 p        2
                                         (ber' kr +
                                                            2
                                                       i bei' kr) r dr           (Al. 41)
                        0
                        2          . 2
            JX   (her• x + be1' x)dx         x(berx ber'x + beix bei'x)          (Al. 42)
                                     .::•
                    2
     Pw      = 2n[c[ PkR(ber kR ber'kR + bei kR bei'kR)                (Al.43)
                          2
Substitute for [c[            from equation (Al.23) and yield;
(Al.45)
(Al. 46)
PD (Al. 4 7)
PD (Al. 48)
or                                                                     (Al.49)
                                                        221
Appendix 2
        R                                                                                    (A2.1)
        m          IJ        IJ A
                        o r
gap is:
                                     t
                                         c
        R                                                                                    (A2. 2)
            mg                           2    2
                        IJ       'll(r -r )
                             0        c w
                        N I
                             c c
        R                                                                                    (A2.   3)
            mw               ~w
                 = 1J or
                      1J             H A (P-jQ)
                                             w
                                                                                             (A2.   4)
                    N I
                     c c
           H =                                                                                 (A2. 6)
                     li-
                           e
to yield;
                                              li-
                                                    e
           R                                                                                   (A2. 7)
               mw
                                              2
                               ll ll n              (P-jQ)
                                   o r        w
the turns is ignored for the time being. The coil can then
   greater than the penetration depth [A2.1), the total flux per unit
                                                                  ~)
   width being:-                                             \)
                                                                                               (/\2. 8)
                               2
                                                          { ,-· .                .'
is the coil perimeter, the coil flux can therefore be defined as;
           k
               r
                    = 0.92
                                    M  c c
                                                                                         (1\2 .10)
                            N I
                             c c
           R
            me
                    =                                                                    (1\2.11)
                             <Pc
to yield;
                           £
                             c
           R        = -~~----;c--c:-                                                     (A2. 12)
               mc       ~
                        o
                             nk     a r (1-j)
                                   r c c
see Figs. 2.1 and 2.2. The length of the return flux path
t       is a function of:                the coil radius; the difference between the coil
    r
and the workpiece lengths; and the workpiece penetration depth.
                                                 .., ,_.,
                                                      '
the difference between the coil and workpiece lengths; and the
and 0.1 for the return flow through air. If the flux returns
range of k will lie between -0.22 and 0.1 [A2.6]. For a coil and
      Jl        = 2r (0.45 + k)                                                          (A2.14)
           r               c
                           2rc(0.45+k)            2(0.45+k)
      R
           mr
                 =                    2
                                                                                         (A2.15)
                               11 11r               11 11r
                                 0    c                0     c
                                        '\ 1:
Appendix 3
configuration shown in Fig. A3.1, [A3.1]. The coil current is known and
Fig. A3.2, where all inductances are mutually coupled to all others.
(A3. 1)
.....--~---- - - -------....---....
                • L
                •   n
(A3. 3)
        Z              -j"'L                                                           (A3. 4)
            cwk                  ck
k l , 2 , 3 , ... ,n
1 l, 2, 3, ... , n
currents I         .
                  w
inductance between such a turn and a solenoid are given by Graver [A3.21.
(A3. 5)
                              b2                  2       2
                                        221     b       b     -l b
              + 3.45         2        + ---1.6112 + 3.2 2 tan    c
                                 c       60     c       c
                   2       b2    . 4            2
                 c               JJ           c
              - - - in(l + - ) + - -   in(l + - ) }                                       (A3. 7)
                     2       2       4        b2
                lob        c     2c
and b and c are the breadth and depth of the segment. in is the
natural logarithm.
(A3. 8)
                                      2 v'r r
                                           1 2
      k                                                                                   (A3. 9)
where
                    3
        c0          8
                                                                              (A3 .11)
                         n          2
and     x2n =
                         LK   (- .!:__) n-p                                   (A3.13)
                        p=O p    A2
where K         = 1.0
          0
First, the segment to be divided into two filaments a and b, then the
and h .         The mutual inductance between the whole solenoid and filament
     2
a is then:
        M                                                                     (1\3.15)
            a
wherG M is the mutual inductance between the filament a and the
       1
upper portion of the solenoid, M is the mutual inductance between
                                2
a .J.nJ the lower portion of the solenoid.   11       and 11       arc the
                                                  1            2
lengths of upper portion and lower portion of the solenoid and
      ~c                                                                        (A3.16)
                                                  -~   :J _l
          1                  a2~
          l1             +   a/ 1 =                                                              (M. 1)
                         J J { (~)
                   1
          Fun =
                   2!1
                             R
                                ax
                                   2             +      c:t> 2 J.ds
                                   ,-2
               +
                   jw
                   2p
                         JJ      ~       .ds-          J J J.~ds                                 (M. 2)
                             R                                 R
(A4. 3)
temperature only.
               d             <lB
              +-                                                         (114. 5)
               3y            dt
      H   ~   H coswt                                                    (A4. 6)
               0
B ~(H)H (114. 7)
Lavers [A4.2]:
where [s] and [T] are the finite element matrices and [!!} is
the billet.
                                                Appendix S
TlH!se probes were used to measure the current density on the surfacR
The entire length of the wires, with the exception of the two
The other two ends of the wires were twisted together and connected
the current in the conductor. The voltage across the probe was
        V           V        +    V.                                                   (AS .l)
                        d          ~
            V.              Nd~                                                        (AS. 3)
                l            dt
and         f       ~
                            B/1                                                        (AS.~)
 where       p      The conductor resistivity
                       d
         V = p£J + dt (BA)                                                (AS. 5)
The area A is very small as the probe was close to the conductor.
V = p£J (AS. 6)
on a cylinder, the thin wire was wrapped tightly around the billet
so that the emf induced on the surface of the cylinder was equal
to that on the probe. As the leads of the probe wer~ twisted together
then the voltmeter measured only the vol tagc on the surface of the
         V   = E£                                                     (AS. 7)
                                                                         ......
E = pJ (A5.8)
V 21fRpJ
1984.
1.8 DENERDASH, N.A .. , MOHAJ11NED, O.A., NEHL, T.W. and MILLER, R.H.:
       1
       'Solution of eddy current problems using three dimensional
Loughborough, 1981.
the finite element method", Geophysics, 1971, Vol. 36, pp. 132-155.
pp. 359-367.
1984.
York), 1967.
5.6 BRIANSKY, D.: "Single analytical formula for the entire B-H
pp. 106-112 0
1\5.5
              BOWOEN, A.L. and 01\VIES, E.J.: "Analytic separation of the
SHORT
LONG
EQUIV
BESSEL
slab
W-P-PO\vER
PROXIMITY
W-SC-FIT
FITTINGl
\v-FC-TEt1P
ALl
Magnetic Material
ii. Publications:
      1.     11
                  Improvements in the design of induction billet heaters .. ,
      END
$INSERT BESSEL
OK,
SLIST LONG
      REAL PS,V,F,DV,ROVV,URV,ALC,DC,ROVC,URC,AKR,P,Q
      REAL BER,BEI,KER,KEI,BERD,BEID,KERD,KEID
      COMPLEX THETA,THY,KO,IO,KOD,IOD
      READ<5,*lPS,V,F,DV,ROVV,URV,ALC,DC,ROVC,URC,AKR
      DDV'503.3*SQRT<ROVV/(F*URVll
      DDC=503.3*SQRT<ROVC/(F*URCll
      X=1000000.0
       BA=DV/(1.414214*DDVl
C       CALCULATION OF BESSEL FUNCTIONS
      IF<BA.GT.8lGO TO 5001
       A1=BER<BAl
      A2=BEI <BA l
      A3=BERD<BAl
      A4=BEID<BAl
      GO TO 5002
5001 A1=REAL<IO<BAll
      A2=AIMAG(IO<BAll
      A3=REAL<IOD<BAl l
      A4=AIMAG<IOD<BAll
5002 CONTINUE
      DEN= ( (A 1 l * (A 1 l l + C<A2 l * CA2 l l
     TA=AG
      EFF=<AP/CAA+APll*100.0
      VRITE<1 ,13lEFF
13    FORMATC1H0,13HEFFICIENICY =,F6.3l
      Z=SQRT<<<AP+AAl**2l+<CTA+TF+TBl**2ll
      PF=<AP+AAl/Z
      VRITE<1 ,15lPF
15    FORMATC1H0,4HPF =,F6.4l
      VA=PS/PF
      TR=Z*SP
      E=SQRTCVA*TRl
      AIN=SQRT<VA/TRl
      TN=<V/El
      WRITE C1 , 1 8 l TN
18    FORMAT<1H0,4HTN =,F7.2l
      AI=AIN/TN
      WRITE C1 , 20 lA I
20    FORMAT<1H0,9HCURRENT =,F10.4l
      STOP
      END
$INSERT BESSEL
OK,
SLIST EQUIV
c       *********************************************************************
c                    COIL OF CONCENTRIC CYLINDERS
c        THIS PROGRAMME APPROXIMATES A BILLET AS N CONCENTRIC CYLINDERS
c       ********************************************************************~
c
        REAL*4 L,K1 ,K,N1 ,I1
         INTEGER S,P
        DIMENSION XPTC100l,EFFC100l,W1 C100l,I1 C100l,Q1 (100),
       *RESC100l,URC100l
         WRITEC1,800l
800    ·FORMATC//'H0\1 MANY CONCENTRIC CYLINDERS ARE REQUIRED'//)
         READC1 ,*lN
         WRITE ( 1 , 15 l
15       FORMAT(//' ENTER 1 IF RANGE OF VALUES REQD'/' ENTER 2 IF ONE •
       *'VALUE REQD'//l
         READ ( 1 , * l S
         IFCS.EQ.1 l\IRITEC1 ,16)
16       FORMATC//' WHICH PARAMETER DO YOU WISH TO VARY'/'ENTER 1 FOR F'
       *I'           2 FOR N1'/'  3 FOR L'l'       4 FOR RB'/
       *'           5 FOR RC'//)
         IFCS.EQ.1 lREADC1 ,*lM
      \/RITE (1, 1 l
1         FORMAT(//'      **** NCOIL ****'//)
C         ***** INPUT RESISTIVITY PROFILE *****
c
      WRITE ( 1 ,803)
803   FORMATC//'ENTER RESISTIVITY OF INNER CYLINDER,RES1 '//)
      READ ( 1 , * l RES ( 1 l
       WRITEC1,5000l
 5000   FORMATC//'ENTER RESISTIVITY OF OUTER CYLINDER,RES2'//l
       READ ( 1 , * JRES C2 l
       WRITEC1,776l
776     FORMAT(//'ENTER INITIAL PERMEABILITY,URCOJ '//)
       READC1,*lURO
        WRITEC1 ,777)
777     FORMATC//'ENTER FIRST CONSTANT OF PERMEABILITY,AF '//)
       READ(1 ,*lAAF
