Chapter 12 : System components

Example 12.1 Page No 290

In [1]:
import math

#Variable initialization
t=4.0               #hr time
Ihp=8.0             #ihp power
Ohp=5.0             #hp power

#CALCULATIONS
Hl=t*2544.0*(Ihp-Ohp)

#RESULT
print(' Total Btu heat loss over a period of 4 hr = %.3f Btu ' %Hl)

 Total Btu heat loss over a period of 4 hr = 30528.000 Btu

Example 12.2 Page No 292

In [3]:
import math

#Variable initialization
t=1.0               #sec time
P=1000.0            #lbf/in^2 pressure 
Q=3.0               #gpm flow rate
Sg=0.85             # density
s=0.42

#CALCULATIONS
H=42.4*((P*Q)/1714)     #Heat Generation Rate
m=8.34*Q*Sg             #mass flow Rate
Tr=H/(m*s)

#RESULTS
print(' Rise in temperature of the fluid = %.2f F' %Tr)

 Rise in temperature of the fluid = 8.31 F

Example 12.3 Page No 296

In [3]:
import math

#Variable initialization
P=1500.0            #lbf/in^2 pressure
d=12.0              #in diameter
V=50.0              #gal volume

#CALCULATIONS
F=P*(math.pi*d**2/4)
S=V*231.0*4.0/(math.pi*d**2)

#RESULTS
print('Weight = %.3f lbf ' %F)
print('\n Stroke length = %.3f in ' %S)

Weight = 169646.003 lb 

 Stroke length = 102.124 in

Example 12.4 Page No 298

In [4]:
import math

#Variable initialization
P=1500.0          #lbf/in^2 pressure
V=5.0             #gal volume
P1=3000.0         #lbf/in^2 pressure
P2=2000.0         #lbf/in^2 pressure

#CALCULATIONS
V2=V*231.0*(P2+14.7)/(P1-P2)
V1=V2*(P1+14.7)/((P+14.7)*231.0)

#RESULT
print('\n Size of accumulator = %.3f gal ' %V1)

 Size of accumulator = 20.049 gal

Example 12.5 Page No 300

In [1]:
import math

#Variable initialization
beta=1.4
p3=2000.0+14.7          #non guage presure
p2=3000.0+14.7          #non guage presure
p1=1500.0+14.7          #non guage presure
deltav=1155.0

#Calculations
v2=(p3/p2)**(1/beta)*(deltav)/(1-(p3/p2)**(1/beta))
v1=v2*(p2/p1)**(1/beta)
perdiff=(v1-4627.25)*100.0/v1

#Results
print(' Volume 2 = %.3f in^3' %v2)
print(' \n Volume 1 = %.3f in^3' %v1)
print(' \n Percentage difference in volume = %.3f percent' %perdiff )

 Volume 2 = 3462.278 in^3
 
 Volume 1 = 5660.756 in^3
 
 Percentage difference in volume = 18.257 percent

Example 12.6 Page No 305

In [6]:
import math

#Variable initialization
Fr=20.0             #gpm flowrate
P=2500.0            #lbf/in^2 pressure
sf=4.0              #safety factor
Ts=55000.0          #lbf/in^2 pressure
V=15.0              #ft/s velocity

#CALCULATIONS
A=Fr*0.3208/V
ID=2.0*math.sqrt(A/math.pi)
Wt=P*ID/(2*(Ts-P))
Wt1=Wt*sf

#RESULT
print('Wall thcikness = %.3f in ' %Wt1)

Wall thcikness = 0.070 in