# Chapter 12 :- Ideal Gas Mixtures: General Considerations¶

## Example 12.1 Page no-560

In [1]:
# Given:-
n1 = 0.08                                                                        # mole fraction of CO2
n2 = 0.11                                                                        # mole fraction of H2O
n3 = 0.07                                                                        # mole fraction of O2
n4 = 0.74                                                                        # mole fraction of N2

# Part(a)
M1 = 44.0                                                                        # molar mass of CO2 in kg/kmol
M2 = 18.0                                                                        # molar mass of H2O in kg/kmol
M3 = 32.0                                                                        # molar mass of O2 in kg/kmol
M4 = 28.0                                                                        # molar mass of N2 in kg/kmol

# Calculations
M = M1*n1 + M2*n2 + M3*n3 + M4*n4                                                # in kg/kmol
# Result
print 'The apparent molecular weight of the mixture in kg/kmol is:  ',M

# Part(b)
mf1 = (M1*n1/M)*100.0                                                             # mass fraction of CO2 in percentage
mf2 = (M2*n2/M)*100.0                                                             # mass fraction of H2O in percentage
mf3 = (M3*n3/M)*100.0                                                             # mass fraction of O2 in percentage
mf4 = (M4*n4/M)*100.0                                                             # mass fraction of N2 in percentage

# Results
print 'The mass fraction of CO2 in percentage is:  ',mf1
print 'The mass fraction of H2O in percentage is:  ',mf2
print 'The mass fraction of O2 in percentage is:  ',mf3
print 'The mass fraction of N2 in percentage is:  ',mf4

The apparent molecular weight of the mixture in kg/kmol is:   28.46
The mass fraction of CO2 in percentage is:   12.3682361209
The mass fraction of H2O in percentage is:   6.95713281799
The mass fraction of O2 in percentage is:   7.87069571328
The mass fraction of N2 in percentage is:   72.8039353479

## Example 12.2 Page no-561

In [2]:
# Given:-
mf1 = 0.1                                                                        # mass fractiion of H2
mf2 = 0.6                                                                        # mass fraction of N2
mf3 = 0.3                                                                        # mass fraction of CO2

# Part(a)
M1 = 2.0                                                                          # molar mass of H2 in kg/kmol
M2 = 28.0                                                                         # molar mass of N2 in kg/kmol
M3 = 44.0                                                                         # molar mass of CO2 in kg/kmol

# Calculations
n1 = (mf1/M1)/(mf1/M1 + mf2/M2 + mf3/M3)                                          # mole fraction of H2
n2 = (mf2/M2)/(mf1/M1 + mf2/M2 + mf3/M3)                                          # mole fraction of N2
n3 = (mf3/M3)/(mf1/M1 + mf2/M2 + mf3/M3)                                          # mole fraction of CO2

# Results
print 'The mole fraction of H2 in percentage is:  ',n1*100
print 'The mole fraction of N2 in percentage is:  ',n2*100
print 'The mole fraction of CO2 in percentage is:  ',n3*100

# Part(b)
# Calculation
M = n1*M1 + n2*M2 + n3*M3                                                         # in kg/kmol
# Result
print 'The apparent molecular weight of the mixture in kg/kmol is:  ',M

The mole fraction of H2 in percentage is:   63.9004149378
The mole fraction of N2 in percentage is:   27.3858921162
The mole fraction of CO2 in percentage is:   8.71369294606
The apparent molecular weight of the mixture in kg/kmol is:   12.7800829876

## Example 12.3 Page no-569

In [3]:
# Given:-
m1 = 0.3                                                                         # mass of CO2 in kg
m2 = 0.2                                                                         # mass of N2 in kg
p1 = 1.0                                                                         # in bar
T1 = 300.0                                                                       # in kelvin
p2 = 3.0                                                                         # in bar
n = 1.25

# Part(a)
# Calculation
T2 = T1*(p2/p1)**((n-1)/n)                                                      # in kelvin
# Result
print 'The final temperature in Kelvin is:  ',T2