       WRITE ( 1 , 77 8 l
778     FORMATC//'ENTER SECOND CONSTANT OF PERMEABILITY,BF '//)
           READ(1 ,*lBBF
           WRITEC1,779l
779         FORMATC//'ENTER THIRD CONSTANT OF PERMEABILITY,CF '//)
           READC1 ,*lCCF
         \/RITE ( 1 , 2 l
2        FORMATC//'ENTER VOLTAGE,U1'//l
         READC1,*JU1
         IFCM.EQ.1 JGO TO 100
         WRITE ( 1 , 3 l
3        FORMAT(//'ENTER FREQUENCY,F'//l
READC1 ,*>F
       IFCM.EQ.2lGO TO      103
100    WRITEC1,4l
4      FORMATC//'ENTER      POWER,N1'//l
       READC1 •* lN1
       IFCM.EQ.3lGO TO      105
103    WRITEC1,7l
7      FORMATC//'ENTER      LENGTH,L'//l
       READC1 ,*lL
       IFCM.EQ.4lGO TO      106
105    WRITEC1,9l
9      FORMATC//'ENTER      BILLET RADIUS,RB'//)
       READC1 ,*lRB
106    WRITEC1,10l
10     FORMATC//'ENTER      COIL RESISTIVITY,RESC'//l
       READ C1 , * l RESC
       WRITE ( 1 , 1 2 l
12     FORMATC//'ENTER      K1'//)
       READC1,*lK1
       IFCM.EQ.5lGO TO      107
       WRITE C1 , 1 3 l
13     FORMATC//'ENTER      COIL RADIUS,RC'//l
       READC1 ,*>RC
107    WRITEC1,14l
14     FORMATC//'ENTER K'//)
       READC1,*>K
       I F CS • E Q • 1 l WR I TE <1 , 1 7 l
17     FORMAT(//' HOW MANY DATA POINTS ARE THERE ,LESS THAN 100 ONLY'//)
       IF CS.EQ.1 lREAD<1 •* lNN
       IF <S • EQ. 1 l WRITE <1 , 1 8 l
18     FORMAT(//' ENTER PARAMETER VALUES,LESS THAN 100 ONLY'//)
       IF< S. EQ. 1 l READ< 1 , * l <XPT CI l , I= 1 , NN l
       IF<S.EQ.1 >GO TO 19
       CALL SUBCU1 ,N,RES,UR,F,N1 ,L,RB,RESC,K1 ,RC,K,EFFC1 l,W1 <1 l
      *, I 1 <1 l , Q 1 ( 1 l , URO , AAF , BBF, CCF l
       GO TO 21
19     CONTINUE
       DO 20 P=1,NN
       IF<M.EQ.1 >CALL SUBCU1 ,N,RES,UR,XPTCPl,N1 ,L,RB,RESC,K1 ,RC
      *,K,EFF<Pl,W1 CPl,I1 CPl,Q1 CPl,URO,AAF,BBF,CCFl
       IF CM.EQ.2 !CALL SUB<U1 ,N,RES,UR,F ,XPTCPl ,L,RB,RESC,K1 ,RC
      *,K,EFF <P>, W1 CPl, !1 CPl ,Q1 CPl ,URO,AAF ,BBF ,CCF)
       IF<M.EQ.3lCALL SUBCU1 ,N,RES,UR,F,N1 ,XPTC
Pl ,RESC,K1 ,RC,K,
      *EFF CPl ,W1 CPl, !1 <Pl ,Q1 (Pl ,URO,AAF ,BBF ,CCF)
       IF<M.EQ.4lCALL SUBCU1 ,N,RES,UR,F,N1 ,L,XPTCPl,RESC,K1 ,RC
      *,K,EFF<Pl,W1 CPl,I1 <Pl,Q1 <Pl,URO,AAF,BBF,CCFl
       IFCM.EQ.5lCALL SUBCU1 ,N,RES,UR,F,N1 ,L,RB,RESC,K1 ,XPTCPl
      *,K,EFF<Pl,W1 CPl,I1 CPl,Q1 C
20     CONTINUE
C      **** RESULTS ****
c
       YR I TE ( 1 , 2 2 l
22     FORMAT!//'              ****RESULTS****'//)
c
21      IF cS . EQ . 1 l GO TO 2 3
        YR I TE ( 1 , 2 4 l
24      FORMAT!//'               EFF          Y1                            I1 '
      *                        Q1    '//)
        YRITEC1 ,25lEFFC1 l,Y1 (1 l,I1 (1 l,Q1 (1 l
25      FORMATC5E14.5l
        GO TO 500
23      YR I TE ( 1 , 26 l
26      FORMAT!//'            XPTC!l      EFF           Y1                                  I1
      *'                  Q1       '//)
        DO 27 I= 1 , NN
        \IRITEC1 ,28lXPTCil,EFFCil,Y1 (!l,I1 Cll,Q1 c I l
28      FORMATC6E13.5l
27      CONTINUE
500     CALL EXIT
        END
c
       SUBROUTINE SUBCU1 ,N,RES,UR,F,N1,
L,RB,RESC,K1 ,RC,K,EFF,
      *W'1 ,!1 ,Q1 ,URO,AAF,BBF,CCFl
c
        REAL*4 L,K1 ,K,N1 ,KERD,KEID,KER,KEI ,KK,I1
       COMPLEX I OT, I ODT, KOT, KODT, RF, Z! 1 0 l , C ( 1 0 l , H ( 1 0 l , JD ( 1 0 l , A,
      *ZAA,ZZ,IO,KO,IOD,KOD,AFC10,10l
        INTEGER P,N,M,T
        DIMENSION IOTC10,10l,IODTC10,10l,KOTC10,10l,
      *RESC10l,URC10l,RRC10l,XC10,10l,AC20,20l
      *, KODT ( 1 0, 1 0 l , RC 1 0 l , SD ( 1 0 l , YT1 ( 1 0, 1 0 l , YT2 ( 1 0, 1 0 l
      * , YT3 C1 0 , 1 0 l , YT4 ( 1 0, 1 0 l , YT5 C1 0 , 1 0 l , YT6 ( 1 0, 1 0 l ,
      * YT7 ( 1 0, 1 0 l , YT8 C1 0, 1 0 l , XD ( 1 0 l , Y ( 1 0 l
c
       D=RB/N
       R ( 1 l =D
       IFCN.EQ.1 lGO TO 802
       DO 802 I=2,N
       RC! l=RCI-1 l+D
802    CONTINUE
       PIE=3.1415926
       U0=4*PIE*1E-7
       Y=2*PIE*F
       PC=0.92
     AM=PIE*CRC**2-RB**2l
        DDRR=CRESC2l-RESC1 ll/CN-1 l
        DO 5001 I=2,N
        RESCI l=RESCI-1 l+DDRR
5001    CONTINUE
C      ***** CALC. OF SKIN DEPTHS *****
c
         TT
E=O
          UR ( 1 l = URO
          DO 1500 JP=1 ,N
1 51 0      DO 807 I=1 ,N
          IF CTTE.EQ.1 lSDCil=SQRTC2.0*RESCil/CW*UO*URCilll
          IF CTTE .NE.1 lSDC I l=SQRTC2.0*RESC I l/ CW*UO*URCJPl l l
807      CONTINUE
         SDC=SQRTC2.0*RESC/CK1*W*UOll
c        ***** CALC. OF XDCil *****
c
         DO 808 I=1 ,N
         XDCil=-SQRTC2.0l/SDCil
808      CONTINUE
         RFCR=PC*RESC*RC*2.0*PIE/CK1*L*SDCl
         RFCI=RFCR
         RF=CMPLXCRFCR,RFCil
         XR=W*UO*PIE*RC/C2.0*(0.45+Kll
         XO=W*UO*AM/L
         DO 806 I=1 ,N
         RRCI l=-RESCI l*2.0*PIE*RCI l/L
806      CONTINUE
c        **
     YT6 ( 1 , 1 l = KE I (X ( 1 , 1 l l
       YT7 (1, 1 l=KERDCXC1, 1 l l _
       YT8 ( 1 , 1 l = KE ID (X ( 1 , 1 l l
       IOTC1 ,1 l=CMPLXCYT1 (1 ,1 l,YT1 (1 ,1 ll
       I ODT ( 1 , 1 l = CMPLX ( YT3 ( 1 , 1 l , YT4 ( 1 , 1 l l
       KOT ( 1 , 1 l = CMPLX ( YT5 ( 1 , 1 l , YT6 ( 1 , 1 l l
       KODT ( 1 , 1 l = CMPLX ( YT7 ( 1 , 1 l, YT8 ( 1 , 1 l l
       GO TO 100
1 01   I OT ( 1 , 1 l = I 0 (X ( 1 , 1 l l
       I ODT ( 1 , 1 l = I OD (X ( 1 , 1 l l
       KOT ( 1 , 1 l =KO (X ( 1 , 1 l l
       KODT ( 1 , 1 l = KOD (X ( 1 , 1 l l
1 00     CONTINUE
       IF CN.EQ.1 JGO TO 620
       DO 102 I=2,N
       IFCXCI,Il.GT.8lGO TO 103
       YT1 CI,Il=BERCXCI,Ill
       YT2 ( I , I l =BE I (X (I , I l l
       YT3 ( I , I l = BERD (X ( I , I l l
       YT4 ( I , I l =BE ID (X (I , I l l
       YT5 ( I , I l =KER (X ( I , I l l
       YT6 (I , I l = KE I (X ( I , I l l
       YT7 (I , I l =
KERD (X <I , I l l
       YT8 ( I , I l = KE ID< X ( I , I l l
        I OT ( I , I l = CMPLX ( YT1 ( I , I l , YT2 ( I , I l l
        I ODT ( I , I l = CMPLX ( YT3 (I , I l , YT4 (I , I l l
       KOTII,Il=CMPLXIYT51I,Il,YT61I,Ill
       KODT<I,Il=CMPLX<YT7CI,Il,YT81I,Ill
       GO TO 104
10 3    I OT ( I , I l =I 0 (X ( I , I l l
        I ODT ( I , I l = I OD (X ( I , I l l
       KOTII,Il=KOIXII,Ill
       KODT (I , I l = KOD (X (I , I l l
104    CONTINUE
        IF (X ( I -1 , I l • GT. 8 l GO TO 1 05
       YT1 ( ( I -1 l , I l = BER (X< ( I -1 l , I l l
       YT2 ( ( I -1 l , I l =BE I (X ( (I -1 l , I l l
       YT3 ( ( I -1 l , I l = BERD (X ( ( I -1 l , I l l
       YT4 ( ( I -1 l , I J =BE ID (X ( (I -1 l , I J l
       YTS ( <I -1 J , I J = KER (X ( ( I -1 J , I J l
       YT6 ( <I -1 J , I J = KE I (X ( (I -1 J , I ) )
       YT711I-1 l,Il=KERDCXIII-1 J,Ill
       YT8!(I-1 l,IJ=KEIDCXIII-1 l,IJJ
        I OT ( (I -1 l , I J = CMPLX <YT1 < <I -1 l , I J , Y
T2 ( (I -1 J , I J l
          I ODT ( <I -1 l , I l =CMPLX ( YT3 ( (I -1 J , I J , YT4 ( <I -1 J , I l J
         KOT ( (I -1 l , I J = CMPLX ( YT5 ( (I -1 l , I J, YT6 ( (I -1 J , I l l
         KODT ( (I -1 l , I l =CMPLX ( YT7 ( (I -1 J , I l , YT8 ( (I -1 J , I l l
         GO TO 102
1 05      I OT ( ( I -1 l , I J =I 0 <X ( ( I -1 J , I l J
          IODT«I-1 l,Il=IOD<X«I-1 l,Ill
        KOT ( CI -1 l , I l =KO (X ( CI -1 l , I l l
        KODT ( (I -1 l , I l = KOD (X ( (I -1 l , I l l
102     CONTINUE
c       *****************************************************************
C                               SOLUTION OF N SIMULTANEOUS EQUATIONS
c       *****************************************************************
c
c        ***************************************************************
C                            SETTING MATRIX OF COEFFICIENTS
c
C        **** SET ARRAY TO ZERO ****
c
        DO 900 I=1 ,MM
        DO 901 J=1 ,MN
        ACJ,Il=O
901     CONTINUE
900     CONTINUE
        A ( 1 , 1 l =I OT ( 1 , 1 l
        A ( 1 , 2 l = - I OT ( 1 , 2 l
        A ( 1 , 3 l =- KOT ( 1 , 2 l
        AC2 ,1 l=IODTC1 ,1 l*YC1 l
        A ( 2 , 2 l =-I ODT ( 1 , 2 l
        A C 2 , 3 l =- KODT ( 1 , 2 l
        A
CMN,NMl=IOTCN,Nl
     ACMN,MNl=KOTCN,Nl
     ACMN,MMl=1
     IFCN.LE.2lGO TO 902
     DO 902 I=2,M
     J=2*I-1
     JI=J-1
     A ( J , J I l =I OT ( I , I l
     A(J, CJI+1 l l=KOTCI ,I l
     A ( J , ( J I + 2 l l =-I OTC I , CI + 1 l l
     ACJ, (JI+3ll=-KOTCI, CI+1 ll
     A ( CJ + 1 l , J I l = I 0 DT ( I , I l * Y ( I l
     A ( ( J + 1 ) , ( J I + 1 ) ) = KODT ( I , I l * Y ( I l
     A ( ( J + 1 l , ( J I +2) l =-I ODT (I , (I+ 1 l l
     AC (J+1 l, CJI+3) )=-KODTCI, CI+1 l l
902  CONTINUE
     DO 700 I=1 ,MN
     DO 61 2 J = 1 , MN
C    ***** EVALUATE PIVOT COEFFICIENT *****
c
  IFCJ.EQ.IlGO TO 612
       AF ( J , I l =A ( J , I l I A ( I , I )
       DO 61 3 J J = 1 , MM
C      *****EVALUATE LINE (J-I) *****
c
        ACJ,JJl=ACJ,JJl-AFCJ,Il*ACI,JJl
613     CONTINUE
612     CONTINUE
        DO 61 ~ J • 1 , MN
        IFCJ.EQ.IlGO TO 614
        ACJ,Il•O
c       ***** THIS IS TO REDUCE ROUND-OFF ERROR ****
614     CONTINUE
700     CONTINUE
.C      ***** CALC. OF CONSTANTS CCil *****
c
        DO 620 I •1 ,MN
        CC I l • A CI , MM l I A (I , I l
 620    CONTINUE
        IFCN.NE.1 lGO TO 750
 c      ***************************************************************
 c          FOR N•1 THERE IS ONLY ONE CONSTANT GIVEN BY:-
        CC1 l•1 .O/IOTC1 ,1 l
        JDC1 l• C-SQRTC2.0l*IODTC1, 1 l*CC1 l l/SDC1 l
        GO TO 210
 c      ***************************************************************
 c
 c      ****************************************************************
 c
                           CALC. OF JDCil AND HCil
 c      ****************************************************************
 c
 750    HC1 l•CC1 l*IOTC1 ,1 l
        JDC1 l• C-SQRTC2.0l*IODTC1 ,1 l*CC1 l l/SDC1 l
        DO 200 I•2,M
        J•2*I-1
        HCI l: C ( J -1 l* I OT ( I , I l +CC J l *KOT CI , I l
        JD CI l • XD CI l * CC CJ -1 l *I ODT CI , I l +CC J l * KODT < I , I l l
 200    CONTINUE
        JDCN l •XDCN l *CC C2*N-2 l*IODTCN ,N l +C C2*N-1 l*KODTCN ,Nl l
         URCJP+1 l•C CAAF/CBBF+HCJPlll+CCFl/UO
 1500      CONTINUE
          IF CTTE.EQ.1 lGO TO 1520
         TTE• 1
         GO TO 1510
 1520      CONTINUE
 c      ****************************************************************
 c                                     CALC. OF IMPEDANCES ZCil
 c      ****************************************************************
 c
        DO 210 I • 1 , M
 21 0   CONTINUE
        ZCNl•RRCNl*JDCNl
        AA•AIMAGCZCNll
        B•REALCZCNl l
        KK•XR/CXR+XO+AAl
        ZB•KK*B
        ZA•CXO+AAl*KK
        ZAA•CMPLXCZB,ZAl
        ZZ•RF+ZAA
        WRITEC1,1755lZZ
 1755   FORMATC//'ZZ•',
2E14.5l
      ZBR=REAL(ZZl
      ZAI=AIMAG(ZZl
      ZM=CABS<ZZl
C     ZM IS MAGNITUDE OF FURNACE IMPEDANCE ZZ
c
C     ZAI=IMAGINARY PART OF ZZ
C     ZBR=REAL PART OF ZZ
c
      V1=(U1/ZMJ•<SQRT(ZBR/N1 ll
      I1=U1/((V1••2l•ZMl
      Q1=(I1••2l•<V1••2l•ZAI
      EFF=100•KK•BI<RFCR+KK•Bl
      RETURN
      END
c
$1 NSERT BESSEL
c
      COMPLEX FUNCTION DIV<Y,Xl
      COMPLEX Y,X
      REAL MODA,MODB,MODX
c
C     ****CALC. OF COMPLEX DIVISION *
***
C     **** CALC. OF Y/X ****
c
      A=REAL<Xl
      B=AIMAG<Xl
      MODA=ABS<Al
      MODB=ABS(Bl
      Z= 1 . 0
      IF<MODA.GT.MODBJGO TO 500
      IF<MODB.LT.0.1 JGO TO 501
502   IF(MODB.LT.1 lGO TO 510
      X=X/10
      MODB=MODB/10
      Z=Z•10
      IF<MODB.GT.1 JGO TO 502
      GO TO 510
501   IF<MODB.LT.0.1 lX=X•10
      Z=Z/10
      MODB=MODB•10
      IF<MODB.LT.0.1 lGO TO 501
      GO TO 510
500   IF<MODA.LT.0.1 lGO TO 503
                                  ------~------------
      END
c
c
       REAL*4 FUNCTION BEICXJ
       X2= CX/8. >**2
       X4=X2*X2
       X6=X2*X4
       Z=16*X2-113.77777774*X6
       X10=X6*X4
       Z=Z+72.81777742*X10
       X14=X10*X4
       Z=Z-10.56765779*X14
       X18=X14*X4
       Z=Z+0.52185615*X18
       X22=X18*X4
       Z=Z-0.01103667*X22
       X26=X22*X4
       Z=Z+0.00011346*X26
       BEI=Z
     . RETURN
       END
c
c
     Z=Z+124.23569650*X10
       X14=X10*X4
       Z=Z-21 .30060904*X14
       X18=X14*X4
       Z=Z+1 .1'l509064*X18
       X22=X18*X4
       Z=Z-0.02695875*X22
       X26=X22*X4
       Z=Z+0.00029532*X26
       KEI=Z
       RETURN
       END
c
c
       REAL*4 FUNCTION BERD!Xl
       X2=!X/8.l**2
     X4=X2*X2
       X6=X4*X2
       X10=X4*X6
       X14=X10*X4
       X18=X14*X4
       X22=X18*X4
       X26=X22*X4
       Z=-4*X2+14.22222222*X6
       Z=Z-6.06814810*X10+0.66047849*X14
       Z=Z-0.02609253*X18+0.00045957*X22
    Z=Z-0.00000394*X26
        BERD=Z*X
        RETURN
        END
c
c
       REAL*4 FUNCTION BEID!Xl
       X2=!X/8.l**2
        X4=X2*X2
        X8=X4*X4
        X12=X8*X4
        X16=X12*X4
        X20=X16*X4
        X24=X20*X4
        Z=0.5-10.66666666*X4
        Z=Z+11.37777772*X8-2.31167514*X12
        Z=Z+0.14677204*X16-0.00379386*X20
        Z=Z+0.00004609*X24
        BEID=Z*X
        RETURN
        END
c
c
        REAL*4 FUNCTION KERDCXl
        Y=X/2.
