Project :
PEDC
Design Calculations for :
Calc by N M Patel
Checked by
500 KL SS Storage Tank
Job No.
Date
Date
22/11/10
DESIGN OF 500 KL SS STORAGE TANK
QTY.
Design Calculation No:
FZD- 01
DATA
Sr
Parameter
Nom
Unit
Value
1 Content
Ethylene glycol
2 Working Capacity
VL
KL/MT
15
3 Vessel Diameter
2.6
4 Height
2.9
5 Specific Gravity
6 Corrosion Allowance
1.2
mm
7 MOC
SS 316
8 Viscosity
9 Ratio ,H/D
Ni
CP
30
1.12
10 Sweep Diameter
Da
0.9
11 TIP Speed
Ts
M/s
12 Blade type
3 Bladed pitch turbine
13 Nos of Blade
Nb
14 Distance Between Impeller
Si
3
M
1.3
15 Ratio of Si/Da
1.50
17 Type of impeller
16 Nos of impeller
multiimpeller
17 Type of Seal
Mechanical seal
LS
3.4
19 Weight of 1st impeller
W a1
Kg
18
20 Weight of 2nd impeller
W a2
Kg
18
18 Length of Shaft
18
Page 1 of 9
Nos of Impeller =
2
3
1 Speed of impeller:
N=
Where Da = Sweep Diameter,M
60 S
= D/3
D a
N=
44.1
33.0
RPM
2 Pumping Rate, Q in M3/sec
Q=N Q ND
a3
Q=
Impeller Type
Nos of Baffles
Propeller
0
Propeller
3 to 8
Turbine,VT blade
0
Turbine,VT blade
4
Pitch Turbine,45
0
Pitch Turbine,45
4
Anchor
0
0.9
Np
0.3
0.33 -0.37
0.93 -1.08
3 to 5
0.7
1.3-1.4
0.28
Where NQ = Pumping Number
12.9
NQ
0.3
0.4-0.55
0.33-0.34
0.7-0.85
0.3
0.6-0.87
0.28
0.6
M3/min
3 Mixing Time,in Min for 100 volume to be change
QS =100
( QV )
Qs =
116
min
4 Reynold Number,Re
ND
Re =
a2
u =Viscosity in Kg.M/hr
108
16524
Hence,Re >10000 Flow is Turbulent
Power Number for Reynold and For 3 bladed pitch turbine is
Np =
5
0.8
Power Required :
PL =Power loss , =
H . P=1. 1
FN P N 3 D
S
735
=
Provided H.P=
+ PL
0.41
H.P
H.P
0.2
0.5
For Mechanical Seal
For Stuffing Box
F= Multi impellerFactor
Pi / Da
0.1 to 0.5
0.5 to 1.0
1.0 to 1.5
1.5 to 2.0
2.0 to 3.0
3.0 to 4.0
>4
Page 2 of 9
1.6
Multi impeller factor
F
1.4
1.5
1.6
1.7
1.8
1.9
2
Page 3 of 9
6 Width of Blade,mm
W=
40
mm
7 Diameter of Shaft,mm
d s =100
J m ( H . P ) L
D a NN i
ds =
Provided ds =
0. 33
Drive length =
0.5
Ls=Length of Shaft ,M =
3.8
Where,Jm =Jamming Factor
EL =Overhung length=
58.2
60
1.5 to 2
2
Jm =
L1+L2
5.55
2.8
L1 =Distance from Centre of Bottom First Impeller to
mm
mm
to Vessel Top Surface
3.4
L2 =Distance from Centre of BottomSecond Impeller to
Vessel Top Surface
=
2.15
8 Thickness of Blade,mm
H.P
t b =555
NN i N b W
tb =
0. 5
+2CA
5.1
mm
mm
Provided tb =
9 Critical Speed,RPM
Nc=
1
1 . 92 W S L
EI
S3
2. 67
+Wa1 L
a1 2
( La 1 + L 1 ) +W a2 L a2 2 ( La 2 + L 1 ) +W c L c2 ( Lc + L 1 )
0. 5
Where E=Modulas of Elasticity =
I=Moment of Inertia,cm4 =
W S=Weight of Shaft,Kg =
=
Bearing Span,L1
=
W C=Weight of Coupling,Kg=
1800000
Kg/cm2
63.6
cm4
[dS2LS/159]
86.0
Kg
0.3
20
Kg
Lc =Distance from Centre of Coupling to tanks Top height
=
NC =
97.6