# Part(b)
Rbar = 8.314                                                                     # universal gas constant in SI units
# Calculations
M = (m1+m2)/(m1/44 + m2/28)                                                      # molar mass of mixture in kg/kmol
W = ((m1+m2)*(Rbar/M)*(T2-T1))/(1-n)                                             # in kj
# Result
print 'The work in kj is:  ',W

# Part(c)
# From table A-23
uCO2T1 = 6939.0                                                                  # internal energy of CO2 on molar mass basis at temperature T1
uCO2T2 = 9198.0                                                                  # internal energy of CO2 on molar mass basis at temperature T2
uN2T1 = 6229.0                                                                   # internal energy of N2 on molar mass basis at temperature T1
uN2T2 = 7770.0                                                                   # internal energy of N2 on molar mass basis at temperature T2
deltaU = (m1/44)*(uCO2T2-uCO2T1) + (m2/28)*(uN2T2-uN2T1)                         # internal energy change of the mixture in KJ

# With assumption, The changes in kinetic and potential energy between the initial and final states can be ignored
Q = deltaU + W
# Result
print 'The heat transfer in kj is:  ',Q

# Part(d)
# From table A-23
sbarT2CO2 = 222.475
sbarT1CO2 = 213.915
sbarT2N2 = 198.105
sbarT1N2 = 191.682
Rbar = 8.314                                                                     # universal gas constant
# Calculation
import math
deltaS = (m1/44)*(sbarT2CO2-sbarT1CO2-Rbar*math.log(p2/p1)) + (m2/28)*(sbarT2N2-sbarT1N2-Rbar*math.log(p2/p1))
# Result
print 'The change in entropy of the mixture in kj/k is:  ',deltaS

The final temperature in Kelvin is:   373.719281885
The work in kj is:   -34.2270009251
The heat transfer in kj is:   -7.81758534069
The change in entropy of the mixture in kj/k is:   -0.0232760033842

## Example 12.4 Page no-571

In [4]:
# Given:-
y1 = 0.8                                                                        # mole fraction of CO2
y2 = 0.2                                                                        # mole fraction of O2
T1 = 700.0                                                                      # in kelvin
p1 = 5.0                                                                        # in bars
V1 = 3.0                                                                        # in m/s
p2 = 1.0                                                                        # in bars

# Part(a)
# From table A-23
sO2barT1 = 231.358
sCO2barT1 = 250.663
# Calculations
import math
RHS = y2*sO2barT1 + y1*sCO2barT1 + 8.314*math.log(p2/p1)
# Using table A-23
LHSat510K = y2*221.206 + y1*235.7
LHSat520K = y2*221.812 + y1*236.575
# Using linear interpolation,
T2 = 510 +((520-510)/(LHSat520K-LHSat510K))*(RHS-LHSat510K)
# Result
print 'The temperature at the nozzle exit in K is:  ',T2

# Part(b)
# From table A-23
sbarO2T2 = 221.667                                                              # in kj/kmol.K
sbarO2T1 = 231.358                                                              # in kj/kmol.K
sbarCO2T2 = 236.365                                                             # in kj/kmol.K
sbarCO2T1 = 250.663                                                             # in kj/kmol.K
# Calculations
deltasbarO2 = sbarO2T2-sbarO2T1-8.314*math.log(p2/p1)                           # in kj/kmol.K
deltasbarCO2 = sbarCO2T2-sbarCO2T1-8.314*math.log(p2/p1)                        # in kj/kmol.K
# Results
print 'The entropy changes of the CO2 from inlet to exit, in KJ/Kmol.K is:  ',deltasbarCO2
print 'The entropy change of the O2 from inlet to the exit in kj/kmol.k is:  ',deltasbarO2

# Part(c)
# From table A-23, the molar specific enthalpies of O2 and CO2 are
h1barO2 = 21184.0
h2barO2 = 15320.0
h1barCO2 = 27125.0
h2barCO2 = 18468.0
# Calculations
M = y1*44.0 + y2*32.0                                                            # apparent molecular weight of the mixture in kg/kmol
deltah = (1.0/M)*(y2*(h1barO2-h2barO2) + y1*(h1barCO2-h2barCO2))
V2 = math.sqrt(V1**2+ 2*deltah*10**3)
# Result
print 'The exit velocity in m/s is:  ',V2

