15-1 A gas mixture consists of 12 kg of CO2, 16 kg of N2, and 10 kg of O2.

Determine:
a. the mass fraction of each component
b. the mole fraction of each component
c. the mixture average molecular weight.

Approach:
We assume the gases are ideal and that each gas in the
mixture behaves as if it alone fills the volume at the mixture
temperature. We will follow Example 15-1.

Assumptions:
1. The mixture behaves as an ideal gas and follows
Dalton’s law.

Solution:
a and b) Mass fraction is defined as X i = mi mm and the mass of each gas, mi, is given. For the mole fraction,
Yi = ni nm , we first need to determine the number of moles of each gas, ni = mi M i . The molecular weight of
each gas is obtained from Appendix A-1. The mixture molecular weight then is M m = mm nm .

Setting up a table to determine all the above information:

Gas mi ( kg ) X i = mi mm Mi ni = mi M i Yi = ni nm
( kmol )
CO2 12 0.316 44.01 0.273 0.236
N2 16 0.421 28.01 0.571 0.494
O2 10 0.262 32.00 0.313 0.270
Mixture mm = 38 1.000 nm = 1.157 1.000
Answer Answer
c) The mixture molecular weight is
m 38 kg kg
Mm = m = =32.84 Answer
nm 1.157 kmol kmol

15- 1
15-2 An ideal gas mixture at 210 kPa, 50 °C has a molar analysis as follows: N2, 35%: CO2, 25%; O2, 40%.
Determine:
a. the mass fractions
b. the partial pressures of each component (in kPa)
c. the volume occupied by 1 kg of the mixture (in m3).

Approach:
We assume the gases are ideal and that each gas
in the mixture behaves as of it alone fills the
volume at the mixture temperature. We can
follow the procedure given in Example 15-1.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.

Solution:
a) Mass fraction is defined as X i = mi / mm . We can evaluate X i by considering 1 kmol of mixture, using
molecular weights from Table A-1.

Gas ni (kmol) Mi mi = ni M i (kg) X i = mi / mm

CO2 0.25 44.01 11.00 0.327
O2 0.40 32.00 12.80 0.381
N2 0.35 28.01 9.80 0.292
1 kmol mm=33.60 kg 1.000
Answers

b) The partial pressure is defined with mole fraction: Pi = Yi P . Above, we assumed kmol of mixture in order to
calculate the mass fraction. Hence, the value in the N i column is also the mole fraction. Therefore:
PCO2 = 0.25(210kPa) = 52.5kPa
PO2 = 0.40(210kPa) = 84.0kPa Answers
PN2 = 0.35(210kPa) = 73.5kPa
mRT
c) Assuming the mixture behaves as an ideal gas, V = .
PM
Because we used 1 kmol to create the table, the entry labeled mm is also the mixture molecular weight. Therefore:
kJ
(1kg)(8.314 )(50 + 273)K
V= kmolK = 0.381 m3 Answer
kN kg 1kJ
(210 2 )(33.6 )( )
m kmol 1kNm

15- 2
15-3 A gas mixture at 20 ºC consists of 0.95 kg O2, 0.82 kg N2, 1.32 kg CO2, and 0.091 kg CO and is contained
in a 0.825-m3 tank. Determine the total pressure (in kPa).

Approach:
This ideal gas equation can be used to calculate
this mixture’s pressure. This total number of
moles can be calculated from the given masses
and the molecular weights of each gas.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.

Solution:
The ideal gas equation is
m RT
P= m
M mV
Assume that each gas behaves as if it alone fills the volume at the mixture temperature.
The total mass of the mixture is:
mm = 0.95 + 0.82 + 1.32 + 0.091 = 3.181kg
The mixture’s molecular weight can be determined using Eq.15-8:
1 1 m
= ∑ Xi where Xi = i
Mm Mi mm
With molecular weights from Appendix A-1.
1 0.95 1 0.82 1 1.32 1 0.091 1
=( )( )+( )( )+( )( )+( )( )
Mm 3.181 32.00 3.181 28.01 3.181 44.01 3.181 28.01
M m = 34.5 kg/kmol
(3.181kg)(8.314 kJ/kmolK)(20 + 273)K
P= = 272.3kPa Answer
1kJ
(34.5 kg/kmol)(0.825 m3 )( )
1kNm

15- 3
15-4 The composition of natural gas varies depending on its source. One sample has the following volume
fraction analysis: methane (CH4), 84%; ethane (C2H6), 14.8%; carbon dioxide (CO2), 0.70%; nitrogen (N2),
0.50%. For the mixture, determine:
a. mass fractions
b. the mixture molecular weight
c. the mixture density if the total mixture pressure is 175 kPa at a temperature of 27 °C.

Approach:
The definition of volume fraction can be combined with the
ideal gas law to determine the mass fraction. Once the mass
fractions are known, then the mixture molecular weight can
be calculated from the individual gas molecular weights.
The mixture density can be obtained from Eq. 15-24 plus the
ideal gas equation.

Assumptions:
1. The mixture behaves as an ideal gas and follows
Dalton’s law.

Solution:
a) The volume fraction is defined as Vi V , and the partial volume is Vi = ni RTm Pi , where P is the total mixture
pressure. Assuming the individual gases and mixtures can be treated as ideal gases.
Vi ni RTm Pm n
= = i = Yi
V nm RTm Pm nm
Thus, the volume fraction equals the mole fraction. Answer
b) The apparent molecular weight of the mixture, using Eq. 15-7, is:
M m = ∑ Ym M i = (0.84)(16.043) + (0.148)(30.020) + (0.007)(44.01)(0.005)(28.01)
= 18.37kJ/kmol Answer
c) The mixture density is obtained from Eq. 15-24:
PM
ρ = ∑ ρi or ρ =
RT
where each individual density is calculated with the ideal gas law using the partial pressure of each gas.
The first approach requires the partial pressure of each gas, Pi = Yi P
PM Y PM (0.84)(175kN/m 2 )(16.043kg/kmol)
Pmethane = i
= i = = 0.94355kg/m3
RT RT (8.314kJ/kmolK)(27 + 273)K
In a similar manner:
(0.148)(175)(30.02)
ρ methane = = 0.3117kg/m3
(8.314)(27 + 273)
(0.007)(175)(44.01)
ρCO2 = = 0.0216kg/m3
(8.314)(27 + 273)
(0.005)(175)(28.01)
ρ N2 = = 0.0098kg/m3
(8.314)(27 + 273)
∑ρ i = 1.289 kg/m3 Answer
Using a second approach that uses the total moles and mixture molecular weight:
PM m (175 kN/m )(18.37 kg/kmol) (1kJ 1kNm )
2

ρ= = = 1.289 kg/m3 Answer

RT ( 8.314 kg/kmolK )( 27 + 273 ) K
which is the same as before.

15- 4
15-5 The pressure and temperature in a rigid tank are 375 kPa and 46 ºC, respectively. The ideal gas mixture
contained in the tank is composed of 0.75 kmol argon (Ar) and 1.35 kmol nitrogen (N2). Determine:
a. the volume of the tank (in m3)
b. the final pressure if the mixture is heated to 127 ºC (in kPa).

Approach:
We assume the gases are ideal and that each gas in the
mixture behaves as if it alone fills the volume at the mixture
temperature. The ideal gas mixture rules are applied.

Assumptions:
1. The mixture behaves as an ideal gas and follows
Dalton’s law.

Solution:
a) The pressure and temperature of the mixture are given, as is the composition of the mixture. If we calculate
the mixture average molecular weight, then we can use the ideal gas equation to calculate the volume.
Mixture average molecular weight is: M m = mm nm
The total number of moles in the mixture is: nm = nAr + nN2 = 0.75 kmol+1.35 kmol=2.10 kmol
The total mass is:
mm = ∑ ni M i = nAr M Ar + nN2 M N2 = ( 0.75 kmol )( 39.94 kg kmol ) + (1.35 kmol )( 28.01kg kmol ) =67.77 kg
Therefore, M m = 67.77 kg 2.10 kmol =32.27 kg kmol
Using the ideal gas law:
m RT ( 67.77kg )(8.314 kJ kmolK )( 46+273) K
V= m = =14.9 m3 Answer
PM m ( 375 kN m 2 ) ( 32.27 kg kmol )(1kJ 1kNm )
Another approach is to use the molar form of the ideal gas equation:
n RT ( 2.10 kmol )( 8.314 kJ kmolK )( 46+273) K
V= m = =14.9 m3
P ( 375 kN m 2
)(1kJ 1kNm )
b) Because this is an ideal gas mixture and the number of moles is fixed:
PV PV T ⎛ 127 + 273 ⎞
n= 1 1 = 2 2 → P2 = 2 P1 = ⎜ ⎟ ( 375 kPa ) =470 kPa Answer
RT1 RT2 T1 ⎝ 46 + 273 ⎠

15- 5
15-6 An ideal gas mixture of 1 lbm air and 1 lbm water vapor is compressed isentropically in a closed system
from 50 lbf/in.2, 250 ºF to 100 lbf/in.2. Determine:
a. the final temperature (in ºF)
b. the work required (in Btu).

Approach:
Assuming the mixture behaves as an ideal gas,
the isentropic relations for an ideal gas can be
used to obtain the final temperature. Once that is
known, work can be calculated with conservation
of energy.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.
2. The process is adiabatic and reversible with
negligible potential and kinetic energy
effects.
3. The specific heats are constant.

Solution:
a) Assuming the mixture is an idea gas with constant specific heats:
T2 = T1 ( P2 / P1 ) k −1/ k where k = c p / cv and k is the mixture property.
We need to evaluate the two specific heats for the mixture. We use Eq. 15-22:
k
c p = ∑ X i c p ,i and c p − cv = R M
i =1

From Appendix B-8, assuming T2 ≈ 850 R, Tavg = ( 710 + 850 ) 2 = 780 R , for air:
c p = 0.243Btu lbmR
cv = 0.243 − (1.986 Btu lbmolR ) ( 28.97 lbm lbmol ) = 0.174 Btu lbmR
k = 0.243 0.174 = 1.393
For water vapor:
c p = c p M = ( 8.303Btu/lbmolR ) (18.02 lbm lbmol ) = 0.461Btu lbmR
cv = 0.461 − (1.986 Btu lbmolR ) (18.02 lbm lbmol ) = 0.351Btu lbmR
k = 0.461 0.351 = 1.315
For the mixture:
⎛ 1 ⎞ ⎛ 1 ⎞ Btu
c p,m = ⎜ ⎟ ( 0.243) + ⎜ ⎟ ( 0.461) = 0.352
⎝ 1 + 1 ⎠ ⎝ 1 + 1 ⎠ lbmR
⎛ 1 ⎞ ⎛ 1 ⎞ Btu
cv , m = ⎜ ⎟ ( 0.174 ) + ⎜ ⎟ ( 0.351) = 0.263
⎝ 1+1⎠ ⎝ 1+1⎠ lbmR
k = 0.352 0.263 = 1.341
(1.341-1) 1.341
⎛ 100 ⎞
Therefore: T2 = ( 710R ) ⎜ ⎟ =846.8 R
⎝ 50 ⎠
b) Applying conservation of energy to the closed system defined above, and assuming adiabatic with no potential
or kinetic energy effects:
∆U + ∆PE + ∆KE = Q − W
⎛ Btu ⎞
W = ∆U = mm c p , m (T2 − T1 ) = ( 2 lbm ) ⎜ 0.263 ⎟ ( 846.8-710 ) R=72.0 Btu Answer
⎝ lbmR ⎠

15- 6
15-7 A pipe connects two tanks with a valve in between. One tank contains 2 kg methane (CH4) at 200 kPa, 10
°C; the other tank contains 4 kg oxygen (O2) at 600 kPa, -10 °C. The valve between the tanks is opened,
and the two gases mix adiabatically. Determine:
a. the final mixture temperature (in °C)
b. the final mixture pressure (in kPa).

Approach:
The closed system energy equation can be used to
calculate the final temperature of the mixture.
We assume the mixture behaves as a mixture of
ideal gases.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.
2. The system is adiabatic, with no work or
potential or kinetic energy effects.
3. Specific heats are constant.

Solution:
a) The closed system energy equation is: ∆E = ∆U + ∆PE + ∆KE = Q − W
The system is adiabatic (Q=0), rigid (W=0), with no potential or kinetic energy effects:
∆U = 0 → ∆U CH 4 + ∆U O2 = 0
Assuming both gases are ideal with constant specific heat: [mcv (T2 − T1 )]CH 4 + [mcv (T2 − T1 )]O2 = 0
[mcvT1 ]CH 4 + [mcvT1 ]O2
Solving for T2 : T2 =
(mcv )CH 4 + (mcv )O2
Evaluate the specific heats at Tavg = (T1 + T2 ) 2 . Assuming T2 270 K, Tavg ,CH 4 278 K, Tavg ,O2 267 K
From Appendix A at the average temperatures:
34.0
cv ,O2 ≈ 0.655 kJ/kgK c p ,CH 4 ≈ = 2.120 kJ/kgK
16.04
cv ,CH 4 = c p ,CH 4 − R M CH 4 = 2.120 − 8.314 16.04 = 1.601 kJ/kgK
(2kg)(1.601kJ/kgK)(283K) + (4 kg)(0.655 kJ/kgK)(263K)
T2 = = 274.0 K Answer
(2 kg)(1.601kJ/kgK) + (4 kg)(0.655 kJ/kgK)
b) The final mixture pressure can be calculated with the ideal gas equation:
m RT
P= m
VM m
The apparent molecular weight is determined from Eq. 15-8:
1 1 2 1 4 1
= ∑ Xi =( )( )+( )( )
Mm Mi 2 + 4 16.04 2 + 4 32.00
M m = 24.03kg/kmol
The total volume is determined from the individual volumes of the gases before they mix:
mRT (2 kg)(8.314 kJ/kmolK)(283K) (4)(8.314)(263)
VCH 4 = = 2
= 1.47 m3 VO2 = = 0.46 m3
PM (200 kN/m )(16.04 kg/kmol) (600)(32.00)
Vtot = 0.46 + 1.47 = 1.93m3
(2 kg + 4 kg)(8.314 kJ/kmolK)(274 K)
Therefore, P= = 296 kPa Answer
(1.93m3 )(24.03kg/kmol) (1kJ 1kNm )

15- 7
15-8 Two gas streams enter a mixer and a single stream exits. The first stream, with a flow rate of 0.2 kg/s, is a
mixture of 13% methane (CH4) and 87% air (mole fractions). The second stream is air. If the mole
fraction of methane must be 5% in the outlet stream, determine:
a. the mass flow rate of the second entering stream of air (in kg/s)
b. the mass flow rate of oxygen in the exiting stream (in kg/s).

Approach:
This is a conservation of mass problem with the
properties evaluated assuming the individual
gases and the mixture behave ideally.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.
2. The system is steady.
3. The air is composed only of O2 and N2 in
the usual proportions.

Solution:
dm
a) For the given control volume, conservation of mass is: = mA + mB - mC
dt
The system is steady, so m B = mC − m A
Methane is conserved (no combustion), so nCH 4 , A = nCH 4 ,C = nCH 4
The entering flow at A is 13% (mole fraction) methane and the exiting flow must be 5% (mole fraction) methane.
Defining molar flow rates in terms of mass flow rates:
mCH 4
nCH 4 =
M CH 4
Likewise, the mole fractions are:
nCH 4 , A nCH 4 ,C
= 0.13 and = 0.05
nm , A nm ,C
mA m
Combining these equations: nCH 4 = 0.13nM , A = 0.05nM ,C → 0.13 = 0.05 C
MA MC
0.13 M C
Therefore, mC = ( )( )mA
0.05 M A
The apparent molecular weights at A and C are found with Eq. 15-7, M m = ∑ Yi M i
The molecular weights are:
M air = (0.21)(32.00) + (0.79)(28.01) = 28.85 kg/kmol
M A = (0.13)(16.04) + (0.87)(28.85) = 27.18 kg/kmol
M C = (0.05)(16.04) + (0.95)(28.85) = 28.21 kg/kmol
Finally,
0.13 28.21 kg kg kg
mC = ( )( )(0.2 ) = 0.54 and m B = 0.54 − 0.2 = 0.34 Answer
0.05 27.18 s s s

b) The molar flow of oxygen at C is:

m mO2 ,C
nO2 ,C = YO2 ,C nm ,C = YO2 ,C C and nO2 ,C =
MC M O2
Combining these equations:
MO 32.00 kg kg
mO2 ,C =YO2 ,C ( 2 )mC = (0.95)(0.21)( )(0.54 ) = 0.122 Answer
MC 28.21 s s

15- 8
15-9 An ideal gas mixture at 500 kPa, 300 K consisting of 20 kg O2 and 16 kg N2 is contained in a piston-
cylinder assembly. Heat is added, and the mixture expands at constant pressure until it reaches a
temperature of 450 K. Determine the heat transfer during this process (in kJ).

Approach:
The closed system energy equation is used to calculate
the heat transfer. The expansion work is calculated
assuming a quasi-equilibrium process.

Assumptions:
1. The process is quasi-equilibrium.
2. The mixture behaves as an ideal gas and follows
Daltons’ law.
3. Potential and kinetic energy effects are negligible.

Solution:
For the control volume defined above, assuming negligible potential and kinetic energy effects, the closed system
energy equation
∆U + ∆PE + ∆KE = Q − W → ∆U = Q − W
Assuming a quasi-equilibrium expansion process, work is calculated with:
W = ∫ PdV = P (V2 − P1 )
Combining this with the energy equation
Q = ∆U + W = ∆U + P∆V = ∆H = ( m∆h )O + ( m∆h ) N
2 2

Using Appendix A-18 for the molar enthalpies:

O2: at 300 K h1 = 8, 736 kJ kmol, at 450 K h2 = 13, 228 kJ kmol
N2: at 300 K h1 = 8, 723kJ kmol, at 450 K h2 = 13,105 kJ kmol
Therefore,
(13,228-8,736 ) kJ kmol (13,105-8,723) kJ kmol
Q = ( 20kg ) + (16kg )
32.0 kg kmol 28.01kg kmol
= 5310 kJ Answer

15- 9
15-10 Hydrogen at 0.2 MPa, 70 ºC and nitrogen at 0.2 MPa, 270 ºC enter an adiabatic mixing chamber in separate
flow streams. The molar flow ratio is 3:1, respectively. The mixture leaves at 0.19 MPa. Using constant
specific heats evaluated at 27 ºC, determine:
a. the exit temperature (in ºC)
b. the entropy generation rate per kg of mixture (in kJ/kg·K).

Approach:
Conservation of energy is used to calculate the
outlet temperature of the mixture. Once that is
known, the entropy generation can be determined
with the entropy balance equation.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.
2. The process is steady, adiabatic with no
potential or kinetic energy effects or work.
3. The specific heats are constant.

