Chapter 10: Hydraulic Conductors and Fittings¶

Example 10.1 pgno:367¶

In [1]:
# Aim:To find minimum inside diameter of pipe
# Given:
# flow-rate through pipe:
Q=30.0; #gpm
# average fluid velocity:
v=20.0; #ft/s

import math
from math import pi

# Solution:
# flow-rate in ft**3/s,
Q_fps=Q/449; #ft**3/s
# minimum required pipe flow area,
A=(Q_fps/v)*144; #in**2
# minimum inside diameter,
D=((4*A)/(pi))**0.5; #in

# Results:
print"\n  Results:  "
print"\n The minimum inside diameter of pipe is in.",round(D,3)

  Results:

The minimum inside diameter of pipe is in. 0.783


Example 10.2 pgno:368¶

In [2]:
# Aim:To find minimum inside diameter of pipe in Metric units
# Given:
# flow-rate through pipe:
Q=0.002; #m^3/s
# average fluid velocity:
v=6.1; #m/s

from math import pi

# Solution:
# minimum required pipe flow area,
A=(Q/v); #m^2
# minimum inside diameter,
D=(((4*A)/(pi))**0.5)*1000; #mm

# Results:
print"\n  Results:  "
print"\n The minimum inside diameter of pipe is  mm.",round(D,1)

  Results:

The minimum inside diameter of pipe is  mm. 20.4


Example 10.3 pgno:371¶

In [3]:
# Aim:To find safe working pressure for the tube
# Given:
# outside diameter of steel tube:
Do=1.250; #in
# inside diameter of steel tube:
Di=1.060; #in
# tensile strength of steel tube:
S=55000.0; #psi
# factor of safety:
FS=8.0;

# Solution:
# wall thickness,
t=(Do-Di)/2; #in
# burst pressure,
BP=(2*t*S)/Di; #psi
# working pressure,
WP=BP/FS/1000; #psi

# Results:
print"\n  Results:  "
print"\n The working pressure of steel tube is  psi.",round(WP,3)
print"\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation"

  Results:

The working pressure of steel tube is  psi. 1.232

The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation


Example 10.4 pgno:380¶

In [4]:
# Aim:Refer Example 10-4 for Problem Description
# Given:
# flow-rate:
Q=30; #gpm
# operating pressure:
p=1000; #psi
# maximum velocity:
v=20; #ft/s
# tensile strength of material:
S=55000; #psi
# factor of safety:
FS=8;
from math import pi
# Solutions:
# flow-rate,
Q=Q/449; #ft^3/s
# minimum required pipe flow area,
Ai=(Q/v)*144; #in^2
# minimum inside diameter,
Di=((4*Ai)/(pi))**0.5; #in
# wall thickness,
t1=0.049; t2=0.065; #in
# tube inside diameter,
D1=0.902; D2=0.870; #in
# burst pressure,
BP1=(2*t1*S)/D1; #psi
# working pressure,
WP1=BP1/FS; #psi
print" \n The working pressure  psi is not adequate (less than  psi) so next case is considered,",WP1,p
# burst pressure,
BP2=(2*t2*S)/D2; #psi
# working pressure,
WP2=BP2/FS; #psi
# ratio of inner diameter to thickness,
r2=D2/t2;
print" \n The working pressure  psi is greater than  psi) ,",WP2,p

# Results:
print"\n  Results:  "
print"\n The ratio of inner diameter to length is .",round(r2,1)
print"\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation"


The working pressure  psi is not adequate (less than  psi) so next case is considered, 746.951219512 1000

The working pressure  psi is greater than  psi) , 1027.29885057 1000

Results:

The ratio of inner diameter to length is . 13.4

The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation


Example 10.5 pgno:389¶

In [5]:
#Aim:Refer Example 10-5 for Problem Description
# Given:
# flow-rate:
Q=0.00190; #m^3/s
# operating pressure:
p=70.00000000000; #bars
# maximum velocity:
v=6.10000000000; #m/s
# tensile strength of material:
S=379.0000000; #MPa
# factor of safety:
FS=8.0000000;
from math import pi

# Solutions:
# minimum required pipe flow area,
A=(Q/v); #m^2
# minimum inside diameter,
ID=(((4*A)/(pi))**0.5)*1000; #mm
# wall thickness,
t1=1.; t2=2.; #mm
# tube inside diameter,
D1=20.; D2=24.0; #mm
# burst pressure,
BP1=(2*(t1/1000)*S)/(D1/1000); #MPa
# working pressure,
WP1=(BP1/FS)*10; #bars
print" \n The working pressure bars is not adequate (less than bars) so next case is considered,",round(WP1,1)
# burst pressure,
BP2=(2*(t2/1000)*S)/(D2/1000); #MPa
# working pressure,
WP2=(BP2/FS)*10; #MPa
# ratio of inner diameter to thickness,
r2=D2/t2;
print" \n The working pressure  bars is greater than bars) ,",round(WP2)

# Results:
print"\n  Results:  "
print"\n The ratio of inner diameter to length is ",r2


The working pressure bars is not adequate (less than bars) so next case is considered, 47.4

The working pressure  bars is greater than bars) , 79.0

Results:

The ratio of inner diameter to length is  12.0