The temperature at the nozzle exit in K is:   517.549113444
The entropy changes of the CO2 from inlet to exit, in KJ/Kmol.K is:   -0.917133196023
The entropy change of the O2 from inlet to the exit in kj/kmol.k is:   3.68986680398
The exit velocity in m/s is:   623.983296128

## Example 12.5 Page no-574

In [5]:
# Given:-
nN2 = 0.79                                                                       # initial moles of nitrogen in kmol
pN2 = 2.0                                                                        # initial pressure of nitrogen in bars
TN2 = 250.0                                                                      # initial temperature of nitrogen in kelvin
nO2 = 0.21                                                                       # initial moles of oxygen in kmol
pO2 = 1.0                                                                        # initial pressure of oxygen in bars
TO2 = 300.0                                                                      # initial temperature of oxygen in kelvin

# Part(a)
MN2 = 28.01                                                                      # molar mass of nitrogen in kg/kmol
MO2 = 32.0                                                                       # molar mass of oxygen in kg/kmol
# Calculations
# With the help of table A-20
cvbarN2 = MN2*0.743                                                              # in kj/kmol.K
cvbarO2 = MO2*0.656                                                              # in kj/kmol.K
T2 = (nN2*cvbarN2*TN2+nO2*cvbarO2*TO2)/(nN2*cvbarN2+nO2*cvbarO2)
# Result
print 'The final temperature of the mixture in kelvin is:  ',T2

# Part(b)
# Calculation
p2 = ((nN2+nO2)*T2)/(nN2*TN2/pN2 + nO2*TO2/pO2)
# Result
print 'The final pressure of the mixture in bar is:  ',p2

# Part(c)
Rbar = 8.314                                                                     # universal gas constant
# Calculations
import math
cpbarN2 = cvbarN2 + Rbar
cpbarO2 = cvbarO2 + Rbar
yN2 = nN2/(nN2+nO2)                                                              # mole fraction of N2
yO2 = nO2/(nN2+nO2)                                                              # mole fraction of O2
sigma = nN2*(cpbarN2*math.log(T2/TN2)-Rbar*math.log(yN2*p2/pN2)) + nO2*(cpbarO2*math.log(T2/TO2)-Rbar*math.log(yO2*p2/pO2))
# Result
print 'The amount of entropy produced in the mixing process, in kJ/K is:  ',sigma

The final temperature of the mixture in kelvin is:   260.571840521
The final pressure of the mixture in bar is:   1.61095419179
The amount of entropy produced in the mixing process, in kJ/K is:   4.94715526048

## Example 12.6 Page no-577

In [6]:
# Given:-
T1 = 32.0                                                                         # temperature of dry air in degree celcius
p1 = 1.0                                                                          # pressure of dry air in bar
AV1 = 100.0                                                                       # volume rate of dry air in m^3/min
T2 = 127.0                                                                        # temperature of oxygen stream in degree celcius
p2 = 1.0                                                                          # pressure of oxygen stream in bar
T3 = 47.0                                                                         # temperature of mixed stream in degree celcius
p3 = 1.0                                                                          # pressure of mixed stream in bar

# Part(a)
Rbar = 8314.0                                                                     # universal gas constant
Ma = 28.97                                                                        # molar mass of air
Mo = 32.0                                                                         # molar mass of oxygen
# From table A-22 and A-23
haT3 = 320.29                                                                     # in kj/kg
haT1 = 305.22                                                                     # in kj/kg
hnotT2 = 11711.0                                                                  # in kj/kmol
hnotT1 = 9325.0                                                                   # in kj/kmol

# Calculations
va1 = (Rbar/Ma)*(T1+273.0)/(p1*10**5)                                             # specific volume of air in m^3/kg
ma1dot = AV1/va1                                                                  # mass flow rate of dry air in kg/min
modot = ma1dot*(haT3-haT1)/((1/Mo)*(hnotT2-hnotT1))                               # in kg/min
# Results
print 'The mass flow rate of dry air in kg/min is:  ',ma1dot
print 'The mass flow rate of oxygen in kg/min is:  ',modot