Solution:
a) For the control volume defined above, and assuming steady, adiabatic, no work and no potential or kinetic
energy effects: 0 = mA hA + mB hB − mC hC
Using conservation of mass: 0 = mA + mB − mC → mC = mA + mB
Substituting the mass equation into the energy equation and simplifying:
0 = mA ( hA − hC ) + mB ( hB − hC )
Note that nH 2 = nA , nN2 = nB , nA nB = 3.0, and n = m M . Also, for an ideal gas with constant specific heat
∆h = c p ∆T . Combining these equations:
nA M A c p , A (TA − TC ) + nB M B c p , B (TB − TC ) = 0 → 3nB M A c p , A (TA − TC ) + nB M B c p , B (TB − TC ) = 0
The molar flow rate cancels, so solving for the exit temperature:
3M A c p , ATA + M B c p , BTB
TC =
3M A c p , A + M B c p , B
From Appendix A-8 at 300K, c p , A = 14.307 kJ kgK and c p , B = 1.039 kJ kgK
3 ( 2.018 )(14.307 )( 343K ) + ( 28.01)(1.039 )( 543K )
TC = =393.3K=120.3o C Answer
3 ( 2.018 )(14.307 ) + ( 28.01)(1.039 )

dS Q
b) The entropy balance equation is: = ∑ + ∑ mi si − ∑ mi si + S gen
dt T in out

For steady and adiabatic: S gen = ∑ mi si − ∑ mi si

out in

We are given the mole fractions, so we can rewrite this equation in terms of molar flow rates and molar entropies.
Let m = nM . Also, we need to keep track of the nitrogen and hydrogen flows separately.
( ) (
S gen = nH 2 M H 2 sH 2 ,C − sH 2 , A + nN2 M N2 sN2 ,C − sN2 , B )
Dividing by the molar flow rate of the mixture:
S gen nH 2 nN
nm
=
nm
( ) (
M H 2 sH 2 , C − s H 2 , A + 2 M N 2 s N 2 , C − s N 2 , B
nm
)
Substituting nm = mm M m and Yi = ni nm
S gen
1 ⎡
=
mm M m ⎣ 2
( ) (
YH M H 2 sH 2 ,C − sH 2 , A + YN2 M N2 sN2 ,C − sN2 , B ⎤⎦ )
The mixture molecular weight is:
M m = YH 2 M H 2 + YN2 M N2 = ( 0.75 )( 2.018 ) + ( 0.25 )( 28.01) = 8.516 kg kmol

15-10
Using Eq. 15-30 for the entropy change of the gases:
⎛ T ⎞ R ⎛ P2,i ⎞
s2,i − s1,i = c p ,i (T ) ln ⎜ 2 ⎟ − ln ⎜ ⎟⎟
⎜
⎝ T1 ⎠ M i ⎝ P1,i ⎠
⎛ kJ ⎞ ⎛ 393.3 ⎞ 8.314 kJ kmolK ⎛ ( 0.75 )( 0.19 ) ⎞ kJ
H2: ∆s = ⎜ 14.307 ⎟ ln ⎜ ⎟- ln ⎜ ⎟ =3.354
⎝ kgK ⎠ ⎝ 343 ⎠ 2.018 kg kmol ⎝ 0.20 ⎠ kgK
⎛ kJ ⎞ ⎛ 393.3 ⎞ 8.314 kJ kmolK ⎛ ( 0.25 )( 0.19 ) ⎞ kJ
N2: ∆s = ⎜ 1.039 ⎟ ln ⎜ ⎟- ln ⎜ ⎟ =0.569
⎝ kgK ⎠ ⎝ 343 ⎠ 28.01kg kmol ⎝ 0.20 ⎠ kgK

S gen 1 ⎡ ⎛ kg ⎞ ⎛ kJ ⎞ ⎛ kg ⎞ ⎛ kJ ⎞ ⎤
= ⎢( 0.75 ) ⎜ 2.018 ⎟ ⎜ 3.354 ⎟ + ( 0.25 ) ⎜ 28.01 ⎟ ⎜ 0.569 ⎟⎥
mm 8.516 kg kmol ⎣ ⎝ kmol ⎠ ⎝ kgK ⎠ ⎝ kmol ⎠ ⎝ kgK ⎠ ⎦
kJ
= 1.064 Answer
kgK

15-11
15-11 Natural gas with a molar analysis of methane (CH4), 70% and ethane (C2H6), 30% is compressed
isothermally from 600 kPa, 77 °C to 2000 kPa. No entropy is generated. Determine the power input per
kmol of mixture (in kJ/kmol).

Approach:
The open system energy equation is used.
However, neither heat transfer rate nor power is
known. The second equation used is the entropy
balance equation.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.
2. The system is isentropic.
3. Potential and kinetic energy effects are
negligible.

Solution:
The open system energy equation, ignoring potential and kinetic energy effects, is:
dE
= Q − W + mA hA − mB hB
dt
The system is steady, and from conservation of mass:
mA = mB = m
Solving for W :
W = Q + m ( hA − hB )
The open system entropy balance equation is:
dS Q
= + mA s A − mB sB + S gen
dt T
Applying the same assumptions as before, and noting S gen = 0 :
Q = Tm ( sB − s A )
Combining the two conservation equations:
W = Tm ( sB − s A ) + m ( hA − hB )
Assume the mixture of ideal gases behaves as an ideal gas. Because the temperature is constant, hA − hB = 0 .
Using Eqs. 15-27 and 15-31 for the entropy difference
⎡ ⎛T ⎞ ⎛ P ⎞⎤ ⎡ ⎛T ⎞ ⎛ P ⎞⎤
∆s = YCH 4 ⎢cP ln ⎜ B ⎟ − R ln ⎜ B ,i ⎟ ⎥ + YC2 H 6 ⎢ cP ln ⎜ B ⎟ − R ln ⎜ B ,i ⎟ ⎥
⎜ P ⎟⎥ ⎜ P ⎟⎥
⎢⎣ ⎝ TA ⎠ ⎝ A,i ⎠ ⎦ CH 4 ⎢⎣ ⎝ TA ⎠ ⎝ A , i ⎠ ⎦ C2 H 6
Because the temperature is constant, TA = TB , and for constant composition PB ,i PA,i = PB PA . Therefore:

( )
∆s = − YCH 4 + YC2 H 6 R ln ( PB PA ) = − R ln ( PB PA )
Note that: ∆s = ∆s M m and nm = mm M m
⎛ ∆s ⎞
W = T ( M m nm ) ⎜ ⎟ = Tnm ∆s
⎝ Mm ⎠
W ⎛P ⎞ ⎛ kJ ⎞ ⎛ 2000 ⎞ kJ
= −TR ln ⎜ B ⎟ = − ( 350K ) ⎜ 8.314 ⎟ ln ⎜ 600 ⎟ = −3503 kmol Answer
n ⎝ PA ⎠ ⎝ kmolK ⎠ ⎝ ⎠

15-12
15-12 Two gas streams are adiabatically mixed. The first stream is methane is at 4 bar, 80 ºC with a flow rate of
8 kg/min; the second stream is air at 1.2 bar, 30 ºC. After mixing, the mixture is at 1 bar, 40 ºC.
Determine:
a. the mass flow rate of the 30 ºC air (in kg/min)
b. the entropy generation rate (in kJ/min·K).

Approach:
Conservation of energy is used to determine the
outlet temperature. The entropy balance equation
is used to calculate the entropy generation note.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.
2. The system is steady, adiabatic, with no work
and negligible potential and kinetic energy.
3. Specific heats are constant.

Solution:
a) The open system conservation of energy equation, ignoring potential and kinetic energy effects, is:
dE
= Q − W + mA hA + mB hB − mC hC
dt
The system is steady, adiabatic, with no work, so that: 0 = mA hA + mB hB − mC hC
Using conservation of mass with the same assumptions: 0 = mA + mB − mC → mC = mA + mB
Substituting into the conservation of energy, and assuming each gas is ideal and the mixture of gases also behaves
as an ideal gas with constant specific heat so that ∆h = cP ∆T :
mA (hA − hC ) + mB (hB − hC ) = 0 = mA c p , A (TA − TC ) + mB c p , B (TB − TC )
−c p , A (TA − TC )
Solving for the flow rate of air: mB = mA
c p , B (TB − TC )
Evaluating the specific heats using Appendix A-8 at the average temperatures:
Air: Tavg = ( 303 + 313) 2 = 308 K, c p = 1.008 kJ/kgK
CH4: Tavg = (353 + 313) 2 = 333K, c p = 37.62 16.043 = 2.345 kJ/kgK
2.345 353 − 313
`mB = −( )( )(8 kg / min) = 74.6 kg / min Answer
1.006 303 − 313
dS Q
b) The entropy balance equation is = + mA s A − mB sB − mC sC + S gen
dt T
The system is steady and adiabatic, so solving for the generation rate and incorporating the conservation of mass:
S gen = mA ( sc − s A ) + mR ( sc − sB )
The entropy change of each gas is obtained with Eq. 15-30: ∆s = c p ln (T2 T1 ) − ( R M ) ln ( P2i P1i )
Because the partial pressures of the gases are needed for the mixture, we need to calculate the mole fractions:
mCH 4 (8 kg/min) 74.6 kg/min
nCH 4 = = = 0.499 kmol/min nair = = 2.575 kmol/min
M CH 4 16.043kg/kmol 28.97 kg/kmol
0.499 2.575
nm = 0.499 + 2.575 = 3.074 kmol/min YCH 4 = = 0.162 Yair = = 0.838
3.074 3.074
313 8.314 kJ / kmolK (0.162)(100)
CH 4: sc − s A = (2.345 kJ/kgK) ln( )− ln[ ] = 1.380 kJ/kgK
353 16.043kg/kmol 400
313 8.314 (0.838)(100)
Air: s B − s A = (1.006) ln( )− ln[ ] = 0.136 kJ/kgK
303 28.97 120
S gen = (8 kg/min)(1.380 kJ/kgK) + (74.6 kg/min)(0.136 kJ/kgK) = 21.2 kJ/minK Answer

15-13
15-13 A mixture of 0.3 lbmol of N2 and 0.2 lbmol of O2 at 15 lbf/in.2, 500 °F is compressed isothermally to 60
lbf/in2. Heat transfer during the process is to the surroundings that are at 40 °F. Determine:
a. the work (in Btu)
b. the heat transfer (in Btu)
c. the entropy produced (in Btu/R).

Approach:
Compression work can be determined with
W = ∫ PdV . An energy balance is used to determine
the heat transfer, and the entropy balance equation can
be used to calculate the entropy produced.

Assumptions:
1. The mixture behaves as an ideal gas and follows
Dalton’s law.
2. Potential and kinetic energy effects are negligible.
3. Specific heats are constant.

Solution:
a) Assuming a quasi-equilibrium process, the compression work is determined with:
W = ∫ PdV
Assuming an ideal gas for the individual gases and for the mixture, we can use the ideal gas law
nRT V V V
W = ∫( )dV = nRT ln( 2 ) = (nN2 + nO2 ) RT ln( 2 ) = nm RT ln( 2 )
V V1 V1 V1
Again using the ideal gas law
PV PV V2 P1
n= 1 1 = 2 2 → =
RT1 RT2 V1 P2
W = (0.5lbmol)(1.986 Btu lbmolR)(960R) ln(15 60) = −1322 Btu Answer
b) The closed system energy equation is:
∆E = ∆U + ∆PE + ∆KE = Q − W
There are no potential or kinetic energy effects, and with constant temperature ∆U = 0 Therefore:
Q = W = −1322 Btu Answer
c) The entropy balance equation for a closed system is:
Q Q Q Q
∆S = + S gen → S gen = ∆S − = ∆S N2 + ∆SO2 − = nN2 ∆sN2 + nO2 ∆sO2 −
T T T T
Assuming constant specific heats, using Eq. 15-31, the entropy change of both components in the mixture:
T P T P
∆S = nN2 [cP ln( 2 ) − R ln( 2i )]N2 + nO2 [cP ln( 2 ) − R ln( 2i )]O2
T1 P1i T1 P1i
But T2 = T1 , and with constant composition P2i / P1i = P2 / P1
Therefore
P P
∆S = −(nN2 + nO2 ) R ln( 2 ) = − nm R ln( 2 )
P1 P1
Therefore,
P nRT ln( P1 / P2 ) P nRT ln( P1 / P2 ) P 1 1
S gen = − nR ln( 2 ) − = nR ln( 1 ) − = nRT ln( 1 )[ − ]
P1 Tb P2 Tb P2 T Tb
1 1 1 1 Btu
= Q[ − ] = (−1322Btu)[ − ] = 1.267 Answer
T Tb 960 R 500 R R

15-14
15-14 A mixture of N2 and CO2 (mole fractions 60 % and 40%, respectively) with a flow rate of 2.5 kg/s is
compressed adiabatically from 150 kPa, 77 ºC to 450 kPa, 207 ºC. Determine:
a. the power required (in kW)
b. the compressor isentropic efficiency.

Approach:
Conservation of energy is used to calculate the
power from the given information. The definition
of isentropic efficiency is used for part b.

Assumptions:
1. The process is steady and adiabatic with no
potential or kinetic energy effects.
2. The mixture behaves as an ideal gas mixture
with constant specific heats..

Solution:
a) Conservation of energy is applied to the control volume defined above. Assuming steady, adiabatic, negligible
potential and kinetic energy effects, and ideal gas mixture with constant specific heats:
W = m ( h1 − h2 )
k
Using Eq. 15-19, h = ∑ Yi hi and h = h M
i =1

W=
mm ⎡
{ (
M m ⎣⎢
Y h1 − h2 )} N2
{ (
+ Y h1 − h2 )}
CO2
⎤
⎦⎥
From Appendix A-1, M N2 = 28.01kg kmol and M CO2 = 44.01kg kmol . Also, from Appendix A-18
N2: at 350 K h1 = 10,180 kJ kmol and at 480 K h2 = 13,988 kJ kmol
CO2: at 350 K h1 = 11,351kJ kmol and at 480 K h2 = 16, 791kJ kmol
The mixture molecular weight is:
M m = YN2 M N2 + YCO2 M CO2 = ( 0.60 )( 28.01) + ( 0.40 )( 44.01) = 34.41kg kmol
( 2.5 kg s )
⎡ kJ kJ ⎤
W= ⎢ ( 0.60 )(10,180-13,988) + ( 0.40 )(11,351-16,791)
34.41kg kmol ⎣ kmol kmol ⎥⎦
= −324 kW Answer

b) The definition of isentropic efficiency is:

h −h W
ηC = 1 2 s = s
h1 − h2 W
For the isentropic process, s1 = s2 s or s1 − s2 s = 0
Using Eq. 15-27, s = ∑ Yi si → ( ) ( )
YN2 sN2 ,1 − sN2 ,2 s + YCO2 sCO2 ,1 − sCO2 ,2 s = 0
and Eq. 15-33 s2,i − s1,i = s − s − R ln ( P2,i P1,i )
o
2, i
o
1, i

Note that with the constant composition, P2,i P1,i = P2 P1 . From Appendix A-18, for N2 at 350 K,
s1o = 196.173kJ kmolK and for CO2, s1o = 219.831kJ kmolK . Therefore, combining the previous two
equations:
⎡ kJ ⎞ ⎛ 450 ⎞ ⎤
( 0.60 ) ⎢⎛⎜196.173
kJ ⎞ o ⎛
⎟ -s2s,N2 + ⎜ 8.314 ⎟ ln ⎜ ⎟
⎣⎝ kmolK ⎠ ⎝ kmolK ⎠ ⎝ 150 ⎠ ⎥⎦
⎡⎛ kJ ⎞ o ⎛ kJ ⎞ ⎛ 450 ⎞ ⎤
+ ( 0.40 ) ⎢⎜ 219.831 ⎟ -s2s,CO2 + ⎜ 8.314 ⎟ ln ⎜ ⎟⎥ = 0
⎣⎝ kmolK ⎠ ⎝ kmolK ⎠ ⎝ 150 ⎠ ⎦
For the isentropic process, once the outlet temperature is fixed, both s o are known. So, solving by trail and error:

15-15
 Guess T2s (K) sNo2 (kJ/kmolK) o
sCO (kJ/kmolK) Left hand side of
2
equation
420 201.499 227.258 2.967
430 202.189 228.252 2.155
450 203.523 230.194 0.579
460 204.170 231.144 -0.189

By interpolation, T2 s = 457.5 K, → h2 s , N2 = 13,327 kJ kmol, h2 s ,CO2 = 15,809 kJ kmol

The ideal work is:
( 2.5 kg s ) kJ
Ws = ⎡( 0.60 )(10,180-13,327 ) + ( 0.40 )(11,351-15,809 ) ⎤⎦
34.41kg kmol ⎣ kmol
= −267 kW
The isentropic efficiency is:
-267 kW
ηC = = 0.823 Answer
-324 kW

15-16
15-15 Hydrogen (H2) at 14.7 psia, 70 °F with a flow rate of 0.1 lbm/s is mixed adiabatically with methane (CH4)
at 14.7 psia, 150 °F to form a mixture that is 50% hydrogen (by volume). Determine:
a. the methane flow rate (in lbm/s)
b. the exit temperature (in °F).

Approach:
The ideal gas mixture laws are used to determine the
required mass flow rate of methane. Once that is
determined, conservation of energy is used to
calculate the out lot temperature.

Assumptions:
1. The mixture behaves as an ideal gas and follows
Dalton’s law.
2. The system is steady and adiabatic with no work
or potential or kinetic energy effects.
3. Specific heats are constant.

Solution:
a) Using the two definitions Yi = ni nm and n = mi M i
YH 2 nH nm nH mH 2 M H 2
= 2 = 2 =
YCH 4 nCH 4 nm nCH 4 mCH 4 M CH 4
YCH 4 M CH 4 0.5 16.043
mCH 4 = ( )( )mH 2 = ( )( (0.1 lbm/s) = 0.795 lbm/s Answer
YH 2 M H2 0.5 2.018)
b) Using conservation of mass, and assuming steady, adiabatic, no work, negligible potential and kinetic energy
effects, each gas is ideal and the mixture of gases also behaves as an ideal gas, and constant specific heats so that
∆h = c p ∆T
0 = mA hA + mB hB − mC hC
Using conservation of mass with the same assumptions
0 = mA + mB − mC → mc = mA + mB
Substituting into the energy equation:
mA (hA − hC ) + mB (hB − hC ) = 0 = mA c p , A (TA − TC ) + mB c p , B (TB − TC )
Solving for the outlet temperature
mA c p , ATA + mB c p , BTB
TC =
mA c p , A + mB c p , B
Evaluating the specific heats using Appendix B-8 at the average temperatures, and guessing TC ≈ 120°F
(70 + 120)
H 2: Tavg = = 95°F, cP = 3.424 Btu/lbmR
2
(150 + 120) 8.552
CH 4: Tavg = = 135°F, cp = = 0.558 Btu/lbmR
2 16.043
(0.1)(3.424)(530 R) + (0.795)(0.558)(610 R)
TC = = 575 R
(0.1)(3.424) + (0.795)(0.558)
= 115°F Answer
Our initial guess of the outlet temperature was close enough so no iteration is required on the specific heats.

15-17
15-16 A rigid tank is divided into two compartments, each of which has a volume of 0.25 m3. One compartment
contains CO2 at 200 kPa, 27 ºC, and the other compartment contains O2 at 100 kPa, 50 ºC. The partition
separating the gases is removed and the gases mix. Heat is lost to the surroundings in the amount of 15 kJ.
Determine:
a. the final mixture temperature (in ºC)
b. the final mixture pressure (in kPa).

Approach:
The closed system energy equation is used to
calculate the final temperature once the two gases
mix. Assuming the gases are ideal and the mixture
behaves as an ideal gas mixture, the ideal gas laws
can be used to calculate the final pressure.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s laws.
2. The specific heats are constant.
3. Potential and kinetic energy effects are
negligible.

Solution:
a) For the control volume defined above, assuming negligible potential and kinetic energy effects and noting that
for a rigid volume W = 0, the closed system energy equation
∆U + ∆PE + ∆KE = Q − W → ∆U = Q → ∆U CO2 + ∆U O2 = Q
With constant specific heats, so that ∆u = cv ∆T : ⎣⎡ mcv (T2 − T1 ) ⎦⎤ CO2 + ⎣⎡ mcv (T2 − T1 ) ⎦⎤ O2 = Q
The individual masses are calculated with the ideal gas equation with molecular weights from Appendix A-1.
PVM CO2 ( 200 kN m )( 0.25 m ) ( 44.01kg kmol )
2 3

mCO2 = = =0.882 kg
RT (8.314 kJ kmolK )( 300K )
mO2 =
PVM O2
=
(100 kN m )( 0.25 m ) ( 32.0 kg kmol ) =0.298 kg
2 3

RT (8.314 kJ kmolK )( 323K )

Assuming T2 ≈ 300K , from Appendix A-8,
cv ,O2 = 0.658 kJ kgK and cv ,CO2 = 0.657 kJ kgK
Therefore,
⎛ kJ ⎞ ⎛ kJ ⎞
( 0.882kg ) ⎜ 0.657 ⎟ (T2 − 300K ) + ( 0.298kg ) ⎜ 0.658 ⎟ (T2 − 323K ) = −15 kJ
⎝ kgK ⎠ ⎝ kgK ⎠
T2 = 286.5K=13.5o C Answer
b) For the final pressure use the ideal gas law:
n RT
P= m
V
The total number of moles is:
⎛m⎞ ⎛m⎞ 0.882 kg 0.298 kg
nm = ⎜ ⎟ + ⎜ ⎟ = + =0.0294 kmol
⎝ M ⎠CO2 ⎝ M ⎠O2 44.01kg kmol 32.0 kmol
nm RT ( 0.0294 kmol )( 8.314 kJ kmolK )( 286.5K )
P= = =140.0 kPa Answer
V ( 0.50 m3 ) (1kJ 1kNm )

15-18
15-17 A mixture with a molar analysis of N2, 83%, CO2, 13%, and O2, 4% with an inlet flow rate of 50 m3/min is
expanded in an adiabatic turbine from 400 kPa, 227 °C to 105 kPa, 77 °C. Determine the power output (in
kW).

