Chapter 7: Energy and Energy Balances¶

Example 7.2-1, page no. 318¶

In [1]:
#Initialization of variables
import math
import numpy
from numpy import linalg

ID=2.0 #cm
Vdot=2.0 #m^3/h

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
u=Vdot*100*100 /(math.pi*math.pow(ID/2,2) *3600)
mdot=Vdot*math.pow(10,3) /3600.
Ek=mdot*math.pow(u,2) /2
print '%s %.3f' %(" \n Ek (J/s) = ",Ek)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook

Ek (J/s) =  0.869
press enter key to exit

Out[1]:
''

Example 7.2-2, page no. 318¶

In [2]:
#Initialization of variables
import math
import numpy
from numpy import linalg

g=9.81 #m/s^2
mdot=15.0 #Kg/s
z2=20.0 #m
z1=-220. #m

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
Power=mdot*g*(z2-z1)
print '%s %d' %(" \n Power (J/s) = ",Power)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook

Power (J/s) =  35316
press enter key to exit

Out[2]:
''

Example 7.3-1, page no. 319¶

In [3]:
#Initilization of variables
Q1=2. #Kcal
Ek1=0#System is stationary
Ep1=0#No vertical displacement
W1=0#No moving boundaries
deltaU1=Q1-W1-Ek1-Ep1
deltaU1J=deltaU1*1000./0.23901
print '%s %.2f' %("Energy gained by the gas in going from 25C to 100C is (J)",deltaU1J)
W2=100. #J
Ek2=0 #The system is stationary at the initial and final states
Ep2=0#Assumed negligible by hypothesis
deltaU2=0#U depends only on T for an ideal gas and T doesn't change
Q2=deltaU2+W2+Ek2+Ep2
print '%s %.1f' %("Heat transferred to the gas while it expands and reequilibrates at 100C is (J)",Q2)
raw_input("Press the Enter key to quit")

Energy gained by the gas in going from 25C to 100C is (J) 8367.85
Heat transferred to the gas while it expands and reequilibrates at 100C is (J) 100.0
Press the Enter key to quit

Out[3]:
''

Example 7.4-1, page no. 322¶

In [4]:
#Initialization of variables
import math
import numpy
from numpy import linalg

U=3800.0 #J/mol
P=1.0 #atm
Vcap=24.63 #L/mol
ndot=250.0 #Kmol/h

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
Hcap=U+P*Vcap*101.3
H=ndot*Hcap*math.pow(10,3)
print '%s %d' %(" \n Specific Enthalpy (J/mol) = ",Hcap)
print '%s %.3E' %("\n Enthalpy of Helium (J/h) = ",H)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook

Specific Enthalpy (J/mol) =  6295

Enthalpy of Helium (J/h) =  1.574E+09
press enter key to exit

Out[4]:
''

Example 7.4-2, page no. 324¶

In [6]:
#Initialization of variables
import math
import numpy
from numpy import linalg

mdot=500.0/3600.#Kg/s
u1=60.0 #m/s
u2=360.0 #m/s
deltaZ=-5. #m
g=9.81 #m/s^2
Qdot= -10000.#Kcal/h
Ws=70.0 #KW

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
Ek=mdot*math.pow(10,-3)*(math.pow(u2,2)-math.pow(u1,2))/2
Ep=mdot*g*deltaZ/math.pow(10,3)
Qdot=Qdot/(0.239*3600.)
Hdot=Qdot-Ws-Ek-Ep
print '%s %f' %(" \n DeltaH (KW) = ",Hdot)
Hcap=Hdot/mdot
print '%s %f' %("\n Specific Enthalpy (Kj/Kg) = ",Hcap)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook

DeltaH (KW) =  -90.365689

Specific Enthalpy (Kj/Kg) =  -650.632958
press enter key to exit

Out[6]:
''

Example 7.5-1, page no. 326¶

In [8]:
#Initialization of variables
import math
import numpy
from numpy import linalg

H0=196.23 #Btu/lbm
H50=202.28 #Btu/lbm
Pfinal=51.99 #psia
Pinitial=18.90 #psia
Vfinal=1.920 #ft^3/lbm
Vinitial=4.969 #ft^3/lbm

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
deltaH=H0-H50
deltaU=deltaH+((Pfinal*Vfinal-Pinitial*Vinitial)*1.987/10.73)
print '%s %.3f' %(" \n change in Specific Enthalpy (Btu/lbm) = ",deltaH)
print '%s %.3f' %(" \n change in Specific Internal Energy (Btu/lbm) = ",deltaU)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook

change in Specific Enthalpy (Btu/lbm) =  -6.050

change in Specific Internal Energy (Btu/lbm) =  -4.956
press enter key to exit

Out[8]:
''

Example 7.5-3, page no. 329¶

In [9]:
#Initialization of variables
import math
import numpy
from numpy import linalg

mdot=2000.0 #Kg/h
P=10.0 #bar

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
print("From Steam tables, ")
Hin=3201. #Kj/Kg
Hout=2675. #Kj/Kg
Ws= -mdot*(Hout-Hin)/3600.
print '%s %d' %("Work delivered by Turbine to surroundings (Kw) = ",Ws)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook
From Steam tables,
Work delivered by Turbine to surroundings (Kw) =  292
press enter key to exit

Out[9]:
''