# Part(b)
nadot = ma1dot/Ma                                                                 # molar flow rate of air in kmol/min
nodot = modot/Mo                                                                  # molar flow rate of oxygen in kmol/min
yo = nodot/(nadot+nodot)                                                          # mole fraction of oxygen
# Results
print 'The mole fraction of dry air in the exiting mixture is:  ',ya
print 'The mole fraction of dry oxygen in the exiting mixture is:  ',yo

# Part(c)
# With the help of tables A-22 and A-23
sanotT3 = 1.7669                                                                  # in kj/kg.K
sanotT1 = 1.71865                                                                 # in kj/kg.K
sbarT3 = 207.112                                                                  # in kj/kmol.K
sbarT2 = 213.765                                                                  # in kj/kmol.K
# Calculations
import math
# Result
print 'The time rate of entropy production, in kJ/K . min is:  ',sigmadot

The mass flow rate of dry air in kg/min is:   114.245377144
The mass flow rate of oxygen in kg/min is:   23.0903984383
The mole fraction of dry air in the exiting mixture is:   0.845326536426
The mole fraction of dry oxygen in the exiting mixture is:   0.154673463574
The time rate of entropy production, in kJ/K . min is:   17.4180498088

## Example 12.7 Page no-584

In [7]:
# Given:-
m =1.0                                                                            # mass of sample in kg
T1 = 21.0                                                                         # initial temperature in degree celcius
psi1 = 0.7                                                                        # initial relative humidity
T2 = 5.0                                                                          # final temperature in degree celcius

# Part(a)
# From table A-2
pg = 0.02487                                                                      # in bar
# Calculations
pv1 = psi1*pg                                                                     # partial pressure of water vapor in bar
omega1 = 0.622*(0.2542)/(14.7-0.2542)
# Result
print 'the initial humidity ratio is:  ',omega1

# Part(b)
# The dew point temperature is the saturation temperature corresponding to the partial pressure, pv1. Interpolation in Table A-2 gives
T = 15.3                                                                          # the dew point temperature in degree celcius
# Result
print 'The dew point temperature in degree celcius is:  ',T

# Part(c)
# The partial pressure of the water vapor remaining in the system at the final state is the saturation pressure corresponding to 5C:
# Calculations
mv1 = 1/((1/omega1)+1)                                                           # initial amount of water vapor in the sample in kg
ma = m-mv1                                                                        # mass of dry air present in kg
pg = 0.00872                                                                      # in bar
omega2 = 0.622*(pg)/(1.01325-pg)                                                  # humidity ratio after cooling
mv2 = omega2*ma                                                                   # The mass of the water vapor present at the final state
mw = mv1-mv2

# Result
print 'The amount of water vapor that condenses, in kg. is:  ',mw

the initial humidity ratio is:   0.0109452159105
The dew point temperature in degree celcius is:   15.3
The amount of water vapor that condenses, in kg. is:   0.00548579192846

## Example 12.8 Page no-586

In [8]:
# Given:-
V = 35.0                                                                          # volume of the vessel in m^3
p1 = 1.5                                                                          # in bar
T1 = 120.0                                                                        # in degree celcius
psi1 = 0.1
T2 = 22.0                                                                         # in degree celcius

# Part(a)
# The dew point temperature at the initial state is the saturation temperature corresponding to the partial pressure pv1. With the given relative humidity and the saturation pressure at 120C from Table A-2
pg1 = 1.985
# Interpolating in Table A-2 gives the dew point temperature as
T = 60.0                                                                          # in degree celcius
# Calculation
pv1 = psi1*pg1                                                                    # partial pressure in bar
# Result
print 'The dew point temperature corresponding to the initial state, in degee celcius is:  ',T

# Part(b)
Rbar = 8314.0                                                                     # universal gas constant
Mv = 18.0                                                                         # molar mass of vapor in kj/kmol
# Interpolation in Table A-2
Tdash = 56.0                                                                      # in degrees
vv1 =((Rbar/Mv)*(T1+273))/(pv1*10**5)                                             # the specific volume of the vapor at state 1 in m^3/kg
# Result
print 'The temperature at which condensation actually begins in degree celcius is:  ',Tdash