Approach:
Conservation of energy is used to calculate the power
from the given information. The effects of the
mixture on the properties must be taken into account.
Assumptions:
1. The mixture behaves as an ideal gas and follows
Dalton’s law.
2. The system is steady and adiabatic with negligible
potential and kinetic energy effects.

Solution:
Conservation of energy and mass applied to the control volume defined above, assuming steady, adiabatic,
negligible potential and kinetic energy effects, and the mixture behaves as a mixture of ideal gases:
W = m(hA − hB )
Using the definitions of mass flow and enthalpy in terms of molar quantities
h −h
W = M m nm ( A B ) = nm (hA − hB )
Mm
Using Eq. 15-19,
W = nm [YN2 (hA − hB ) N2 + YCO2 (hA − hB )CO2 + YO2 (hA − hB )O2 ]
Using the ideal gas equation, PV = nRT or in terms of flows:
kN m3 1min
(400 2 )(50 )( )
PV m min 60s = 0.080 kmol
nm = =
RT kJ s
(8.314 )(500K)
kmolK
Using Appendix A-18 for the molar enthalpies:
Gas hA (at 500 K) hB (at 500 K)
N2 14,581 10,180
O2 14,770 10,213
CO2 17,678 11,351

kmol kJ
W = (0.080 )[(0.83)(14,581 − 10,180) + (0.13)(17, 678 − 11,351) + (0.04)(14, 770 − 10, 213)]
s kmol

= 373 kW Answer

15-19
15-18 A mixture of 40% carbon dioxide and 60% nitrogen (by weight) is compressed isentropically from 100
kPa, 27 °C to 400 kPa. Assume the specific heats are constant at the inlet temperature. Determine:
a. the outlet temperature (in °C)
b. the work required (in kJ/kg).

Approach:
Assume the mixture behaves as an ideal gas with
constant specific heat. The final temperature can
be determined from the isentropic relations for an
ideal gas, once the mixture’s properties are
determined. Work can be calculated using
conservation of energy.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.
2. The system is steady, adiabatic, with
negligible potential and kinetic energy.
3. Specific heats are constant.

Solution:
a) For an isentropic process of an ideal gas with constant specific heat:
P k −1 cp
T2 = T1 ( 2 ) k where k =
P1 cv
Using Eq. 15-20 and Eq. 15-22 and from Appendix A-8 at 300K,
c p , N2 = 1.039 kJ/kgK, cv , N2 = 0.743kJ/kgK, and c p ,CO2 = 0.846 kJ/kgK, cv ,CO2 = 0.657 kJ/kgK.
c p , m = (0.60)(1.039) + (0.4)(0.846) = 0.962 kJ/kgK
c v,m = (0.60)(0.743) + (0.4)(0.657) = 0.708 kJ/kgK
c p,m 0.962
km = = = 1.358
cv , m 0.708
−1
400 1.358
T2 = (300K)( ) 1.358 = 432.5 K = 159.5°C Answer
100
b) Applying conservation of energy and mass, and assuming steady, adiabatic, negligible potential and kinetic
energy effects, and ideal gas mixture with constant specific heat:
W = m(h1 − h2 ) = mc p , m (T1 − T2 )
W
= (0.962 kJ/kgK)(300 − 432.5)K = −127.5 kJ/kg Answer
m

15-20
15-19 Two gas streams are mixed in an insulated device. The first stream is nitrogen (N2) at 20 psia, 120 ºF with
a volume flow rate of 300 ft3/min. The second stream is oxygen (O2) at 20 psia, 200 ºF with a mass flow
rate of 50 lbm/min. The exiting stream is at 17 psia. Determine the exit temperature (in ºF).

Approach:
Conservation of energy is used to calculate the
outlet temperature.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.
2. The system is steady and adiabatic with
negligible potential and kinetic energy effects.
3. Specific heats are constant.

Solution:
The open system conservation of energy equation, ignoring potential and kinetic energy effects, is:
dE
= Q − W + mA hA − mB hB
dt
The system is steady, adiabatic with no work, so that:
0 = mA hA + mB hB − mC hC
Using conservation of mass with the same assumptions:
0 = mA + mB − mC → mC = mA + mB
Substituting into the conservation of energy equations, and assuming each gas is ideal and the mixture behaves as
an ideal gas with constant specific heats, so that ∆h = cP ∆T :
mA ( hA − hC ) + mB ( hB − hC ) = 0 = mA cP , A (TA − TC ) + mB cP , B (TB − TC )
Solving for the outlet temperature:
m c T + mB cP , BTB
TC = A P , A A
mA cP , A + mB cP , B
The mass flow rate of nitrogen is obtained with the ideal gas equation written in terms of flows.
⎛ lbf ⎞ ⎛ ft 3 ⎞ ⎛ lbm ⎞ ⎛ in.2 ⎞
⎜ 20 in.2 ⎟ ⎜ 300 min ⎟ ⎜ 28.01 lbmol ⎟ ⎜ 144 ft 2 ⎟
PVM ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ = 27.0 lbm
mA = =
RT ⎛ ft-lbf ⎞ min
⎜ 1545 lbmR ⎟ ( 580R )
⎝ ⎠

Using Appendix B-8 to evaluate the specific heats at average temperatures, assume TC 170°F :
N 2 : at Tavg = (120 + 170 ) 2 = 145°F c p = 0.248 Btu lbmR
O2 : at Tavg = ( 200 + 170 ) 2 = 185°F c p = 0.222 Btu lbmR

( 27.0 lbm min )( 0.248 Btu lbmR )(120° F ) + ( 50 lbm min )( 0.222 Btu lbmR )( 200°F )
TC =
( 27.0 lbm min )( 0.248 Btu lbmR ) + ( 50 lbm min )( 0.222 Btu lbmR )
TC = 169.9 °F Answer
This answer is close enough to our assumed value of outlet temperature, so no iteration is required.

15-21
15-20 A mixture of N2, CO2, and H2O at 100 kPa, 27 ºC is compressed isentropically in a steady-flow compressor
to 600 kPa. The molar ratio is 4:1:1, respectively. Using constant specific heats evaluated at the inlet
temperature, determine:
a. the outlet temperature (in ºC)
b. the required work per kmol (in kJ/kmol).

Approach:
Assume the mixture behaves as an ideal gas with
constant specific heat. The outlet temperature can
be determined from the isentropic relations for an
ideal gas, once the mixture properties are
determined. Work can be calculated using
conservation of energy.

Assumptions:
1. The gases are ideal with constant specific
heats and the mixture follows Dalton’s law.
2. The process is steady, adiabatic, and reversible
with negligible potential and kinetic energy
effects.

Solution:
a) For an isentropic process for an ideal gas with constant specific heat: T2 = T1 ( P2 / P1 ) k −1/ k where
k = c p / cv and k is the mixture property. Using Eq. 15-21 and Eq. 15-23,

c p = ∑ Yi c p ,i and cv = ∑ Yi cv ,i Note that c p = c p / M

4 1
The mole fractions are: YN2 = = 0.667 YH 2O = YCO2 = = 0.167
4 +1+1 4 +1+1
kg kJ kJ
From Appendix A-8 at 300K, cP , N2 = 1.039 kJ kg iK → c p ,N2 = (28.01kg )(1.039 ) = 29.10
kmol kgiK kmoliK
kJ kJ
cv , N2 = (0.743)(28.01) = 20.81 c p ,CO2 = (0.846)(44.01) = 37.23
kmol•K kmoliK
kJ
cv ,CO2 = (0.657)(44.01) = 28.91
kmoliK
kJ kJ
c p , H 2 O = 33.67 and cv = c p − R = 33.67 − 8.314 = 25.36
kmoliK kmoliK
kJ
Therefore, c p , m = (0.667)(29.10) + (0.167)(37.23) + (0.167)(33.67) = 31.32
kmoliK
kJ
cv , m = (0.667)(20.81) + (0.167)(28.91) + (0.167)(25.36) = 22.94
kmoliK
k = c p / cv = 31.32 / 22.94 = 1.365
T2 = (300K)(600 100)( )
1.365 −1 1.365
= 484.4 K = 211.4 °C
By applying conservation of energy and mass, and by assuming steady, adiabatic, negligible kinetic and potential
energy effects, and ideal gas mixture with constant specific heat:
W = m(h1 − h2 ) = mic p , m (T1 − T2 )
Using the definitions, n = m / M and c p = c p / M → W = nc p , m (T1 − T2 ) → W n = c p , m (T1 − T2 )
W kJ kJ
= (31.32 )(300 − 484.4)K = −5775 Answer
n kmoliK kmol

15-22
15-21 A rigid tank contains 1 kg of CO2 at 100 kPa, 27 °C. A second tank contains 0.8 kg O2 at 500 kPa, 127 °C.
The valve in the pipe connecting the two tanks is opened, and the gases are allowed to mix until an
equilibrium temperature of 77 °C is obtained. Determine:
a. the volume of each tank (in m3)
b. the final pressure (in kPa)
c. the heat transfer to or from the gases (in kJ).

Approach:
Assuming each gas and the mixture are ideal
gases, the ideal gas law can be used to
determine the volumes. The energy equation is
used to calculate the heat transfer rate. The
final pressure can be determined using mixture
properties in the ideal gas law.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.
2. Potential and kinetic energy effects are
negligible and there is no work.
3. Specific heats are constant.

Solution:
a) Assuming both gases are ideal:
m RT (1kg )(8.314 kJ kmolK )( 300 K )
VA = A A = = 0.567 m3 Answer
PA M A ( )
100 kN m ( 44.01kg kmol )
2

( 0.8)(8.314 )( 400 )
VB = = 0.166 m3 Answer
( 500 )( 32.00 )

b) The final pressure is obtained assuming the mixture is an ideal gas:

m RT
P2 = m m
M mVm
The mixture molecular weight is:
1 1 ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 0.8 ⎞⎛ 1 ⎞ kg
= ∑ Xi = + → M m = 37.72
Mm M i ⎜⎝ 1 + 0.8 ⎟⎠ ⎜⎝ 44.01 ⎟⎠ ⎜⎝ 1 + 0.8 ⎟⎜ ⎟
⎠⎝ 32.00 ⎠ kmol
(1.8 kg )(8.314 kJ kmolK )( 350 K )
P2 = = 189.4 kPa Answer
( 37.72 kg kmol )( 0.567 + 0.166 ) m3
c) Assuming negligible potential and kinetic energy effects and no work, conservation of energy gives:
Q = ∆U = ∆U CO2 + ∆U O2
For an ideal gas with constant specific heats:
Q = ⎡⎣ mcv (T2 − T1 ) ⎤⎦ CO + ⎡⎣ mcv (T2 − T1 ) ⎤⎦ O
2 2

From Appendix A-8 at the average temperatures:

CO2: at Tavg = ( 300 + 350 ) 2 = 325K, cv = 0.682 kJ kgK
O2: at Tavg = ( 400 + 350 ) 2 = 375K, cv = 0.675 kJ kgK
⎛ kJ ⎞ ⎛ kJ ⎞
Therefore, Q = (1kg ) ⎜ 0.682 ⎟ ( 350-300 ) K+ ( 0.8 kg ) ⎜ 0.675 ⎟ ( 350-400 ) K
⎝ kgK ⎠ ⎝ kgK ⎠
= 34.1kJ-27.0 kJ=7.1kJ Answer

15-23
15-22 A mixture of N2 and CO2 (mole fractions 70 % and 30%, respectively) with a flow rate of 1.5 kg/s is
compressed adiabatically from 125 kPa, 27 ºC to 450 kPa. The compressor has an isentropic efficiency of
83%. Determine:
a. the exit temperature (in ºC)
b. the power required (in kW).

Approach:
Conservation of energy and the definition of
isentropic efficiency are used to calculate the
compressor power.

Assumptions:
1. The process is steady and adiabatic with
negligible potential and kinetic energy
effects.
2. The mixture behaves as an ideal gas.

Solution:
The definition of isentropic efficiency is:
W
ηC = s
W
The power for isentropic compression, Ws , can be determined from conservation of energy. For the control
volume defined above, assuming steady and adiabatic with negligible potential kinetic energy effects and an ideal
gas mixture:
W = m ( hA − hB )
k
Using Eq. 15-19, h = ∑ Yi hi and h = h M
i =1

m
M m ⎢⎣
{ (
W = m ⎡ Y hA − hB )} N2
{ (
+ Y hA − hB )}
CO2
⎤
⎥⎦
From Appendix A-1, M N2 = 28.01kg kmol and M CO2 = 44.01kg kmol .
The mixture molecular weight is:
M m = YN2 M N2 + YCO2 M CO2 = ( 0.70 )( 28.01) + ( 0.30 )( 44.01) = 32.81kg kmol
Also, from Appendix A-18
N2: at 300 K hA = 8, 723kJ kmol , s Ao = 191.682 kJ kmol K
CO2: at 300 K hA = 9, 431kJ kmol , s Ao = 213.915 kJ kmol K
For the isentropic process, s A = sB , s or s A − sB , s = 0
Using Eq. 15-27, s = ∑ Yi si → ( ) ( )
YN2 sN2 , A − sN2 , B , s + YCO2 sCO2 , A − sCO2 , B , s = 0
and Eq. 15-33 s2,i − s1,i = s − s − R ln ( P2,i P1,i )
o
2, i
o
1, i

Note that with the constant composition, P2,i P1,i = P2 P1 . Therefore, combining the previous two equations:
⎡ kJ ⎞ ⎛ 450 ⎞ ⎤
( 0.70 ) ⎢⎛⎜191.682
kJ ⎞ o ⎛
⎟ -sB,s,N2 + ⎜ 8.314 ⎟ ln ⎜ ⎟⎥
⎣⎝ kmolK ⎠ ⎝ kmolK ⎠ ⎝ 125 ⎠ ⎦
⎡⎛ kJ ⎞ o ⎛ kJ ⎞ ⎛ 450 ⎞ ⎤
+ ( 0.30 ) ⎢⎜ 213.915 ⎟ -sB,s,CO2 + ⎜ 8.314 ⎟ ln ⎜ ⎟⎥ = 0
⎣ ⎝ kmolK ⎠ ⎝ kmolK ⎠ ⎝ 125 ⎠ ⎦
For the isentropic process, once the outlet temperature is fixed, both s o are known. So, solving by trail and error:

Guess T2s (K) sNo2 (kJ/kmolK) o

sCO (kJ/kmolK) Left hand side of
2
equation
550 209.461 239.135 -9.363
500 206.63 234.814 -6.085

15-24
 420 201.499 227.258 0.22
The LHS of the equation is close enough to zero, so additional iterations are not required. Therefore,
→ hBs , N2 = 12, 225 kJ kmol, hBs ,CO2 = 14, 206 kJ kmol
The ideal work is:
(1.5 kg s ) kJ
Ws = ⎡⎣( 0.70 )( 8,723-12,225 ) + ( 0.30 )( 9,431-14,206 ) ⎤⎦
32.81kg kmol kmol
= −178 kW
Ws -178 kW
Therefore, W = = =-214 kW Answer
ηC 0.83

15-25
15-23 A 10-m3 tank, initially filled with O2 at 40 kPa, 27 ºC, is connected to a pipeline that contains N2 at 200
kPa, 27 ºC. Nitrogen is allowed to flow into the smaller tank until the pressure reaches 110 kPa. Heat
transfer to the surroundings results in an isothermal process. Determine:
a. the mass of nitrogen that enters the tank (in kg)
b. the heat transfer (in kJ).

Approach:
This is a transient (filling) problem. The open
system conservation of energy equation can be
used to determine the heat transfer. The mass of
nitrogen added can be determined with the given
information, ideal gas laws, and mixture
condensations.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.
2. The system has no work and negligible
potential and kinetic energy effects.

Solution:
a) Assuming each gas and the mixture are ideal, i i = mi RTi M i = ni RTi
PV
mO2 RT1
Initially, only oxygen is present 1 1 =
PV = nO2 RT1
M O2
After the nitrogen is added, P2V2 = nm RT2 = (nO2 + nN2 ) RT2
Combining these two equations and canceling terms
P1 nO2 P
= → nN2 = nO2 ( 2 − 1)
P2 nO2 + nN2 P1
PV (40kN/m 2 )(10m3 )
nO2 = 1 1
= = 0.160 kmol
RT1 (8.314kJ / kmolK)(300K)
110
nN2 = (0.160kmol)( -1)=0.281kmol
40
mN 2 = M N 2 n N 2 = (28.01kg / kmol)(0.281kmol) = 7.86 kg
Answer
mO2 = (32.0)(0.160) = 5.12 kg
b) Applying the open system energy equation to the control volume defined above, and assuming: no work, ideal
dU
gases, mixture is ideal gas, negligible potential and kinetic energy effects: = Q + mN2 hN2
dt
dm
From conservation of mass: = mN 2
dt
Substituting into the energy equation, integrating with respect to time, and noting that hN2 is constant
∆U = Q + mN2 hN2
Solving for the heat transfer
Q = ∆U − mN2 hN2 = (m2 u2 − m1u1 ) − mN2 hN2
Using Eq.15-15
Q = (mN2 u N2 ,2 + mO2 uO2 ,2 ) − mO2 ,1uO2 ,1 − mN2 hN2 ,2 → Q = mN2 (u N2 ,2 − hN2 ,2 ) + mO2 (uO2 ,2 − uO2 ,1 )
For an ideal gas, internal energy is function of temperature only so uO2 ,2 − uO2 ,1 = 0 . Note h = u + pv , so:
Q = mN2 (− P2 v2 ) N2 = − PN2 ,2V = −YN2 P2V
0.281
= −( )(110kN/m 2 )(10m3 ) = −701kJ Answer
0.281 + 0.160

15-26
15-24 A mixture of 1 kg methane (CH4) and 15 kg air initially at 27 ºC, 101.3 kPa is compressed isentropically to
one tenth of its initial volume. Determine the final temperature (in ºC) and pressure (in kPa).

Approach:
Assuming the mixture behaves as an ideal gas, the
ideal gas equation and expressions for entropy
change of an ideal gas mixture can be used to
calculate the final temperature and pressure.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.
2. The process is isentropic.

Solution:
For the mixture, the ideal gas equation is:
PV1 1 = n1 RT1 and P2V2 = n2 RT2
Dividing these two equations and solving for the final pressure:
T V ⎛ T ⎞⎛ 1 ⎞
P2 = 2 1 P1 = ⎜ 2 ⎟ ⎜ ⎟ (101.3kPa ) (1)
T1 V2 ⎝ 300 ⎠ ⎝ 0.1 ⎠
We need to evaluate T2.
For the isentropic process: s1 = s2 or s1 − s2 = 0 . Therefore, using Eq. 15-27, s = ∑ Yi si
⎡⎣Y ( s2 − s1 ) ⎤⎦ air + ⎡⎣Y ( s2 − s1 ) ⎤⎦ CH = 0
4

and Eq. 15-31 with constant specific heats:

⎛T ⎞ ⎛P ⎞
s2,i − s1,i = c p ,i (T ) ln ⎜ 2 ⎟ − R ln ⎜ 2,i ⎟
⎜P ⎟
⎝ T1 ⎠ ⎝ 1,i ⎠
The composition is constant, so P2,i P1,i = P2 P1 .
Using Appendix A-8 at 300K, c p , air = c p , air M air = (1.005 )( 28.97 ) = 29.11kJ kmolK
Likewise, c p ,CH 4 = c p ,CH 4 M CH 4 = 35.81kJ kmolK
For the moles
15 kg 1 kg
nair = =0.518 kmol nCH 4 = =0.0623kmol
28.97 kg kmol 16.04 kg kmol
The mole fractions are:
0.518
Yair = = 0.893 YCH 4 = 1 − Yair = 0.107
0.518 + 0.0623
The entropy change equation is:
⎡ kJ ⎞ ⎛ T2 ⎞ ⎛ kJ ⎞ ⎛ P2 ⎞ ⎤
( 0.893) ⎢⎛⎜ 29.11 ⎟ ln ⎜ ⎟ − ⎜ 8.314 kmolK ⎟ ln ⎜ 101.3 ⎟ ⎥
⎣ ⎝ kmolK ⎠ ⎝ 300 ⎠ ⎝ ⎠ ⎝ ⎠⎦
(2)
⎡⎛ kJ ⎞ ⎛ T2 ⎞ ⎛ kJ ⎞ ⎛ P2 ⎞ ⎤
+ ( 0.107 ) ⎢⎜ 35.81 ⎟ ln ⎜ − ⎜ 8.314 ⎟ ln =0
⎣⎝
⎟
kmolK ⎠ ⎝ 300 ⎠ ⎝ kmolK ⎠ ⎜⎝ 101.3 ⎟⎠ ⎥⎦
The two equations (1) and (2) are solved for T2 and P2. Doing so, we obtain:
T2 = 730.3K and P2 = 2466 kPa Answer

15-27
15-25 A mixture with a molar analysis of CO2, 50%, CO, 34%, and O2, 16% with a flow rate of 2 kg/s is
adiabatically compressed from 100 kPa, 27 °C to 550 kPa, 227 °C. The inlet velocity is 40 m/s and the
outlet velocity is 80 m/s. Determine the required power input (in kW). How does the power change if
velocity is not taken into account?

