Download - Pipe Sizing and Pressure Drop Calculations
PIPE SIZING AND PRESSURE DROP CALCULATIONS
• Compressible fluid
• Incompressible fluid
• Steam Service
• Equivalent Length
NETWORK SOLUTION BY HARDY CROSS METHOD
A. Compressible fluid
Density changes with change in pressure
Line sizing
Q = AVVP
QTd 157.1
Where, d = Calculated ID in mm. Q = Flow in Nm3/hr., T = Fluid temperature in ° K P = Fluid pressure in Kg/cm2a ( Kg/cm2g + 1.0332) V = Velocity in m/s
SL.Nu. Service Velocity1 Compressed air 102 Nitrogen 103 Argon 104 Oxygen 85 Acetylene 66 LPG(Gas) 87 BF,Coke Oven,Mixed Gas 12
Recommended velocities for compressible fluids
Conversion of m3/hr. to Nm3/hr.
22
2
1
11
)273
100V
t
RHPvP
T
VP
Where, V1 = Nm3/hr P1 = Atmospheric pressure = 1.0332 Kg/cm2a T1 = 0° C = 273 ° K V2 = m3/hr.
P2 = System pr. Kg/cm2g +1.0332 t2 = Fluid temperature in °C RH = Relative humidity Pv = Vapor pressure at temperature t2 °C
Tempr. Pr. Tempr. Pr. Tempr.°C kg/
cm2(a)°C kg/
cm2(a)°C kg/
cm2(a)0 0.00622 27 0.0363 54 0.1531 0.00669 28 0.0385 55 0.16072 0.00719 29 0.0408 56 0.16843 0.00772 30 0.0433 57 0.17674 0.00829 31 0.0459 58 0.1855 0.00889 32 0.0485 59 0.194056 0.00953 33 0.05135 60 0.20317 0.01021 34 0.0542 62 0.2338 0.01093 35 0.0574 64 0.2449 0.0117 36 0.0606 66 0.26710 0.01251 37 0.0641 68 0.29111 0.01338 38 0.0676 70 0.31812 0.01429 39 0.0714 72 0.34613 0.01526 40 0.0752 74 0.37714 0.01629 41 0.0794 76 0.4115 0.01738 42 0.0836 78 0.44516 0.01853 43 0.0882 80 0.48317 0.01975 44 0.0928 82 0.52318 0.02103 45 0.0978 84 0.56719 0.02239 46 0.1028 86 0.61320 0.02383 47 0.1083 88 0.66221 0.0254 48 0.1138 90 0.71522 0.027 49 0.1198 92 0.77123 0.0286 50 0.1258 94 0.83124 0.0304 51 0.1323 96 0.89425 0.0323 52 0.1388 98 0.96226 0.0343 53 0.1459 100 1.033
Vapour Pressure Table
Pressure Drop Calculation
i) Weymouth equation (for gas transmission lines)
5.33
2
dPx
Q x L x 0.3 P
xSxT
Where, ∆P = Pressure drop in kgf/cm2
Q = Discharge in Nm3/hrP = Average line pressure, in kgf/cm2a = (P1 + P2)/2
P1 = Initial pressure in kgf/cm2 aP2 = Final pressure in kgf/cm2aL = Equivalent length in meterD = Pipe bore in mmS = Sp. gr. of gasT = Actual Gas temp. oK
ii) Fritzsche Equation
5
857.11.875
dPx
Q x L x 0.15 P
xTxS
Where, ∆P = Pressure drop in kgf/cm2
Q = Discharge in Nm3/hrP = Average line pressure, in kgf/cm2a = (P1 + P2)/2
P1 = Initial pressure in kgf/cm2 aP2 = Final pressure in kgf/cm2aL = Equivalent length in meterD = Pipe bore in mmS = Sp. gr. of gasT = Actual Gas temp. oK
B. Incompressible fluid
Density does not changes with change in pressure
Line sizing
V
Qd 38.1128
Where, d = Calculated ID in mm., Q = Flow in lit/min., V= Velocity in m/s
V
Qd 8.18
OR
Where, d = Calculated ID in m, Q = Flow in m3/hr., V = Velocity in m/s
SL.No. Service Velocity1 Feed Water 12 LPG (Liquid) 1.53 Condensate(Pressure) 24 Condensate(Gravity) 1.55 Hydraulic Fluid(7Kscg) 25 Hydraulic Fluid(70 Kscg) 2.56 Hydraulic Fluid(140 Kscg) 37 Hydraulic Fluid(210 Kscg) 3.58 Roll Coolant 29 Oil Lubricant 2
10 Grease 211 Fuel oil/LSHS 1.512 Water-Discharge line/General service 213 Water-Pump suction line 1 to 0.514 Water-Pump suction line(Boiling