        X2=CX/8.l**2
        X4=X2*X2
    X6=X2*X4
         X10=X6*X4
         X14=X10*X4
         X18=X14*X4
         X22=X18*X4
         X26=X22*X4
         Z=-ALOGCYl*BERDCXl-BERCXl/X+0.25*3.1415926*BEIDCXl
         ZZ=-3.69113734*X2+21 .42034017*X6
         ZZ=ZZ-11 .36433272*X10+1 .41384780*X14
         ZZ=ZZ-0.06136358*X18+0.00116137*X22
         ZZ=ZZ-0.00001075*X26
         ZZ=ZZ*X
         KERD=Z+ZZ
         RETURN
         END
c
c
         REAL*4 FUNCTION KEIDCXl
        Y=X/2.
         X2= CX/8. >**2
         X4=X2*X2
         X8=X4*X4
         X12=X4*X8
         X16=X12*X4
         X20=X16*X4
         X24=X20*X4
         Z= -ALOGCYl*BEIDCXl-BEICXl/X-0.25*3.1415926*BERDCXl
         ZZ=0.21139217-13.39858846*X4
         ZZ=ZZ+19.41182758*X8-4.65950823*X12
      - ZZ=ZZ+0.33049424*X16-0.00926707*X20
         ZZ=ZZ+0.00011997*X24
         ZZ=ZZ*X
         KEID=Z+ZZ
         RETURN
         END
c
c
        COMPLEX FUNCTION THETACXl
        REAL*4 K,J
        Y=8./X
        Y2=Y*Y
        Y3=Y2*Y
        Y4=Y3*Y
        Y5=Y4*Y
        Y6=Y5*Y
    K=0.0110486*Y
        K=K-0.0000906*Y3
        K=K-0.0000252*Y4
        K=K-0.0000034*Y5
        K=K+0.0000006*Y6
        J=-0.0110485*Y-0.3926991
        J=J-0.0000901*Y3-0.0009765*Y2
        J=J+0.0000051*Y5+0.0000019*Y6
        THETA=CMPLXCK,Jl
        RETURN
        END
c
c
        COMPLEX FUNCTION THYCXl
        REAL*4 K,J
        Y2=Y*Y
        Y3=Y2*Y
        Y4=Y3*Y
        Y5=Y4*Y
        Y6=Y5*Y
        K=0.7071068-0.0625001*Y
        K=K-0.0013813*Y2+0.0000005*Y3
        K=K+0.0000346*Y4+0.0000117*Y5
        K=K+0.0000016*Y6
        J=0.7071068-0.0000001*Y
        J=J+0.0013811*Y2+0.0002452*Y3
        J=J+0.0000338*Y4-0.0000024*Y5
        J=J-0.0000032*Y6
        THY=CMPLXCK,Jl
        RETURN
        END
c
c
    COMPLEX FUNCTION KO!Xl
    COMPLEX CN,Z2;F,K,THETA
    Y=SQRT!3.1415926/!2*Xll
    W=SQRT( 1./2. l
    Z=W
    CN=CMPLX ( -W, -Z l
    K=THETA!-Xl
    Z2=CN*X+THETA!-Xl
    F=CEXP !Z2 l
    KO=Y*F
    RETURN
    END
c
c
    COMPLEX FUNCTION IO!Xl
    COMPLEX CN,ZX,F,FY,B,D,DF,THETA,KO
    REAL M,N
    Y=1/(SQRT<2*3.1415926*Xll
    W=SQRT!1./2. l
    CN=CMPLX!W,Zl
    ZX=CN*X+THETA!Xl
    F=CEXP!ZXl
    FY=F*Y
    B=KO!Xl
    M=AIMAG!Bl
    N=REAL!Bl
    D=CMPLX!-M,Nl
    DF=D/3.1415926
    IO=FY+DF
    RETURN
    END
c
c
    COMPLEX FUNCTION KOD!Xl
    COMPLEX CN,ZX,F,FY,FX,THETA,THY
    Y=SQRT!3.1415926/!2*Xll
    W=1/SQRT<2.0l
         Z=W
         CN= CMPLX ( -W, -Z l
         ZX=CN*X+THETA<-Xl
         F=CEXP<ZXl
         FY=F*Y
         FX= -FY*THY <-X l
         KOD=FX
         RETURN
         END
c
c
         COMPLEX FUNCTION IOD<Xl
         COMPLEX CN,ZX,F,FY,FX,B,D,DX,THETA,THY,KOD
         REAL*4 M,N
      Y=SQRT(1/(2*3.1415926*Xll
        W=SQRT<1./2. l
        Z=W
        CN=CMPLX<W,Zl
        ZX=CN*X+THETA<Xl
        F=CEXP<ZXl
        FY=F*Y
        FX=FY*THY<Xl
        B=KOD<Xl
        M=AIMAG(Bl
        N=REAL<Bl
        D= CMPLX C-M, N l
        DX=D/3.1415926
        IOD=FX+DX
        RETURN
        END
OK,
pr slab.fortran
c THE SLAB PROGRAM- this has been written by Ali Al-Shaikhli on his Ph.D
c                   project.It calculates the magnetic field ,power
c                   density and power distribution in a
c                   non magnetic rectangular slab.It also calculates the
c                   total power induced in the slab.
        print• (a,/l' ,'You will be asked to input values for HAG(the surface
       1 magnetic field strength'
parameter(mx=9 l
parameter(my=10l
parameter(mr=4l
slab
parameterCn=mx*myl
parameterCmac=mX**2l
parameter(nbc=1000l
parameterCmq=15l
         real hxCmxl,hyCmyl,muCnl,rotab,roxaCnl
         real rox
Cnl,royCnl,muo,pgCml,bCm),q(mql
      real pCml,mgCnl,hag,freq,max
      complex hCnl,acCmacl,bcCnbcl,ca,cb,cc
logical pringo
data hx/.075,.07,.039,.025,.018,.009,.006,.005,.003/
data sl
bthx/.5/
data slbthy/.1/
data rotab/2.824e-8/
c      FREQ is
    the supply rrequency CHzl
pringo=(q(7l.eq.O.Ol
100        xa=O.O
           do1001i=1,mx
           xa=xa
+hx Ci l
           i r ( hx ( i ) .1 t. 1 e -1 0) then
           print*,'HORIZONTAL MESHCesl IS/ARE TOO SMALLCless than 1e-10 ml'
           stop
           endir
1001       continue
         if(absCxa+xa-slbthxl.gt.slbthx*1e-5lthen
         print*,'TWICE THE SUM OF HORIZONTAL MESHES IS NOT CONSISTENT WITH
        1 DIMENSION SLBTHX'
         stop
         endir
           xa=O.O
           do 10015 i=1 ,my
           xa=xa+hyCil
           ir Chy ci l .1 t .1 e-1 o lthen
           print*,'VERTIC
           irCabsCxa+xa-slbthyl.gt.slbthy*1e-5lthen
           pri
nt*,'twice the sum of vertical meshes not consistent with
      1 dimension slbthy'
      ,print*,'TWICE THE SUM OF VERTICAL MESHES NOT CONSISTENT
      1 WITH DIMENSION SLBTHY'
       stop
       end if
        call setup(pg,b,h,hag,hx,hy,rox,roy,mu,ac,bc,
       1rotab,roxa,mg,pringo,max,
       2freq,q,mq,m,mx,my,mac,nbc,n,p)
         stop
         end
         real hx(mxl,hy(myl,mu(n),rotab,roxa(nl
         real rox(nl,royCnl,muo,pg(mJ,b(mJ,q(ml
         real p(ml,mg(nl,hag,freq,max
         complex h(nl,ac(macl,bc(nbcl,ca,cb,cc
logical pringo
10       format(10f7.2,6xl
15       format(9f8.0,2xl
20       format(9f8.0,xl
         mxpi =mx+1
         mslab=mxpi*my+mxpi
         if(m.gt.mslab.
        1 and.mac.ge.mx*mxl
        2 goto 101
print*,'
         xa=8*atan(1 .0)
         omega=xa*freq
         muo=xa*2e-7
          ifC.not.pringolgoto 1051
          print*,' TH
 1051     xa=2/omega
          do 1 055 j =my, 1 , -1
          ia=j*mX
          xb=xa/(hyCj l*hy(j))
          do 1054 i=mx,1,-1
          xc=xa/ Chx ( i l*hx Ci l)
          if Cia .eq.1 )then
          royCial=xb*roxCial
          else
          roy Cia l=xb* Croxa Cia l+roxa ( ia-1))
          end i f
          if ( i .eq.1 lroy Cia l =xb*roxa Cia l
          if Cj .eq.1 l goto 1052
          roxCial=xc*CroxaCial+roxaCia-mxll
          goto 1053
 1052     roxCial=xc*roxaCial
 1053     ia=ia
 -1
 1054     continue
 1055     continue
          ifC.not.pringolgoto 1084
          print*,'THE RESISTANCE VECTORS ROX AT EACH MODE (array roxl'
          print • C4 e 1 6 . 7 l ' , Crox Cj l , j = 1 , n l
          print*,'THE RESISTANCE VECTORS ROY AT EACH NO
DE (array ray l •
      print• (4e16.7l', (roy(j J ,j•1 ,nl
1084        j•n-mx
            do 1 087 i • 1 , j
            h(il•O.O
1087        continue
            do 1088 i•mx,j,mx
            h(i l•rox<i l*hag
1088        continue
            j : j +1
            do 1089 i•j ,n
            h ( i J=roy ( i )*hag
1089        continue
            h(nl=h(nl+rox(nl*hag
            if(.not.pringolgoto 2000
            print*,'THE NODE ARRAY MU'
            print • ( 4e 1 6. 7 l • , (mu (j l , j • 1 , n l
121         ic=my-1
            do 130 j=1 ,ic
            ia=j*mx
            ib=ia-mx
            do125i=1,mxmi
            ca=h(ia+i+1 l-hUb+i l
            cb=h(ia+i l-h<ib+i+1 l
            cc=<ca+cbl/hy(j)
            c b = ( c a- c b l I hx ( i l
            rox<ib+il=roxa<ib+il*(cc*conjg(ccl+cb*conjg(cbll*0.03125
125         continue
            ca=hag-h<ia l
            cb=h<ia+mxl-hag
            cc=(ca+cbl/hy(j)
            cb=(ca-cbl/hx<mxl
            rox<ial=roxa<ial*(Cc*conjg(ccl+cb*conjg(cbll*0.03125
·130        continue
            ia=mx*my-rnx
            do 140 i=1,mxmi
            ca=hag-h<ia+il
            cb=hag-h<ia+i+1 l
            cc=(ca+cbl/hy(myl
            cb= <ca-cbl/hx<i l
            rox<ia+il=roxa(ia+i l*(CC*Conjg(ccl+cb*conjg(cbll*0.03125
140         continue
            cc=<hag-h(mx*myll/hy<myl
            cb=(hag-hCmx*myll/hx<mxl
            rox(mx*myl=roxa<mx*mYl*(Cc*conjg(ccl+Cb*conjg(cbll*0.03125
            pg<1 l=rox(1 l
            pg(mxpil=rox<mxl
            pg(mxpi*my+1 l=rox <mx*my-mxmi l
            pg<mslabl=rox<mx*myl
            do 150 i=2,mx
            pg(i l=rox<i-1 l+rox(i l
            pg<mslab-mxpi+i l=rox Cmx*my-mx+i-1 l+rox<mx*my-mx+i l
150         continue
    do 160 j=2,my
      ia=mxpi*j-mxpi
      ib=mx*j-mx
      pg<ia+1 l=rox<ib+1 l+rox(ib+1-mxl
      pg(ia+mxpil=rox<ib+mxl+rox(ibl
      do 155 i=2,mx
      pg ( i a+ i l =ro x ( i b + i l +ro x ( i b +i -1 l +ro x ( i b + i - mx l +rox ( i b + i - mx- 1
155   continue
                         ------    -----~-~---------------------------------------------
160 continue
        i a =msl ab+1
        do 161 i=ia,m
        pg(i )=0.0
161     continue
c      the following section calculates the power density within each mesh
c      element of the quarter section slab
         ia=O
         ib=O
         do 1 64 j =1 , my
         do 162 i=1 ,mx
         b<ia+i )=4/hy<jl/hx<i l*rox<ib+i l
162      continue
         ia=ia+mxpi
         ib=ib+mx
         b(ial=O.O
164      continue
         ia=mslab-mx
         do 165 i=ia,mslab
         b(i )=0.0
165      continue
         print   15,<b<jl,j=1,9l
         print   15,(b(j),j=11,19l
         print   15, (b (j l ,j=21 ,29 l
         print   15,<b<jl,j=31,39l
         print   15,<b<jl,j=41,49l
         print   15,<b<jl,j=51,59l
         print   15,<b<jl,j=61,69l
         xa=O.O
         do 1665 i=1 ,mslab
         xa=xa+pg<il
1665     continue
         xa=4*xa
     OF THE MESH'
          print 10,(pgCjl,j=1,10l
          print 10, CpgCjl,j=11,20l
          print 10,CpgCjl,j=21,30l
          print _1 0 , ( pg ( j ) , j = 31 , ll 0 l
          print 10, (pgCj l ,j=ll1 ,50)
          print 10, Cpg(j l ,j=51 ,60)
          print 10, CpgCj l ,j=61 ,70l
          print 10, CpgCj l ,j=71 ,BOl
          print 10, Cpg(j l ,j=B1 ,90)
          print 1 0 , ( pg ( j l , j = 91 , 1 00 l
          print 10,(pg(jl,j=101,110l
            max=b(1 l
            do 166 k=1,m
            if(b(kl.gt.maxlthen
            max=b(
            do 167 k=1 ,m
            pCkl=(b(kl/maxl*100
 167        continue
            print    10 , ( p ( j l , j =1 , 9 l
            print    10 , ( p ( j l , j =1 1 , 19 )
            print    1 0, ( p ( j l , j = 21 , 2 9 l
            print    1 0, ( p ( j l , j = 31 , 3 9 )
            print    10, (p(j l ,j=ll1 ,49l
            print    10, (p(j l ,j=51 ,59l
            print    10,(p(jl,j=61,69l
            print    10,(p(jl,j=71,79l
            print    10,(p(jl ,j=B1,89l
            print    10,(pCjl,j=91,99l
            if(.not.pringolgoto 169
            print*,'MAGNETIC FIELD STRENGTHS WRITTEN AS COMPLEX NUMBERS'
            print' ( lld 1 6 • 6 l ' , ( h ( j l , j = 1 , n l
           print       2 0, ( mg (j l , j = 1 , 9 l
           print       20, ( mg ( j l , j = 1 0, 1 8 l
           print       20,(mg(jl,j=19,27l
           print       20,CmgCjl,j=28,36l
           print       20,Cmg!jl,j=37,45l
           print       20,Cmg(jl,j=46,54l
           print       20, <mg(j l,j=55,63l
           print       20,Cmg!jl,j=64,72l
           print       20, <mg(j l,j=73,81 l
           print       20,Cmg(jl,j=82,90l
           subroutine gauss(rx,ry,mu,h,nxy,a,na,b,nb,mx,myl
           complex hCnxyl,a(nal,b(nbl,ca
           real rx<nxyl,ryCnxyl,muCnxyl
logical ntgo
           mxmi=mx-1
           mxmx=mX*ffiX
           mxy=mx*mY
           if(na.ge.mxmx.