Approach:
Conservation of energy and mass can be used to
calculate the required power. The effects of the
mixture properties must be taken into account. We
can follow the approach given in Example 15-3.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.
2. The system is steady and adiabatic with
negligible potential and kinetic energy effects.
3. The specific heats are constant.

Solution:
Conservation of mass and energy applied to the control volume defined above, assuming steady, adiabatic,
negligible potential energy effects, and the mixture behaves as an ideal gas mixture:
mA = mB = mm = ∑ mi
VA2 VB2 V 2 − VB2 V 2 − VB2
W = mm [(hA − hB ) + ( − )] = mm (hA − hB ) + mm ( A ) = ∑ mi ∆hi + mm ( A )
2 2 2 2
V 2 − VB2
= ∑ ni ∆hi + mm ( A )
2
k
Using Eq.15-19, h = ∑ Yi hi , and assuming ideal gases with constant specific heats, so that ∆h = c p ∆T :
i =1

VA2 − VB2
W = ∑ ni c p ,i ∆T + mm ( )
2
Factor out the total molar flow rate and use the definition of mole fraction:
n V 2 − VB2 V 2 − VB2
W = nm ∑ i c p ,i ∆T + mm ( A ) = nm ∑ Yi c p ,i ∆T + mm ( A )
nm 2 2
mm VA2 − VB2
=
Mm
∑ Yi c p,i ∆T + mm ( 2
)

kg
The mixture molecular weight is: M m = ∑ Yi M i = ( 0.50 )( 44.01) + ( 0.34 )( 28.011) + ( 0.16 )( 32.0 ) = 36.65
kmol
Noting that c p = c p M and using Appendix A-8 at the average temperature, Tavg = (300 + 500) / 2 = 400K
⎛ kJ ⎞ ⎛ kg ⎞ kJ
c p ,CO2 = ⎜ 0.939 ⎟ ⎜ 44.01 ⎟ =41.3 kgK
⎝ kgK ⎠⎝ kmol ⎠
⎛ kJ ⎞ ⎛ kg ⎞ kJ
c p ,CO = ⎜ 1.047 ⎟ ⎜ 28.011 ⎟ = 29.3 kgK
⎝ kgK ⎠⎝ kmol ⎠
⎛ kJ ⎞ ⎛ kg ⎞ kJ
c p ,O2 = ⎜ 0.941 ⎟ ⎜ 32.0 ⎟ =30.1 kgK
⎝ kgK ⎠⎝ kmol ⎠
kJ
∑Y ci p ,i = (0.50)(41.3) + (0.34)(29.3) + (0.16)(30.1) = 35.4
kmolK
kg 35.4 kJ kmolK kg (40 m s) 2 − (80 m s) 2
W = (2.0 )( )(300 − 500)K+(2.0 )( )
s 36.65 kg kmol s 2(1000kgm kNs 2 )
= −386 kW − 4.8 kW = −391kW Answer
The velocity added only 4.8 / 391× 100% = 1.2% of the power. Answer
15-28
15-26 A mixture of oxygen (O2) and nitrogen (N2) (mass fractions 60% and 40%, respectively) is expanded
adiabatically from 500 kPa, 450 °C to 150 kPa. The flow rate is 2.5 kg/s, and the isentropic efficiency is
80%. Determine:
a. the exit temperature (in °C)
b. the output power (in kW)

Approach:
Conservation of energy and the definition of
isentropic efficiency, combined with mixture
property information, are used to calculate the
turbine power. Power is calculated first, then the
outlet temperature is calculated.

Assumptions:
1. The mixture behaves as an ideal gas and
follows Dalton’s law.
2. The system is steady and adiabatic with
negligible potential and kinetic energy effects.

Solution:
b) The definition of isentropic efficiency for a turbine is: ηT = W Ws → W = ηT Ws
The turbine power for an isentropic process, Ws , can be determined form conservation of energy and mass for the
control volume defined above, assume steady adiabatic, isentropic, negligible potential or kinetic energy effects,
and an ideal gas mixture. Ws = m(hA − hB , s )
Using Eq. 15-18, h = ∑ X i hi , and assuming constant specific heats, so that ∆h = c p ∆T :
W = m[{ Xc p (TA − TB , s )}O2 + { Xc p (TA − TB , s )}N2 ] = m[( Xc p )O2 + ( Xc p ) N2 ](TA − TB , s )
We need to determine TB , s . For an isentropic process S A = S B or S A − S B = 0
TB R P
Using Eq. 15-26, s = ∑ X i si and Eq. 15-30: sB ,i − s A,i = c p ,i ln( )− ln( B ,i )
TA Mi PA,i
The composition is constant, so, PB ,i / PA,i = PB / PA therefore,
TB , s R P T R P
X O2 [c p ,O2 ln( )− ln( B )] + X N2 [c p , N2 ln( B , s ) − ln( B )] = 0
TA M O2 PA TA M N2 PA

Solving for TB , s :
2 2
(
⎡ X O M O + X N M N R ln ( PB PA ) ⎤
TB , s = TA exp ⎢
2 2
⎥
)
⎢ X O2 c p ,O2 + X N2 c p , N2 ⎥
⎣ ⎦
The specific heats are evaluated using Appendix A-8 at the average temperature. Assuming TB , s ≈ 500 K ,
Tavg (500 + 723) / 2 = 611K . So, using 600 K, c p ,O2 = 1.003kJ/kgK , c p , N2 = 1.025kJ/kgK
0.60 0.40 kg kJ 150
( + ) i(8.314 )iln( )
32.00 28.01 kmol kgmolK 500
TB , s = (723K) exp[ ] = 521K
kJ kJ
(0.60)(1.003 ) + (0.40)(1.025 )
kgK kgK
This is close enough to our assumed value, so no iteration is required. Therefore:
Ws = (2.5 kg s)[(0.60)(1.003) + (0.40)(1.025)](kJ kgK)(723 − 521)K = 510 kW
W = ηT Ws = (0.80)(510 kW) = 408 kW Answer
a) For the actual outlet temperature, use the definition of isentropic efficiency in terms of temperatures
W h −h c p (TA − TB ) T −T
ηT = = A B = = A B
Ws hA − hB , s c p (TA − TB , s ) TA − TB , s
TB = TA − ηT (TA − TB , s ) = 723K- ( 0.80 )( 723-521) K=561.4 K Answer

15-29
15-27 Sometimes on a winter day, eyeglasses fog up when a person enters a building from the outdoors.
Assuming an outdoor air temperature of 5 ºC and an indoor air temperature of 25 ºC, determine the
maximum relative humidity the building can have for a person’s eyeglasses not to become fogged.

Approach:
For fog to form on the glasses, the air mixture would
need a humidity whose dewpoint is 5oC. We can use
the definition of relative humidity, Eq. 15-39 to
determine the maximum relative humidity for the
glasses not to fog.

Assumptions:
1. Air and water vapor are ideal gases, and their
mixture behaves as an ideal gas.

Solutions:
Relative humidity is defined as:
P
φ= v
Pg
Using Appendix A-10, Pg is evaluated at the mixture temperature, TDB = 25°C
Pg (TDB ) = 3.169 kPa
The vapor pressure, Pv, when the mixture reaches its dewpoint temperature of 5oC is:
Pv = Pg (TDP ) = 0.8721 kPa
Therefore:
0.8721
φmax = = 0.275 or 27.5% Answer
3.169
Any relative humidity greater than this will cause the eyeglasses to fog.

15-30
15-28 Air at a pressure of 101.3 kPa with a flow rate of 150 m3/min enters a humidifier with a dry-bulb
temperature of 38 °C and a wet-bulb temperature of 20 °C. The air exits the cooler with a dry-bulb
temperature of 25 °C and a relative humidity of 65%. Without using the psychrometric chart, determine the
make-up water flow rate (in kg/min).

Approach:
Conservation of mass, applied to the defined control
volume, can be used to determine the make-up water
flow rate.

Assumptions:
1. Air and water vapor are ideal gases, and their
mixture behaves as an ideal gas.

Solution:
Assuming steady flow and tracking the water and the air separately, we obtained:
Air: ma1 = ma 2 = ma
Water: mv1 + m f 3 = mv 2
Using the definition of specific humidity in terms of flow rates, ω = mv ma , we rewrite the water balance
equation:
ω1ma + m f 3 = ω2 ma → m f 3 = (ω2 − ω1 )ma
Assuming the air and water vapor behave as ideal gases, the inlet specific humidity can be determined with Eq.
15-46 and the saturation tables for water.
0.622 Pg (TWB ) 0.622(2.339 kPa) kg water
ω* = = =0.0147
P − Pg (TWB ) 101.3kPa-2.339 kPa kg dry air
c p , a (TWB − TDB ) + ω * h fg (TWB )
ω1 =
hg (TDB ) − h f (TWB )
(1.005 kJ kg K)(20 − 38)K + (0.0147 kg water kg dry air)(2454.4 kJ kg) kg water
= = 0.00723
(2570.7 − 83.96) kJ kg kg dry air
The outlet specific humidity is determined with Eq. 15-42.
0.622φ2 Pg (TDB ) (0.622)(0.65)(2.339kPa) kg water
ω2 = = = 0.00948
P − φ2 Pg (TDB ) 101.3kPa − (0.65)(2.339kPa) kg dry air
The dry air flow rate, assuming air is ideal gas:
PM ( P − Pv ) M ( P − φ1 Pg1 ) M
ma = ρ aV = a V = V= V
RT RT RT
Using Eq. 15-42
ωP (0.00723)(101.3)
φ1 = = = 0.498
(0.622 + ω ) Pg (0.622+0.00723)(2.339)
kN kg m3
[101.3 − (0.498)(2.339) (28.97 )(150 )
ma = m 2
kmol min = 168.3 kg
kJ min
(8.314 )(38 + 273)K
kmolK
kg kg
m f 3 = (0.00948 − 0.00723)(168.3 ) = 0.379 Answer
min min

15-31
15-29 Outside air at 30 °F, 60% relative humidity is to be conditioned and delivered to a building at 70 °F, 50%
relative humidity. The volume flow rate is 7500 ft3/min. Using the psychrometric chart, determine the
required water addition per lbm of dry air.

Approach:
Conservation of mass applied to defined control
volume is used to determine the make-up water flow
rate.

Assumptions:
1. Air and water are ideal gases, and their mixture
is an ideal gas.
2. The system is steady.

Solution:
Assuming steady flow, and tracking the water and air separately, we obtain:
Air: ma1 = ma 2 = ma
Water: mv1 + m f 3 = mv 2
Using the definition of specific humidity in term, of flow rates, ω1ma + m f 3 = ω2 ma → m f 3 = (ω2 − ω1 )ma
The air mass flow rate is:
V
ma = ρ aV =
va
Assuming the air and water vapor behave as ideal gases, from the pyschrometric chart (assuming 1 atm pressure):
lbm water ft 3
ω1 ≈ 0.002 v1 ≈ 12.4
lbm dry air lbm dry air
ω2 ≈ 0.00775
Therefore,
ft 3
7500
ma = min = 605 lbm
ft 3 min
12.4
lbm
lbm
m f 3 = (0.00775 − 0.002)(605) = 3.48 Answer
min

15-32
15-30 When feed corn is harvested, it must be dried before it is stored so that it does not mildew and rot.
Consider corn that enters a dryer at 40% moisture by mass (solid plus moisture). Dry air enters the dryer at
80 ºC at a rate of 16 kg/kg of corn entering. Moist air exits the dryer at 100 kPa, 35 ºC, and 55% relative
humidity. Determine the moisture content by mass of the corn leaving the dryer.
Approach:
Conservation of mass applied to the control volume
defined above is used to determine the moisture
content of the corn.
Assumptions:
1. The air and water vapor are ideal gases, and
their mixture behaves as an ideal gas.

Solution:
We define the moisture in the corn as X = mw, c mcorn
where mw,c is the water flow rate in the corn and mcorn is the total (solid plus water) mass flow rate of corn.
Applying conservation of mass to the control volume and assuming steady flows:
Air: ma , A = ma , B = ma
Water: mv , A + mw,c ,1 − mv , B − mw,c ,2 = 0
Because dry air enters at A1 , mv , A = 0. Solving for the exiting corn moisture flow and dividing by the corn mass
flow at 2:
mw, c ,2 mw,c ,1 − mv , B mw,c ,1 mv , B
X2 = = = −
mcorn,2 mcorn ,2 mcorn,2 mcorn,2
The corn mass flow at 2, mcorn ,2 , is the difference between the corn entering at 1 and the water leaving at B,
Assuming the air and water vapor behave as ideal gases.
ωB ma
mcorn ,2 = mcorn ,1 − mv , B = mcorn,1 − ωB ma = mcorn ,1 (1 − )
mcorn,1
Substituting into the main equation
mw,c ,1 mv , B m 1 m 1
X2 = − = w,c ,1 − v, B
ma ma mcorn ,1 m mcorn ,1 ma
mcorn,1 (1 − ωB ) mcorn ,1 (1 − ωB ) (1 − ωB a
) (1 − ωB )
mcorn ,1 mcorn,1 mcorn ,1 mcorn ,1

mw, c ,1 1 mv , B 1 m 1 ma ( mv , B ma ) 1
X2 = ( )− ( ) = w,c ,1 ( )− ( )
mcorn ,1 ma mcorn ,1 ma mcorn ,1 ma mcorn ,1 ma
1 − ωB ma ( ) 1 − ωB 1 − ωB 1 − ωB
mcorn ,1 ma mcorn ,1 mcorn ,1 mcorn ,1
⎛ ⎞
⎜ ⎟⎛ m ⎞
=⎜
1 ⎟ ⎜ w, c ,1 − maωB ⎟
⎜ ma ⎟ ⎜⎝ mcorn,1 mcorn ,1 ⎟⎠
⎜ 1 − ωB m ⎟
⎝ corn ,1 ⎠

Using Eq. 15-42

0.622φB Pg (0.622)(0.55)(5.628)
ωB = = = 0.0199
P − φB Pg 100 − (0.55)(5.628)
⎡ 1 ⎤
X2 = ⎢ ⎥ [ 0.40 − 16(0.0199) ] = 0.120 (12.0%) Answer
⎣ 1 − (0.0199)(16) ⎦

15-33
15-31 Dry air (0% relative humidity) at 14.7 lbf/in.2, 150 ºF is used to dry a damp fabric that enters the dryer with
a 45% moisture content by mass (solid plus moisture) and leaves with a 5% moisture content. The mass
flow rate of the exiting fabric (solid plus moisture) is 10 lbm/min. Moist air exits at 14.7 lbf/in.2, 130 ºF,
and 65% relative humidity. Determine the required volumetric flow rate of the entering air (in ft3/min).

Approach:
Conservation of mass is combined with
psychrometric principles to determine the required
air flow rate.

Assumptions:
1. The air and water vapor are ideal gases, and
their mixture is an ideal gas.
2. The system is steady.

Solution:
Assuming air and water vapor are ideal gases, the ideal gas equation in terms of flow rates is:
m RT
Va = a = ma va
Pa M a
To obtain the air flow rate, apply conservation of mass to the control volume defined above. For steady flow:
Air: ma1 = ma 2 = ma
Water: mw, FA + mv1 − mw, FB − mv 2 = 0
where “w,F” represents the water in the fabric. Because dry air enters at 1, mv1 = 0 . Solving for the vapor flow
at 2 and dividing by the air flow rate:
mv 2 m m
= ω2 = w, FA − w, FB
ma ma ma
Solving for the air flow rate:
ma = (1 ω2 ) ( mw, FA − mw, FB )
mw, F mw, F
We define X = = where “s” represents the solid and “w” represents the water. At B,
mF ms , F + mw, F
mFB = 10 lbm min . Therefore, solving for mw, FB
mw, FB = XmFB = 0.05 (10 lbm min ) = 0.5lbm min
and ms , F = mFB − mw, FB = 10 − 0.5 = 9.5lbm min
For location A, solve the definition of X for the water flow rate in the fabric:
⎛ XA ⎞ ⎛ 0.45 ⎞⎛ lbm ⎞ lbm
mw, FA = ⎜ ⎟ ms , F = ⎜ ⎟⎜ 9.5 ⎟ = 7.78
⎝ 1− X A ⎠ ⎝ 1 − 0.45 ⎠⎝ min ⎠ min
Using Eq. 15-42
0.622φ2 Pg 21 (0.622)(0.65)(2.225) lbm water
ω2 = = = 0.0679
P − φ1 Pg 2 14.7 − (0.65)(2.225) lbm dry air
1 lbm
ma = ( 7.78 − 0.5) = 107.2
0.0679 min
RT (1545ft-lbf/lbmolR)(150 + 460)R
va = = = 15.4 ft 3 /lbm
Pa M a (14.7 lbf/in. ) (28.97 lbm/lbmol (144in. 1ft )
2 2 3

V = (107.2 lbm min ) (15.4 ft 3 lbm ) =1,650 ft 3 min Answer

15-34
15-32 The design specifications for a new room humidifier are: air volume flow rate 0.015 m3/s, inlet relative
humidity 40%, outlet relative humidity 60%, dry bulb temperature fixed at 21 ºC, and barometric pressure
of 101.3 kPa. Water is to be added only once per day. If the humidifier is to run continuously for 24 hrs
before the water supply (assume it is at 21 ºC) must be refilled, determine the required supply tank volume
(in cm3).

Approach:
Conservation of mass is used on two control
volumes. An unsteady problem is obtained for this
water supply tank; the equation is integrated with
respect to time to find the required volume. A steady
control volume is used around the humidifier itself.

Assumptions:
1. The humidifier is steady.
2. The air and water vapor are ideal gases, and
their mixture behaves as an ideal gas.

Solution:
Applying conservation of mass to the water supply tank
dm dt = − m f 3
Separating variables, assuming the water flow note is constant, and integrating with respect to time
0 t

∫ dm = − ∫ m f 3 dt
m 0
→ − m = − m f 3t

With m = ρ wV → V = m f 3t ρ w
The water flow rate is obtained by applying conservation of mass to the control volume around the humidifier.
Assuming steady flow:
Air: ma1 = ma 2 = ma
Water: mv1 + m f 3 = mv 2
Using the definition of specific humidity in terms of flow rates, ω = mv ma , we rewrite the water balance equation:
ω1ma + m f 3 = mv 2 ⇒ m f 3 = (ω2 − ω2 )ma
Note that ma = V va .
Assuming the air and water are ideal gases, and using the psychrometric chart:
ω1 0.0062 kg water/kg dry air
va 0.842 m 3 /kg dry air
ω2 0.0095 kg water/kg dry air
Therefore,
(0.015m3 /s)
ma = = 0.0178kg/s
(0.842m3 /kg)
m f 3 = (0.0095 − 0.0062)(0.0178kg/s) = 5.88 × 10 −5 kg/s
(5.88 × 10 −5 kg/s)(24 hr)(3600s 1hr)
V= = 0.00508 m3
1000 kg/m3
100 cm 3
= (0.00508 m3 )( ) = 5080 cm3 Answer
m

15-35
15-33 Air at 14 lbf/in.2, 75 ºF, and 75% relative humidity is compressed to 70 lbf/in.2, 280 ºF. The air then passes
through a water-cooled heat exchanger and exits as saturated air at 68 lbf/in.2, 100 ºF. Determine:
a. the specify humidity and relative humidity at the compressor discharge
b. the water vapor condensed in the heat exchanger for each lbm of dry air flowing through it.

