HT24
HT24
Installation Details
for Self-Regulating
Heating Cable
                                                                                                                          Design Guide and Installation
    Self-Regulating                                                                                                       Details for Self-Regulating
    Heating Cables                                                                                                        Heating Cable
Electrical Specifications 2703-1 2703-2 2705-1 2705-2 2708-1 2708-2 2710-1 2710-2 2305-1 2305-2 2310-1 2310-2 2315-1 2315-2
      Service Voltage (Volts)                     120   240   120      240      120   240   120        240          120      240   120     240     120      240
      Maximum Circuit Length (Feet)               330   660   270      540      210   420   180        360          240      480   180     280     135      200
      Thermal Rating at 50-F (Watts/FT.)           3     3     5           5     8     8    10          10           5        5     10     10       15       15
      Temperature Rating
        Maximum Maintain (Deg. F)                 150   150   150      150      150   150   150        150          250      250   250     250     250      250
        Maximum Exposure (Deg. F)                 185   185   185      185      185   185   185        185          366      366   366     366     366      366
Alternate Voltages
RSCC 240 VAC self-regulating heating
cables can be operated at alternative
                                                                                                         o
voltages. The chart below compares
heating cable power output with prod-
uct rating.
Example:
Thermal output of 2705-2 5 Watts/Ft.
at 50°F, powered at 208 VAC = 5
Watts/Ft. x .86 = 4.3 Watts/Ft.
    To determine the pitch and amount of      In some circumstances it may be desired                         Determine Heater Pitch
    RSCC self-regulating heating cable        to use a heater with less power output                          To determine the required pitch (P),
    required to heat trace a pipe, you’ll     per foot of heater than the calculated                          select value from Tables 3A and 3B for
    need to know the pipe temperature to      heat loss per foot of pipe. In these                            calculated value of (Rp) and pipe size.
    be maintained, minimum ambient tem-       cases, the heater can be spiralled
    perature, pipe size and insulation type   onto the pipe to achieve the required
                                                                                                              Calculate Required
    and thickness.                            power output per foot of pipe. A
                                                                                                              Heater Length
                                              developed power ration and heater
                                                                                                              To determine required heater length
    Calculating Heat Loss                     pitch will need to be determined.
                                                                                                              (Lh), multiply length of pipe (Lp) by (Rp).
    1. First determine temperature differ-                                                                    Lh = Lp x Rp.
       ence (⌬T) between temperature to       Calculate Developed
       be maintained (Tm) and minimum         Power Ratio
       ambient temperature (Ta).              To calculate developed power ratio
       ⌬T = Tm - Ta.                          (Rp) divide heat loss from pipe (Qp)
    2. Select insulation K factor from        by heater power output (Qh).
       Table 1 (Ki) and divide by .021 to     Rp = Qp ÷ Qh.
       determine conductivity ratio (Rk).
       Rk = Ki ÷ .021.                        Table 2A — Heat Loss (Qa) from Insulated Pipe (Watts/Foot-°F).
    3. Determine heat loss from Table 2A        Pipe              Pipe
       (Qa) by selecting pipe size and                                                                          Insulation Thickness
                                                Size              O.D.
       insulation thickness. If piping is       (IPS)           (Inches)          1/2"           1"        1-1/2"          2"        2-1/2"          3"           4"
       indoors multiply (Qa) by 0.9.            1/2              0.840              .05          .04          .03         .03           —            —            —
                                                3/4              1.050              .06          .04          .03         .03           —            —            —
    4. Calculate heat loss from pipe (Qp)
                                                  1              1.315              .07          .05          .04         .03          .03           —
       by multiplying ⌬T by Rk and Qa.         1-1/2             1.900              .09          .06          .04         .04          .03           —            —
       Qp = ⌬T x Rk x Qa.                         2              2.375              .11          .07          .05         .04          .04           —            —
                                               2-1/2             2.875              .13          .08          .06         .05          .04           —            —
                                                  3              3.500              .16          .09          .07         .05          .05           —            —
    Determine Heater                           3-1/2             4.000              .18          .10          .07         .06          .05          .05           —
                                                  4              4.500              .20          .11          .08         .06          .06          .05          .04
    Power Output                                  6              6.625              .28          .15          .11         .09          .07          .06          .05
    From Graph I (page 3) select the heater       8              8.625              .35          .19          .13         .10          .09          .08          .06
    with the power output (Qh) which meets       10              10.750             .44          .23          .16         .13          .10          .09          .07
                                                 12              12.750             .51          .27          .19         .14          .12          .10          .08
    or exceeds the heat loss (Qp) from           14              14.000             .56          .29          .20         .16          .13          .11          .09
    the pipe. For non-metal pipe multiply        16              16.000             .64          .33          .23         .18          .15          .12          .10
    the power output Qh from the chart by        18              18.000             .71          .37          .25         .20          .16          .14          .11
                                                 20              20.000             .79          .41          .28         .21          .18          .15          .12
    0.7 before selecting the heater.             24              24.000             .94          .48          .33         .25          .21          .18          .14
                                              Values given above are heat loss for metal pipe in units of Watts/Foot of pipe per °F temperature difference from pipe to
                                              ambient temperature fiberglass insulation.