          1 and.nxy.ge.mxy.
          2 and.mx.gt.2.
          3 and.my.gt.2. lgoto 100
           print• (6(a,i6/ll',
          1 • mx = '
,mx
          2, 'my=' , my
          3,'mxy=' ,mxy
          4, 'nxy=' , nxy
          5, 'mxrnx=', mxmx
          6,'na=' ,na
           return
100        mxpi=mx+1
           mxt=2*mx
           mxmxmm=mxmx-mx
c     i
f ntgo is true then the subroutine gauss will not give the
c   right solution because the array b is not large enough
        ntgo=nb.lt.mxmx*<my-1 >
        i f <ntgo )then
        print*,'NBC IN THE MAIN PROGRAM IS NOT LARGE ENOUGH FOR THE MATRIX
       1 TO BE SOLVED'
        print*,'nbc=' ,nb
        stop
        end if
        kb=mx
        if <ntgol kb=O
        ka=-kb
  ib=i*mxpi
      a<ib+1 )=cmplx<rx<il+rx<i+1 )+ry<i+1 l,mu<i+1 ))
      a(ib+2l=-rx<i+1)
      a<ibl=-rx<il
240   continue
      a Ci b +mx +1 ) = - r x Cmx -1
      a(ib+mx+2)=cmplxCrx<mx-1 l+rx<mxl+ry(mx),mu(mx))
       ie=mxy-mxt+1
        do 600 i=1,ie,mx
        ia=1
        ka=ka+kb
        ca=1/a(1 >
        do 410 j=2,mx
        b(ka+j-1 >=ca*a(j)
410     continue
        b(ka+mxl=-ca*ry<i>
        h<i l=hCi )*ea
        ib=i
        do 430 J=mxpi,mxmx,mx
        do 420 k=1,
mxmi
         a (k+j-mx-1 l=a Ck+j )-a Cj l*b(ka+k)
420      continue
         a (j-1 l=-a (j l*b<ka+mx >
         ib=ib+1
         hCibl=hCibl-aCjl*h<i>
430      continue
  do 440 k=1,mx
       aCmxmxmm+kl=ryCil*b(ka+kl
c      print*,Ch(jl,j=1,90l
440    continue
       a Cmxmxl=a Cmxmxl +cmplx Cry ( i l+rx Ci +mx )+ry ( i +mx) ,mu Ci +mx))
       h ( i + mx ) = h Ci +mx ) +r y ( i ) * h ( i )
       ib=i+1
           ic=i +mxmi
           do 550 ia=ib,ic
           ka=ka+kb
           ca=1 /a (1 l
506        do 510 j=2,mx
           bCka+j-1 l=ca*a Cj)
510        continue
           bCka+mxl=-ca*ryCial
           hCial=hCiahca
           id=ia.
           do 530 j=mxpi,mxmx,mx
           do 520 k= 1 ,mxmi
           a Ck+j-mx-1 l=a Ck+j )-a (j l*b (ka+k)
520        continue
           aCj-1 l=-a <J l*bCka+mx)
           id
= i d+1
           hCidl=hCidl-aCjl*hCial
530        continue
           do 540 k=1 ,mx
           a(mxmxmm+kl=ryCial*bCka+kl
540        continue
           a(mxmxmml=aCmxmxmml-rxCia+mx-1
           a(mxmx-1 l=a(mxmx-1 l-rx(ia+mx-1
           a Cmxmx l=a (mxmx l+cmplx Cry ( i a) +rx ( ia+mx-1 ) +rx ( ia+mx l +ry Cia+mx)
          1 ,mu Cia+mxl)
545        hCmx+ial=hCmx+ial+ryCial*hCial
550        continue
600        continue
            ja=O
            i=2
            ic=mxpi
            id=mxy-1
            ie=mxy-mxmi
            do 640 ia=ie,id
            ca=1 /a Cja+i-1)
           do 610 j=i,mx
            aCj+jal=ca*a(j+jal
610         continue
            h Cia l =ca*h ( ia
           ib=ia
           do 630 j=ic,mxmx,mx
           do 620 k=i,mx
           a Ck+j-!+1 l=a Ck+j-i+1 l-a (j l*a (ja+k l
620        continue
           ib=ib+1
           hCibl=hCibl-a Cj l*hCia l
630        continue
           ja=ja+mx
       h(mxyl=h(mxyl/a(mxmx)
       i =mx
       ja=mxmx-mxt
       do 700 ia=id,ie,-1
       do 690 j=1 ,mx
       h(ia )=h( ia )-h(ia-i +1 +j )*a (ja+j)
690    continue
       i = i -1
       ja=ja-mx
700    continue
       id=mxy-mx
810    do 850 ia=id,1 ,-1
       do 830 j=1 ,mx
       h(ial=h(ial-h(ia+j)*b(ka+j)
830    continue
       ka=ka-kb
850    continue
       return
       end
C         IIRITEC1,12lN
C12       FORMATC1H ,3HN =,I8l
C         IIR I TE ( 1 , 61 l F
C61       FORMAT!1H ,11HFREQUENCY =,F15.4l
C         IIRITE!1,90lAIC
C90       FORMAT!1H ,14HCOIL CURRENT =,F9.3,4X,'AMP' l
C         IIRITE!1,62lROII
C62       FORMATC1H ,5HROII =,E15.7,4X,'OHM-M' l
C         IIRITE(1,63lUR
C63       FORMATC1H ,4HUR =,F15.3l
C           IIRITE!1,9l
C9        FORMAT!1H ,2X,'Z' ,13X,'CURRENT DENSITY' ,10X,'POIIER DENSITY')
C         IIRITEC1,7l
C7        FORMAT!1H ,1X,•---• ,12X,•---------------• ,10X,•-------------• l
          IND=O
            D!LSl=AL!LSl/(N!LSl-1 l
                 AF!LSl=AL!LSl/700.0
          Z=O.O
5         IND=IND+1
          XZCLS, INDl =Z
          SUM=O.O
                 NL=NCLSl
          DO 10 K=1 ,NL
          I=K-1
          SUM1=1 ./(C(LSl+Z**2+2.0*Z*D!LSl*I+DCLSl*DCLSl*I*Il
          SUM=SUM+SUM1
10        CONTINUE
       PI=3.14159265
         U0=4.0*PI/10000000.0
         AXA=UR!LSl*FlLSl
         XAX=PhUO*AXA
         DD!LSl=!ROW/XAXl**0.5
       AJ!LS,INDl=B!LSl*SUM
       AP!LS,INDl=ROW*AJ!LS,INDl*AJ!LS,INDl/!DD!LSll
       DIS=ABS!Zl
C      WRITE!1,6lDIS,AJ!LS,INDl,AP!LS,INDl
C6     FORMAT!1H ,F6.3,5X,F15.6,14X,E10.4l
         ROW=ROW+DR
       Z=Z-AF!LSl
       E=Z+AL!LSl
       XZ!LS,INDl=ABS!XZ!LS,INDll
       IF !E.GE.O.OlGO TO 5
1000       CONTINUE
   AL1=0.0
        YB=O.O
        YET=-9999.0
        YEV=-9999.0
            DO 1500 MM=1,LM
        DO 2000 J=1,IND
        IF!YET.LT.AP!MM,JllYET=AP!MM,Jl
        IF!YEV.LT.AJ!MM,JllYEV=AJ!MM,Jl
           IF!AL1 .LT.AL!MMllALI=AL<MMl
2000       CONTINUE
1500     CONTINUE
        IF!LH!LSl.EQ.3lCALL C1051N
        IF!LH!LSl.EQ.1 lCALL TREND
   '    IF!LH!LSl.EQ.2lCALL 55600
         CALL DEVPAP!210.0,297.0,1 l
       CALL MOVT02(0.0,0.0l
       CALL LINT02!0.0,297.0l
       CALL MOVT02!210.0,0.0l
       CALL LINT02!210.0,297.0l
      CALL WINDOW!2l
        CALL ERRMAX!100l
        CALL AXIPOS!I ,40.0,90.0,150.0,1 l
        CALL AXIPOS!1 ,40.0,90.0,180.0,21
 CALL DEVPAPC210.0,297.0,1 l
       CALL MOVT02CO.O,O.Ol
       CALL LINT02C0.0,297.0l
       CALL MOVT02C210.0,0.0l
       CALL LINT02C210.0,297.0l
       CALL WINDOWC2l
       CALL ERRMAXC100l
        CALL AXIPOSC1 ,40.0,90.0,150.0,1)
        CALL AXIPOSC1 ,40.0,90.0,180.0,2)
        CALL AXI SCA C1 , 9, 0. 0, AL 1 , 1 l
        CALL AXISCAC1 ,9,0.0,YET,2l
        CALL AXIDRAC2,1 ,1)
        CALL AXIDRAC-2,-1 ,2l
        CALL MOVT02C25.0,285.0l
           CALL CHAHOLC'PDCW/SQ ml*.' l
       'CALL MOVT02C190.0,78.0l
        CALL CHAHOLC'ZCml*.' l
         CALL MOVT02C40.0,70.0l
      CALL CHAHOLC'Fig.4 • . POWER DENSITY DISTRIBUTION ALONG THE LOAD *·' l
        CALL MOVT02C95.0,62.0l
     WRITE<1 ,66lAIC
66      FORMAT!1H ,14HCOIL CURRENT =,F15.3,4X,'Amp' l
        WRITE!1,62lROW
62      FORMAT!1H ,18HLOAD RESISTIVITY =,E10.4,4X,'Ohm.m• l
        WRITE(1 ,63lUR
63      FORMAT!1H ,14HPERMEABILITY =,F10.1 l
        AF=AL/100
16      ASS=O.O
        D=AL/ <N-1 l
        Z=O.O
5       SUM=O.O
        DO 10 K=1 ,N
        I=K-1
        SUM1=1 ./(C+Z**2+2.0*Z*D*I+D*D*I*Il
        SUM=SUM+SUM1
10      CONTINUE
        AJ=B*SUM
       AP=ROW*AJ*AJ/!2.0*DDl
         AAP=AP*AF
         AAP=ASS+AAP
         Z=Z-AF
         E=Z+AL
         IF <E.GE.O.OlGO TO 5
         IF <AAP-PWl28,25,25
28       N=N+1
         GO TO 16
25     . WRITE!1,22lN
22       FORMAT!1H ,17HNUMBER OF TURNS =,I10,4X,'Turns• l
         WRITE< 1 , 7 lAAP
7        FORMAT<1H ,22HPOWER PER UNIT WIDTH =,F15.3,4X,'Watts/meter• l
         ER=100.0*<AAP-PWl/PW
         WRITE!1 ,53lER
53       FORMAT!1H ,12HDIFFERENCE =,F5.2,'%' l
         WRITE!1 ,77l
77       FORMAT!1H ,•---------------------------------------------------• l
         STOP
         END
SLIST \1-SC-FIT
C    THE SYMBOLS ARE
c                         N   NUMBER OF POINTS
c                         X   DISTANCE FROM SINGLE TURN CMMl
c                         Y   VOLTAGES FROM SINGLE TURN CUVl
c                      All    CURRENT FOR SINGLE TURN CAMPl
c                      AI2    CURRENT FOR THE COIL CAMPl
c                       NT    NUMBER OF TURNS
c                         D   COIL PITCHE CMMl
c                       AL    LOAD LENGTH CMMl
c                      R0\1   LOAD RESISTIVITY COHM.Ml
c                      ROL    LOAD RADIUS OR SLAB'S PROBE LENGTH <MMl
c                       UR    LOAD RELATIVE PERMEABILITY
c                         F   FREQUENCY CHzl
c                       LH    CHOICE OF THE DEVICE,TREND=1 ,S5600=2,C1051N=3
c                      NSC    CHOICE OF SLAB=1 ,CYLINDER=2
c
c
       REAL BER,BEI,KER,KEI,BERD,BEID,KERD,KEID
       COMPLEX THETA,THY,KO,IO,KOD,IOD
       DIMENSION AZC999l,AVC999l,PDC999l,AJC999l
      1,SAJC999l,HRMC999l
 REAL*8 XC50l,Y(50l,AC50l,REF
        INTEGER N,NT
      READ CS,* lN, <X< I l, Y< I l, I= 1 ,N l ,AI 1 ,AI2 ,NT ,D,AL ,R0\1 ,ROL ,F, UR,NSC,LH
      CALL E02ACFCX,Y,N,A,5,REFl
C     \IR I TE ( 1 , 3 0 l <A ( I l , I =1 , 5 l
C30     FORMATC1H ,5X,E12.4l
        SLO= CYCN)-Y(N-1 l l/CXCNl-XCN-1 l l
        CON=Y<Nl-CSLO*X<Nll
          CL=D* <NT-1 l
          R=-<CL/2.0l
C     \IR I TE C1 , 9 l
C9      FORMATC1H ,'DISTANCE' ,7X,'VOLTAGE' l
C     \IRITEC1,90l
C90     FORMATC1H ,•--------• ,?X,'                        •l
        ML=1
1000      SUM=O.O
          RA=AI2/AI1
          DO 200 J=1,NT
          I=J-1
          ZA=R+CI*Dl
          Z=ABSCZAl
    IFCZ.GT.XCNllGO TO 300
        SUMO=AC1 l+CAC2l*Zl+CAC3l*CZ**2ll+CAC4l*(Z**3ll+CAC5l*<Z**4ll
         SAMO=O.O
          SUM1=SUMO+SAMO
         GO TO 400
300       SUM1=CSLO*Zl+CON
400     IFCSUM1 .LT.O.OlSUM1=0.0
          SUM= ( SUM+SUM1 l
200     CONTINUE
        SUM2=SUM*RA
       ZZ=R+ CCL/2 .0 J
         AZCMLl=ZZ
          AV CML J =SUM2
          ML=ML+1
         R=R+0.25
         IFCZZ.LE.ALJGO TO 1000
         K=ML-1
         DO 500 I=1 ,K
         AVCIJ=AVCIJ/1000000.0
500      CONTINUE
         IFCNSC.EQ.1JROL=ROL/1000.0
         IFCNSC.EQ.2JR=ROL
       PIY=3.141592654
       U0=4.0*PIY/10000000.0
       U=UO*UR
       11=2.0*PIY*F
         R=R/1000.0
       PER=2.0*R*PIY
       RL=ROII*PER
       D2=C2.0*RO\Il/CII*Ul
       D1=SQRTCD2l
     BA=R*SQRTC2.0l/D1
C        CALCULATION OF BESSEL FUNCTIONS
      IFCBA.GT.8lGO TO 5001
       A1=BERCBAl
      A2=BEICBAl
      A3=BERDCBAl
      A4=BEIDCBAl
      GO TO 5002
5001 A1=REALCIOCBAJl
      A2=AIMAGCIOCBAll
      A3=REALCIODCBAJl
      A4=AIMAGCIODCBAlJ
5002 CONTINUE
      DEN=CCA1 l*CA1 ll+CCA2l*CA2ll
      B1 = CCA3 >* CA3 J J + CCA4 l * CA4 J l
      B2 = CCA 1 l * CA3 J l + CCA2 l * CA4 J l
      C1=B1/DEN
      C2=B2/DEN
IFCNSC.EQ.2JGO TO 550
        DP=D1/C2.0*ROII*ROL*ROLJ
        DO 510 I= 1 , K
        PDCIJ=DP*AVCII*AVCIJ
510     CONTINUE
        GO TO 999
550       CONTINUE
      CPD=U*PIY*F*R*C2/2.0
   DO 11 0 I= 1 , K
       AJ<Il=AV<Il/RL
       SAJ<I l=AJ(I l*AJ(I l
       HRM<Il=D2*SAJ<Il/C1
       PD!Il=CPD*HRM!Il
11 0   CONTINUE
999        CONTINUE
c        WRITE ( 1 , 1001 l
C1001      FORMAT<1H ,2X,'DISTANCE' ,7X,'VOLTAGE' ,6X,'POWER DENSITY' l
c        WRITE <1 , 1 1 0 0 l
C1100      FORMAT<IH ,2X,'--------' ,7X,'-------• ,6X,•-------------' l
c        DO 120 I=l ,K
c        WRITE <1 , 1 30 l AZ <I l , AV< I l , PD <I l
C130        FORMAT<1H ,3X,F6.1 ,5X,E12.5,5X,E12.5l
C120           CONTINUE
YB=O.O
          YET=-9999.0
          YEV=-9999.0
          DO 2000 J=I,K
          IF!YET.LT.PD<JllYET=PD(Jl
          IF<YEV.LT.AV<JllYEV=AV(Jl
2000        CONTINUE
             YET= I .1 *YET
             YEV=1 .I*YEV
          IF<LH.EQ.3lCALL C1051N
          IF<LH.EQ.1 lCALL TREND
          IF<LH.EQ.2lCALL S5600
           CALL DEVPAP<210.0,297.0,1 l
         CALL MOVT02<0.0,0.0l
         CALL LINT02!0.0,297.0l
 CALL MOVT02!210.0,0.0l
       CALL LINT02!210.0,297.0l
      CALL WINDOW<2l
        CALL ERRMAX!IOOl
        CALL AXIPOS(I ,40.0,90.0,150.0,1 l
        CALL AXIPOS<I ,40.0,90.0,180.0,2)
        CALL AXISCA(I ,9,0.0,AL,1 l
        CALL AXISCA<I ,9,0.0,YEV,2l
        CALL AXIDRA!2,1,1 l
        CALL AXIDRA<-2,-1,2)
        CALL MOVT02(25.0,285.0l
          CALL CHAHOL<'V <VOLTl*.' l
        CALL MOVT02(182.0,78.0l
        CALL CHAHOL('Z(mml*.' l
       CALL GRACUR<AZ,AV,Kl
           CALL MOVT02!50.0,70.0l
        CALL CHAHOL!'SURFACE VOLTAGE DISTRIBUTION ALONG THE LOAD*·' l
          CALL MOVT02!95.0,60.0l
         IF<NSC.EQ.1 lCALL CHAHOL<'ALUMINUM SLAB*.' l
            IF<NSC.EQ.2lCALL CHAHOL('ALUMINUM CYLINDER*.' l
         CALL MOVT02(40.0,50.0l
        CALL CHAHOL('RESISTIVITY=2.8E-8 Ohm.m *·' l
        CALL MOVT02(140.0,50.0l
CALL CHAHOLC'RELATIVE PERMEABILITY=!*.' l
  CALL MOVT02C40.0,40.0l
  CALL CHAHOLC'N=    Turns   d=   mm    g=  mm *·' l
  CALL MOVT02C113.0,30.0l
  CALL LINT02C125.0,30.0l
  CALL MOVT02C126.0,30.0l
  CALL CHAHOLC'PREDICTION BY SUPERPOSITION*.' l
  CALL MOVT02C145.0,40.0l
  CALL CHAHOLC'o PRACTICAL READINGS*.' l
  CALL MOVT02C40.0,30.0l
   CALL CHAHOL('I=    Amp    F=    Hz*.'l
  CALL PICCLE
    CALL DEVEND
    IFCLH.EQ.3JCALL C1051N
    IFCLH.EQ.2lCALL S5600
    IFCLH.EQ.1 >CALL TREND
      CALL DEVPAPC210.0,297.0,1 l
   CALL MOVT02CO.O,O.Ol
   CALL LINT02C0.0,297.0l
   CALL MOVT02C210.0,0.0l
   CALL LINT02C210.0,297.0l
   CALL WINDOWC2l
   CALL ERRMAXC100l
    CALL AXIPOSC1 ,40.0,90.0,150.0,1 l
    CALL AXIPOSC1,40.0,90.0,180.0,2l
    CALL AXISCAC1,9,0.0,AL,1 l
    CALL AXISCAC1,9,0.0,YET,2l
    CALL AXIDRAC2,1 ,1 l
   CALL AXIDRAC-2,-1,2)
    CALL MOVT02C25.0,285.0l
       CALL CHAHOLC'PDCW/SQ ml*.' l
    CALL MOVT02C190.0,78.0l
    CALL CHAHOLC'ZCmml*.' l
    CALL GRACURCAZ,PD,Kl
     CALL MOVT02C50.0,70.0l
  CALL CHAHOLC'POWER DENSITY DISTRIBUTION ALONG THE LOAD *·' l
    CALL MOVT02C95.0,60.0l
   IFCNSC.EQ.1 >CALL CHAHOLC'ALUMINUM SLAB*.' l
      IFCNSC.EQ.2lCALL CHAHOLC'ALUMINUM CYLINDER*.' l