Approach:
The specific humidity remains constant through the
compression, so at its exit we know two independent
properties once the inlet specific humidity is
determined. The condensate flow rate from the heat
exchange is obtained with conservation of mass.

Assumptions:
1. The air and water vapor are ideal gases, and
their mixture behaves as an ideal gas.
2. The system is steady.

Solution:
a) The specific humidity is constant, so ω2 = ω1 . Eq. 15-42 is used to determine first ω1 , and then φ2 .
0.622φ1 Pg1
ω1 =
P1 − φ1 Pg1
From Appendix B-10, Pg1 (75°F) = 0.435 lbf/in.2
(0.622)(0.75)(0.435)
ω1 = = 0.0148lbm water/lbm dry air
14 − (0.75)(0.435)
Now
ω2 P2
φ2 =
(0.622 + ω2 ) Pg 2
From Appendix B-10, Pg 2 (280°F) = 49.18lbf/in.2
(0.0148)(70)
φ2 = = 0.033 = 3.3%
(0.622 + 0.0148)(49.18)
b) Assuming steady flow for the control volume defined around the heat exchanges, and tracking the air and
water separately, conservation of mass gives:
Air: ma 2 = ma 3 = ma
Water: mv 2 − mv 3 − m f 4 = 0 ⇒ m f 4 = mv 2 − mv 3
Using the definition of specific humidity in terms of flow rates:
m f 4 = ω2 ma − ω3 ma = (ω2 − ω3 )ma
mf 4
= ω2 − ω3
ma
For ω3 from Appendix B-10, Pg 3 (100°F) = 0.9503 lbf/in.2
(0.622)(1.00)(0.9503)
ω3 = = 0.00882
68 − (1.00)(0.9503)
mf 4 lbm water
= 0.0148 − 0.00882 = 0.00598 Answer
ma lbm dry air

15-36
15-34 Air at 101.3 kPa, 20 °C, and 55% relative humidity flows from a humidifier into a room. From equipment
and people in the room, 6 kW of heat is added to the air; assume no water vapor is added. If the exit
temperature of the air cannot exceed 26 °C, determine:
a. the inlet volumetric flow rate of air required (in m3/hr)
b. the percent error if the water vapor in the air is ignored.

Approach:
Conservation of energy and mass are used to
determine the required air flow rate. Note that
because no water vapor is added to the air in the
room, the specific humidity entering and leaving the
room is constant.

Assumptions:
1. Air and water vapor are ideal gases, and their
mixture is an ideal gas.
2. The system is steady with no work and
negligible potential and kinetic energy effects.

Solution:
a) For the control volume defined above, apply conservation of energy and mass. Assume steady, no work,
negligible potential and kinetic energy effects, and ideal gas mixture, therefore,
m
Q = m(h2 − h1 ) = ma (ha 2 − ha1 ) + mv (hv 2 − hv1 ) = ma [(ha 2 − ha1 ) + v (hv 2 − hv1 )]
ma
Assuming constant specific heats and using ω = mv ma and ma = Vva , we can solve this equation for volume
flow rate:
Q
V=
va [c p , a (T2 − T1 ) + ω c p ,v (T2 − T1 )]
From Appendix A-7 at 23oC, c p , a = 1.005 kJ/kgK, c p ,v = 1.867 kJ/kgK
RT RT RT
va = = =
2 Pa M a ( P − Pv ) M a ( P − φ Pg ) M a
At 20oC, Pg = 2.339 kPa → Pa = 101.3kPa − 0.55(2.339 kPa) = 100.0 kPa
(8.314 kJ)(20 + 273)K
va = = 0.841m3 /kg
(100 kN/m 2 )(28.97 kg/kmol)
0.622φ Pg 0622(0.55)(2.339) kg water
ω= = = 0.0080
P − φ Pg 100.0 kg dry air
(6000W) (1kJ 1000J )
V=
(0.841m /kg)[(1.005kJ/kgK) + (0.0080)(1.867 kJ/kgK)](26 − 20)K
3

= 1.166 m3 /s = 4, 200 m3 /hr Answer

b) If we ignore the water vapor,
6000 (1 1000 )
V= = 1.183m3 /s = 4,260 m3 /hr
(0.841)(1.005)(26 − 20)
4260 − 4200
% error = × 100 = 1.4% Answer
4200

15-37
15-35 Shown in the figure below is a heating/humidifier unit. Air enters at 101.3 kPa, -5 ºC, and 90% relative
humidity. The preheating coil raises the temperature to 2 ºC. The air is then humidified with water whose
temperature is held at the dew-point temperature of the air exiting from the complete unit. Reheating is
used to set the outlet to 22 ºC and 35% relative humidity. The make-up water enters at 4 ºC. Using the
psychrometric chart, determine:
a. the water added per kg of dry air
b. the total heat transfer (preheater plus spray water heater plus reheat) to the air per kg of dry air
(kJ/kg dry air).

Approach:
Conservation of mass applied to the overall control volume
is used to determine the required make-up water flow rate.
Heat transfer in the three heaters is determined from
conservation of energy.

Assumptions:
1. Air and water vapor are ideal gases, and their mixture
behaves as an ideal gas.
2. The system is steady with no work or potential or
kinetic energy effects.

Solution:
a) Applying conservation of mass to the overall control volume, assuming steady and ideal gas mixture, and
tracking the air and water separately:
Air: ma1 = ma 4 = ma
Water: mv1 + m5 = mv 4 → m5 = mv 4 − mv1 = (ω4 − ω1 )ma
kg water kg water
Using the psychrometric chart: ω1 ≈ 0.0024 and ω4 ≈ 0.00557
kg dry air kg dryair
m5 kg water
= 0.00557 − 0.0024 = 0.00317 Answer
ma kg dry air
b) An energy balance on the overall control volume that encompasses all three heaters, assuming steady, no work,
and negligible potential and kinetic energy effects:
0 = Q + ma1h1 − ma 4 h4 + m5 h5
Q mh
= h4 − h1 − 5 5 = h4 − h1 − (ω4 − ω1 )h5
ma ma
From the psychrometric chart:
h1 ≈ 1.0 kJ/kg, h4 ≈ 36.2 kJ/kg
From Appendix A-10, h5 = h f (4°C) = 16.8 kJ/kg
Q
= 36.2 − 1.0 − (0.00557 − 0.0024)(16.8) = 35.1 kJ/kg dry air Answer
ma

15-38
15-36 For energy conservation and indoor air quality reasons, fresh air and recirculation air from a building are
mixed before being used again in the building. Consider the following mixing situation. Air exits from the
cooling coil in an air conditioning system at 12 ºC and 100% relative humidity and is adiabatically mixed
with fresh air at 36 ºC and 30% relative humidity. The fresh air has a mass flow rate 30% of that exiting
from the cooling coils. Determine the final temperature and relative humidity of the mixture.

Approach:
For this mixing problem, we uses ideal gas mixture
properties in the conservation of mass and energy to
determine the outlet conditions.
Assumptions:
1. The mixture behaves as an ideal gas and follows
Dalton’s law.
2. The system is steady, adiabatic, with no work
and negligible potential and kinetic energy.

Solution:
Applying conservation of mass to this control volume, assuming steady and ideal gas mixture, and tracking the air
and water separately, we obtain:
Air: ma ,1 + ma ,2 = ma ,3 (1)
Water: mv ,1 + mv ,2 = mv ,3 → ω1ma ,1 + ω2 ma ,2 = ω3 ma ,3 (2)
Applying conservation of energy and assuming adiabatic, no work, and negligible potential and kinetic energy
effects:
0 = ma ,1ha ,1 + mv ,2 hv ,2 + ma ,2 ha ,2 + mv ,2 hv ,2 − ma ,3 ha ,3 − mv ,3 hv ,3
= ma ,1 (ha ,1 + ω1hv ,1 ) + ma ,2 (ha ,2 + ω2 hv ,2 ) − ma ,3 (ha ,3 + ω3 hv ,3 ) (3)
Combining Eqs. (1) and (2)
ma ,2 ω1 − ω3 ω1 + ω2 ( ma ,2 ma ,1 )
= → ω3 = (4)
ma ,1 ω3 − ω2 1 + ma ,2 ma ,1
ma ,2 (ha ,1 + ω1hv ,1 ) − (ha3 + ω3 hv ,3 )
Combining Eqs. (1) and (3) =
ma ,1 (ha 3 + ω3 hv ,3 ) − (ha ,2 + ω2 hv ,2 )
(ha ,1 + ω1hv ,1 ) + (ha ,2 + ω2 hv ,2 ) ( ma ,2 ma ,1 )
ha ,3 + ω3 hv ,3 = (5)
1 + ma ,2 ma ,1
Note that m1 = ma ,1 + mv ,1 = (1 + ω1 )ma ,1 and m2 = ma ,2 + mv ,1 = (1 + ω2 )ma ,2
So that:
ma ,2 (1 + ω1 )m2
= (6)
ma ,1 (1 + ω2 )m1
Once we calculate ω1 and ω2 , Eq. (6) is used to determine the dry air mass ratio, Eq. (4) is used to calculate ω3, and
Eq. (5) must be solved for the dry bulb temperature.
From Appendix A-9 ha ,1 (12°C ) = 285.12 kJ/kg ha ,2 (36°C ) = 309.24 kJ/kg
From Appendix A-10 hv ,1 = hg (12°C ) = 2523.4 kJ/kg Pg ,1 (12°C ) = 1.419 kPa
hv ,2 = hg (36°C) = 2567.1 kJ/kg Pg ,2 (36°C) = 5.929 kPa
0.622φ1 Pg1 (0.622)(1.00)(1.419) kg water
ω1 = = = 0.00884
P1 − φ1 Pg1 101.3 − (1.00)(1.419) kg dry air
(0.622)(0.30)(5.929) kg water
ω2 = = 0.0112
101.3 − (0.30)(5.929) kg dryair

15-39
 ma ,2 (1 + 0.00884)
= (0.30) = 0.2993
ma ,1 (1 + 0.0112)
0.00884 + (0.0112)(0.2993) kg water
ω3 = = 0.00938
1 + 0.2993 kg dry air
[285.14 + (0.00884)(2523.4)] + [309.24 + (0.0112)(2567.1)](0.2993)
ha ,3 + ω3 hv ,3 = = 314.5 kJ/kg
1 + 0.2993
Iterate on this last equation by guessing T3, evaluating ha ,3 and hv ,3 , and calculating the LHS of the equation.
Doing so:
T3 (°C) ha ,3 (kg/kg) hg (T3 ) (kJ/kg) LHS [kJ/kg] Pg 3
20 293.17 2538.1 316.98
19 292.16 2536.3 315.95
17.5 290.56 2533.5 314.3 2.022kPa
The calculated LHS is close enough to the value on the RHS to stop iterating. Therefore,
ω3 P (0.00938)(101.3)
φ3 = = = 0.744
(0.622 + ω3 ) Pg 3 (0.622 + 0.00938)(2.022)
= 74.4% Answer

15-40
15-37 Air with a volume flow rate of 600 m3/min enters the cooling/dehumidifying coils of an air conditioner at
101.3 kPa, 40 °C, 80% relative humidity. The design specification is for the air to leave the coil as
saturated air at 25 °C. Determine:
a. the water removal rate (in kg/min)
b. the heat transfer rate (in kW).

Approach:
Conservation of mass and the equations for the
air/water vapor properties can be used to determine
the water removal rate. Conservation of energy is
used to calculate the heat transfer rate.

Assumptions:
1. Air and water vapor are ideal gases, and their
mixture is an ideal gas.
2. The system is steady with no work or potential
or kinetic energy effects.

Solution:
a) Assuming steady flow, and tracking the air and water separately, conservation of mass gives:
Air: ma1 = ma 2 = ma
Water: mv1 − mv 2 − m f 3 = 0 ⇒ m f 3 = mv1 − mv 2
Using the definition of specific humidity in terms of flow rates:
m f 3 = ω1ma − ω2 ma = (ω1 − ω2 )ma
Using Eq. 15-42 to evaluate the specific humidities with Pg1 (40°C) = 7.384 kPa and Pg 2 (25°C) = 3.169 kPa .
0.622φ1 Pg1 (0.622)(0.80)(7.3984) kg water
ω1 = = = 0.0385
P1 − φ1 Pg1 101.3 − (0.80)(7.384) kg dry air
(0.622)(1.00)(3.169) kg water
ω2 = = 0.0200
101.3 − (1.00)(3.169) kg dry air
The air mass flow rate is:
V V1 ( P1 − φ1 Pg1 ) M a (600 m3 min)[(101.3 − (0.80)(7.384)]kN m 2 (28.97 kg kmol) kg
m= 1 = = = 637
va1 RT (8.314 kg kmol K)(40 + 273)K min
kg kg
m f 3 = (0.0385 − 0.020)(637 ) = 11.7
min min
b) For the control volume defined above, apply conservation of energy and mass. Assume steady, no work,
negligible potential and kinetic energy effects, and an ideal gas mixture:
0 = Q + ma ha1 + mv1hv1 − ma ha 2 − mv 2 hv 2 − m f 3 h3
Solving for the heat transfer rate, factoring out air flow rate, and using the definition of specific humidity:
Q = ma [(ha 2 − ha1 ) + ω2 hv 2 − ω1hv1 ] + m f 3 h3
From Appendix A-10 hv1 = hg1 (T1 ) = 2524.1kJ/kg
hv 2 = hg 2 (T2 ) = 2547.2 kJ/kg
h3 = h f 3 (T2 ) = 104.8 kJ/kg
Note that water condensation will be at the moist air outlet temperature
From Appendix A-9 ha1 (T1 ) = 2524.1kJ/kg
ha 2 (T2 ) = 298.17 kJ/kg
kg kJ kg kJ kJ
Q = (637 )[(298.17 − 313.25) + (0.020)(2547.2) − (0.0385)(2574.1)] + (11.7 )(104.89 ) = −39060
min kg min kg min
Q = −651 kW Answer

15-41
15-38 A warm-air furnace/humidifier unit is used to heat 1000 ft3/min of air from 14.5 lbf/in.2, 60 ºF, and 60%
relative humidity to 120 ºF and 12% relative humidity. Liquid water is supplied at 65 ºF. Determine:
a. the liquid water flow rate (in lbm/hr
b. the required heat transfer rate (in Btu/hr).

Approach:
Conservation of mass can be used to determine the
liquid water flow rate. Once that is known,
conservation of energy can be used to determine the
heat transfer rate.

Assumptions:
1. The air and water vapor are ideal gases, and
their mixture is an ideal gas.
2. The system is steady with no work or potential
or kinetic energy effects.

Solution:
a) Assuming steady flow, and tracking the water and air separately, conservation of mass gives us:
Air: ma1 = ma 3 = ma
Water: mv1 + m f 3 = mv 2
Using the definition of specific humidity in terms of flow water:
ω1ma + m f 3 = ω2 ma → m f 3 = ma (ω2 − ω1 )
Using Eq. 15-42 to evaluate the specific humidities with Pg (60°F) = 0.2563lbf/in.2 and Pg (120°F) = 1.695lbf/in.2
0.622φ1 Pg1 (0.622)(0.60)(0.2563) lbm water
ω1 = = = 0.00667
P − φ1 Pg1 14.5 − (0.60)(0.2563) lbm dry air
(0.622)(0.12)(1.695) lbm water
ω2 = = 0.00885
14.5 − (0.12)(1.695) lbm dry air
The dry air mass flow rate is:
V VPM V1 ( P − φ1 Pg1 ) M
ma = 1 = 1 a a =
Va1 RT RT
(1000 ft 3 /min)[14.5 − (0.60)(0.2563)]lbf/in.2 (28.97 lbm/lbmol)
= = 74.5lbm/min
(1545ft-lbf/lbmolR)(60 + 460)R (1ft 3 144in.2 )
m f 3 = (74.5lbm/min)(0.00885 − 0.00667) = 0.162 lbm/min Answer
b) For the control volume defined above, apply conservation of mass and energy. Assume steady, no work,
negligible potential and kinetic energy effects, and an ideal gas mixture:
0 = Q + ma ha1 + mv1hv1 + m f 3 h f 3 − ma ha 2 − mv 2 hv 2
Dividing out the air mass flow rate, combining like terms, and using the definition of specific humidity:
Q = ma [(ha 2 − ha1 ) + ω2 hv 2 − ω1hv1! + (ω2 − ω1 )h f 3 ]
From Appendix B-10, hv1 = hg1 (T1 ) = 1087.7 Btu/lbm
hv 2 = hg 2 (T2 ) = 1113.5 Btu/lbm
h f 3 = h f (T3 ) = 33.09 Btu/lbm
From Appendix B-9 ha1 = 124.27 Btu/lbm
ha 2 = 138.66 Btu/lbm

Q = (74.5lbm/min) [ (138.66 − 124.27) + (0.00885)(1113.5) − (0.00667)(1087.7) + (0.00885 − 0.00667)(33.09) ] Btu/lbm

= 1, 271Btu/min = 76, 270 Btu/hr Answer

15-42
15-39 Products of combustion with a molar analysis of N2, 74%, H2O, 14%, CO2, 7%, and O2, 5% at 14.7 psia,
800 ºF with a volumetric flow rate of 300 ft3/min enter a counterflow heat exchanger and exit at 200 º F.
On the other side of the heat exchanger moist air enters at 14.7 psia, 60 ºF, and 30% relative humidity and
exits at 100 ºF. Determine the mass flow rate of the moist air (in lbm/min).
Approach:
Conservation of mass and energy are applied to the
control volume. Mixture properties must be used.

Assumptions:
1. The air and water vapor are ideal gases, and their
mixture behaves as an ideal gas.
2. The system is steady and adiabatic with no work or
potential or kinetic energy effects.
3. The combustion gases have constant specific heat.

Solution:
Apply conservation of mass to the two fluids separately. Assume steady and ideal gas mixtures:
m1 = m2 and m3 = m4 ⇒ ma 3 = ma 4 = ma
Note that because the specific humidity is constant on the air side ω3 = ω4 and m3 = (1 + ω3 )ma
Apply conservation of energy to the control volume surrounding the adiabatic, no work, and negligible potential
and kinetic energy effects. Using the moist air enthalpy: 0 = m1h1 − m2 h2 + ma h3 − ma h4
Using conservation of mass and solving for the dry air flow rate: ma = m1 ⎡⎣( h1 − h2 ) ( h4 − h3 ) ⎤⎦
For the moist air, h3 = ha 3 + ω3 hv 3 and h4 = ha 4 + ω4 hv 4
From Appendix B-9 ha ,3 (520 R) = 124.270 Btu/lbm ha ,4 (560 R) = 133.86 Btu/lbm
From Appendix B-10
hv 3 = hg (60°F) = 1087.7 Btu/lbm, hv 4 = hg (100°F) = 1105.0 Btu/lbm, Pg3 (60°F) = 0.2563 lbf/in.2
0.622φ3 Pg 3 (0.622)(0.30)(0.2563) lbm water
ω3 = = = 0.00327
P3 − φ3 Pg 3 14.7 − (0.30)(0.22163) lbm dry air
h3 = 124.27 + (0.00327)(1087.7) = 127.83Btu/lbm h4 = 133.86 + (0.00327)(1105.0) = 137.47 Btu/lbm
For the combustion gases; assuming constant specific heats:
1 1 ∆T
∆h =
Mm
∑ Yi ∆h = M ∑ Yi c p ,i ∆T = M ∑ Yi c p,i where c p = c p M
m m

From Appendix B-1 we obtain M i and from Appendix B-8 we obtain c p ,i at the average temperature

Gas Yi c p (Btu/lbmR) M c p (Btu/lbmolR) Yi c p ,i (Btu/lbmolR) Yi M i

N2 0.74 0.248 28.01 6.95 5.143 20.73
H2O 0.14 0.473 18.02 8.53 1.194 2.52
CO2 0.07 0.247 44.01 10.87 0.761 3.08
O2 0.05 0.235 32.00 7.52 0.376 1.60
∑ = 7.474 ∑ = M m = 27.93
(800 − 200)R
Therefore, ∆h = (7.474 Btu/lbmR) = 160.0 Btu/lbm
27.93lbm/lbmol
P1 M m (14.7 lbf/in.2 )(27.93lbm/lbmol)(144in.2 1ft 2 )
m1 = ρ1V1 = V1 = (300 ft 3 min )=9.11 lbm/min
RT1 (1545ft lbf lbmolR )(800+460)R
160.6
ma = (9.11 lbm/min)( ) = 151.8 lbm/min
137.47 − 127.83
m3 = (1 + 0.00327)(151.8) = 152.3 lbm/min Answer

15-43
15-40 To remove dust particles and odors from air, an air washer is often used. This device is composed of a
chamber in which water is sprayed and liquid water is removed from the bottom, taking the dust and odors
with it. Air with a flow rate of 5000 ft3/min enters the device shown below at 14.7 lbf/in.2, 65 ºF, and 70%
relative humidity and leaves as saturated air at 60 ºF. Water at 70 ºF with a flow of 30 lbm/min is sprayed
into the air. For the water exiting the chamber, determine:
a. the flow rate (in lbm/min)
b. its temperature (in ºF).