Example 7.6-1, page no. 330¶

In [10]:
#Initialization of variables
import math
import numpy
from numpy import linalg

m1=120.0 #kg
m2=175.0 #kg
m3=295.0 #kg
ID=6.0 #cm
P=17.0 #bar
H1=125.7 #Kj/Kg
H2=271.9 #Kj/Kg
H3=2793.0 #Kj/kg

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
deltaH=m3*H3-m1*H1-m2*H2
print("From tables, Vdot=0.1166 m^3/kg")
Vdot=0.1166
A=math.pi*math.pow(ID/2,2) /math.pow(10,4)
u=m3*Vdot/(A*60)
Ek=m3*math.pow(u,2) /(2*math.pow(10,3))
Qdot=deltaH+Ek
print '%s %.3E' %("Heat required (Kj/min) = ",Qdot)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook
From tables, Vdot=0.1166 m^3/kg
Heat required (Kj/min) =  7.673E+05
press enter key to exit

Out[10]:
''

Example 7.6-2, page no. 331¶

In [11]:
#Initialization of variables
import math
import numpy
from numpy import linalg

basis=1.0 #Kg/s
x=0.6 #ethane
T1=150.0 #K
T2=250.0 #K
P=5.0 #bar
Hout1=434.5 #KJ/Kg
Hout2=130.2 #KJ/Kg
Hin1=314.3 #KJ/Kg
Hin2=30.0 #KJ/Kg

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
Qdot=basis*(x*Hout1+(1-x)*Hout2-x*Hin1-(1-x)*Hin2)
print '%s %.3f' %(" \n Heat required (KJ/Kg) = ",Qdot/basis)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook

Heat required (KJ/Kg) =  112.200
press enter key to exit

Out[11]:
''

Example 7.6-3, page no. 332¶

In [12]:
#Initialization of variables
import math
import numpy
from numpy import linalg

m3=1150.0 #Kg/h
H3=2676.0 #KJ/Kg
H2=3074.0 #KJ/Kg
H1=3278.0 #KJ/Kg

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
print("Mass balance on Water,")
print("m3+m1=m2")
print("Energy balance,")
print("m3*H3+m1*H1=m2*H2")
A=([[1,-1],[H2,-H1]])
b=([[m3],[m3*H3]])
C=numpy.dot(linalg.inv(A),b)
#here we solved two linear equations simultaneously.
m2=C[0,0]
m1=C[1,0]
print '%s %.3f' %(" Input flowrate,m1 (Kg/h) = ",m1)
print '%s %.3f' %(" \n Output flowrate, m2 (Kg/h) = ",m2)
print("From tables,Vdot=3.11 m^3/Kg")
Vdot=3.11
print '%s %.3f' %(" Volumetric input flowrate (m^3/h) = ",m1*Vdot)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook
Mass balance on Water,
m3+m1=m2
Energy balance,
m3*H3+m1*H1=m2*H2
Input flowrate,m1 (Kg/h) =  2243.627

Output flowrate, m2 (Kg/h) =  3393.627
From tables,Vdot=3.11 m^3/Kg
Volumetric input flowrate (m^3/h) =  6977.681
press enter key to exit

Out[12]:
''

Example 7.7-1, page no. 334¶

In [13]:
#Initialization of variables
import math
import numpy
from numpy import linalg

Vdot=20.0 #L/min
P2=1.01325*100000. #atm
ID1=0.5 #cm
ID2=1.0 #cm
g=9.81 #m/s^2
deltaZ=50.0 #m

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
u1=Vdot*math.pow(10,4) /(math.pow(10,3) *60*math.pi*math.pow(ID1/(2),2))
u2=Vdot*math.pow(10,4) /(math.pow(10,3) *60*math.pi*math.pow(ID2/(2),2))
deltaP=-((u2*u2-u1*u1)/2 +g*deltaZ)*1000.
P1=P2-deltaP
print '%s %.3f' %(" \n P1 (Pa) = ",P1)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook

P1 (Pa) =  456730.088
press enter key to exit

Out[13]:
''

Example 7.7-2, page no. 335¶

In [14]:
#Initialization of variables
import math
import numpy
from numpy import linalg

deltaZ= -2.5 #ft
u1=0.0
D=50.0 #lbm/ft^3
F=0.80 #ft.lbf/lbm
V=5.0 #gal
g=32.174 #ft/s^2
ID=0.25 #in

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
u2=math.sqrt(2*32.174*(-F-g*deltaZ/32.174))
Vdot=u2*math.pi*math.pow(ID/(2),2) /144.
t=V*0.1337/(Vdot*60.)
print '%s %.3f' %("Total time taken (min) = ",t)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook
Total time taken (min) =  3.125
press enter key to exit

Out[14]:
''

Example 7.7-3, page no. 336¶

In [15]:
#Initialization of variables
import math
import numpy
from numpy import linalg

Ws=1000000. #N.m/s
deltaP= -83.*1000. #N/m^2
g=9.81 #m/s^2
deltaZ= -103 #m
D=1000. #kg/m^3

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
mdot= -Ws/(deltaP/D + g*deltaZ)
print'%s %.3f' %(" \n Water flow rate (kg/s) = ",mdot)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook

Water flow rate (kg/s) =  914.553
press enter key to exit

Out[15]:
''