water)0.5
15 Water- Gravity line 0.315 Water-Sewere system 1 to 0.5
Recommended velocity
Pressure Drop Calculation (Incompressible fluid)
i) D’Arcy Equation
gd
fLvhf
2
4000 2
ORHgR
fLvhf
2
1000 2
Where, d = ID in mm. ; hf = Pressure drop in MLC,
L = Equivalent length in m., V= Velocity in m/s , g = 9.81 m/s2
f = Friction factor ; RH = Hydraulic Radius (See table below)
Shape Hydraulic Radius
Circle, dia= d d/4
Square, side = D D/4
Rectangle, side a & b
Table for Hydraulic Radius (RH)
)2 ba
ab
Table for Hydraulic Radius (RH)
NETWORK SOLUTION BY HARDY CROSS METHOD
Line NBID (d) Direction
Flow (Q) V L hf hf Q1
No. mm mm m3/hr. m/s m m Q m3/hr m3/hr
ab 100 106 + 50 1.57 50 1.32 0.0264 -0.2 49.8bc 50 53.6 + 25 3.08 25 5.07 0.2026 -0.2 24.8cd 50 53.6 + 25 3.08 25 5.07 0.2026 -0.2 24.8de 50 53.6 + 21 2.58 25 3.67 0.1747 -0.2 20.8ef 50 53.6 + 21 2.58 25 3.67 0.1747 -0.2 20.8bh 50 53.6 - 25 3.08 25 -5.1 0.2026 -0.2 -25.2hi 50 53.6 - 25 3.08 50 -10 0.4053 -0.2 -25.2if 50 53.6 - 21 2.58 25 -3.7 0.1747 -0.2 -21.2fg 100 106 + 42 1.32 50 0.96 0.0228 -0.2 41.8dj 25 27.8 - 4 1.83 10 -1.7 0.4163 -0.2 -4.2ik 25 27.8 + 4 1.83 10 1.67 0.4163 -0.2 3.804
0.88 2.419
Q
hf hf
i) v = (4 x 7 x 106 x Q)/( 3600 x 22 x d2)
Working formulae :
852.1
63.2
6.77511)
Cd
QLhfii
Qhf
hfiii
85.1)
Procedure :i) Number each circuit from left to right and then downward.ii) Put diameter, direction of flow, flow and equivalent length of each line.iii) Calculate as per the table below iv) Manipulate the data to make the desired flow as desired point so that the value of the last but before column becomes nearer to zero.
iv)Q1=Q+
Friction factor can be found out from Moody Diagram or by Colebrook & White version equation :
Pressure Drop Calculation (Incompressible fluid), Contd’
a. Moody Chart :
Where, e = Linear measurement of absolute roughness, which can be found out from the table :
Pipe Quality Roughness factor (e)
Cast Iron : 0.2
Commercial black steel : 0.046
Steel (Smooth) : 0.025
Steel (light rust) : 0.25
Steel(Galvanized) : 0.15
Plastic : 0.002
Pressure Drop Calculation (Incompressible fluid), Contd’
b. Colebrook & White version equation
fRd
e
f
255.1
7.3log4
110
VdVd
ForceViscous
ForceInertiaR
Where,Density ; V= Velocity of fluid, m/s
xstokeityVisKinematic 001.cos
d= ID, m
•If RN < 2000, the flow is laminar and then f =16/R
•If R > 3000, the flow is turbulent and f can be calculated by Colebrook & White Version Equation as given below :
If 3000> RN > 2000, the flow is in Transition zone and pressure drop calculation result are better, if turbulent condition is assumed.
Pressure Drop Calculation (Incompressible fluid), Contd’
ii) William’s and Hazen Formula
852.1
63.2
6.77511
Cd
QLhf
Where, hf=head loss in metre Q=flow in m3/hr d=pipe bore in mm L=equivalent length in metre C=a co-efficient representing the roughness of the interior surface of the pipe
Value of C C = 140 for “extremely smooth and straight pipes” with “continuous interior” and
welded or coupled joints, such as new brass, copper, lead, tin, new cast iron, new welded or seamless steel, smooth cement - lined cast iron or steel pipe.