Approach:
Conservation of mass is used to determine the water
flow rate leaving. Conservation of energy is used to
determine its temperature.
Assumptions:
1. Air and water vapor are ideal gases, and their
mixture is an ideal gas.
2. The system is steady and adiabatic with no work
or potential or kinetic energy effects.

Solution:
a) Assuming steady flow for the control volume defined above, and tracking the air and water separately,
conservation of mass gives:
Air: ma1 = ma 2 = ma
Water: mv1 + m3 − mv 2 − m4 = 0
Solving for m4, and using the definition of specific humidity in terms of flow rates:
m4 = ω ma − ω2 ma + m3 = (ω1 − ω2 )ma + m3
Using Eq. 15-42 to evaluate the specific humidities, and from Appendix B-10, Pg1 (65°F) = 0.3098lbf/in.2
and Pg 2 (60°F) = 0.2563 lbf/in.2
0.622φ1 Pg1 (0.622)(0.70)(0.3098) lbm water
ω1 = = = 0.00931
P1 − φ1 Pg1 14.7 − (0.70)(0.3098) lbm dry air
(0.622)(1.00)(0.2563) lbm water
ω2 = = 0.0110
14.7 − (1.00)(0.2563) lbm dry air
V1 V1 ( P1 − φ1 Pg1 ) M a
The air mass flow rate is ma = =
va1 RT
(5000ft 3 /min)[14.7 − (0.70)(0.3098)]lbf/in.2 (28.97 lbm lbmR)
ma = = 372 lbm/min
(1545ft-lbf lbmolR)(65 + 460)R(1ft 2 144in.2 )
m4 = (0.00931 − 0.0110)(372 lbm/min) + 30 lbm/min = 29.4 lbm/min Answer
b) For the same control volume, apply conservation of energy and mass. Assume steady, adiabatic, no work,
negligible potential and kinetic energy effects, and an ideal gas mixture:
0 = ma ha1 + mv1hv1 + m3 h3 − ma ha 2 − mv 2 hv 2 − m4 h4
Solving for h4, using the definition of specific humidity, and combining like terms:
h4 = (1 m4 ) [ma {ha1 − ha 2 + ω1hv1 − ω2 hv 2 } + m3 h3 ]
From Appendix B-10,
hv1 = hg1 (65°F) = 1089.9 Btu/lbm hv 2 = hg 2 (60°F) = 1087.7 Btu/lbm h3 = h f 3 (70°F) = 38.09 Btu/lbm
From Appendix B-9 ha1 (65°F) = 125.47 Btu/lbm ha 2 (60°F) = 124.27 Btu/lbm
h4 = (1 29.4 ) ⎡⎣(372) {125.47 − 124.27 + (0.00931)(1089.9) − (0.0110)(1087.7)} + (30)(38.09) ⎤⎦
= 31.05 Btu/lbm
By interpolating in Appendix B-9, T4 = 73.0°F Answer

15-44
15-41 Moist air enters an adiabatic compressor at 95 kPa, 20 ºC, and 60% relative humidity with a volumetric
flow rate of 20 m3/min and leaves at 200 kPa, 100 ºC. Determine:
a. the relative humidity at the exit
b. the power input (in kW).

Approach:
The specific humidity remains constant, so at the exit
we know two independent properties once the inlet
specific humidity is determined. Power input can be
determined with application of the energy equation.

Assumptions:
1. The air and water vapor are ideal gases, and
their mixture behaves as an ideal gas.
2. The system is steady and adiabatic with
negligible potential and kinetic energy effects.

Solution:
a) The specific humidity is constant, so ω2 = ω1 . Equation 15-42 can be used to determine first ω1 and then φ2 .
0.622φ1 Pg1
ω1 =
P1 − φ1 Pg1
From Appendix A-10, Pg1 (20°C) = 2.339 kPa
(0.622)(0.60)(2.339)
ω1 = = 0.00933
95 − (0.60)(2.339)
ω2 P2
Now φ2 =
(0.622 + ω2 ) Pg 2
From Appendix A-10, Pg 2 (100°C) = 101.3kPa
(0.00933)(200)
φ2 = = 0.0292 = 2.92%
(0.622 + 0.00933)(101.3)
b) Apply conservation of energy and mass to the control volume defined above. Assume steady, adiabatic,
negligible potential and kinetic energy effects, and an ideal gas mixture.
ma1 = ma 2 and mv11 = mv 2 and ω1 = ω2
W = ma (ha1 − ha 2 ) + mv (hv1 − hv 2 )
Dividing by the air mass flow rate and using the definition of specific humidity
W = ma [(ha1 − ha 2 ) + ω (hv1 − hv 2 )]
For and ideal gas ma = V1 va1
RT1 RT1 (8.314 kg/kmolK)(20 + 273) K
va1 = = = = 0.898 kg/m3
Pa1 M ( P1 − φ1 Pg1 ) [95 − (0.60)(2.339)]kN/m 2 (28.97 kg/kmol)
20 m3 min
ma = = 22.3kg/min = 0.371kg/s
0.898 kg/m3
From Appendix A-10, hv1 = hg (20°C) = 2538.1kJ/kg, hv 2 = hg (100°C) = 2676.1 kJ/kg
From Appendix A-9 ha1 = 293.16 kJ/kg, ha 2 = 373.70 kJ/kg
W = (0.371kg/s)[(293.16 − 373.70) + (0.00933)(2538.1 − 2676.1)]kJ/kg
= −30.4kW Answer

15-45
15-42 An air duct is routed through an uninsulated space in a building. The air enters at 14.7 psia with a dry-bulb
temperature of 82 ºF, a wet-bulb temperature of 68 ºF, and a flow rate of 10 lbm/min. The air exits the duct
at 62 ºF. Using the psychrometric chart, determine:
a. the relative humidity at the duct inlet
b. the relative humidity at the duct outlet
c. the heat transfer rate (in Btu/hr).

Approach:
The relative humidity at the inlet can be picked off
the pyschrometric chart using the given information.
For the duct outlet we assume the outlet specific
humidity is equal to the inlet specific humidity; we
need to check this. Conservation of energy is used to
determine the heat transfer rate.

Assumptions:
1. The air and water vapor are ideal gases, and
their mixture behaves as an ideal gas.
2. The system is steady with no work or potential
or kinetic energy effects.

Solution:
a) Using the psychrometric chart, φ1 ≈ 50% Answer
lbm water
b) If no condensation occurs, ω2 = ω1 . From the psychrometric chart ω1 ≈ 0.0115
lbm dry air
Therefore at ω2 and T2 , φ2 ≈ 0.96 = 96%. Because the mixture did not reach the saturation line, our assumption
ω2 = ω1 is valid.
c) Applying conservation of energy to the control volume, assuming steady, no work, negligible potential and
kinetic energy effects, and an ideal gas mixture, and using mixture properties.
Q = M a (h2 − h1 )
Using the psychrometric chart to evaluate the moist air enthalpies:
h1 ≈ 32.2 Btu/lbm dry air
h2 ≈ 27.3Btu/lbm dry air
The total flow rate is 10 lbm/min which is the sum of the dry air and water vapor flows:
m
m = ma + mv = ma (1 + v ) = ma (1 + ω )
ma
Therefore
m 10 lbm/min
ma = = = 9.886 lbm/min
1+ ω 1 + 0.015
Q = (9.886 lbm/min)(27.3 − 32.2) Btu/lbm = −48.4 Btu/min
= −2907 Btu/hr Answer

15-46
15-43 Air enters an industrial dryer at 100 kPa, 25 ºC, and 65% relative humidity and is heated to 140 ºC at
constant pressure. The hot air is passed over the material to be dried and exits the dryer at 60 ºC and 30%
relative humidity. Water must be removed from the material at a rate of 20 kg/min. Determine:
a. the dry air mass flow rate (in kg/min)
b. the heat transfer rate (in kW).

Approach:
Apply conservation of energy and mass to control
volume I to determine the required heat transfer per
unit mass dry air. Conservation of mass applied to
control volume I + II is used to determine the
required air mass flow rate.

Assumptions:
1. The air and water vapor are ideal gases, and
their mixture behaves as an ideal gas.
2. The system is steady with no work or potential
or kinetic energy effects.

Solutions:
a) Applying conservation of mass to control volume I + II, assuming steady flow and ideal gas mixture, and
tracking the air and water separately:
Air: ma1 = ma 2 = ma 3 = ma
Water: mv1 + m4 − mv 3 = 0
Using the definition of specific humidity in terms of flow rates:
ω1ma + m4 − ω3 ma = 0 → ma = m4 (ω3 − ω1 )
Using Eq. 15-42 to calculate the specific humidities, and from Appendix A-10, Pg1 (25°C) = 3.169 kPa
and Pg 3 (60°C) = 19.94 kPa.
0.622φ1 Pg1 (0.622)(0.65)(3.169) kg water
ω1 = = = 0.0131
P1 − φ1 Pg1 100 − (0.65)(3.169) kg dry air
(0.622)(0.30)(19.94) kg water
ω3 = = 0.0396
100 − (0.30)(19.94) kg dry air
20 kg/min
ma = = 755 kg/min Answer
0.0396 − 0.0131
b) Applying conservation of energy to control volume I, and assuming steady, no work, negligible potential and
kinetic energy effects, and an ideal gas mixture:
0 = Q + ma ha1 − ma ha 2 + mv1hv1 − mv 2 hv 2
Using the definition of specific humidity in terms of flow rates and combining like terms
Q = ma [ha 2 − ha1 + ω2 hv 2 − ω1hv1 ] note that ω1 = ω2
From Appendix A-10 hv1 = hg1 (25°C) = 2547.2 kJ/kg
hv 2 = hg 2 (140°C) = 2733.9 kJ/kg
From Appendix A-9 ha1 (25°C) = 298.17 kJ/kg
ha 2 (140°C) = 414.16 kJ/kg
1min
Q = (755kg/min)( )[414.16 − 298.17 + (0.0131)(2733.9 − 2547.2)]kJ/kg
60s
= 1490 kW Answer

15-47
15-44 Air at 14.7 lbf/in.2, 50 ºF, and 40% relative humidity enters a heater/humidifier device and exits at 70 ºF
and 50% relative humidity. This is a constant pressure process, and liquid water at 45 ºF is used.
Determine per lbm of dry air:
a. the amount of water injected (in lbm water/lbm dry air)
b. the heat transfer rate (in Btu/lbm dry air).

Approach:
Conservation of mass can be used to determine the
make-up water flow rate. Once that is known,
conservation of energy is used to determine the
required heating rate.

Assumptions:
1. The air and water vapor are ideal gases, and
their mixture behaves as an ideal gas.
2. The system is steady with no work or potential
or kinetic energy effects.

Solution:
a) Assuming steady flow for the control volume shown above, and tracking the air and water separately,
conservation of mass gives:
Air: ma1 = ma 2 = ma
Water: mv1 + m f 3 − mv 2 = 0
Using the definition of specific humidity in teams of flow rates:
ω1ma + m f 3 − ω2 ma → m f 3 ma = ω2 − ω1
Using Eq. 15-42 to evaluate the specific humidities, and from Appendix B-10, Pg1 (50°F) = 0.178lbf/in.2
and Pg 2 (70°F) = 0.363 lbf/in.2
0.622φ1 Pg1 (0.622)(0.40)(0.178) lbm water
ω1 = = = 0.00303
P1 − φ1 Pg1 14.7 − (0.40)(0.178) lbm dry air
(0.622)(0.50)(0.363) lbm water
ω2 = = 0.00778
14.7 − (0.50)(0.363) lbm dry air
mf 3 lbm water
= 0.00778 − 0.00303 = 0.00475 Answer
ma lbm dry air
b) For the same control volume, apply conservation of energy and mass. Assume steady, no work, negligible
potential and kinetic energy effects, and an ideal gas mixture:
0 = Q + ma ha1 + mv hv1 + m f 3 h3 − ma ha 2 − mv 2 hv 2
Solving for the heat transfer rate, diving by air flow rate, and combining like terms:
Q mf 3
= (ha 2 − ha1 ) + ω2 hv 2 − ω1hv1 − h3
ma ma
From Appendix B-10 hv1 = hg1 (T1 ) = 1083.3Btu/lbm hv 2 = hg 2 (T2 ) = 1092.0 Btu/lbm
h3 = h f 3 (T3 ) = 13.04 Btu/lbm
From Appendix B-9 ha1 (T1 ) = 121.88 Btu/lbm ha 2 (T2 ) = 126.66 Btu/lbm
Q
= (126.66 − 121.88 ) + (0.00778)(1092.0) − (0.00303)(1083.3) − (0.00475)(13.04)
ma
Btu
= 9.94 Answer
lbm dry air

15-48
15-45 For energy conservation and air quality purposes, designers of building air handling systems mix fresh air
from outside with recycled air from inside. Consider two air streams, both at 14.7 lbf/in.2, that are mixed in
a steady flow adiabatic mixer. The first stream has a flow of 650 ft3/min and is at 55 °F and 10% relative
humidity. The second stream has a flow of 900 ft3/min and is at 85 °F and 80% relative humidity.
Determine the outlet relative humidity, temperature, and specific humidity.

Approach:
For this mixing problem, we use ideal gas mixture
properties in the conservation of mass and energy to
determine the outlet conditions.

Assumptions:
1. Air and water are ideal gases, and their
mixture is an ideal gas.
2. The system is steady and adiabatic with no
work or potential or kinetic energy effects.

Solution:
Applying conservation of mass to the control volume, assuming steady and ideal gas mixture, and tracking the air
and water separately, we obtain:
Air: ma1 + ma 2 = ma 3 (1)
Water: mv1 + mv 2 = mv 3 → ω1ma1 + ω2 ma 2 = ω3 ma 3 (2)
Applying conservation of energy, and assuming adiabatic, no work and negligible potential and kinetic energy
effects:
0 = ma1ha1 + mv1hv1 + ma 2 ha 2 + mv 2 hv 2 − ma 3 ha 3 − mv 3 hv 3
= ma1 (ha1 + ω1hv1 ) + ma 2 (ha 2 + ω2 hv 2 ) − ma 3 (ha 3 + ω3 hv 3 ) (3)
Combining Eq. (1) and (2):
ma 2 ω1 − ω3 ω + ω2 ( ma 2 ma1 )
= → ω3 = 1 (4)
ma1 ω3 − ω2 1 + ma 2 ma1
Combining Eq. (1) and (3):
ma 2 (ha1 + ω1hv1 ) − (ha 3 + ω3 hv 3 )
=
ma1 (ha 3 + ω3 hv 3 ) − (ha 2 + ω2 hv 2 )
(ha1 + ω1hv11 ) + (ha 2 + ω2 hv 2 ) ( ma 2 ma1 )
ha 3 + ω3 hv 3 = (5)
1 + ma 2 ma1
Once we calculate ω1 and ω2 , Eq. (6) is used to determine the dry air mass ratio, Eq. (4) is used to
calculate ω3 , and Eq. (5) must be solved iteratively for the dry bulb temperature.
From Appendix A-9 ha1 (55°F) = 121.88 Btu lbm ha 2 (85°F) = 135.25 Btu lbm
From Appendix A-10 hv1 = hg (55°F) = 1085.5 Btu lbm Pg1 (55°F) = 0.2172 lbf in.2
hv 2 = hg (85°F) = 1098.6 Btu lbm Pg 2 (85°F) = 0.603lbf in.2
0.622φ1 Pg1 (0.622)(0.10)(0.2172) lbm water
ω1 = = = 0.000920
P1 − φ Pg1 14.7 − (0.10)(0.2172) lbm dry air
(0.622)(0.80)(0.603) lbm water
ω2 = = 0.0211
14.7 − (0.80)(0.603) lbm dry air
The flow rates are:
ft 3 lbf lbm
(650 )[14.7 − 0.10(0.2171)] 2 (28.97 )
V1 V1 ( P1 − φ1 Pg1 ) M min in. lbmol = 50.0 lbm
ma1 = = =
va1 RT1 ft-lbf 1ft 2 min
(1545 )(55 + 460)R( )
lbmolR 144in.2
(900)[14.7 − (0.80)(0.603)](28.97) lbm
ma2 = = 63.4
(1545)(85 + 460)(1 144) min

15-49
 ma 2 63.4
= = 1.267
ma1 50
0.00092 + (0.0211)(1.267) lbm water
ω3 = = 0.0122
1 + 1.267 lbm dry air
[121.88 + (0.00092)(1085.5)] + [130.26 + (0.0211)(1098.6)](1.267) Btu
ha 3 + ω3 hv 3 = = 139.96
1 + 1.267 lbm
Iterate on this last equation by guessing T3, evaluating ha 3 and hv 3 , and calculating the LHS of the equation. Doing
so:
T3 = 70°F ha 3 = 126.66 Btu lbm hg (T3 ) = 1092.0 Btu lbm Pg 3 = 0.3532 lbf in.2
LHS of equation = 139.98 Btu lbm
This value of temperature results in the LHS of the equation being close enough to the value on the RHS so no
further iteration is needed. Therefore,
ω3 P (0.0122)(14.7)
φ3 = = = 0.779 = 77.9% Answer
(0.622 + ω3 ) Pg 3 (0.622 + 0.0122)(0.3632)

15-50
15-46 In an air conditioning system, air with a volume flow rate of 25 m3/min enters the dehumidifying section at
101.3 kPa, 28 ºC, and 70% relative humidity and leaves as saturated moist air; some water vapor
condenses, and it leaves the system at the same temperature as the saturated moist air. The air then is
reheated to 24 ºC and 40% relative humidity. Using the psychrometric chart, determine:
a. the temperature of the moist air leaving the dehumidifier (in ºC)
b. the water condensation rate (in kg/min)
c. the heat transfer rate in the cooling coil (in kW)
d. the heat transfer rate in the reheat section (in kW).

Approach:
The condensate flow rate is determined from a mass
balance on the whole system. Conservation of
energy applied to the dehumidifying and reheat
sections is used to determine the respective heat
transfer rates.

Assumptions:
1. The air and water vapor are ideal gases, and
their mixture behaves as an ideal gas.
2. The system is steady with no work or potential
or kinetic energy effects.

Solutions:
a) For the temperature at location 2, we know φ2 . The second property is obtained by noting that ω2 = ω3 . (No
water is added between points 2 and 3). From the pyschrometric chart:
ω3 ≈ 0.0075 kg water kg dry air
Now at ω2 = ω3 and φ2 = 100%, TWB = TDB 10°C Answer
b) Applying conservation of mass to the whole system, and assuming steady and an ideal gas mixture, and
tracking the air and water separately:
Air: ma1 = ma 2 = ma 3 = ma where ma = V1 va1
Water: mv1 − m4 − mv 3 = 0 → m4 = (ω1 − ω3 )ma
From the pyschrometric chart, ω1 ≈ 0.0168 kg water kg dry air , va1 ≈ 0.878 m3 kg
25 m3 /min
ma = = 28.5 kg/min
0.878 m3 /kg
m4 = (0.0168 − 0.0075)(28.5 kg/min) = 0.265 kg/min Answer
c) Applying conservation of energy to the dehumidifying section, and assuming no work and negligible potential
and kinetic energy effects:
0 = Q + ma h1 − ma h2 − m4 h4 → Q = (h2 − h1 )ma + m4 h4
From the pyschrometric chart: h1 ≈ 71.5 kJ/kg, h2 ≈ 29.0 kJ/kg
From Appendix A-10, h4 = h f (10°C) = 42.01 kJ/kg
Q = (29.0 − 71.5)kJ/kg(28.5 kg/min) + (0.265 kg/min)(42.01kJ/kg) = −1200 kJ/min
= (−1200 kJ/min)(1min 60s) = −20.0 kW Answer
d) Using a similar analysis on the reheat section
Q = (h3 − h2 )ma = (43.7 − 29.0)kJ/kg(28.5 kg/min)(1min 60s) = 6.98 kW Answer

Comment:
The signs on the heat transfer in the dehumidifying and reheat sections are consistent with the sign convention.
Note the significantly larger heat transfer rate in the dehumidifying section compared to the reheat section.
Condensing water vapor takes very large amounts of energy compared to sensible heating of air.