       CALL MOVT02Cqo.0,50.0l
      CALL CHAHOLC'RESISTIVITY=2.8E-8 Ohm.m *·' l
      CALL MOVT02(1q0.0,50.0l
      CALL CHAHOLC'RELATIVE PERMEABILITY=1*.' l
CALL MOVT02C113.0,30.0l
      CALL LINT02C125.0,30.0l
      CALL MOVT02C126.0,30.0l
      CALL CHAHOLC'PREOICTION BY SUPERPOSITION*.' l
      CALL MOVT02C1q5.0,qO.Ol
      CALL CHAHOLC'o PRACTICAL READINGS*.' l
      CALL MOVT02cqo.o,qO.Ol
      CALL CHAHOLC'N=      Turns     d=    mm g=   mm *·' l
        CALL MOVT02Cq0.0,30.0l
           CALL CHAHOLC'I=       Amp    F=    Hz*.'l
      CALL PICCLE
          IFCLH.NE.2lCALL DEVENO
      STOP
      END
$INSERT BESSEL
OK,
SLIST FITTING1
       REAL*B XC50l,YC50l,AC50l,REF,D,AI1 ,AI2,AL
       INTEGER N,NT
      READ C5 , * l N, CXC I l , Y CI l , I =1 , N) , AI 1 , AI 2 , NT, D, AL
      CALL E02ACFCX,Y,N,A,5,REFl
      WRITE C1 , 30 l CA CI l , I= 1 , 5 l
  30   FORMATC1H ,5X,E12.4l
       SLO= CYCNl-YCN-1 l l/CXCNl-XCN-1) l
       CON=YCNl-CSLO*XCNll
        CL=D*CNT-1 l
        R=-CCL/2.0l
      WRITE C1 , 9 l
9     FORMATC1H ,'DISTANCE• ,?X,'VOLTAGE' l
      WRITEC1 ,90l
90    FORMATC1H ,•--------• ,?X,'                                 'l
1000    SUM=O.O
        RA=AI2/AI1
         DO 200 J=1 ,NT
         I =J -1
           Z=ABSCZAl
           IFCZ.GT.XCNllGO TO 300
           S UMO =A C1 l + CA C2 l * Z l + CA C3 l * CZ* * 2 l l + CA C4 l * CZ**3 l l + CA C5 l * CZ* *4 l l
             SAMO=O.O
             SUM1 =SUMO+SAMO
             GO TO 400
300          SUM1=CSLO*Zl+CON
400        IFCSUM1 .LT.O.OlSUM1=0.0
             SUM= CSUM+SUM1 l
200        CONTINUE
           SUM2=SUM*RA
           ZZ=R+CCL/2.0)
         WRITEC1 ,6lZZ,SUM2
6        FORMATC1H ,F6.1,10X,E15.7l
           R=R+10.0
           IFCZZ.LE.ALlGO TO 1000
         CALL EXIT
         STOP
         END
OK,
SLIST \1-FC-TEMP
c                                      NON UNIFORM COIL
c
c                                    ------------------
c
C      THE SYMBOLS ARE
C                    CD            CONDUCTOR DIAMETER CMMl
C                    AG            AIR GAP CMMl
C                      N           NUMBER OF POINTS
C                      X           DISTANCE FROM SINGLE TURN (MMl
C                      Y           VOLTAGES FROM SINGLE TURN CUVl
C                   AI1            CURRENT FOR SINGLE TURN CAMP>
C                   AI2            CURRENT FOR THE COIL <AMPl
c                            NT NUMBER OF TURNS
c                          D<I> COIL PITCHES <MMl
c                            AL LOAD LENGTH <MMl
c                           R0\1 LOAD RESISTIVITY <OHM.Ml
c                              R LOAD RADIUS
c                            UR LOAD RELATIVE PERMEABILITY
c                              F FREQUENCY <Hzl
c                           THC THERMAL CONUCTIVITY CW/M Kl
c                           ALL NUMBER OF CYLINDERS
c                           SSS NUMBER OF DISCS
c
c
        REAL BER,BEI,KER,KEI,BERD,BEID,KERD,KEID
        COMPLEX THETA,THY,KO,IO,KOD,IOD
        DIMENSION Z<120l,AVC120l,PD<120l,D<50l,AXC120l,AJ<120l
       1 , SAJ ( 1 20 l , HRM ( 120 l , AR ( 1 20 l , HR C1 20 l , SRR ( 120 l , BR ( 120 l , ARM ( 120 l
       2 , BI ( 1 20 l , SBR ( 1 20 l , SB I ( 1 20 l , XHR ( 1 2 0 l , XHRZ ( 1 20 l , SXZ ( 1 20 l , I TIT ( 80 l
         REAL*B X(9l,YC9l,A(9l,REF,D
       1 ,AX,AZ,V,V1 ,RA
         REAL T
         COMMON T ( 11 , 11 l , WX ( 242 l
         DIMENSION \IXLC4l,WXHC4l,WYLC4l,WYHC4l
         DATA WXLC1 l,WXL(2l,WXLC3l,\IXLC4l/20.0,140.0,20.0,140.0/
         DATA WXH ( 1 l , WXH ( 2 l, \IXH ( 3 l , \IXH ( 4 l /1 40.0, 260.0, 1 40.0, 260.0/
         DATA WYLC1 l,WYLC2l,WYLC3l,\IYLC4l/110.0,110.0,20.0,20.0/
         DATA WYH ( 1 l , WYH ( 2 l , WYH ( 3 l , \IYH ( 4 l /200.0,200.0, 11 0. 0, 11 0. 0/
         READC5,*lCD,AG,N, CXCI l, Y(l l, 1=1 ,Nl ,All ,AI2,NT,AL, (DC I l, 1=1 ,NTl
       1 ,RO\I,R,UR,F,THC,ALL,SSS
         WRITE(1 ,3000l
3000     FORMATC1H ,34X,'W-PDC• l
         WRITE<1 ,3001 l
3001     FORMATC1H ,34X,'*****' l
         WRITEC1 ,3002 l
3002    FORMAT<1H ,//22X,'COIL AND LOAD SPECIFICATIONS' l
         WRITEC1 ,3003 l
3003     FORMATC1H ,21X,•----------------------------' l
         WRITE<1 ,6001 lCD,R
6001     FORMATC1H ,5X,'CONDUCTOR DIAMETER= • ,F6.1 ,1X,'MM',
       1 20X, 'LOAD RADIUS = ' , F4. 1 , 1 X, • MM' l
   WRITE Cl ,60021AG,ROW
6002 FORMATIIH ,16X,'AIR GAP = ',F6.1,1X,'MM'
      1,7X,'LOAD RESISTIVITY= ',E9.4,1X,'OHM.M' I
       WRITE Cl ,60031NT,UR
6003 FORMATIIH ,BX,'NUMBER OF TURNS= • ,I5,1X,'TURNS'
      1,15X,'LOAD PERMEABILITY = ',F6.01
          WRITE Cl ,60041AI2,F
6004 FORMATIIH ,11X,'COIL CURRENT= ',F6.0,1X,'AMP' ,20X,'FREQUENCY =
      1,F6.0,1X, 'Hz' I
          WRITEI1 ,46501THC
4650      FORMATI1H ,3X,'THERMAL CONDUCTIVITY= • ,F5.1 ,IX,'W/M K' I
       WRITEI1 ,6005 I
6005 FORMATI1H ,//30X,'THE RESULTS' I
         WRITEI1,60061
6006      FORMATI1H ,29X,•-----------',/3Xl
       CALL E02ACFIX,Y,N,A,5,REFl
       WRITEI1,10l
 10       FORMATIIH ,14X,'THE CONSTANTS OF THE SINGLE TURN EQUATION' l
         WRITEI1,77l
77       FORMATIIH ,14X,'-----------------------------------------')
       WRITE I 1 , 3 0 lA I 1 l , A I 2 l , A I 3 l , A I 4 I , A I 5 l
 30      FORMATIIH ,51E10.3,4Xll
         SLO= IYINl-YIN-1 l l/ IXINl-XIN-1 l l
         CON=YINI-ISLO*XINll
         RA=AI2/AI1
           S=AL/SSS
         AK=AL/S
         K=AK+I
         PIY=3.141592654
         U0=4.0*PIY/10000000.0
         U=UO*UR
         W=2.0*PIY*F
           R=R/1000.0
         PER=2.0*R*PIY
         RL=ROW*PER
         D2=12.0*ROYl/IW*Ul
   D1=SQRTID2l
      BA=R*SQRTI2.0l/D1
C       CALCULATION OF BESSEL FUNCTIONS
      IFIBA.GT.BlGO TO 5001
       AI=BERIBAl
      A2=BEIIBAl
      A3=BERDIBAl
      A4=BEIDCBAl
      GO TO 5002
5001 AI=REALCIOIBAll
      A2=AIMAGIIOIBAll
    A3=REALIIODIBAll
      A4=AIMAGIIODIBAll
5002 CONTINUE
      DEN= I IAI >* IAI l l+ I IA2 >* IA2 l l
      B1 = I I A3 >* I A3 l l + I I A4 l * I A4 l l
      82 = I I A1 l * I A3 l l + I I A2 l * I A4 l l
      CI=BI/DEN
      C2=B2/DEN
     AX C1 J =DC 1 J
         DO 80 I=2,NT
         AXCI l=AXCI-1 l+DCI J
80       CONTINUE
      WRITEC1 ,9000)
9000   FORMATC1H ,//28X,'THE COIL PITCHES' J
       WRITEC1 ,9001 J
9001   FORMATC1H ,27X,'----------------' J
       WRITE C1 , 40 J
40     FORMATC1H ,20X,'DISTANCE',10X,•THE COIL PITCHES' J
       DO 20 I= 1 , NT
      WRITE C1 , 6 0 J AX CI J , I , D CI J
60    FORMATC1H ,20X,F6.1,14X,2HDC,I2,3HJ =,F6.1J
20       CONTINUE
 ZC1 l=O.O
       DO 70 I=2,K
       ZCI l=ZCI-1 l+S
70     CONTINUE
       DO 90 I=1 ,K
       V1=0.0
       DO 1 00 I I = 1 , NT
       ABC=ZCIJ-AXCIIJ
       AZ=ABSCABCJ
       IFCAZ-XCNJJ200,200,300
200    V=AC1 l+CAC2l*AZJ+CAC3l*AZ**2l+CAC4l*AZ**3l
      *+CAC5l*AZ**4l
       GO TO 400
300    V=CSLO*AZJ+CON
400    IFCV.LT.O.OJV=O.O
       V1=V1+V
100    CONTINUE
      AVCIJ=V1*RA/1000000.0
90    CONTINUE
      CPD=U*PIY*F*R*C2/2.0
      DO 110 I=1 ,K
      AJCIJ=AVCIJ/RL
      SAJCI l=AJCI hAJCI l
      HRMCIJ=D2*SAJCIJ/C1
      PDCIJ=CPD*HRMCIJ
110   CONTINUE
c     WRITEC1 ,9003J
C9003   FORMATC1H ,//23X,•VOLTAGE AND POWER DENSITY' J
c     WRITEC1 ,9004J
C9004 FORMAT ( 1 H , 22X, • ------------------------- • J
c       WRITEC1 ,1000)
C1000 FORMATC1H ,2X,'DISTANCE' ,7X,'VOLTAGE' ,6X,'POWER DENSITY' J
c
C1100     FORMAT<1H ,2X,•--------• ,7X,'-------' ,6X,•-------------' l
c         DO 120 I=1 ,K
c          WRITE ( 1 , 13 0 l Z <I l , AV( I l , PD <I l
C130      FORMAT<1H ,3X,F6.1 ,5X,E12.5,5X,E12.5l
C120       CONTINUE
           YB=O.O
           YET=-9.999 .0
           YEV=-9999.0
           DO 2000 J=1 ,K
           IF<YET.LT.PD<JllYET=PD(J)
           IF<YEV.LT.AV<JllYEV=AV(Jl
2000       CONTINUE
        LLL=ALL+1
        DR=R/ALL
        H1=R/(2.0*THCl
        H2=BA/R
         . H3=B2*BA
        AR <1 l =0. 0
   HR<1l=O.O
       SRR <1 l =0. 0
       DO 4100 I=2,LLL
       AR <I l =AR <I - 1 l +OR
       HR <I l =H2 * AR <I l
       SRR <I l = <AR <I h AR <I l l I ( R* R l
41 00 CONTINUE
C      CALCULATIONS OF BESSEL FUNCTIONS
       DO 4200 I=1 ,LLL
       IF<HR<Il.GT.8lGO TO 4250
       BR<Il=BER<HR<Ill
       BI ( I l =BE I <HR <I l l
       GO TO 4200
4250 BR!Il=REAL<IO<HR!Illl
       BI <I l=AIMAG<IO<HR<I l l l
4200 CONTINUE
       DO 4400 I=1,LLL
       SBR<I l=BR<I l*BR<I l
       SBI <I l=BI (I l*BI <I l
      XHRZ<Il=XHR<Il/H3
      SXZ<Il=H1*<SRR<Il-XHRZ<Ill
4400 CONTINUE
       DO 4050 I=1 ,LLL
       ARM<Il=1000.0*AR<Il
4050   CONTINUE
c      WRITE!1,4000l
C4000   FORMAT!1H ,//23X,'TEMPERATURE DISTRIBUTION' l
c      WRITE<1 ,4350)
C4350   FORMAT<1H ,22X, '--------------------~---' l
      DO 4500 J=1 ,K
     CALL MOVT02<20.0,15.0l
       CALL CHAARR<ITIT,80,1l
       CALL DATE
       CALL ISOFRA<Ol
       IF<IVIEY.EQ.4lGO TO 85
       CALL I SOPRJ ( K, 1 • 0, ANX, LLL, 1 • 0, ANY, T, I VI EY, NY, YX l
89     CALL PICCLE
       PRINT 82
82     FORMAT('1=PLOT AGAIN, 2=STOP' l
       READ ( 1 , * l I PLOT
       IF<IPLOT.EQ.1 lGO TO 63
       IF<IPLOT.EQ.2lGO TO 99
85     DO 11 I= 1 , 4
       IVIEY=I-1
       CALL YIND02(YXL<I>,YXHCil,YYLCil,YYHCill
11     CALL ISOPRJ<K,1.0,ANX,LLL,1.0,ANY,T,IVIEY,NY,YXl
     GO TO 89
99    CALL DEVEND
      STOP
      END
          SUBROUTINE DATE
          INTEGER ADATE<8l
           CALL DATE$A<ADATEl
          CALL MOVT02(240.0,200.0l
            CALL CHAARR<ADATE,8,2l
        RETURN
          END
$INSERT BESSEL
OK,
                          ALl
      .   '                                                                                                                     i>l
                                                                                                                                                Lnl;;_;..
                                                                                                                                                                        ';'                            \.    !"
                                                                                                                                                                                                            ·. :,_ ·
  ' " ' .'1/'Y ._, 1·                                 A.r:.r-1. AL-::-::h.J.ikhli                             UHll_
  '('L: ;, ,:,1,.~;:,! '... \ I '                       _                                                     j);\ I ur.~
                                                                                                                                                ~ ;·     ,i ;_;,: ·•-         t·:-.,,.:;~_:\\1:-.L\TlU:-:
 ~-'1~<0~~~-k~~~~~~~~-\;lcU~'<~~---r~~~~~--~~~~~~-~[)~A~f~l-~~~
: LOC          COD~:           K!.Y                    (..<li\1Mt'Nr~                 l_OC      CODE.       KE.Y            COMMf:·r\tfs--
  ADR          KODE            lAST£                   flt.r,U~H.IUJNGCN              Af.JH     KODE        TASTE           tiCMERKUNGEN
  ADR          CODF            TOUCHF                  ('OMMENlAHlF<;                 ADR       CODE        TOUCHE          COMMENTAtRF''•
1 __ u _____                 LRN _ ...                                  __ __         5    000              0     3                   I
     1   76                  2ND Lbi                                                  5     65                  X
 ---~     "      ll                     A                                             5     03                  3
          3      42                 STO                                               5 .•  93
   --:;-boo                         cil 1                       z                     5     02                  2
          •, ~-- 91                 R/S                                               6     04                4
  ····---(J    .. 4-3
                                    RCL                                               6     Ol                1
  -;.boo'                           Ill 4                                             6          06           6
          !! 33-                        x2                                            6          52           EE
       0     85                         +                                             6          05           5
  ---1()' -- 4 3                    RCL                                               6          94           +I-
      1:       bo o                 0 1                                               6          95
 -- y;           33                  x2                                               6          33           x2
    lJ
1--,C   -·
           95                                                                         6          65           X
    1·1    34                       ;;:                                               6          02           2
 -- 1;, 65                              X                                             7          93
    le 43                           RCL                                               7          02          2
 l?- 00(                     - ·o     6                        a                      7          07          7