Table 3A — Pitch in Inches of Heater Wrap on Pipe for Given                                                     Example: (2700 Heater)
           Heat Loss/Developed Power Ratios of 1.1-2.0
                                                                                                                Ta = -20°F
  Pipe          Pipe                                                                                            Tm = 40°F
                                                    Heat Loss/Developed Power Ratio                             Insulation — Calcium Silicate
  Size          O.D.
  (IPS)       (Inches)       1.1      1.2       1.3      1.4      1.5      1.6       1.7      1.8   1.9   2.0   Pipe Material — Metal
                                                                                                                Insulation Thickness — 2"
  1/2          0.840           6        4         3        3        3         2        2      —     —     —
  3/4          1.050           7        5         4        4        3         3        2       2     2     2    Length of Pipe — 100'
   1           1.315           9        6         5        4        4         3        3       3     2     2
 1-1/4         1.660          11        8         6        5        5         4        4       3     3     3
                                                                                                                Step I     ⌬T = Tm-Ta
 1-1/2         1.900          13        9         7        6        5         5        4       4     4     3                  = 40 - (-20)
    2          2.375          16       11         9        7        6         6        5       5     4     4                  = 60
 2-1/2         2.875          20       14        11        9        8         7        6       6     5     5
    3          3.500          24       17        13       11       10         9        8       7     7     6    Step II    Rk = K ÷ .021
 3-1/2         4.000          27       19        15       13       11        10        9       8     8     7                  = .031 ÷ .021
    4          4.500          31       21        17       14       13        11       10       9     9     8
    5          5.563          38       26        21       18       16        14       13      12    11    10                  = 1.48
    6          6.625          45       31        25       21       18        17       15      14    13    12    Step III   Qa from Table 2A for 6 IPS
    8          8.625          59       41        32       27       24        22       20      18    17    15
   10          10.750         74       51        41       34       30        27       25      23    21    19               pipe and 2" thick insulation
   12          12.750         87       60        48       41       36        32       30      27    25    23               is .09.
   14          14.000         96       66        53       45       39        35       32      29    27    25
   16          16.000        110       76        61       51       45        40       37      34    31    29    Step IV Qp = T x Rk x Qa
   18          18.000        123       89        68       58       51        45       41      38    35    33               = 60 x 1.48 x .09.
   20          20.000        137       95        76       64       56        50       46      42    39    36
   24          24.000        164      114        91       77       67        60       55      50    47    43
                                                                                                                           = 8.0
                                                                                                                Step V     From Graph 1, at 40°F the 2708
                                                                                                                           heater produces Qi of 8.5 watts
Table 3B — Pitch in Inches of Heater Wrap on Pipe for Given                                                                per foot. Select the 2708 heater
           Heat Loss/Developed Power Ratios of 2.2-4.0                                                                     for this application.
  Pipe          Pipe
                                                                                                                           Alternate Heater by
                                                    Heat Loss/Developed Power Ratio                                        Spiralling
  Size          O.D.
  (IPS)       (Inches)       2.2      2.4       2.6      2.8      3.0      3.2       3.4      3.6   3.8   4.0              Assume that for the above example
    2          2.375           4         3        3        3        3         2        2       2     2     2               you wish to use a 2705 heater.
 2-1/2         2.875           5         4        4        3        3         3        3       3     2     2
    3          3.500           6         5        5        4        4         3        3       3     3     3
                                                                                                                           The Qi, from Graph 1, at
 3-1/2         4.000           6         6        5        5        4         3        3       3     3     2               40°F for the 2705 heater is
    4          4.500           7         6        6        5        5         5        4       4     4     4               5.5 watt per foot.