15-51
15-47 Air at 101.3 kPa, 12 °C, and 70% relative humidity flows into a heater/humidifier and leaves at 38 °C, and
60% relative humidity. Determine per kg of dry air:
a. the amount of water to be added (in kg water/kg dry air)
b. the heat transfer required if the water is supplied at 18 °C (in kJ/kg dry air)
c. the percent error if only the heat transfer required by the dry air (water vapor and make-up water
ignored) is taken into account (in kJ/kg dry air).

Approach:
Conservation of mass can be used to determine the
make-up water flow rate. Once that is known,
conservation of energy can be used to determine the
required heating rate.

Assumptions:
1. Air and water vapor are ideal gases, and their
mixture is an ideal gas.
2. The system is steady with no work or potential
or kinetic energy effects.

Solution:
a) Assuming steady flow, and tracking the water and air separately, conservation of mass gives us
Air: ma1 = ma 2 = ma
Water: mv1 + m f 3 = mv 2
Using the definition of specific humidity in terms of flow rates:
ω1ma + m f 3 = ω2 ma → m f 3 ma = ω2 − ω1
Using Eq. 15-42 to evaluate specific humidiities with Pg (12°C) = 1.419 kPa and Pg (38°C) = 6.682 kPa
0.622φ1 Pg1 (0.622)(0.70)(1.419)
ω1 = = = 0.00616
P1 − φ1 Pg1 101.3 − (0.70)(1.419)
(0.622)(0.60)(6.682)
ω2 = = 0.02563
101.3 − (0.60)(6.682)
mf 3 kg water
= 0.02563 − 0.00616 = 0.0195 Answer
ma kg dry air
b) For the control volume defined above, apply conservation of mass and energy. Assume steady, no work, and
negligible potential and kinetic energy effects. The gases form an ideal gas mixture.
0 = Q + ma ha1 + mv1hv1 + m f 3 h f 3 − ma ha 2 − mv 2 hv 2
Dividing by the dry air flow rate, combining like terms, and using the definition of specific humidity:
Q
= (ha 2 − ha1 ) + ω2 hv 2 − ω1hv1 − (ω2 − ω1 )h f 3
ma
From Appendix A-10 hv1 = hg1 (T1 ) = 2523.40 kJ/kg hv 2 = hg 2 (T2 ) = 2570.70 kJ/kg h f 3 = h f (T3 ) = 75.57 kJ/kg
From Appendix A-9 ha1 = 285.14 kJ/kg ha 2 = 311.25 kJ/kg
Q kJ
= 311.25 − 285.14 + (0.02563)(2570.7) − (0.00616)(2523.4) − (0.0263 − 0.00616)(75.57) = 75.0
ma kg dry air
c) To heat only the dry air
Q kJ
= 311.25 − 285.14 = 26.11 Answer
ma kg dry air
Hence, we would be low by a factor of 3 if we ignored the contributions by the water vapor and the make-up
water.

15-52
15-48 The air conditioning system for a large building that has a cooling load of 1.5 MW is shown in the figure
below. Refrigerant R-134a enters the compressor as a saturated vapor at 360 kPa and leaves at 1.2 MPa;
the flow rate is 6 kg/s. The refrigerant leaves the condenser as a saturated liquid. The water flowing
through the condenser can have an 8 ºC temperature rise. The pressure rise across the pump is 75 kPa. The
compressor isentropic efficiency is 0.70 and that of the pump is 0.60. Using appropriate assumptions and
approximations, determine:
a. the compressor power (in kW)
b. the condenser heat transfer rate (in kW)
c. the water flow rate through the condenser (in kg/min)
d. the required make-up water flow rate (in kg/min)
e. the pump power (in kW).

Approach:
Conservation of energy is used to calculate each of the
desired quantities, along with basic concepts. An
approximation is needed to obtain the make-up water
flow rate.

Assumptions:
1. The compressor and pump are steady and adiabatic
with negligible potential and kinetic energy effects.
2. The liquid water is ideal with constant specific heat.
3. The condenser is steady with no work and negligible
potential and kinetic energy effects.

Solution:
a) Apply conservation of mass and energy to a control volume around the compressor. Assume steady, adiabatic,
and negligible potential and kinetic energy effects. Using the definition of isentropic efficiency for a compressor:
Win = mR (h2 − h1 ) = mR ( h2 s − h1 ) ηC
From Appendix A-15 h1 = hg ( P ) = 250.58 kJ/kg s1 = sg ( P ) = 0.9160 kJ/kgK
h2 s − 270.99 0.9160 − 0.9023
Because s2 = s1 , we obtain h2 s ( P2 , s2 ) by interpolation = → h2 s = 275.4 kJ kg
275.52 − 270.99 0.9164 − 0.9023
⎛ kg ⎞ ⎛ 275.4 − 250.58 ⎞
Win = ⎜ 6 ⎟ ⎜ ⎟ = 213kW Answer
⎝ s ⎠⎝ 0.70 ⎠
b). Applying conservation of energy to the complete steady cycle:
Qcond = Qevap + Win = 1500 kW + 213kW = 1713kW Answer
c) Applying conservation of energy to water flowing through condenser, and assuming steady, no work,
negligible potential and kinetic energy effects, and an ideal liquid with constant specific heat so that ∆h = c ρ∆T .
Qcond
Qcond = mw (h7 − h6 ) = mw c p , w (T7 − T6 ) → mw =
c p , w (T7 − T6 )
From Appendix A-6, c p , w ≈ 4.18 kJ kgK
1713kW kg
mw = = 51.2 Answer
(4.18 kJ kgK)(8°C) s
d) The heat transferred to the water in the condenser is dissipated in the cooling tower. Some of the water
evaporates (mmw ). We assume all the heat is removed by evaporating water. This is an approximation.
Therefore: Qcond ≈ mmw h fg → mmw ≈ Qcond h fg
1713kW kg
From Appendix A-10 at 25oC, h fg = 2442.3kJ kg → mmw ≈ = 0.70
2442.3kJ kg s
e) Applying conservation of energy to the pump with assumptions as before, and assuming incompressible liquid:
V ∆P mw ∆P (51.2 kg s)(75 kN m 2 )
WP = mw (h6 − h5 ) = w = = = 6.4 kW Answer
ηP ρ wη P (1000 kg m3 )(0.60)

15-53
15-49 In an air conditioning system, air at 14.7 psia, 84 °F, and 70% relative humidity flows over the cooling coil
with a flow of 15,000 ft3/min. The air is cooled and dehumidified, and the liquid water leaves the system at
50 °F. The air then flows through the reheat section and exits at 75 °F and 40% relative humidity. Using
the psychrometric chart, determine:
a. the condensate flow rate (in lbm/hr)
b. the heat transfer rate in the cooling coil (in Btu/hr)
c. the heat transfer rate in the reheat section (in Btu/hr).

Approach:
Conservation of mass applied to the overall control
volume is used to determine the condensation flow
rate. Heat transfer to the cooling coil and from the
reheat coil are determined with conservation of
energy over control volumes encompassing just the
appropriate duct sections.

Assumptions:
1. Air and water vapor are ideal gases, and their
mixture is an ideal gas.
2. The system is steady with no work and
negligible potential and kinetic energy effects.

Solutions:
a) Applying conservation of energy to the overall control volume, assuming steady and ideal gas mixture, and
tracking the air and water separately:
Air: ma1 = ma 2 = ma 3 = ma
Water: mv1 − m4 − mv 3 = 0 → m4 = mv1 − mv 3 = (ω1 − ω3 )ma
The air mass flow rate is ma = V va1 . From the psychrometric chart:
lbm water ft 3 lbm water
ω1 ≈ 0.0178 , va1 ≈ 14.1 , ω3 ≈ 0.0075
lbm dry air lbm lbm dry air
(15000 ft 3 min)(60min 1hr) lbm
ma = 3
= 63,830
14.1ft lbm hr
m4 = (0.0178 − 0.0075)(63830 lbm hr) = 657 lbm hr Answer
b) Applying conservation of energy to the section containing the cooling coil, and assuming no work and
negligible potential and kinetic energy effects:
0 = Q + ma h1 − ma h2 − m4 h4 → Q = (h2 − h1 )ma + m4 h4
From the psychrometric chart:
h1 ≈ 39.3Btu lbm
At point 2, the air is saturated (φ2 = 100%) and the wet bulb temperature is the same as the condensation
temperature, 50oF. Therefore, h2 ≈ 20.1Btu lbm
From Appendix B-10, h4 = h f (50°F) = 18.06 Btu lbm
Btu lbm lbm Btu Btu
Q = (20.1 − 39.3) (63,830 ) + (657 )(18.06 ) = −1.214 × 106 Answer
lbm hr hr lbm hr
c) Using a similar analysis on the reheater
Q = (h3 − h2 )ma
Btu lbm Btu
= (26.3 − 20.1) (63,830 ) = 3.96 × 105 Answer
lbm hr hr

15-54
15-50 Cooling towers often are operated so that the air leaving the tower is not saturated and no fog forms.
Consider a tower in which warm water enters at 38 ºC with a flow rate of 900 kg/min and leaves at 24 ºC.
Air enters the cooling tower at 101.3 kPa, 20 ºC, and 40% relative humidity with a flow rate of 600 m3/min.
Make-up water at 22 ºC is added at a rate of 15 kg/min. Does fog form? That is, determine the dry-bulb
temperature (in ºC) and the relative humidity of the air leaving the cooling tower. Use the psychrometric
chart where appropriate.

Approach:
Conservation of mass and energy are used to
determine the outlet air temperature and relative
humidity.

Assumptions:
1. Air and water vapor are ideal gases, and their
mixture behaves as an ideal gas.
2. The system is steady and adiabatic with no work
or potential or kinetic energy effects.

Solution:
Apply conservation of mass to the control volume. Assuming steady and ideal gas mixture, and tracking air and
water separately:
Air: ma1 = ma 2 = ma
Water: mv1 + m3 + m5 − mv 2 − m4 = 0
Note that m3 = m4 . Using the definition of specific humidity in terms of flow rates:
m5 = mv 2 − mv11 = (ω2 − ω1 )ma → ω2 = ω1 + m5 ma
Applying conservation of energy, and assuming adiabatic, no work, and negligible potential and kinetic energy
effects:
0 = ma ha1 + mv1hv1 + m3 h3 + m5 h5 − ma ha 2 − mv 2 hv 2 − m4 h4
Using the definition of specific humidity and combining like terms:
0 = ma [ha1 − ha 2 + ω1hv1 − ω2 hv 2 ] + m3 (h3 − h4 ) + m5 h5 = ma (h1 − h2 ) + m3 (h3 − h4 ) + m5 h5
m3 (h3 − h4 ) + m5 h5
Solving for h2 = h1 +
ma
From the psychrometric chart at T1 = 20°C, φ1 = 40%
kg water
ω1 0.0058 h1 34.6kJ/kg va1 0.893m3 /kg
kg dry air
ma = V1 va1 = 600 m3 /min 0.893m3 /kg = 715 kg/min
ω2 = 0.0058 + 15 715 = 0.0268 kg water kg dry air
From Appendix A-10 h3 = h f 3 (38°C) = 152.2 kJ/kg
h4 = h f 4 (24°C) = 100.7 kJ/kg h5 = h f 5 (22°C) = 92.3kJ/kg
(900 kg/min)(152.2 − 100.7) kJ/kg + (15 kg/min)(92.3kJ/kg)
h2 = 34.6 kJ/kg + = 101.4 kJ/kg
715 kg/min
From the pyschrometric chart at ω2 and h2
φ2 ≈ 80% T2 ≈ 33.5°C Answer
Because the relative humidity is less than 100%, fog will not form.

15-55
15-51 In a cooling tower, 400 gal/min of liquid water enters at 120 ºF. Air enters at 1 atm, 80 ºF dry-bulb
temperature, and 45% relative humidity, and exits at 100 ºF and 95% relative humidity. Make-up water is
available at 70 ºF. Determine the mass flow rates of the dry air and the make-up water (in lbm/min).
Approach:
The outlet temperature of the cooling water is not
specified, because of this, a range of air and make-up
water flow rates are possible. Conservation of mass
and energy applied to the control volume are used to
determine the flow rates.

Assumptions:
1. The air and water vapor are ideal gases, and
their mixture behaves as an ideal gas.
2. The system is steady and adiabatic, with no
work or potential or kinetic energy effects.

Solution:
Apply conservation of mass to the control volume, and assume steady and an ideal gas mixture. Tracking the air
and water separately:
Air: ma1 = ma 2 = ma
Water:: mv1 + m3 + m5 − mv 2 − m4 = 0
Note that m3 = m5 , and use the definition of specific humidity in terms of flow rates:
m5 = mv 2 − mv1 = (ω2 − ω1 )ma
Use conservation of energy and assume adiabatic, no work, and negligible potential and kinetic energy effects:
0 = ma ha1 + mv1hv1 + m3 h3 + m5 h5 − ma ha 2 − mv 2 hv 2 − m4 h4
Again, using the definition of specific humidity, combining like terms, and incorporating conservation of mass:
0 = ma [ha1 − ha 2 + ω1hv1 − ω2 hv 2 ] + (ω2 − ω1 )ma h5 + m3 (h3 − h4 )
m3 (h4 − h3 )
Solving for the air flow rate: ma =
ha1 − ha 2 + ω1hv1 − ω2 hv 2 + (ω2 − ω1 )h5
From Appendix B-10, hv1 = hg (80°F) = 1096.4 Btu/lbm Pg1 (80°F) = 0.5073lbf/in.2
hv 2 = hg (100°F) = 1105.0 Btu/lbm Pg2 (100°F) = 0.9503lbf/in.2
h5 = h f (70°F) = 38.09 Btu/lbm h3 = h f (120°F) = 88.0 Btu/lbm
From Appendix B-9 ha1 (80°F) = 129.06 Btu/lbm ha2 (100°F) = 133.86Btu/lbm
0.622φ1 Pg1 (0.622)(0.45)(0.5073) lbm water
ω1 = = = 0.00981
P1 − φ1 Pg1 14.7 − (0.45)(0.5073) lbm dry air
(0.622)(0.95)(0.9503) lbm water
ω2 = = 0.0407
14. − (0.95)(0.9503) lbm dry air
m3 = ρV3 = (61.7lbm/ft 3 )(400gal/min)(0.1337ft 3 /gal) = 3,300 lbm/min
The make-up water flow rate and the required air flow rate depend on the outlet temperature of the water at 4
(3,330lbm/min)(h4 − 88.0)Btu/lbm
ma =
[129.06 − 133.86 + (0.00981)(1906.4) − (0.0407)(1105.0) + (0.0407 − 0.00981)(38.09)]Btu/lbm
= −87.21 (h4 − 88.0)
Therefore, over a range of temperatures at 4:
T4 (°F) h4 ( Btu lbm ) ma ( lbm min ) m5 ( lbm min )
80 48.09 3480 108
85 53.08 3045 94.1
90 58.07 2610 80.6
95 63.06 2175 67.2

15-56
15-52 Warm water at a flow rate of 100,000 kg/hr enters a cooling tower at 40 °C and leaves at 25 °C. Air enters
the cooling tower at 98 kPa, 20 °C, 40% relative humidity and leaves saturated at 35 °C. Determine the
required 20 °C make-up water flow rate (in kg/hr) and the air flow rate (in kg/hr).

Approach:
Conservation of mass is used to determine the make-
up water flow rate per unit mass of dry air.
Conservation of energy is used to determine the air
flow rate.

Assumptions:
1. Air and water vapor are ideal gases, and their
mixture behaves as an ideal gas.
2. The system is steady and adiabatic with no work
or potential or kinetic energy effects.

Solution:
Apply conservation of mass to the control volume assuming steady and ideal gas mixture, and tracking air and
water separately:
Air: ma1 = ma 2 = ma
Water: mv1 + mw3 + mm − mw 4 − mv 2 = 0
Note that when we add make-up water, mw3 = mw 4 = mw . Using the definition of specific humidity in terms of flow
rates:
mm = mv 2 − mv1 = ω2 ma − ω1ma = (ω2 − ω1 )ma
Applying conservation of energy to the control volume, and assuming no work, adiabatic, and negligible potential
and kinetic energy effects:
0 = ma ha + mv1hv1 + mw3 h3 + mm h5 − ma ha 2 − mv 2 hv 2 − mw 4 h4
Using the definition of specific humidity in terms of flow rates, and combining like terms:
0 = ma (ha1 − ha 2 + ω1hv1 − ω2 hv 2 ) + mm h5 + mw (h3 − h4 )
Incorporating conservation of mass:
0 = ma [ha1 − ha 2 + ω1hv1 − ω2 hv 2 ] + (ω2 − ω1 )ma h5 + mw (h3 − h4 )
Solving for air flow rate:
mw (h4 − h3 )
ma =
ha1 − ha 2 + ω1hv1 − ω2 hv 2 + (ω2 − ω1 )h5
From Appendix A-10 hv1 = hg1 (20°C) = 2538.1kJ/kg
hv 2 = hg 2 (35°C) = 2565.3kJ/kg h3 = h f 3 (40°C) = 167.57 kJ/kg
h4 = h f 4 (25°C) = 104.89 kJ/kg h5 = h f 5 (20°C) = 83.96 kJ/kg
Pg1 (20°C) = 2.339 kPa Pg2 (35°C) = 5.628 kPa
From Appendix A-9 ha1 (20°C) = 293.16 kJ/kg ha 2 (35°C) = 308.23kJ/kg
0.622φ1 Pg1 (0.622)(0.40)(2.339) kg water
ω1 = = = 0.00600
P − φ1 Pg1 98 − (0.40)(2.339) kg dry air
(0.622)(1.00)(5.628) kg water
ω2 = = 0.0379
98 − (1.00)(5.628) kg dry air
(100, 000 kg hr)(104.89 − 167.57) kJ kg
ma =
[293.16 − 308.23 + (0.0060)(2538.1) − (0.0379)(2565.3) + (0.0379 − 0.006)(83.96) kJ kg
ma = 66, 410 kg hr
mm = (0.0379 − 0.006)(66, 410 kg hr) = 2,120 kg hr Answer

15-57
15-53 The heat rejected to the cooling water in the condenser of a Rankine cycle power plant is 850 MW. The
cooling water enters the cooling tower at 40 ºC and exits at 25 ºC. Air enters the cooling tower at 101.3
kPa, 10 ºC, and 35% relative humidity and exits at 30 ºC and 95% relative humidity. Make-up water is
supplied at 10 ºC. Determine the mass flow rate of the air, cooling water, and make-up water (in kg/s).
Approach:
There are three unknowns in the this problem, the
cooling water flow rate, m3 , the make-up water flow
rate, m5 , and the air flow rate, ma . Conservation of
mass and energy applied to the cooling tower control
volume, and conservation of mass and energy to a
control volume encompassing only the condensed are
used to determine the three unknowns.

Assumptions:
1. The air and water vapor are ideal gases, and
their mixture behaves as an ideal gas.
2. The condenser is steady with no work or
potential or kinetic energy effects.
3. The cooling tower is steady and adiabatic with
no work or potential or kinetic energy effects.

