# Chapter 4:- Control Volume Analysis Using Energy¶

## Example 4.1 Page no-125

In [1]:
# Given:-
# At inlet 1:-
p1= 7.0                                        # pressure in bar
T2= 200.0                                      # temperature in degree celcius
m1dot= 40.0                                    # mass flow rate in kg/s

# At inlet 2:-
p2= 7.0                                        # pressure in bar
T2= 40.0                                       # temperature in degree celcius
A2= 25.0                                       # area in cm^2

# At exit:-
p3= 7.0                                        # pressure in bar
AV3= 0.06                                      # Volumetric flow rate through wxir in m^3/s

# From table A-3
v3 = (1.108)*(10**(-3))                        # specific volume at the exit in m^3/kg

# from table A-2
v2= (1.0078)*(10**(-3))                        # specific volume in state 2 in m^3/kg

# Calculation:-
m3dot= AV3/v3                                  # mass flow rate at exit
m2dot = m3dot-m1dot                            # mass flow rate at inlet 2
V2= (m2dot*v2)/(A2*(10**(-4)))

# Results:-
print '-> The mass flow rate at the inlet 2 is', round(m2dot,2),' kg/s.'
print '-> The mass flow rate at the exit is', round(m3dot,2),' kg/s.'
print '-> The velocity at the inlet is ', round(V2,2),'m/s.'

-> The mass flow rate at the inlet 2 is 14.15  kg/s.
-> The mass flow rate at the exit is 54.15  kg/s.
-> The velocity at the inlet is  5.7 m/s.

## Example 4.3 Page no-135

In [2]:
# Given:-
p1= 40.0                                     # pressure in bar
T1= 400.0                                    # temperature in degree celcius
V1= 10.0                                     # velocity m/s

# At exit:-
p2= 10.0                                     # pressure in bar
V2= 665.0                                    # velocity in m/s
mdot= 2.0                                    # mass flow rate in kg/s

# From table A-4
h1= 3213.6                                   # snpecific enthalpy in kJ/kg
v2 = 0.1627                                  # specific volume at the exit in m^3/kg

# Calculation:-
h2 = h1 + ((V1**2-V2**2)/2)/1000             # snpecific enthalpy in kJ/kg
A2=(mdot*v2)/V2                              # Exit area

# Results:-
print '-> The exit Area of the nozzle is', round(A2,4),'m^2.'

-> The exit Area of the nozzle is 0.0005 m^2.

## Example 4.4 Page no-138

In [3]:
# Given:-
m1dot = 4600.0                                    # mass flow rate in kg/h
Wcvdot= 1000.0                                    # turbine power output in kv
p1= 60.0                                          # pressure in bar
T1=400.0                                          # temperature in degree celc
V1= 10.0                                          # velocity in m/s

# At exit:-
p2= 0.10                                          # pressure in bar
q2= 0.90                                          # quality
V2= 50.0                                          # velocity in m/s

# From table A-2 and A-3:-
h1= 3177.2                                        # specific enthalpy at inlet in kJ/kg
hf2= 191.83
hg2= 2584.63

# Calculation:-
h2 = hf2+q2*(hg2-hf2)                             # specific enthalpy at exit in kJ/kg
Qcvdot = Wcvdot + m1dot*((h2-h1)+(V2**2- V1**2)/(2*1000))/3600

# Results:-
print '-> The rate of heat transfer between the turbine and surroundings is',round(Qcvdot,2),'kW.'

-> The rate of heat transfer between the turbine and surroundings is -61.39 kW.

## Example 4.5 Page no-140

In [4]:
# Given:-
p1=1.00                                         # pressure in bar
t1= 290.00                                      # temperature in kelvin
A1= 0.1                                         # area in m^2
V1= 6.00                                        # velocity in m/s

# At exit:-

p2=7.00                                         # pressure in bar
t2= 450.00                                      # temperature in kelvin
V2= 2.00                                        # velocity in m/s
Qcvdot= -180.0                                  # heat transfer rate in kJ/min
R= 8.314                                        # universal gas constant in SI units

# from table A-22

h1= 290.16                                      # specific enthalpy in kJ/kg
h2= 451.8                                       # specific enthalpy in kJ/kg

# Calculations:-

v1 =  (R*1000*t1)/(28.97*p1*10**5)              # specific volume
mdot=(A1*V1)/v1                                 # mass flow rate
Wcvdot = Qcvdot/60 + mdot*((h1-h2)+(V1**2-V2**2)/(2*1000))

# Results:-

print '-> The power input to the compressor is ',round(Wcvdot,2),'kW.'