                                     ,--        -··                               17             06          6
                                    RCL                                               7 ·\       0~          5
                             --it       4                      h                      7 .•       02          2
                                    X                          X                      7 .•       08          8
                                    RCL                 a/cosa                        7     06               6
                                    0 1                                               7     52               EE
                                    X
                                      2                                               7 .. 04                4
                                                        '2 .. - ·-· -
                                    =                   z   a/cosa                    8 <: 94                +I-
                                    +                              +                  8    _95
                                    RCL                                               8          41          R/S
                                    0       4                                                                LRN
                              . x2
                                                                                                                  B         trhe values
    30
 -----
                 75                                                                                          STO              of the con j-,
     31          43                 RCL                                                                         0     2       stants and'
~__3 CX)c 5                         0 5                                                                         I             current to
     33          33                     .)(2                                                                 STO              be stored
     34          95                                     h 2 -a 2 +     a                                     (2\ 3
1-~---'-j--=:.--'---         ·-      - ..                        __ -=co"-=-saooi'l                                           here,outsi f>
~-~~-5                            l/X                   1/ :-- - -                                 ...          h             the progra
 36 42                              STO                                                                      STO              so that it
                                                  ··      · 2 az 2
 ~!: 00--Cf,                        0       7
                                                     fl-/ (he eo sa                                          (2\ 4           does not
     3S         43                  RCL                                                                         a            affect the
_:__]~         00 c                 0 4                                                     •                STO             program wh f'
 __ 51_0 33                          x2                                                                      (2\ 5           the change
    4 1 75                                                                                                      a
    4 2 43
        --
                                  . - ---
                                    RCL                                                   ··>                STO
    43 00 0                         0 5                                                     .                0 6
-44 -33                                 x2
-------        -----~.
                                                  .
   4;           95
   4 r;         34             rx
 _3 ;~         _65_           ... X
 __40 43_                         HCL                                       2
                                                               2        z
    4 9 00 0·                     0 7                   l(h +a--                                                                             f-_iL___ _l_ _ ___[ _ _ _ _ _ _ _ _ _
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~ROGHAMt.~EUH                                                                                             DA ll                                    1-'EClLLE                      P!~     i'Etie;E.. ~;-,1.'-</-.'fl(L'~
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                                                                                                                                                                                                       COMMl.Ni1~'
                                                                                                                                                                                                       HEMF..RIO:UNGEN
  ADR        CODF         TOUCHE                 COMMENTA1RE5               ADR          CODE          TOUCHE            lOMMfNTA!RFS             r~DR               CODE            TOUCHE            COMMENTAIRE
        tJ   _____ __ LRN ___                    -------·-----             ___ _5 -~- __ 43...       ___ RCL __                                   11  _76 2ND Lbl
        1    _76. _2ND__ Lbl .                         v-.                    5 ,. 02 0                      27                                   11. 15      E                                                PD
        2    _1~-      ]li!? .. l<.                   __,1_                   5 i 95                                                              11   43    RCL
        3     42            STO                                             _ _?_H _42_                _STO                                       11 01 0     14
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        s     91       -R/s.....                            .....               6ll       os              5
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                                                                                                                                                                                         LRN
                                        1
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~- __91_                    R/S                                  . .            6S        43             RCL                                                                             BF             the resist vil
1-1...2_ _Q2. ...... 2.                      ..                             ..6/ 01 0               ..... 12                                                                             STO            of the mat rL
   12 43. --·-'--·-· --· · - - - -                                         _6..!_ _95_ r---"=··                                                                                          28             to. be stor
   13 08 ____8_ _ - · - - - -                                                   6   835 _J/X ......                                                                                      CF             outside. th
e-l-.:_~Q.l ... ____ 1___           ___ ·--                                1-6~      _65_ ... X                                                                                          STO            program
~r-93.. c__....1._____ -····-·· ... ...                                     _..]_n . 43__     _RCL                                                                                       27
   16 04             4       ---···--··----                                1-_21_. oLQ __26 _____ _                                       . ··-
                                                                                                                                                         .                               p
    17 91.. .        7    -- - - · - - · - - -                              _73. _9.5.... ......E-----···                                                                                STD
      1e       o7__ _         7____ --·----------                          1--1__:;_ __3..4__ _          lx                                                                                  26.
      19       03    3    - ·--·-········ ---·                             f..-?.~. -~~- -~-0                      ...
                                                                                                                                                         :_l
                                                                                                                                                         •...
      20       65    X                                                          7 5 01 0                 l3                           0
   -z 1        4j __:R~E.       ::~.~~--==                                 -·'[(,         4~- .. RCL · ·.·
      22 0 01                 P.).... f....-··-···--··                      .. 7 /01.0.                  11 .                                            ··.                                                                    ..
      2 3 95                                                                _18_ ..]3_ _ x 2                                                                                                                                ......
    4
    4 9
        a ;;                  ~~i- ------·-·- i6-~b!io
                       ··-··-··-- ... ·-·-·---·                                           R/S
                                                                                                    ..   J2
                                                                                                                                                          ·.
   5 1        43           RCL                                              10      r,    14             D
                                     -·-·-----------
   5 2        2 0             29                                            10 ;          43         RCL
   5 3
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              95
              85
                                                                            10           rn    0     13
                                                                                                                                                                    TLXAS INSIIHJMLNTS
                              +                                                 1o        91         R/S
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                                                                                                                                                     1\L   v    cnt::\r-.~t   N 1   s
                                                                                                                                                      Ti\SH:    BCMEl~KVNGEN
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i---'.'~1----        __I,!lN___            .. -~--~-   -~5-~              7r          SilR                                    J,l        _'!~
_ : .J.§~- __l_!!!>_Lb1                                 5   i             25          CLR                                     11        33
____! _],1__ --~~!'_____                                5                 43          ll::L                                  ~~i1       34
_ _2__ 42_ ~-!:i'rQ                                    S        1 Xl       0I         ~ 1                                      11       42              STD
_4_ oo_ o_ ___ P_l                                     _5 "~ 75 ~ _ -:                                                        11       D1 <5            15
---~ _9_1 -~               R(S _                       6 ''   43     RCL                                       ~~ ~--~~   ~-  11        7l              SilR
     '•    43              RCL                         6     D2 <1 ~ 21                                                        11       25              CLR
=+,oD_ _o___---__~0x-_ _• 4 __ ___                   _6 2   95                                                                11    43                  RCL
       , 33         _ ~ 2 ~~~                       _§:':~_)3       x2                                                        11   DD                   01
 _2~ _ls_ -~-~- ____                                  6 ·'~ 34                        rx              . ---~- . _______       u     75
e-J-r:>_ 4}_           RCL _ ~ _                      6     42    . STO                                                       12    43                  RCL
 __ ;I-1_ ~--D       ~ p2 5 ~~                      . 6 ' Dl c 5 ~15                                                          12 D2 D)                  26
 ~~~ 3_3__ ---~)(__                                   6 / __71_ . __$BR _                             _____ __ ___            12    95
J~ 95_______=:__~- __                                 6 >< 25       CLR                                                       12    33                  x2                               .