    5          5.563           9         8        7        7        6         6        5       5     5     4
    6          6.625          11        10        9        8        7         7        6       6     6     5    Step VI Rp = Qp ÷ Qh
    8          8.625          14        12       11       10        9         9        8       8     7     7               = 8.0 ÷ 5.5
   10          10.750         17        15       14       13       12        11       10      10     9     8
   12          12.750         20        18       17       15       14        13       12      12    11    10               = 1.46
   14          14.000         22        20       18       17       15        14       13      13    12    11    Step VII The pitch, P, in inches from
   16          16.000         26        23       21       19       18        16       15      14    14    13
   18          18.000         29        26       23       21       20        18       17      16    15    14             Table 3A for 6 IPS and
   20          20.000         32        29       26       24       22        21       19      18    17    16             Rp = 1.46 is 14 inches
   24          24.000         38        35       31       29       27        25       23      22    20    19
                                                                                                                Step VIII Lh = Lp x Rp
Heat Loss/Developed Power Ratios should be rounded to the next highest value.                                                = 100' x 1.46
Heat Loss/Developed Power Ratios less than 1.1, run the heating cable parallel to the pipe.
                                                                                                                             = 146' cable length required.
    To determine the amount of RSCC self-         Table 2B — Heat Loss (Qb) from Insulated Valves (Watts/°F)
    regulating heating cable required to
    heat trace a valve, you’ll need to                Valve
                                                                                       Insulation Thickness
                                                       Size
    know the pipe temperature to be main-           (Inches)        1/2"          1"         1-1/2"           2"          3"          4"
    tained, minimum ambient
    temperature, valve size and insulation            1/2             .30         .16         .11           .09           .07         .04
                                                      3/4             .31         .16         .12           .10           .07         .05
    type and thickness.                                 1             .35         .18         .13           .10           .08         .05
                                                     1-1/2           .44          .23         .16           .13           .10         .06
    Calculating Heat Loss                               2             .49         .26         .18           .14           .11         .07
                                                     2-1/2           .56          .29         .21           .16           .12         .08
    1. First determine temperature difference           3             .64         .34         .24           .19           .14         .09
       (⌬T) between temperature to be main-          3-1/2           .71          .37         .26           .21           .16         .10
       tained (Tm) and minimum ambient                  4             .77         .41         .29           .23           .17         .11
                                                        6            1.06         .56         .40           .31           .23         .16
       temperature (Ta). ⌬T = Tm - Ta.                  8            1.33         .71         .50           .40           .29         .19
                                                       10            1.67          .88        .62            .49          .37         .24
    2. Select insulation K factor from                 12            2.07         1.09         .77           .61          .46         .30
       Table 1 (Ki) and divide by .021 to              14            2.32         1.23         .86           .69          .51         .34
       determine conductivity ratio (Rk).              16            2.61         1.44        1.01           .80          .60         .40
       Rk = Ki ÷ .021.                                 18            2.99         1.65        1.16           .92          .69         .46
                                                       20            3.24         1.86        1.31          1.04          .78         .51
    3. Determine heat loss from Table 2B               24            3.98         2.28        1.61          1.28          .96         .63
       (Qb) by selecting valve size and
       insulation thickness. If valve is
                                                  Calculate Required                             Step II      Rk = K ÷ .021
       indoors multiply (Qb) by 0.9.                                                                              = .031 ÷ .021
                                                  Heater Length
    4. Calculate heat loss from pipe (Qv)         To determine required heater length                             = 1.48
       by multiplying ⌬T by Rk and Qb.            (Lh), multiply number of valves (Nv) by        Step III     Qa from Table 2B for 6"
       Qv = ⌬T x Rk x Qb.                         (Rp). Lh = Nv x Rp.                                         valve and 2" thick insulation
                                                                                                              is .31.
    Determine Heater                                                                             Step IV      Qv = ⌬T x Rk x Qb
                                                  Example: (2708 Heater)
    Power Output                                                                                                  = 60 x 1.48 x .31.