-> The power input to the compressor is  -119.52 kW.

## Example 4.6 Page no-141

In [5]:
# Given:-
# At Entry:=
t1=20.0                                          # Temperatue in deg celcius
p1=1.0                                           # pressure in atm
AV1= 0.1                                         # volumetric flow rate in litre/s
D1=2.5                                           # Diameter of th hose in cm

# At Exit:=
t2=23.0                                          # temperatuer in deg celcius
p2=1.0                                           # pressure in atm
V2=50.0                                          # Velocity in m/s
Z2=5.0                                           # elevation in m
g= 9.8                                           # acceleration due to gravity in m/s^2

# from table A-2 and A-19:-

v= (1.0018)*((10.0)**(-3))                       # specific volume in m^3/kg
c= 4.18

# Calculation:-
mdot = (AV1/1000)/v                              # mass flow rate in kg/s
V1= (AV1/1000)/(3.14*(D1/(2*100))**2)            # Entry velocity in m/s
deltah = c*(t2-t1)+v*(p2-p1)
Wcvdot= ((mdot*10)/9)*(-deltah+(V1**2-V2**2)/(2*1000)+g*(0-Z2)/1000)

# Results:-
print '-> The power input to the motor is', round(Wcvdot,2),'KW.'

-> The power input to the motor is -1.53 KW.

## Example 4.7:- Page no-144

In [6]:
# Given:-
# Entering:-
p1=0.1                                        # pressure in bar
x1= 0.95                                      # Quality
p2= 0.1                                       # pressure in bar
t2= 45.0                                        # temperature in deg celcius
t3=20.0                                         # temperature of cooling entry in deg cel
t4=35.0                                         # temperature of cooling exit

# From table A-3
hf= 191.53                                    # Enthalpy in KJ/kg
hg= 2584.7                                    # Enthalpy in KJ/kg
h2=188.45                                     # Assumption at states 2,3 and 4, h is approx equal to hf(T), in kJ/kg
deltah4_3= 62.7                               # Assumption 4, in kJ/kg

# Calculations:-
h1= hf + x1*(hg-hf)
ratio= (h1-h2)/(deltah4_3)
QRate= (h2-h1)                                # Part B

# Results:-
print '-> The rate of the mass flow rate of the cooling water to the mass flow rate of the condenstaing stream is (m3dot/m1dot)',round(ratio,2)
print '-> The rate of energy transfer from the condensing steam to the cooling water of the steam passing through the condenser is',round(QRate,2),'kJ/kg.'

-> The rate of the mass flow rate of the cooling water to the mass flow rate of the condenstaing stream is (m3dot/m1dot) 36.31
-> The rate of energy transfer from the condensing steam to the cooling water of the steam passing through the condenser is -2276.59 kJ/kg.

## Example 4.8:- Page no-146

In [7]:
# Given:-
T1 = 293.0                                            # In kelvin
P1= 1.01325 * (10**5)                                 # In pascal
V1max= 1.3                                            # maximum velocity of entering air in m/s
T2max= 305.0                                          # maximum temperature at the exit in kelvin
pec= -80.0                                            # power received by electronic components in watt
Pf= -18.0                                             # Power received by fan in watt
R= 8.314                                              # Universal gas constant
M= 28.97*(10**(-3))                                   # Molar mass of air in kg
Qcvdot=0                                              # Heat transfer from the outer surface of the electronics enclosure to the surroundings is negligible.
Cp= 1.005*(10**3)                                     # in j/kg*k
pi=3.14

# Calculations:-

Wcvdot = pec +Pf                                      # total electric power provided to electronic components and fan in watt
mdotmin=  (-Wcvdot)/(Cp*(T2max-T1))                   # minimum mass flow rate
v1= ((R/M)*T1)/P1                                     # specific volume
A1min = (mdotmin*v1)/V1max
D1min = (4*A1min/(pi))**(0.5)

# Results:-
print '-> The smallest fan inlet diameter is',round(D1min*100,2),'cm.'

-> The smallest fan inlet diameter is 8.13 cm.