# Part(c)
# From table
vf2 = 1.0022e-3
vg2 = 51.447
vv2 = vv1                                                                         # specific volume at final state
# Calculations
mv1 = V/vv1                                                                       # initial amount of water vapor present in kg
x2 = (vv2-vf2)/(vg2-vf2)                                                          # quality
mv2 = x2*mv1                                                                      # the mass of the water vapor contained in the system at the final state
mw2 = mv1-mv2
# Result
print 'The amount of water condense in kg is:  ',mw2

The dew point temperature corresponding to the initial state, in degee celcius is:   60.0
The temperature at which condensation actually begins in degree celcius is:   56.0
The amount of water condense in kg is:   3.14710226995

## Example 12.9 Page no-587

In [9]:
# Given:-
V = 35.0                                                                        # volume of vessel in m^3
p1 = 1.5                                                                        # initial pressure in bar
T1 = 120.0                                                                      # initial temperature in degree celcius
psi = 0.1
T2 = 22.0                                                                       # in degree celcius
Rbar = 8314.0                                                                   # universal gas constant
Ma = 28.97                                                                      # molar mass of air
pv1 = 0.1985                                                                    # in bar, from example 12.8
mv2 = 0.681                                                                     # in kg, from examples 12.8
mv1 = 3.827                                                                     # in kg, from example 12.8
mw2 = 3.146                                                                     # in kg, from example 12.8
# evaluating internal energies of dry air and water from Tables A-22 and A-2, respectively
ua2 = 210.49                                                                    # in kj/kg
ua1 = 281.1                                                                     # in kj/kg
ug2 = 2405.7                                                                    # in kj/kg
uf2 = 92.32                                                                     # in kj/kg
ug1 = 2529.3                                                                    # in kj/kg

# Calculations
ma =( ((p1-pv1)*10**5)*V)/((Rbar/Ma)*(T1+273))                                  # mass of dry air in kg
Q = ma*(ua2-ua1) + mv2*ug2 + mw2*uf2 - mv1*ug1

# Result
print 'The heat transfer during the process, in kJ is:  ',Q

The heat transfer during the process, in kJ is:   -10602.7454057

## Example 12.10 Page no-595

In [10]:
# Given :-
AV1 = 150.0                                                                       # entry volumetric flow rate in m^3/min
T1 = 10.0                                                                         # entry temperature in degree celcius
psi1 = 0.8
T2 = 30.0                                                                         # exit temperature in degree celcius
p = 1.0                                                                           # in bar

# Part(a)
Rbar = 8314.0                                                                     # universal gas constant
Ma = 28.97                                                                        # molar mass of air
# The specific enthalpies of the dry air are obtained from Table A-22 at the inlet and exit temperatures T1 and T2, respectively:
ha1 = 283.1                                                                       # in kj/kg
ha2 = 303.2                                                                       # in kj/kg
# The specific enthalpies of the water vapor are found using hv  hg and data from Table A-2 at T1 and T2, respectively:
hv1 = 2519.8                                                                      # in kj/kg
hv2 = 2556.3                                                                      # in kj/kg
# From table A-2
pg1 = 0.01228                                                                     # in bar
# Calculations
pv1 = psi1*pg1                                                                    # the partial pressure of the water vapor in bar
pa1 = p-pv1
va1 = (Rbar/Ma)*(T1+273)/(pa1*10**5)                                              # specific volume of the dry air in m^3/kg
madot = AV1/va1                                                                   # mass flow rate of the dry air in kg/min
omega = 0.622*(pv1/(p-pv1))                                                       # humidity ratio
Qcvdot = madot*((ha2-ha1)+omega*(hv2-hv1))                                        # in kj/min
# Result
print 'Rate of heat transfer, in kJ/min is:  ',Qcvdot

# Part(b)
# From Table A-2 at 30C
pg2 = 0.04246                                                                     # in bar
# Calculations
pv2 = pv1
psi2 = pv2/pg2                                                                    # relative humidity at the exit
# Result
print 'The relative humidity at the exit is:  ',psi2

Rate of heat transfer, in kJ/min is:   3716.99116097
The relative humidity at the exit is:   0.231370701837