                                                  Ta = -20°F
    From Graph I determine heater power                                                                           = 27.5
    output for pipe temperature to be main-       Tm = 40°F
                                                  Insulation — Calcium Silicate                  Step V       Qh from Graph 1 for 40°F
    tained (Qh). If valve is non-metal multiply                                                               required temperature is 8.5.
    value of Qh from graph by 0.7.                Valve Size — 6 IPS
                                                  Insulation Thickness — 2"                      Step VI      Rp = Qv ÷ Qh
    Calculate Developed                           Number of Valves — 2                                            = 27.5 ÷ 8.5
    Power Ratio                                                                                                   = 3.24 feet per heater
                                                  Step I     ⌬T = Tm-Ta
                                                                                                                    valve
    To calculate developed power ratio                          = 40 - (-20)
    (Rp) divide heat loss from valve                            = 60                             Step VII     Lh = Nv x Rp
    (Qv) by heater power output (Qh).                                                                             = 2 x3.24
    Rp = Qv ÷ Qh.                                                                                                 = 6.5'
To determine the amount of RSCC self-regu-       Example: for 60°F temperature                   Pipe flanges, fittings and hangers act as
lating heating cable required to heat trace an   difference:                                     heat sink devices in a heat trace system.
insulated pipe flange, fitting or hanger, sim-                                                   Allowances must be made for these
                                                 (2) 10" flanges (4" heater per 10 degrees
ply find the size on the verticle axis, read                                                     devices to maintain a consistent and opera-
                                                     difference);     2 x 4 x 60/10 = 48.0"
across to the appropriate device, then read                                                      tional system.
down to the horizontal axis to determine the     (1) 10" fitting (4" heater per 10 degrees
amount of cable required per 10°F tempera-           difference);      1 x 4 x 60/10 = 24.0"     For pipe flanges and fittings under two
ture difference. Multiply the temperature dif-                                                   inches in size use four inches of heater per
                                                 (4) 7" wide hangers (4" heater per 10 degrees
ference by this value and divide by ten to                                                       device. For hangers under two inches in
                                                     difference);       4 x 4 x 60/10 = 96"
get the inches of cable to use per device.                                                       size use six inches of heater per device.
Hanger sizing is determined by the width of                                        Total 168"
the hanger.                                                Total Allowance Required = 14.0'
    Spiral Installation
    of Heating Element
    Note:
    1. If ratio of heater footage to pipe
       is greater than 1.5 — use two
       parallel heaters or select higher
       wattage heater. If ratio is less than
                                                                                 *Glass Tape P/N 1528-01017, 1528-01019
       1.0 — use one parallel heater.
    2. When installing the heater on
       non-metal pipe secure the heater to
       the pipe with aluminum tape. Refer
       to pitch chart on isometric drawings
       for proper pitch length.
      Note:
      1. For more specific details and full
         materials list refer to installation
         instruction sheet packed with
         connection kit
                                                                                                            P/N 1535-02080
                End Seal Kit
              P/N 1548-50300                                      P/N 1548-50211
Note:
This detail is shown as an illustration of a
method of taking advantage of the shape of
a piping configuration to attain good pipe
contact. To simply trace the inside radius of
the corner would not be considered correct.
Although a tee-splice might also be used to
trace the third leg of the tee. The objective of
this detail is to emphasize that it is advisable
to get more heater on any area where the
thermal insulation might not be fitted as well
as on straight pipe. This method is intended
to be used on other fittings besides tees.
         Notes:
         1. Exact configuration may vary per valve type.
         2. For removable valve bodies leave a loop of
            tracing of the proper length when tracing the
            pipe.
         3. See installation chart for correct amount of
            tracing per valve size.
         4. Take care to keep the flat side of the heater in
            as good physical contact with the valve body
            as possible.                                                     *Glass Tape P/N 1528-01017, 1528-01019
         5. Fully insulate and weather protect.
                                         Notes:
                                         1. See National Electrical Code paragraph 427-12(E).
                                         2. Fully insulate and weatherproof (if outdoors).
                                         Note:
                                           Heater must be pulled thru flexible conduit to avoid splicing —
                                           if necessary to splice heater a junction box will be required.
                                         Note:
                                         Use fiberglass or aluminum tape to hold tracer in place on pump body.
                                         Note:
                                         Fully insulate and weather protect.
     Note:
     All forms of rigid pipe supports directly
     in contact with the pipe surface act
     as a heat sink. Heat tracing should be
     doubled over at these points and the
     supports should be insulated as much
     as practicable to limit heat loss.
                                      Note:
                                      Insulate as much of shoe support as
                                      possible and weather protect all openings.
                    Note:
                    Fully insulate and weather protect
                    pipe support if outdoors.
                                            * Glass Tape or
                                              Cable Ties (typical)
0419-32000