                         =
                     -:xr-                               9 :! 43         RCL                                      14      34                            rx
    4o 34 ~ - -~rx--~-                 -- ---          ~9    ~'~ DO  c   9l  1                                    14      42     STD
                                                         9 c 75          ~                                        15 D1 0        15
~ 42 ----sTo- -                                          9 G 25          RCL                                      15. 7l         SilR
    42 1 Q<               15    -- f.--- - ---- - _9 _, o2~<:ij                   -24___
                                                                                       ------~-~---~
                                                                                                                  15·     25     CLR
 -~- -               f--- ---
  _j~ ;2!._ _I'BR____ -                                  9 [; 95                                                  15 ' 43        RCL
~ 25                     CLR                             9 ~- 33                                                  15 . 00 0     .)!\ 1
    45 43                 RCL                            100 34                                                   155 75
  46      oo<
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                      i~l'fl -                           1D1      42     STD                                      15      25     RCL
                    ----H·-~- -----
                                                       _102      01  Op  15                                       15    D2   0   29
    48 43                 RCL                            1o:• 71         SBR                                      15 .. 95
                c
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                                                         10-: 25
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    50         95
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                                                                         RCL
                                                                                                                                                                               ,;            I
 19771r-A•S Instruments
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                                                                                                                                                                     1 AS T E        BCMEHKUNGEN
                                                                                                                                                                     TOUCHE          C'.OMMENTAIRE
  Hi~:-    .3L          _7x __ ___                               2J : 2 2 7        ___n;v LR
  16 --1
 r----      42            STO                                          ''               LRN
'1_():     Q1 0           15
'16:l         71    ___SBR                                             (!               S_                Constant                                                                                         -
                                                                       -;
 16 ~         25         CLR                                                            STO
 16 ,; 43                RCL                                           ,,      -
                                                                                        ~2
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 16,- 33                 x2                                            :.:              STO
 16" 65               X                                                :1
                                                                                      -~ 3
 16 u 43              RCL                                              _,
                                                                                       h                 _Height_
 17<j 01 Qj 10                                                                        STO
.!:._7 ' 95                                                                           ~ 4
 17l 42               STO                                                            _a                  Cond, Radi
 17 3 01 0            11                                                           ____STO __
             .. --·-------             ---------~---
 :!} 4_ 14 3_         RCL       __ _                                                    ~     5
                                                                       ,_,
1-2.:"     01 0          14                                                            _a                Constant                     ---
 17 6    91     ____R/_'§_____ ____H _                                                __ STO
1
~?_ _76_- 2l'lJ?_~b1 -                                           ----"-
                                                                       2
                                                                                        ~ 6-             ------ - ------
£_B__1?_ ---~---                   PD                                  3
                                                                 ------1 - -
                                                                                        pp               T]1__e _c_OJ15t<!J
17 9 43             RCL                                                4                STO              of the PD
__:_:c -- --'--
18 0 01 0           11                                             ---- c               11
181 91              R/S                                                 f·              Zl               he distanc
                                                                         ;
T82 7&                  2ND Lbl        The starti                                       STO               of the
 ·----- -----·-
~_?_5                     CLR          of the                                           19               turns from
118 4 43      R,CL   sul:u;~_t:i,rl_"_                                                  22               t!'_.,__z er2 ___
1185 00 () -·-a-;4~-                                                   0               _STQ____ point .C'l'l _
~_8 6 1--_3 3_ 1--- _X                                                                  20----- the load
                                                                                   ··-··--.
                                                                                                                                          -
 18 7 85             +                                                                  Z3
 -----·---·-----
1188 43              RCL                                                                STO
 189101            o,    15
                         x2
                                   I      -
                                                                       4                21
 19;;1· 33 '                                                                            Z4
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                                                             I
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                                                                       0
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 200          55                                                       '-•
201 25··                 RCL                                                                                                              --
--
- -----
                                                                                                                                                                                 •uuCoH.•,   (.0MI\INFE~
 IHJ>MOvt:Ht:NTS IN 11ft:   DESIGN Of' INOU<...TION BlLI.Y.T
 llf!';\TERS.
that the extrusion speed of the ram and product                              '                         '
temperature through the die be constant throuqhout       Table 1:     The pitches of the coil
the extrusion cycle, thus avoiding continual adjust-
~nt of the press,   This isothermal effect may be
•      CONCLUSION                                                 10     DAVIES, E J 'Induction Heating Handbook'
                                                                         and         (HcGraw-Hill, 1979)
The practical investiqations show that the super-                        SIMPSON, P1
position method can be applied on loads of different
ahapea.      The non-uniform flow of current through              APPENDIX
the conductor. did not affect the applicability of
this .ethod within the ranqe of the teat carried                  The Relationship Between Surface Current and Powec
out.                                                              Densities and the Temperature of the Load
Applying this .-thod to a practical application,                  In induction heating, the heat flows almost in the
i.e. the production of             a linear power density taper   radial direction only because of the high power
along the length of a billet, proved highly success-              u.ually uaed in this kind of heatin9. The one
ful.                                                              dimension solution to the problea of heat flow in a
                                                                  slab and cylinder heated by induction is shown be1dw.
The computer proqr. . h relatively BIMll, i.e. it
does not need a larqe atoraqe capacity or conatne                 a)     Slab heated fro. one side
long computing time. This ..ethod is therefore suit-
able for desk top computers which are cheap and                   The temperature {a) at a depth {x) in a slab of total
readily avialable within even the smallest             c~ny.      thickness {b) heated on ono side at a constant power
                                                                  density {PO) is 9iven by (9j.
Although the results are shown for tests on auaina
frequency and non-magnetic loads, further work now
being carried out to assess the effect of frequency,                   a •                                                  1       4    <
proxilllity of laainations packs and non•uniform                                                                        - l- 2           L
                                                                                                                                    •   n•l
current distribution in conductor. Work also vas
carried out on magnetic material to assess the
effect of the permeability and the possible use of                                                                                             (lA.)
this method in the oft-the-bar forging applications.
REFERENCES
1      BAKER, R M:      'Heating of Non-magnetic Electric         The firse term is the final temperature of the slab.
                        Conductors by Magnetic Induction -        The S\mll!l&tion term is the transient temperature rise
                        Longitudinal Flux 1 , AIEE Tra.ns,        and can be ignored for valued of t 31- 0.25. The tra.n-
                        1944, ~· pp 273-279.                      aient period represents a small part of the heatinq
2   BAKER, R M:         'Design and calculation of Induction      :r~~i!:r.~=~~~e 1 ~ ::r~::~==s~t~ 7 ;~c;~ds                                 foe
                        Heating Coils', AIEE Trans, 1957,         Equation        {i~)   then   s~plifies          to
                        ,!!   (pt II), pp 31-40.
•   VA.UGHAN, J T
    and
                                'Design of Induction Heating
                               Coils for cylindrical Non-         i.e. the temperature at any depth (x} is a function
    WILLIA.MSON, J        W;   .aqnetic Loads 1 , AIEE Trans,     of the surface power density which is in turn a
                               1945, !!· pp 587-592.              function of the surface current density llo) and
                                                                  it is given by equation (4A.} below
'   VAUGHA.N, J T
    and
                      'Design of Induction Heating
                      Coils foLCylindrical Magnetic               - pD •
                                                                                  2 •
                                                                                 .J 2
    WILLIA.MSON, J W: Loads', AIEE Trans, 1946, 65,
                               pp 887-892   0             -
                                                                  This shows that the temperature at any depth is a
                                                                  function of the surface current density •
•   REICHERT,      K: 'A    Nwnerical Hethod for calculating
                         Induction Heating Installations',        b)     cylinder of radius R
                         Electrowarme International, 26 (4),
                         1968, pp 113-123, Englhh Tri'iisla-      The temperature (a) at a radius (r) in a cylinder
                         tion by the Electricity Council,         of radius (R) heated by induction la 9iven by [9]:
                         overseas Activities, Translation
                         Service.
7   DEHERDASH, N A, 'Solution of Eddy CUrrent
                                                                       a • (PO)~ {2t + r    ,-
                                                                                            2R
                                                                                                2
                     1
  [ber (•) ber           (•) + bet (•) bei 1 (•) J
                                                                (7A)
             bn
                    12
                         (•) +   bep (•)
Where   ..    lz    R
               '
As in the case of slab the temperature at any radius
is a function of the surface current density.
                                                                                            ...
                                                                                             I
               I         2         l
                                                                                [          -a••-
                                               N
        . .   <!>
                         ~        <!> 0 0
                                   :
                                               <!>0                                                  -
        -1•...-!
        0
                         0
                         0         0            :                                   ~
                                                                                                                  24
        :--- z, ____,              0
                                   0
                                                0
                                                0
                                   0
                    •• _.,                      ''
                              ... _____..,                                      ,./
                                   0
        I
        0
          --                                                            water po.
        0
                                                                                        ~
                                                                                   ,/
                                                                        sol.i4 copper
          •
                                                                                                 "'
                                                                                                 160
                                                                                            ~140
                                                                                            :!
                                                                                            ~120
                                                                                                 "
                                       eo           120             160 •       200
                                        Disto.nce otonq   t~   load. z. (!ne)
40
.                                                                                     -;;
                                                                                      'i
                                                                                             l6
)2
;."
" s                                                                                   .... "
'"'
 ~
;;
                                                                                      "c
                                                                                      ~
                                                                                            "
~                                                                                      ~
c
c     '                                                                               •
                                                                                      c
                                                                                      .!l
                                                                                            20
j                                                                                           16
':l
      )
                                                                                      ~
                                                                                      ~
 ~
]                                                                                     ':l 12
                                                                                      ~c
                                                                                            '
                         40           .          110
                              Oistunu along the toad. z, lmml
                                                                   ,.,          200         ''         20    c.o    60     60    100   120   140
                                                                                                              OiStanu along the load ,z, {«wnl     "'      110      200
                                                                                                      120
             SI
                                                                                                      110
             "'
     e       41
                                                                                            -. ""
                                                                                             ~
     •-.'    "                                                                              ... "
     "" "
     ..;
      ~
             J6
             D
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                                                                                            d'
                                                                                            ~
     ·;;
             ))
                                                                                            ."
                                                                                             ~
     0
      ~
     'E      "24                                                                            0
                                                                                             ~
                                                                                             ~ so
                                                                                                      .
      i                                                                                     ~
                                                                                                      ..
      V      21
                                                                                            'Z
      'Z 13                                                                                  ~
      •
      ~
      ~
             ~
                                                                                            ~
                                                                                            ~
                                                                                                      30
             12
              9                                                                                   ,.
             6
                                                                                                  10
                                                ..
             3
                                                                                                  "
                                                                                        ,
                                                                                        ~
                                                                                        "e
                                                                                                  70
                                                                                       -.. "
      "                                                                                 "
                                                                                        a·
                                                                                        ~
                                                                                        ~         so
                                                                                        ~
                                                                                        0
k
~
    • 54                                                                                •"
                                                                                        ~
14842                                                                                   ~
                                                                                        'Z 30
                                                                                         ~
~,   ]6
                                                                                        ~
                                                                                        <
v30
                                                                                                 20
~    24
~
.=   18
      12                                                                                         10
                             ..            .,
                                  Distance o.long
                                                                  160                            OL-~-c.~~~~~~~..-~~
                                                                                                  o 20 "Distance
                                                                                                             "    eo 1 m u ~ ~ ~
                                                                                                                 along the loo.d, z ,fmm).
           fig. 9.      Current density distribution o.bng the alu11iniue          Fig 10.        Power densitr distribution along the olucniniu111 cylinder
                        cylinder due to 600A tlowinq in single con!llctor
                        at air gap of l01t1111.                                                   dut to 1t turns coil at air gap of Jhun, non uniforM
                                                                                                  coil pitch and o current of 600A.
Indu•trial designers of induction heatinCJ work-coils             The effect of the different parameters in the
depend largely on relatively simple equivalent                    equivalent circuit depends on the case under con-
circuit techniques supplemented J:Jy empirical data               sideration. To asseslthe influence of the external
acCUIDUl&ted over many years. on the other hand                   reactance Xr and the empirical factors k and krr
academics and other research workers have developed               three computer proqrama were developed J:Jaeed on
highly sophisticated numerical techniques to solve                three different equivalent circuits. A specific
particular problesu. Exiatinq methods of work-c()il               vork co!l/no~etic billet configuration vas
design almost universally assume that tha load ia                 asataned. and calculations were carried out for
subjected to a uniform maqnetic field.                            frequencies of SO Bz: and 3 kHz. The required
                                                                  number of turns obtained from the three circuits is
I t 1a very difticult, therefore. to adapt theae                  vithin 8\ difference from each other for the SO Bz
methods to solve the probl&IU of non-unifoX'III. power            frequency, this difference increases to J:Jecome 10\
density distribution alonq a workpleoe.                           when the frequency 1s 3 kHz. Varying k J:Jetween
                                                                  -o.22 and O.l chanqed. the required number of turns
This paper explains the principles   bahi ~ each form             by up to 10\. The maximuaa effect of chanqifl9 kr
of work-coil desiqn and assesses the ra.nqe .of their             fr0111 1.0 to 1.5 vas 20\ difference in the required
applicability. SUqqestions are made to improve the                numJ:Jttr ot turns.
existinq work coil deaiqn techniques and details
are viven of a nev Mthod of desiqn mora readily                   The equivalent circuit Mthod suffers, "nudnly, from
a.pp11ad. when non-uniform surface power denalty                  three tmperfections.
distriJ:Jutions are requ11'114-:
                                                                  (1) The need of empirical data. The values of these
2. EQUIVALENT CIRCUIT METHOD                                     .factors are not necessarily known for every
                                                                  application.