## Example 12.11 Page no-600

In [11]:
# Given:-
T1 = 30.0                                                                       # in degree celcius
AV1 = 280.0                                                                     # in m^3/min
psi1 = 0.5                                                                      # relative humidity at the inlet
T2 = 10.0                                                                       # in degree celcius
p = 1.013                                                                       # pressure in bar

# Part(a)
# From table A-2
pg1 = 0.04246                                                                   # in bar
Rbar = 8314                                                                     # universal gas constant
Ma = 28.97                                                                      # molar mass of air
# Calculations
pv1 = psi1*pg1                                                                  # in bar
pa1 = p-pv1                                                                     # partial pressure of the dry air in bar
madot = AV1/((Rbar/Ma)*((T1+273)/(pa1*10**5)))                                  # common mass flow rate of the dry air in kg/min
# Result
print 'The mass flow rate of the dry air in kg/min is:  ',madot

# Part(b)
# From table A-2
pv2 = 0.01228                                                                    # in bar
# Calculations
omega1 = 0.622*(pv1/(p-pv1))
omega2 = 0.622*(pv2/(p-pv2))
# Result
print 'The rate at which water is condensed, in kg per kg of dry air flowing through the control volume is:  ',mwdotbymadot

# Part(c)
# From table A-2 and A-22
ha2 = 283.1                                                                      # in kg/kj
ha1 = 303.2                                                                      # in kg/kj
hg1 = 2556.3                                                                     # in kg/kj
hg2 = 2519.8                                                                     # in kg/kj
hf2 = 42.01                                                                      # in kg/kj
# Calculations
Qcvdot = madot*((ha2-ha1)-omega1*hg1+omega2*hg2+(omega1-omega2)*hf2)             # in kj/min
# Result
print 'The required refrigerating capacity, in tons is:  ',Qcvdot/211

The mass flow rate of the dry air in kg/min is:   319.348473885
The rate at which water is condensed, in kg per kg of dry air flowing through the control volume is:   0.00568197500137
The required refrigerating capacity, in tons is:   -52.4650572692

## Example 12.12 Page no-602

In [12]:
# Given:-
T1 = 22.0                                                                         # entry temperature of moist air in degree celcius
Twb = 9.0                                                                         # wet-bulb temperature of entering moist air in degree celcius
madot = 90.0                                                                      # mass flow rate of dry air in kg/min
Tst = 110.0                                                                       # temperature of injected saturated water vapor in degree celcius
mstdot = 52.0                                                                     # mass flow rate of injected saturated water vapor in kg/h
p = 1.0                                                                           # pressure in bar

# Part(a)
# By inspection of the psychrometric chart
omega1 = 0.002
# Calculation
# Result
print 'The humidity ratio at the exit is:  ',omega2

# Part(b)
# The steady-state form of the energy rate balance can be rearranged as
# (ha + omega*hg)2 = (ha + omega*hg)1 + (omega2-omega1)*hg3
# On putting values in the above equation from tables and figures, temperature at the exit can then be read directly from the chart
T2 = 23.5                                                                         # in degree celcius
# Result
print 'The temperature at the exit in degree celcius is:  ',T2

The humidity ratio at the exit is:   0.0116296296296
The temperature at the exit in degree celcius is:   23.5

## Example 12.13 Page no-604

In [13]:
# Given:-
T1 = 38.0                                                                         # temperature of entering air in degree celcius
psi1 = 0.1                                                                        # relative humidity of entering air
AV1 = 140.0                                                                       # volumetric flow rate of entering air in m^3/min
Tw = 21.0                                                                         # temperature of added water in degree celcius
T2 = 21.0                                                                         # temperature of exiting moist air in degree celcius
p = 1.0                                                                           # pressure in atm