## Example 4.9 Page no-148

In [8]:
# Given:-
P1 = 20.0                             # pressure in supply line in bars
P2 = 1.0                              # exhaust pressure in bar
T2 = 120.0                            # exhaust temperature in degree celcius

# from table A-3 at 20 bars
hf1 = 908.79                        # Enthalpy in kj/kg
hg1 = 2799.5                        # Enthalpy in kj/kg

# from table A-4, at 1 bar and 120 degree celcius
h2 = 2766.6                         # in kj/kg
h1 = h2                             # from throttling process assumption

# Calculations:-
x1 = (h1-hf1)/(hg1-hf1)

# Results:-
print '-> The quality of the steam in the supply line is',round(x1,2)

-> The quality of the steam in the supply line is 0.98

## Example 4.10 Page no-150

In [9]:
# Given:-
P1 = 1.0                      # pressure of industrial discharge in bar
T1 = 478.0                    # temperature of industrial discharge in kelvin
m1dot = 69.78                 # mass flow rate of industrial discharge in kg/s
T2 = 400.0                    # temperature of exit products from steam generator in kelvin
P2 = 1.0                      # pressure of exit products from steam generator in bar
P3 = 0.275                    # pressure of water stream entering the generator in Mpa
T3 = 38.9                     # temperature of water stream entering the generator in degree celcius
m3dot = 2.079                 # mass flow rate of water stream entering in kg/s
P5 = 0.07                     # exit pressure of the turbine in bars
x5 = 0.93                     # quality of turbine exit

# Part (a)
m2dot = m1dot                 # since gas and water streams do not mix
m5dot = m3dot                 # --DO

# from table A-22, A-2 and A-3:-
h1 = 480.3                    # in kj/kg
h2 = 400.98                   # in Kj/kg
h3 = 162.9                    # assumption: h3 = hf(T3), units in Kj/kg
hf5 = 161.0                   # in kj/kg
hg5 = 2571.72                 # in kj/kg

# Part (b)
P4 = P3                       # from the assumption that there is no pressure drop for water flowing through the steam generator
T4 = 180                      # in degree celcius

# Calculations:-
h5 = hf5 + x5*(hg5-hf5)
Wcvdot = m1dot*h1 + m3dot*h3 - m2dot*h2 - m5dot*h5
h4 = h3 + (m1dot/m3dot)*(h1 -h2)    # from steady state energy rate balance
# interpolating in table A-4, with these P4 and h4
# Results:-
print '-> The power developed by the turbine is ',round(Wcvdot,2),'kJ/s.'
print '-> Turbine inlet temperature is',round(T4,2),'degree celcius.'

-> The power developed by the turbine is  877.84 kJ/s.
-> Turbine inlet temperature is 180.0 degree celcius.

## Example 4.11 Page no-153

In [10]:
# Given:-
V = 0.85                               # volume of tank in m^3
T1 = 260.0                             # initial temperature of the tank in degree celcius
X1 = 0.7                               # initial quality

# from table A-2
uf1 = 1128.4                           # in kg/kg
ug1 = 2599.0                           # in kg/kg

vf1 = 1.2755e-3                        # in m^3/kg
vg1 = 0.04221                          # in m^3/kg

# for final state, from table A-2,
u2 = 2599.0                             # units in KJ/kg
v2 = 42.21e-3                           # units in m^3/Kg
he = 2796.6                             # units in KJ/kg

# Calculations:-
u1 = uf1 + X1*(ug1-uf1)                 # in kj/kg
v1 = vf1 + X1*(vg1-vf1)                 # in m^3/kg
m1 = V/v1                               # initial mass in kg
m2 =  V/v2                              # final mass in kg
U2 = m2*u2                              # final internal energy in KJ
U1 = m1*u1                              # initial internal energy in KJ
Qcv = (U2-U1) - he*(m2-m1)

# Results:-
print '-> The amount of heat transfer is',round(Qcv,2),'KJ.'

-> The amount of heat transfer is 14162.16 KJ.

## Example 4.12 Page no-155

In [11]:
# Given:-
Pv = 15.0                                            # pressure in the vessel in bar
Tv = 320.0                                           # temperature in the vessel in degree celcius
Vt = 0.6                                             # volume of a tank in m^3
Tt = 400.0                                           # temperature in the tank in degree celcius when the tank is full

# Since the tank is initially empty:-
m1 = 0
u1 = 0

# From table A-4, at 15bar and 400 degree celcius:-
v2 = 0.203                                           # Volume in m^3/kg
m2 = Vt/v2                                           # mass within the tank at the end of the process in kg
hi = 3081.9                                          # in kj/kg
u2 = 2951.3                                          # in kj/kg

# Calculations:-
deltaUcv = m2*u2-m1*u1
Wcv = hi*(m2-m1)-deltaUcv

# Results:-
print '-> The amount of work developed by the turbine is ',round(Wcv,2),'kJ.'

-> The amount of work developed by the turbine is  386.01 kJ.