C = 130 for “very smooth” pipes, such as Welded or seamless steel with “continuous interior” in “fair” conditions, New Welded-Steel Pipe with riveted girth joint, new cast iron, old brass, copper, lead, tin.
C = 120 for “smooth” pipes, such as smooth wooden pipes or wood-stave pipes ordinary concrete.
C = 110 - 130 for “new full reveted” steel or wrought-iron pipe, depending on thickness of plate and extent to which rivets are counter sunk.
C = 110 for old cement-lined pipe, or vitrified-crock sewers in good condition.
C = 100 for old cast-iron or “old continuous interior” steel pipes where the carrying capacity over a long period of years is somewhat impaired through tuberculation or sedimentation. For sizes below 6 in, somewhat lower valves should be used.
C. Steam
Line sizing
V
vQd
*8.18 Where, d = Calculated ID in mm, Q = Flow in Kg/hr., V= Velocity in m/s
kgmlumeSpecificVo /3,Recommended velocity for steam
serviceSL.No.
Service Velocity
1 Medium pressure saturated(20 to 40KScg) 272 Low pressure superheated(12Kscg & below) 303 High pressure saturated(100 KSCG) 364 High pressure superheated(100 KSCG) 435 Medium pressure saturated(20 to 40KScg) 306 Medium pressure superheated(20 to 40KScg) 36
Pressure Drop Calculation (Steam)
∆P = 1 91.44 X 6.95 x V x L x W2
d d5
Where, ∆P = Pressure drop in kgf/cm2
d = Pipe Bore in mmL = Equivalent length of pipe in meter
W = Mass flow rate in kg/hrV = Specific volume in m3/kg
D. Equivalent Length
i) Equivalent Length of Fittings and Flanges
Fittings Le/d Fittings Le/d90° S.R. Welding Elbow 20 Flange Joint 690° L.R. Welding Elbow 15 25%Expander 303 D 10 25% Reducer 1590° Standard screwed Elbow 30 50%Expander 20180° return bend 65 50% Reducer 12
Tee with flow through 20 75 % Reducer 12
Tee with flow through branch 65 75% Expander 790° Mitre Bend- 3 cut 16 Vessel outlet 3345 ° Elbow 15 Vessel inlet 6645 ° Mitre 15 5D bend 8
Ex. : Equivalent length of 1 no. 100 NB Flange joint = 6 x 100/1000 = 0.6 M
ii) Equivalent Length of Valves and strainer
Equivalent Length (Contd’)
Valve
Type
Equivalent straight length in m. for NB valve size
25 40 50 80 100 125 150 200 250 300
Gate valve
(open)
0.17 0.25 0.35 0.5 0.67 0.84 1.0 1.33 1.67 2.0
Gate valve
(1/4th Close)
1.0 1.5 2.0 3.0 4.0 5.0 6.0 8.0 10 12
Globe valve
(1/2 close)
5.0 7.5 10 15 20 25 30 40 50 60
Globe Valve
(3/4 close)
20 30 40 60 80 100 120 160 200 240
Globe (Open) 8.3 12.5 16.7 25 33 41.7 50 66.7 83.3 100
Swing Check 2.07 3.12 4.17 6.5 8.25 10.4 12.5 16.6 20.8 25
Foot Valve 0.62 0.93 1.25 1.87 2.47 3.12 3.15 5.0 6.24 7.5
Strainer
( Basket)
1.2 1.4 1.5 1.8 2.4 3.0 3.3 4.2 4.8 5.5
Globe Check 3.8 5.75 7.68 11.5 15.2 14.2 23 30.7 38
Equivalent Length (Contd’)
iii) Equivalent Length of composite network
CA
d1, l1
d4, l4
d3, l3
d5, l5d2, l2
B
Grid AB
A
de1, l2
d2, l2
16
3
1
211
l
ldde
= Eq. dia of l2 length
Now, eq. dia of single l2 long line becomes16
3
3
8
23
8
101
ddd e
Grid BC
Cde3, l3
d3, l3
de4, l3
B16
3
4
343
l
ldd e
16
3
5
354
l
ldd e&
Now equivalent dia of single l3 line becomes,16
3
3
8
43
8
33
8
302
ee dddd
CA
d01, l2 d02, l3B
The equivalent length of section BC in terms of d01
3
16
02
013
d
dlLeq
So, the whole system becomes equivalent to d01 dia and (l2+Leq) length pipe.