0.8
RPM
Hence,Operating Speed Should be
1.3*NC
126.9
RPM
OR
0.7*N
68.3
RPM
Page 4 of 9
18
Wa1
18
Wa2
10
Deflection,mm:
A Deflection due to 1st Impeller Weight
Weight of 1st impeller W a1 =
[ ][
1=
Wa1 X
6 EI
2 L a1 L1 +3 L a1 X X 2 ]
2.24225
= 0.0224225
18
Kg
X=Distance from centre of lower bearing to distance
at which deflection is to be measured
=
3.4
M
340
Deflection at the end of 1st impeller ,X=La1
cm
cm
M
3.4
Bearing Span,L1 =
=
340
0.3
30
M
cm
cm
B Deflection due to 2nd Impeller Weight
Weight of 2nd impeller W a2 =
2 L a 2 L1 + 3 Lalignl
Wa2 X
[ X X 2]
2=
6 EI
0.59370
cm
0.005937
=
2.15
M
215
Deflection at the end of 2nd impeller ,X=La2
2.15
Bearing Span,L1 =
=
C Deflection due to Shaft Weight
s=
WsL
8 EI
3.69
cm
0.03693
D Deflection due to coupling weight
3=
[ ][
WcX
6 EI
2 Lc L1 +3L c X X 2 ]
0.00000
E Critical Speed at Overhung,Nc
30
g
N C=
1 + 2 + 3 + s
=
Kg
X=Distance from centre of lower bearing to distance
at which deflection is to be measured
a2
18
117
0.5
Rpm
Page 5 of 9
215
0.3
30
M
cm
cm
cm
11 Critical speed of shaft when shaft supported at two end
1 Deflection due to Shaft Weight
S=
5WS L
S3
5.371
384 EI
mm
2 Deflection due to 1st Impeller Weight
1=
W1 L
a 12
( Ls La 1 )
0.255163
3 EIL S
mm
3 Deflection due to 2nd Impeller Weight
2=
W2 L
a2 2
( LS La 2 )
3 EIL s
1.736142
mm
4 Deflection due to 3rd Impeller Weight
3=
W3 L
a3 2
( L S La 3 )
3 EIL S
mm
0.882918
5 Critical Speed ,Nc
30
g
N C=
1 + 2 + 3 + s
NC =
329.55
0.5
RPM
Hence,Operating Speed Should be
1.3*NC
428.4
RPM
OR
0.7*NC
230.7
RPM
Page 6 of 9
12 Design of Hub and Key
Part Description
Yield stress Shear stress
MOC
y in MPA
in MPA
Crushing stress
c in MPA
Shaft
SS-304
175
43.75
87.5
Hub
CS
180
45
90
Key
CS
180
45
90
Coupling
CS
180
45
90
bolt
CS
180
45
90
A Design of Hub
1 Outer Diameter of Hub,Dh in mm
Dh =2 d s
120
mm
2 Length of Hub,Lh in mm
Lh =1 .5 d s
90
mm
3 Induced shear stress in Hub
16TD h
s=
D 4 d 4
Torque
s =
T = 215.981
0.679
Mpa
1 Width of key,W
15
mm
2 Thick of Key,t
15
mm
<
45
ok
B Design of Key
For Square Key
3 Shaft strength reductin factor,e
e=10 .2
W
h
1. 1
d
d
e=
0.8125
4 Length of Key,Lk
94
mm
5 Induced shear stress
is =
Torque
2T
LK Wd s
is =
T=
4.14
215.98
Mpa
<
45
ok
Mpa
<
90
ok
6 Induced Crushing Stress
ic =
2T
LK tds
ic =
4.14
Page 7 of 9
N.M
N.M
Page 8 of 9
13 Design of Rigid coupling
1 Design of flange
Hub Diameter,D
120
MM
Thick of flange,tf
30
mm
Checking for induced shear stress in flange
if =
2T
D 2 t f
if = 0.318444 Mpa
<
45
ok
2 Design for Bolts
PCD of bolts,D1 = 3d
180
OD of Flange,D2 = 4d
mm
240
mm
3.36
mm
Nos of bolts
n=
Bolt design under shear load
8T
db=
4 b nD 1
0. 5
db =
Bolt size 4nos of M6 size
Page 9 of 9
