INPUT DATA
Parameters Value unit
Pipe Diameter(d) 0.193675 m
Density (p) 81.6942 kg/m3
Viscosity(u) 0.0000144 N/s
Max Pressure(P) drop 3661.1161 N/m2 per 30.48m of pipe
Pipe Roughness (ϵ) 4.57E-05 m
Flow rate(Q) 0.097575905 m3/s
Area(A) 0.0294641094 m2 0.02946
Max Velocity 5 m/s
OUTPUT DATA
Velocity Calculated(v) 3.3116868953 m/s
Criteria Check YES
Reynold's Number 3638744.53609786 Turbulent Flow
Friction Factor f
Churchill 0.014523
Friction Factor f
Colebrook 0.014462
churchhill
1/f^0.5+2log[{(ϵ/D)/3.7} + (7/Re)^0.9]=0
Churchill
A = I/f^0.5 8.29788
B = [{(ϵ/D)/3.7} + (7/Re)^0.9] 7.1E-05
C=2log[{(ϵ/D)/3.7} + (7/Re)^0.9] -8.29779
A+C=0 9.13E-05
Colebrook
1/f^0.5+2log[(ϵ/3.7D) + {2.51/(Rf^0.5)}=0
U = I/f^0.5 8.315566
V = (ϵ/3.7D) 6.38E-05
W= 2.51/(Rf^0.5) 5.74E-06
V+W 6.95E-05
Y=2log[(ϵ/3.7D) + {2.51/(Rf^0.5)}] -8.31556
U+Y=0 4.97E-06
Calculating Pressure Drop for 30.48m(1ft) length of pipe
Using Darcy-Weisbach equation
Using Churchill Equation friction factor
Δp=f*L/D*pv^2/2
h=f* L/D* V^2/2g
Pressure Drop using Churchill Equation friction factor gives us 1023.922 N/m2
0.148507 psi
Pressure Drop using Colebrook Equation friction factor gives us 1019.571 N/m2
0.147876 psi
Step 4 : Selected Pipe Size Check
Is Fluid Velocity < Max. Limit YES
Is Pressure Drop < Max. Limit YES
Criteria Check YES
Line Sizing Calculation
3661.116 N/m^2 /30.48 m Pa/30.48 m
Exercise 1
Model a water pipeline with hand Calculations
Use the data in the table and assume the flow is isothermal.
Perform calculations to determine the delivery pressure of
the pipeline using single-phase flow theory
NOTE: You must have a hand calculator or spreadsheet to
complete this exercise.
Pipeline data
Property Symbol Value Unit
Diameter D 3.068 in 0.26 f
Length L 20025 f 1
Elevation Change Z 1000 f
Horizontal Distance x 20000 f 2
Ambient Temperature Tamb 60 deg F
Inclination Angle q 2.862 deg F 0.04996 radians 3
Roughness ε 0.0018 in
Relative Roughness ε/D 0.0005867 in
Fluid data
Water viscosity mw 1.2 cp 8.06E-04 lb/f-s
Water density rw 63.7 lbm/f3
Operating data 4
Source temperature Tinlet 60 deg F
Inlet pressure Pin 1200 psia
Water flow rate Qw 6000 BPD 0.39 f3/s
Constants 5
Gravitational g 32.2 f/s2
velocity 0
Chart Title
0.0600
0.0500
Chart Title
0.0600
P (psia) T (F) 0.0500
783 20 0.0500
620 100 0.0100 0.0400
421 180 0.0056
0.0300
0.0200
0.0100
0.0000
400 450 500 550 600 650 700
Calculate the following Parameters
Water Velocity v 7.342657 f/s
Reynold's number Re 1.51E+05 Turbulent flow
Friction Factor using Churchill Equation
1/f^0.5+2log[{(ϵ/D)/3.7} + (7/Re)^0.9]=0 A + 2LOG B = 0
Friction Factor f 0.034449
A 1/f^0.5 5.387799
B [{(ϵ/D)/3.7} + (7/Re)^0.9] 2.02E-03
A + 2LOG B = 0 0.0002086351
Evaluate the frictional pressure term fpv^2/2gd Pressure loss form
ΔP/L=f * p/2 * v^2/D 227.52017952 psf/f
Frictional drop per 20025 f of pipe 32762.9058508838 psi/f
Pressure loss form
Total pressure drop in pipe 656077189.663949 psi
Evaluate the elevational pressure term
pgsinφ
sin 0.0499392192 psi
0.0499305532
0.0499714286 0.0499506335
Chart Title
Chart Title
500 550 600 650 700 750 800 850
Turbulent flow
Unit converters
Linear
inches to meters
inch m mm cm f
120 3.048 3048 304.8 10 0.193675 m
velocity(v)
f/s m/s f/hr
5 1.524 18000
Volumetric Flow Rate(Q)
f /s
3
m3/s f3/hr
1 f
Area(A)
Volume(V)
Density(p)
Pressure(P)
Mass(m)
Mass flow rate
Viscosity(u)