The basic a .. embly of induction billet heater
consists of a vater cooled coil aurroundinq a met-                (2) The aa~tion of uniform material properties
allic workpiace. A relatively larqe air qap between·              which is not· correct aa the resistivity and
the coil and the workpiece 18 required ao that the                permeability of the workpiece chanqe with the
billet can IIIOYe: freely through the heater and, in              teatperature end magnetic field strength. Thia deficit
addition, thermal inwlation ia provided to rad.ue.                can partially be overCODe by considering the work-
the radiated heat losses froaa the billet to the                  piece aa consisting or a number of concentric
coil. 'nle maqnetic flux produced J:Jy the work coil              cylinders as it will be seen in the next section.
has alternative pa.rellel paths through either the
workpiece, the coil or the air qap as shown in                    Cl) The assumption of uniform magnetic field along
Fiq.l.                                                            the length of the workpiece hinders this ~~ethod
                                                                  from being applied to applications which require
An expression for the total ampere-turn requirement               nonunifom power densities in the load.                This
probleaa can be aolved by. aployinq tha nlaU.vely       determine the induced current.a, power distribution
new auperp0.1tion met.hod--(2).·-                       dd mechanical forces produced by any tvo dJ.Junaion-
                                                        al, linear an4 axiaymmetric induction Mating prob-
2.1. Improvement on the Equivalent Circuit Method       leaa. The limitationa of the ~Nthod are that u.q-
                                                        netio aatarials cannot be included in the problem
One of the Umitationa of the equivalent circuit         9ecmeb:'y and that the number of ele~Nnts into which
~~~t~tho4 la that it doea not tAke the nonuniforaitY    the conductors can be divided is limited by the aize
of the vorkpiece into account. This can ba overcome     of the mat.J:ix which . .y be solved. This number la
by a ..Wilinq the vorkpiece to conaht of ~e than        of the order of 300 unknovna tor the IBM 3033 cOCD-
one concentric cylinder, each with different            putar (10) which haa a memory capability of approx-
physical properties.                                    imately one MCJabyte. Difficult!. . alae arise when
                                                        dealing with curved surfaces and vhen the penetra-
riq. 3 shova a typical billet ~atinq application        tion c!epth la Maall in comparison vith the lCHid
in which the coil of ra4.ius r._ .urrounda a bUlet of   dblendcns. The nwnber of unknown• can partly be
radius rv• The billet la ahoWn divided into n           reduced by finely .ubdividing only the reqion of
concentric cylinders. The impedance of the ccmpodte     goreatest interest, typically bewteen a diatance
load can ba datemined by UIPloyinq the complex          0.1 to 0.33 timea the akin depth G (LO) fra~ the
PQyntinq' a vector (4) to deriw an expre .. ion for     .urface and using Larger aubdiviaicns elaewhere.
the tmp.dance, Zx, on radius x within a cylinder        'l'hia, of courn, requires more complicated software.
A computer proqram vaa developed for a work coil/       In this Mthod, the field reqion t.a subdivided into
multicyllndar billet. 'l'vO caHs v.re considered.       a finite number of discrete sized aubreqicns or
The first is, a load of 8 concentric cylinders with     finite elements. '1"he unknovn quantitiea at each
ditfarant. physical properti... The HCond is tor a      element are presented by suitable interpolation
unifon. load with physical propertiea equal to the      tunctiona that contain the node values of each
average properties of th.a first can. The differ-       element.
ence between the required number of turns for the
two loada vaa found to ba 23\. 'l'bia shOVII the        L'ral the . .thematical point of viav the finite
importance of the load's nonunUormity in the            element Mthod ia based on integral formulations.
calculations.                                           By vay of caaparUon the older finite difference
                                                        methoda are usually based on 41ffarent1al formula-
3. NUMERIOL METHODS OF WOIUC COIL DESIGN                tions. Finite element inteqral fcra11,1lations are
                                                        obtained by tvo different procedure•• Variational
Three distinct nuaerical approaches have been           fcrllllll&tions (11, 12) and weiqhted reaidual formu-
applied to induction Matinq problems a finite           lationa (13). Both of these techniques generate
c!itferenca, finite element and mutually coupled        th8 final assembly of algebraic equation• that must
circuit aethods. In the finite difference and           be .alnd fozo the unknown nodal par&~~etera.
finite element methods the vorkpiece is divided into
subraqions and the relevant non-linear partial          It JaUat ba borne in mind that when employing the
differential ~ationa are replaced by a aet of           finite el-.nt methods. to. 110lve induction heatinq
algebraic ~ations to ba solved in each .ubreqion        prcbleiiUI, one muat (10}
by m.ans of iteratiw procedures. The llll.ltually
coupled circuit Mthod (7) divides the .ystea into       (i) Include an •exterior :element• to represent the
a numba:r of aubconductors INtually coupled with        reqion outside the con.
each othar and applias ICirchcff'• Lava to obtain a
aya~ of linear equations deacribinq the prcblell        (U) Use a aparse matrix aclution routine.
which in turn are aol v.d by procedure a siailar to
those uaed in otlwr nu.rical  •thoi•.     The           Typically, induction heating problem& involving the
application of trulae numerical methods to induction    coraputation of the electrcmaqnetic and thermal
he&tlft9' problems will now be discussed in c!etail.    fields at. 500 nodes requires approximately o. 75
                                                        lllltCJabyt&s et computer memory (10) even vhe.n usinq
3.1. MutualLY Coupled Circuit Method                    a sparse matrix solution routine. Problems involv-
                                                        ift9' 500. to 2QCX) unknovns can be run at a reasonable
Thia method vas developed to solva problems in          coilt ·on an IBM 3033 which compares very favourably
induction heatinq and melting applications by           with the 111\ltually coupled circuit method IIW!ntioned
Kolbe an4 Rei .. (7) and later developed by Duc!ley     previously.
and Burk•· (8). '1'he reqion to be 1nvest1qated U
subdivided into sobconductors and using Jcnovn          3.3. Finite Differences
expressions (9) the reaistance and self inductance
of each subccnductor, toqether with tlw JaUtual         'l'hi• vas the firat numerical technique applied to
inductances between each subconductor are determined.   induction heating problems (14) and ia still in
The expre aaions are usually preaentad in the fcna of   use today. The technique provides a relatively
an impedance matrix and the application~of Xirchoff'•    atraiqhtforvard means of formulation and aolvinq
LAva had to a system of linear equation• with com-      two dimenlional problems.
plex coefficients of the form
                                                        'nie bada of the finite difference method is the
 rzJ   (I)    rvJ                                       conv.rsion of the 9overninq differential equations
                                                        of the problem into a aet of algebraic eqU.ations by
where [z] ia the aquare matrix compriainq all of        the use of the Taylor aeriea approximations, The
the coil and billet self and .utual impedances,         equations are solved at every point on a qrid
[I] is the column matrix of unknown coil and billet     constructed. over the required space domain. The
segment currents and [v] is the column matrix ot        mathematical operations are performed not upon
driving voltaqes.                                       continuous function· but rather in terms of
                                                        equation• about one c!iacrete point. By thia
The mutually coupled circuit method can be u..d to      approach operation& such as c!ift'erentiation and
1nte9rat1ora . .Y b4l reduced to silllpla ariU..tio         bave also included the effect of conductor size,
torm. and can then be conveniently .alved ua1nq             operatinq frequsncy, aaqnetic permeability and
di9ital computers.                                          shape of workpiece, and current distribution within
                                                            the work conductor itself,
A proqru has been written to predict the maqnatic
field and power density distributions in a
haaocJeneou.e noru~~&qnetic rectanqul.az: wor1cp1eca
subjected to a   unlfo~   .urface maqnatio field            1. BOBSON,L. Guide to Induction Heating Equipment,
atreJ\9tho                                                  BNCE, London, 1984, 66 paqes.
Tf\e ~-;ts:;·-iteid'' ard power- densttr- 4tiib:UiQtl.On:   2. AL-SBAIKBLI,A,X,K. and UOBSON,L.   ·~provementa
vi thin an aluainl\111 voxkplece Of croea-Hction            1n the Desian of Induction Billet Beaters•, 19th
500 - by 150 • auhjecUd to a .urface UtJU;tio.              UPEC, university of Dundee, April, 1984.
tidd atzenvth of LoS Arl at a frequency.of ~so.. n
are eon t.D F19•• 4 Ul4 5 reepectlnly. . . . ..             3. BAD!R,Jl.M. "Desiqn and calculation of Induction
                                                            Heat.f.n; Coib". AIEB 'l'rans, 76 1 Pt II, pp.3l-40,
The prQ91'. . v&a relatiwly simple to· blpl...nt an4        ~(arch, 1957.
qava aatiefactory re.ulte when applied to a at=pla
two dimensional problUio                                    4. REXCHERT,J:., "The calculation of coreless
                                                            Furnace• vith Electrically Conductinq crucibles•,
4. CONCLUSIONS AND RECOMMENDATIONS                          Electrotechnik, 49, 6, pp.376-397, February, 1964.
An accurate knovledc)a of the power density and             5. VJUJGBAN',J.T. an4 WILLIMSON,J.W. "Dedqn of
temperature dbtr1hut1on 1n an inductively heated            tnductton Heating Coils for cylindrical Nonmaqnetic
load la of qraat importance. Analytical .xpreaeiona         Lo&da", AIEB Trana, 64, pp.587-589, August, 1945.
are available for simple one ~atonal qeomatriaa
auch ae a cylinder or a dab (15) 'but for the               6. VAUGBAK,.:r ,T, and WILLIAMSON,.:r .w. "Desiqn of
solution of even the .tm:pl.. t two 4.1mauilonal            Induction Beatinq Coiia for Cylindrical Maqnetic
g~trlea numerical technique. are requu.d.                   Loads", AIEl!: Trane, 65, pp.SB7-892, 1946.
AIR GAP
ODD DD D COIL
                                                                           •
          •.
    ,r--1'----.\
V                                                       •.
•                             X                                    .
                              •                                    z}·
                                                                       w
••
                    1
                ·--·--·-
       1~1
0·6
0·6
0·4
0·2
 0
             1                  3                   s         6        7         8
                   Fig. 4   Variation of .aqnetic field distribution (&a a fraction of the
                            surface fidd) with depth y.
POWER (W)
140
120
100
80
60
40
                       Fig. 5       Power distribution down the centre line of the cross-sectional face
       ILLUSTRATING ELECTROMAGNETIC$ USING AN INDUSTRIAL PROCESS
ABSTRACT
extrusion.
            ILLUSTRATING ELECTROMAGNETICS USING AN INDUSTRIAL PROCESS
SYMBOLS
                                                                       2
wp      =      The power density induced on point P (Wm- )
6
               I    P
                   nf~
                             = Skin depth (m)
p Resistivity (0m)
~ Permeability (llm-l)
                                                       (i)                              L.llobson
1.   INTRODUCTION
metallic structure then eddy currents are induced in the metal which
Induction billet heaters are made with many turns of water cooled
copper tubing and the design of these coils requires knowledge of the
induced currents from not one conductor but many in close proximity to
                                      -1-
                                                                          L.Hobson
However this situation is far from those seen in practice.      The
prior to extrusion.
2. SINGLE CONDUCTOR
and carrying current -I. Both the conductor and the image will
to
             I
     H                                                                ( 1)
                                   -2-
                                                                         L.Hobson
 The axial components Hz of the conductor's magnetic field strength and
 its image will aid each other, while the radial components 11                              will
                                                                                        r
 cancel each other.                              The resultant field on point P will therefore
be:-
       11                2 11                                                                  (2)
            p                   z
but 11                                                 h
                     11 cos8                11
                                                 .{2 + z 2
                ~                       ~
z (3)
Hence
       11                                                                                     (4)
         p
                                                   2
    11          ~
                           I                 - a
        p                                                                                     (5)
   J            ~   11          12                  -1
                                                 (Am )
       p             p              0                                                         (6)
   w                                                                                          (7)
       p
                                                             -3-
                                                                                                   L.llobson
        If the metallic workpicce is a long cylinder then it can, to a
circular turn of 6rnrn diameter water cooled copper tubing was wrapped
gaps between the coil and workpiece of 30rnrn and SOrnrn, and a coil
the surface of the billet was measured. The measurements were made
density probes were positioned along the length of the billet, Fig. 2.
insulated from the billet. The output of the probe was measured by a
high impedance voltmeter shielded from the high magnetic field. The
voltage reading was directly related to the current density within the
                       .   (5) (6)
surface of the workp1ece             , as:-
                V
        J =                                                              (8)
                p~
Fig. 3 shows the results obtained and whilst the shape of the curves
    J
        p
            =                                                            (9)
                                          -4-
                                                                              L.Hobson
     They are functions of the effective height h ,                              A is a straight
                                                 e
line represented by
     A = 10-2 h                                                                            ( 10)
                   e
transformer on the high current output was used to measure the work
3, MULTIPLE CONDUCTORS
with the surface current density distribution along the length of the
billet.
                                                 -5-
                                                                                                   L.Hobson
surrounded by an 11 turn work coil made from 6rnrn diameter water cooled
copper tubing. The air gap used was SOrnrn and the coil pitch was
distribution along the billet was measured and the results correlated
4.   TEMPERATURE TAPER
                                      (1)
     Research work at Loughborough           has produced a novel computer
aided technique for the design of induction billet heating work coils
turn work coil of arbitrary pitch. The values of the surface current
density of 30mm air gap single turn coil, Fig. 3, were used to
                                       -6-
                                                                             L.Hobson
different work coil configurations using the procedure outlined
previously.
coil with the spacings given in Table 1 should produce a linear temp-
          .   'b ut1on
erature d 1str1    .   (l) usua 11 y assoc1a
                                          . t e d w1t
                                                   . h a 1 um1n1um
                                                             . .         .
                                                                   extrus1on.
Pitch 1 2 3 4 5 6 7 8 9 10
Spacing (mm) 7 8 11 23 23 23 25 25 25 32
test the theory because of the capital cost and the limitations of the
                                            -7-                                 L.Hobson
staff trainees of the Electricity Supply Industry involved in
5. CONCLUSIONS
                                   -8-                                L.Hobson
 REFERENCES
April 1984
1984
pp.273-278
1984
                                       -9-
                                                                           L.Hobson
LIST OF FIGURES
current of 600 A.
current of 600 A.
UI/PBl/3.12.84
                                         -10-
                                                                             L.Hobson
                             z
7 7 7 /stab 7 7
                   I
                  C'!Ymage
Ia l
           Ibl
            8
EQUATION 9
                                                                                 EQUATION
                                                                                                AIR GAP OF
            7                                                                                   30 mm
                                                                             0   MEASUREMENT)
                                                                                 EQUATION 7 )
                                                                                              AIR GAP OF
                                                                                            )
                                                                                              So mm
                                                                             0   MEASUREMENT)
            6
30 mm AIR GAP
11'10
     ~
     ~
     ....
            4
                "'   '\. .   \
                              \
      ~                  \       \    D
     ..,                     \\. \
     ~      3
     H
     "'z~                            \
     "'z                                  \
     ~                                        \
     D
     u
            2
                                                  ~
                                                    \.
                                                         '\.
            1                                                  '    -..;::
·- -~----
                                                                                                             L.Hobson
                                             current (10,000: 10)
                                             transformer
                                  r----'"VY'I"'-------r- b          INDUCTIVELY
415V
                                                                    HEATED ALUMINIUM
60A
                                                                    CYLINDRICAL
50Hz                                                                WORK PIECE.
                                           0     MEASUREMENTS
"'<:
I
           40
    ~
"'o
    ..-<
    ..,X   30
      -
    !::H
    [(I
    z
    1:!
    I'<
    z
           20
    ~
    D
    u
10
            0
                0   40            80           120              160    200
                                                                      L.Hobson
            40
                                                        SUPERPOSITION
            35
                                                  o     MEASUREMENTS
30
N           25
I
     Ei
    .0:
    ~
IIlo
    .....
    ..,X 20
       .                                                                0
    ~
    ....
    "'~z
    E-<
            15
     z
    ~
    p
    u
10
            0
                 0   40           80             120          160           200
                                                                            L.Hobson
     8
SUPERPOSITION
6 0 MEASUREMENTS
IS   5
8
     0
         0   40           80            120          160       200
L.Hobson