Chapter 15 - Basic Flow equations¶

Example 1 - Pg 407¶

In [1]:
#calcualte the final temperature and pressure of the gas
#Initialization of variables
print '%s' %("From Table B-4,")
h=1187.2 #Btu/lbm
t=328. #F
#calculations
p2=100 #psia
u2=1187.2 #Btu/lbm
t2=540. #F
dt=t2-t
#results
print '%s %d %s' %("Final temperature of steam =",t2,"F")
print '%s %d %s' %("\n Final pressure =",p2,"psia")
print '%s %d %s' %("\n Change in temperature =",dt,"F")

From Table B-4,
Final temperature of steam = 540 F

Final pressure = 100 psia

Change in temperature = 212 F


Example 3 - Pg 409¶

In [2]:
#calculate the work done in the process
#Initialization of variables
import math
p1=100. #psia
p2=14.7 #psia
k=1.4
T1=700. #R
R=10.73/29
V=50.
cv=0.171
cp=0.24
R2=1.986/29.
#calculations
T2=T1/ math.pow((p1/p2),((k-1)/k))
m1=p1*V/(R*T1)
m2=p2*V/(R*T2)
Wrev= cv*(m1*T1 - m2*T2) - (m1-m2)*(T2)*cp
#results
print '%s %d %s' %("Work done in case 1 =",Wrev," Btu")

Work done in case 1 = 572  Btu


Example 4 - Pg 420¶

In [3]:
#calculate the friction of the process
#Initialization of variables
import math
p1=100. #psia
p2=10. #psia
n=1.3
T1=800. #R
cv=0.172
R=1.986/29.
T0=537. #R
cp=0.24
#calculations
T2=T1*math.pow((p2/p1),((n-1)/n))
dwir=cv*(T1-T2)
dwr=R*(T2-T1)/(1-n)
dq=dwr-dwir
#results
print '%s %.1f %s' %("The friction of the process per pound of air =",dq,"Btu/lbm")

The friction of the process per pound of air = 18.6 Btu/lbm


Example 5 - Pg 421¶

In [4]:
#calculate the friction in the process
#Initialization of variables
ms=10 #lbm
den=62.3 #lbm/ft^3
A1=0.0218 #ft^2
A2=0.00545 #ft^2
p2=50. #psia
p1=100. #psia
gc=32.2 #ft/s^2
dz=30. #ft
T0=537. #R
T1=620. #R
T2=420. #R
#calculations
V1=ms/(A1*den)
V2=ms/(A2*den)
df=-144/den*(p2-p1) - (V2*V2 -V1*V1)/(2*gc) - dz
#results
print '%s %.1f %s' %("Friction =",df,"ft-lbf/lbm")

Friction = 72.9 ft-lbf/lbm


Example 6 - Pg 432¶

In [5]:
#calculate the efficiency of the cycle and the loss of available energy
#Initialization of variables
cp1=0.25
T=3460 #R
T0=946.2 #R
T00=520 #R
dG=1228 #Btu/lbm
cp=0.45
#calculations
dqa=cp1*(T-T0)
w=cp*dqa
dg=489.
eff=w/dg*100
dI=-dg+w
#results
print '%s %.1f %s' %("\n Efficiency of cycle =",eff," percent")
print '%s %.1f %s' %("\n Loss of available energy =",dI,"Btu/lbm")

 Efficiency of cycle = 57.8  percent

Loss of available energy = -206.2 Btu/lbm


Example 7 - Pg 434¶

In [6]:
#calculate the engine efficiency and effectiveness. Also, calculate the loss of available energy
#Initialization of variables
p1=400. #psia
t1=600. #F
h1=1306.9 #Btu/lbm
b1=480.9 #Btu/lbm
p2=50 #psia
h2=1122 #Btu/lbm
h3=1169.5 #Btu/lbm
b3=310.9 #Btu/lbm
#calculations
print '%s' %("All the values are obtained from Mollier chart,")
dw13=h1-h3
dw12=h1-h2
dasf=b3-b1
etae=dw13/dw12*100
eta=abs(dw13/dasf)*100
dq=dw13+dasf
#results
print '%s %.1f %s' % ("Engine efficiency =",etae,"percent")
print '%s %.1f %s' %("\n Effectiveness =",eta,"percent")
print '%s %.1f %s' %("\n Loss of available energy  =",dq,"Btu/lbm")

All the values are obtained from Mollier chart,
Engine efficiency = 74.3 percent

Effectiveness = 80.8 percent

Loss of available energy  = -32.6 Btu/lbm