# Part(a)
# From table A-2
pg1 = 0.066                                                                       # in bar
# The specific volume of the dry air can be evaluated from the ideal gas equation of state. The result is
va1 = .887                                                                        # in m^3/kg
cpa = 1.005
# From table A-2
hf = 88.14
hg1 = 2570.7
hg2 = 2539.94
# Calculations
pv1 = psi1*pg1                                                                    # the partial pressure of the moist air entering the control volume in bar
omega1 = 0.622*(pv1/(p*1.01325-pv1))
omega2 = (cpa*(T1-T2)+omega1*(hg1-hf))/(hg2-hf)
madot = AV1/va1                                                                   # mass flow rate of the dry air in kg/min
mwdot = madot*60*(omega2-omega1)                                                  # in kg/h
# Result
print 'The mass flow rate of the water to the soaked pad in kj/h is:  ',mwdot

# Part(b)
pv2 = (omega2*p*1.01325)/(omega2+0.622)                                           # in bars
# At 21C, the saturation pressure is
pg2 = 0.02487
psi2 = pv2/pg2
# Result
print 'The relative humidity of the moist air at the exit to the evaporative cooler is:  ',psi2

The mass flow rate of the water to the soaked pad in kj/h is:   66.4756578168
The relative humidity of the moist air at the exit to the evaporative cooler is:   0.714165693403

## Example 12.14 Page no-607

In [14]:
# Given:-
AV1 = 142.0                                                                       # in m^3/min
T1 = 5.0                                                                          # in degree celcius
omega1 = 0.002
AV2 = 425.0                                                                       # in m^3/min
T2 = 24.0                                                                         # in degree celcius
psi2 = 0.5
p = 1.0                                                                           # in bar

# Part(a)
# From the psychrometric chart, Fig. A-9.
va1 = 0.79                                                                        # in m^3/kg
va2 = 0.855                                                                       # in m^3/kg
omega2 = 0.0094
# Calculations
ma1dot = AV1/va1                                                                  # in kg/min
ma2dot = AV2 /va2                                                                 # in kg/min
omega3 = (omega1*ma1dot+omega2*ma2dot)/(ma1dot + ma2dot)
# Result
print 'The humidity ratio is:  ',omega3

# Part(b)
# Reduction of the energy rate balance gives
# (ha + omega*hv)3 = [ma1dot*(ha + omega*hv)1 + ma2dot*(ha + omega*hv)2]/(ma1dot+ma2dot)
# With (ha + omega*hv)1 = 10kj/kg and (ha + omega*hv)2 = 47.8kj/kg from figure A-9
LHS = (ma1dot*10+ma2dot*47.8)/(ma1dot + ma2dot)

# This value for the enthalpy of the moist air at the exit, together with the previously determined value for omega3, fixes the state of the exiting moist air. From inspection of Fig. A-9,
T3 = 19.0                                                                         # in degree celcius
# Result
print 'The temperature of the exiting mixed stream in degree celcius is:  ',T3

The humidity ratio is:   0.0074347493219
The temperature of the exiting mixed stream in degree celcius is:   19.0

## Example 12.15 Page no-610

In [15]:
# Given:-
T1 = 38.0                                                                       # in degree celcius
m1dot = 4.5e7                                                                   # in kg/h
T2 = 30.0                                                                       # in degree celcius
m2dot = 4.5e7                                                                   # in kg/h
T3 = 25.0                                                                       # in degree celcius
psi3 = 0.35
T4 = 35.0                                                                       # in degree celcius
psi4 = 0.9
T5 = 20.0                                                                       # in degree celcius

# Analysis
# The humidity ratios omega3 and omega4 can be determined using the partial pressure of the water vapor obtained with the respective relative humidity
omega3 =0.00688
omega4 = 0.0327
# From tables A-2 and A-22
hf1 = 159.21
hf2 = 125.79
ha4 = 308.2
ha3 = 298.2
hg4 = 2565.3
hg3 = 2547.2
hf5 = 83.96
# Calculations
madot = (m1dot*(hf1-hf2))/(ha4-ha3+omega4*hg4-omega3*hg3-(omega4-omega3)*hf5)   # in kg/h
m5dot = madot*(omega4-omega3)                                                   # in kg/h
# Results
print 'The mass flow rate of dry air in kg/h is:  ',madot
print 'The mass flow rate of makeup water in kg/h is:  ',m5dot

The mass flow rate of dry air in kg/h is:   20270180.9849
The mass flow rate of makeup water in kg/h is:   523376.07303