Solution:
Apply conservation of mass and energy to the control volume around the condenser. Assume steady, no work,
negligible potential and kinetic energy effects:
m3 = m4 and 0 = Q + m4 h4 − m3 h3 → m3 = Q ( h4 − h3 ) = Qin ( h4 − h3 )
From Appendix A-10 h3 = h f (40°C) = 167.57 kJ/kg h4 = h f (25°C) = 104.89 kJ/kg
850, 000 kW
m3 = = 13,560 kg/s Answer
(167.57 − 104.89)kJ/kg

Apply conservation of mass to the control volume around the cooling tower. Assume steady and ideal gas
mixture, and track air and water separately:
Air: ma1 = ma 2 = ma
Water: mv1 + m3 + m5 − mv 2 − m4 = 0
Note that m3 = m5 , and use the definition of specific humidity in terms of flow rates:
m5 = mv 2 − mv1 = ω2 ma − ω1ma = (ω2 − ω1 )ma
Applying conservation of energy to the cooling tower control volume, and assuming steady, adiabatic, no work,
negligible potential and kinetic energy effects:
0 = ma ha1 + mv1hv1 + m3 h3 + m5 h5 − ma ha 2 − mv 2 hv 2 − m4 h4

Using the definition of specific humidity in terms of flow rates, combining like terms, and incorporating
conservation of mass:
0 = ma [ha1 − ha 2 + ω1hv1 − ω2 hv 2 ] + (ω2 − ω1 )ma h5 + m3 (h3 − h4 )
Solving for the air flow rate
m3 (h4 − h3 )
ma =
ha1 − ha 2 + ω1hv 2 − ω2 hv 2 + (ω2 − ω1 )h5
From Appendix A-10 hv1 = hg1 (10°C) = 2519.9 kJ/kg Pg1 = 1.228 kPa
hv 2 = hg 2 (30°C) = 2556.3 kJ/kg Pg 2 = 4.246 kPa
h5 = h f (10°C) = 42.01 kJ/kg
From Appendix A-9 ha1 (10°C) = 283.14 kJ/kg ha 2 (30°C) = 303.21 kJ/kg

15-58
 0.622φ1 Pg1 (0.622)(0.35)(1.228) kg water
ω1 = = = 0.00265
P − φ1 Pg1 101.3 − (0.35)(1.228) kg dry air
(0.622)(0.95)(4.246) kg water
ω2 = = 0.0258
101.3 − (0.95)(4.246) kg dry air
(13560kg/s)(104.89 − 167.57)kJ/kg
ma =
[283.14 − 303.21 + (0.00265)(2519.8) − (0.0258)(2556.3) + (0.0258 − 0.00265)(42.01)]kJ/kg
= 10,840 kg/s Answer
m5 = (0.0258 − 0.00265)(10840 kg/s) = 251kg/s Answer

Comment:
Note that the make-up water flow rate is substantial (66 gal/s = 3990 gal/min) and the main water flow rate is
3590 gal/s = 215, 300 gal/min, so the make-up water flow rate is about 1.9% of the main flow.

15-59
15-54 For the combustion of octane (C8H18) with air, determine the air-fuel ratio for this reaction on both a molar
and a mass basis assuming:
a. stoichiometric air
b. 150% excess air.

Approach:
The chemical equation for complete combustion is balanced first. We
then rebalance the equation for 250% excess air.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion products only CO2,
H2O, O2, and N2.

Solution:
a) Assuming complete combustion and that air is composed only of oxygen and nitrogen in the usual proportions
the reaction equation is:
A(C8 H18 ) + B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 )
Balances on the elements give:
C: 8 A = E
H: 18 A = 2 D
N2: 3.76 B = C
O: 2 B = D + 2 E
Arbitrarily using A=1 (one mole of fuel), the remaining balance equations give:
E = 8, D = 9, B = 12.5, C = 47.0
C8 H18 + 12.5(O 2 + 3.76N 2 ) ⇒ 47.0(N 2 ) + 9(H 2 O) + 8(CO 2 )
nair 12.5(1 + 3.76) moles air
AFR = = = 59.5
n fuel 1 moles fuel
Using Eq. 15-53 and M fuel = 8(12.01) + 18(2.018 2) = 114.2
M air 28.97 kg air
AFR = AFR = 59.5( ) = 15.09 Answer
M fuel 114.2 kg fuel
b) For 150% excess air = 250% theoretical air, the new balance equation is:
A(C8 H18 ) + 2.5 B (O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D (H 2 O) + E (CO 2 ) + F (O 2 )
Again set A=1. The factor of 2.5 in front of the air component represents the theoretical air of 250%. We use
B=12.5 from the stoichiometric balance equation.
The balance equations are:
C: 8A = E
H: 18 A = 2 D
N2: 2.5 B (3.76) = C
O: 2.5 B (2) = D + 2 E + 2 F
With A=1 and B=12.5: C = 117.5, D = 9, E = 8, F = 18.75
C8 H18 + 31.25(O 2 + 3.76N 2 ) ⇒ 117.5(N 2 ) + 9(H 2 O) + 8(CO 2 ) + 18.75(O 2 )
nair 31.25(1 + 3.76) moles air
AFR = = = 148.8
n fuel 1 moles fuel
M air 28.97 kg air
AFR = AFR = 148.8( ) = 37.7 Answer
M fuel 114.2 kg fuel
Comment:
For part b, the AFR from part a could have been multiplied by 2.5(15.09) = 37.7 kg air/kg fuel

15-60
15-55 Assume gasoline can be treated as octane (C8H18) with a specific gravity of 0.68, and it is to be burned with
air at 101.3 kPa, 27 °C. Determine the theoretical volume of air required to burn a liter of gasoline (in m3).

Approach:
The mass of octane to be combusted is determined from the
specific gravity. The chemical equation for complete
combustion is used to determine the amount of air required
for each mole of fuel. The amount of air is used in the ideal
gas equation to determine the volume.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion products
only CO2, H2O, and N2.

Solution:
Assuming complete combustion and that air is composed only of oxygen and nitrogen in the usual proportions, the
reaction is:
A(C8 H18 ) + B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 )
Balances on the elements give:
C: 8A = E
H: 18 A = 2 D
N2: 3.76 B = C
O: 2 B = D + 2 E
Arbitrarily using A=1 (for one mole of fuel), the remaining balance equations give:
E = 8, D = 9, B = 12.5, C = 47.0
C8 H18 + 12.5(O 2 + 3.76N 2 ) ⇒ 47.0(N 2 ) + 9(H 2 O) + 8(CO 2 )
Therefore, the molar air-fuel ratio is:
n 12.5(1 + 3.76) moles air
AFR = air = = 59.5
n fuel 1 moles fuel
The mass of fuel is:
kg 10−3 m3
m f = ρ f V = SG ρ wV = (0.68)(1000 )(1L)( ) = 0.68 kg
m3 1L
Number of moles of fuel:
mf 0.68 kg
nf = = = 0.00595 kmol
M f [8(12.01) + 18( 2.018 )] kg
2 kmol
nair = (0.00595kmol)(59.5) = 0.354 kmol
Assuming air is an ideal gas
nRT (0.354 kmol)(8.314 kJ kmol K)(27 + 273)K
V= =
P 101.3kN m 2
= 8.72 m3 Answer

15-61
15-56 Dodecane (C12H26) is burned with 50% excess air. Determine:
a. the molar and mass air-fuel ratios
b. the dew-point temperature of the combustion products at 101.3 kPa (in °C).

Approach:
The chemical equation for 50% excess air is balanced.
From the total moles in the products of combustion, the
water vapor partial pressure and its corresponding
saturation temperature can be determined.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, O2,and N2.

Solution:
a) Assume air is composed of oxygen and nitrogen in their usual proportions, and note that 50% excess air is
150% theoretical air. Therefore, the chemical balance equation for stoichiometric conditions is:
A(C12 H 26 ) + B (O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 )
Balances on the elements give:
C: 12A = E
H: 26 A = 2 D
N2: B (3.76) = C
O: B (2) = D + 2 E
Arbitrarily setting A = 1, the remaining balance equations gives:
E = 12, D = 13, B = 18.5, C = 69.56
For excess air, the new reaction equation is:
A(C12 H 26 ) + (0.5 + 1) B (O2 + 3.76 N 2 ) ⇒ C ( N 2 ) + D( H 2 O) + E (CO2 ) + F (O2 )
We use A = 1and B = 18.5 from the stoichiometric balance.
Balances on the elements give:
C: 12 A = E → E = 12
H: 26 A = 2 D → D = 13
N2: (0.5 + 1) B (3.76) = C → C = (0.5 + 1)(18.5)(3.76) = 104.34
O: (0.5 + 1) B(2) = D + 2 E + 2 F → F = 9.25
The molar air-fuel ratio is
n (0.5 + 1)(18.5)(1 + 3.76) kmol air
AFR = air = = 132.1
n fuel 1 kmol fuel
2.018
M fuel = 12(12.01) + 26( ) = 170.35
2
M air 28.97 kg air
AFR = AFR = 132.1( ) = 22.46
M fuel 170.35 kg fuel
b) Assuming the products behave as an ideal gas mixture, the vapor pressure of the water vapor is:
13
Pi = Yi P = (101.3kPa) = 9.502 kPa
104.34 + 13 + 12 + 9.25
From Appendix A-11 at this pressure: TDP 44.8°C Answer

15-62
15-57 Assume gasoline can be treated as octane (C8H18), and it is to be burned with 100% theoretical air at 101.3
kPa. Determine:
a. the partial pressure of the water vapor (in kPa) and the dew-point temperature of the products (in °C)
for stoichiometric combustion
b. the partial pressure (in kPa) and dew-point temperature (in °C) for 65% excess air.

Approach:
Initially, the stoichiometric reaction equation is
balanced, and then a second equation with 65%
excess air is balanced. From the equation, the moles
of each component in the products are used to
determined the water vapor partial pressure and
dewpoint temperature.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for
each mole of O2.
2. The products of combustion behave as an ideal
gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, O2, and N2.

Solution:
a) Assume air is composed of oxygen and nitrogen in their usual proportions. The chemical balance equation for
stoichiometric conditions is:
A(C8 H18 ) + B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 )
C: 8 A = E
H: 18 A = 2 D
N2: 3.76 B = C
O: 2 B = D + 2 E
Arbitrarily using A=1, the remaining balance equations give:
B = 12.5, C = 47.0, D = 9, E = 8
C8 H18 +12.5(O 2 +3.76N 2 )Þ42.0(N 2 )+9(H 2 O)+8(CO 2 )
Assuming the products of combustion behave as an idea gas mixture, the partial pressure of the water vapor is:
9
PH 2O = YH 2 O P = (101.3kPa) = 14.25 kPa Answer
47.0 + 9 + 8
At this pressure, the saturation temperature of water is the dew point temperature. From Appendix A-11 ,
TDP ≈ Tsat (14.25 kPa ) = 52.8°C Answer

b) For the excess air, the new reaction equation is:

A(C8 H18 ) + (0.65 + 1) B (O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
We use A=1 and B=12.5 from the stoichiometric balance. Balances on each element give:
C: 8 A = E → E = 8
H: 18 A = 2 D → D=9
N2: (0.65 + 1) B(2) = C → C = (0.65 + 1)(12.5)(3.76) = 77.55
O: (0.65 + 1) B (2) = D + 2 E + 2 F → F = 8.125
Assuming the products of combustion behave as an ideal gas mixture, the partial pressure of the water vapor is:
9
PH 2O = YH 2O P = (101.3kPa) = 8.88 kPa
77.55 + 9 + 8 + 8.125
TDP = Tsat (8.88 kPa) ≈ 43.3°C Answer

15-63
15-58 An Orsat analysis of the products from the combustion of propane (C3H8) and air yields the following mole
fractions: CO2, 11.5%; O2, 2.7%, CO, 0.7%. Water vapor and nitrogen are also present. Determine the
reaction equation and the percent theoretical air for the reaction.

Approach:
The chemical reaction equation is balanced using the
given mole fractions. The stoichiometric balance is also
obtained. Comparison of the two equations will give the
percent theoretical air.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, O2, and N2.

Solution:
The chemical reaction equation, assuming stoichiometric conditions and air is composed of oxygen and nitrogen
in the usual proportions:
A(C3 H8 ) + B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 )
Balances on the elements give:
C: 3 A = E
H: 8 A = 2 D
N2: 3.76 B = C
O: 2 B = D + 2 E
Arbitrarily setting A=1, the remaining balance equations gives:
B = 5, C = 18.8, D = 4, E = 3
C3 H8 + 5(O 2 + 3.76N 2 ) ⇒ 18.8(N 2 ) + 4(H 2 O) + 3(CO 2 )
For the given information and assuming 100 moles in the products of combustion the balance equation is:
A(C3 H8 ) + B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + 11.5(CO 2 ) + 2.7(O 2 ) + 0.7(CO)
Balances on the elements give:
C: 3 A = 11.5 + 0.7 → A = 6.1
H: 8 A = 2 D → D = 24.4
N2: 3.76 B = C → C = 100.58
O: 2 B = D + 11.5(2) + 2.7(2) + 0.7 → B = 26.75
Normalizing so that 1 mole of fuel is used (divide all coefficients by 6.1), the reaction equation is:
C3 H8 + 4.38(O 2 + 3.76N 2 ) ⇒ 16.49(N 2 ) + 4(H 2 O) + 1.89(CO 2 ) + 0.44(O 2 ) + 0.11(CO)
Comparing the coefficients on the air:
5 stoichiometric air 1
= =
4.38 stoichiometric air + excess air 1 + excess air
excess air = −0.124 = −12.4% or 87.6% theoretical air
A second approach:
AFR actual ( 4.38 1)
Theoretical air = = = 0.876 = 87.6% Answer
AFR stoichiometric ( 5 1)
Comment:
Note that this is incomplete combustion.

15-64
15-59 When a fuel is burned with atmospheric air, water vapor is present in addition to the dry air. Consider a
gaseous fuel mixture that has the following molar analysis: CH4, 72%; N2, 14%; H2, 9%; CO2, 3%; O2, 2%.
The fuel is burned with moist air to form gaseous products at one atmosphere; only CO2, H2O, and N2 are
present in the products. If the dew-point temperature of the products is 60 °C, determine the amount of
water vapor present in the combustion air (in kmol/kmol fuel).

Approach:
The chemical reaction equation can be written with an
unknown amount of water vapor in the combustion air.
The amount of water vapor in the products can be
determined from the dewpoint temperature. Using the
water vapor amount in the products, the chemical reaction
equation can be solved.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, O2, and N2.

Solution:
The chemical balance equation based on 1 mole of fuel is written. Air is assumed to be composed of oxygen and
nitrogen in the usual proportions.

0.72(CH 4 ) + 0.14(N 2 ) + 0.09(H 2 ) + 0.03(CO 2 ) + 0.02(O 2 ) + B(O 2 + 3.76N 2 ) + C (H 2 O) ⇒ D(N 2 ) + E (CO 2 ) + F (H 2 O)

Balances on the elements give:
C: 0.72 + 0.03 = E → E = 0.75
H: 0.72(4) + 0.09(2) + 2C = 2 F → C = F − 1.53 or F = C + 1.53
N2: 0.14 + 3.76 B = D
O: 0.03(2) + 0.02(2) + 2 B + C = 2 E + F = 2 E + (C + 1.53)
From the oxygen balance, B=1.465
From the nitrogen balance, D=5.648
The value of F is obtained using the dewpoint temperature. Assuming the products of combustion behave as an
ideal gas mixture:
F Y (D + E)
PH 2O = YH 2 O Ptot → YH 2O = → F=
D+E+F 1− Y
From Appendix A-10 at TDP = 60°C, PH 2 O = 19.94 kPa
19.94
YH 2O = = 0.1968
101.3
0.1968(5.648 + 0.75)
F= = 1.568
1 − 0.1968
moles H 2 O
Therefore C=1.568-1.53 = 0.038 Answer
mole fuel

15-65
15-60 At an oil refinery, an impure gas mixture is combusted for use in process heating. The gas at 20 lbf/in.2,
300 ºF has the following volumetric analysis: CH4, 20%; C2H6, 40%; N2, 30%, H2O, 10%. Air is supplied
at 15 lbf/in.2, 100 ºF at a rate of 115% theoretical air. Determine:
a. the volume flow of air supplied per volume flow of fuel
b. the dew-point temperature of the products of combustion if the products exit at 15 lbf/in2.

Approach:
Using the chemical reaction equation, we can
determine the air fuel ratio. Assuming both gases are
ideal, the given information is used to calculate the
volume air required per volume of fuel.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for
each mole of O2.
2. The products of combustion behave as an ideal
gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, O2, and N2.

Solution:
a) Assume air is composed of oxygen and nitrogen in the usual proportions. Using the fact that volume fractions
equal to mole fractions, the chemical balance equation with 115% theoretical air for 1 mole fuel is:

0.20(CH 4 ) + 0.40(C 2 H 6 ) + 0.30(N 2 ) + 0.10(H 2 O) + (1.15) B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 ) .

Balances on each element gives:
C: 0.2 + 0.4(2) = E → E = 1.00
H: 0.2(4) + 0.4(6) + 0.1(2) = 2 D → D = 1.700
N2: 0.3 + 1.15 B(3.76) = C
O: 0.1 + 1.15 B (2) = D + 2 E + 2 F
Note that by inspection (See Example 15-10), F = B × (% excess air) so solving the oxygen balance, B=1.80, and
then C = 8.083 and F = 0.27
n (1.15)(1.8)(1 + 3.76)
AFR = air = = 9.853
n fuel 1
Assuming the air and fuel behave as ideal gases
nRT
( ) air naTa Pf
Vair 100 + 460 20 ft 3 air
= P = = (9.853)( )( ) = 9.68 3
V fuel nRT n f T f Pa 300 + 460 15 ft fuel Answer
( ) fuel
P

b) Assuming the products of combustion behave as an ideal gas mixture, the vapor pressure of the water vapor is:
1.700 lbf lbf
PH2 O = YH2 O P = (15 2 ) = 2.307 2
8.083 + 1.700 + 1.00 + 0.27 in. in.
From Appendix A-10 at this pressure:
TDP = TSAT 131.2°F Answer

15-66
15-61 Decane (C10H22) is burned with 25% excess air, which is at 98 kPa, 30 °C, and 75% relative humidity. The
combustion products are cooled to 98 kPa, 30 °C. Determine the amount of water vapor that condenses (in
kg water/kg fuel).

Approach:
The chemical reaction equation is first balanced for stoichiometric
conditions using dry air, and then with 25% excess air. The moles of
water vapor in the reactants can be determined with psychrometric
equations, and these moles are added directly to both sides of the
equation. If the products are cooled until water vapor condenses,
then the relative humidity in the products is 100%. Again using
psychrometric equations, the amount condensed can be determined.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion products only
CO2, H2O, O2, and N2.

Solution:
Assume dry air is composed of oxygen and nitrogen in the usual proportions. The chemical balance equation for
the stoichiometric condition is:
C: 10 A = E → E = 10
H: 22 A = 2 D → D = 11
N2: 3.76 B = C
2B = D + 2E → B = 15.5 → C = 58.28
For 25% excess dry air, the new reaction equation is:
A(C10 H 22 ) + (0.25 + 1) B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
We use A=1 and B=15.5 from the stoichiometric balance
C: 10 A = E → E = 10
H: 22 A = 2 D → D = 11
N2: (0.25 + 1) B (3.76) = C → C = (1.25)(15.5)(3.76) = 72.85
O: (0.25 + 1) B (2) = D + 2 E + 2 F → F = 3.875
The water vapor in the reactants is added directly to both sides. Assuming the products and reactants behave as
ideal gases: ( )
PH 2O = YH 2O Ptot = nH 2O ntot Ptot and PH 2O = φR Pg
o
From Appendix A-10 at 30 C, Pg = 4.264 kPa, so PH 2O = (0.75)(4.246) = 3.185 kPa
Using the coefficients from the chemical balance equation with excess air:
nH 2O nH 2 O PH O (92.225) (3.185)(92.225)
PH 2O = Ptot = Ptot → nH 2 O = 2 = = 3.098 kmol
(1.25) B(1 + 3.76) + nH 2O 92.225 + nH 2O Ptot − PH 2O 98 − 3.185
This number of moles of water vapor is added to both sides of the reaction equation so that is now reads:
C10 H 22 + 19.375(O 2 + 3.76N 2 ) + 3.098(H 2 O) ⇒ 72.85(N 2 ) + (11 + 3.098)(H 2 O) + 10(CO 2 ) + 3.875(O 2 )
The products are the saturated (φ p = 100%) if water vapor is condensed. Therefore,
nH 2O
PH 2O = YH 2O Ptot = Ptot where nH 2O is the maximum water vapor possible in the products.
72.85 + nH 2 O + 10 + 3.875
(4.246)(72.85 + 10 + 3.875)
Solving for nH 2 O = = 3.928kmol
98 − 4.246
For this amount of water vapor to exist in the products, the amount condensed per kmol of fuel is:
11 + 3.098 − 3.928 = 10.17 kmol water condensed kmol fuel
kmol water 18 kg water kmol water kg water
On a percent mass basis: 10.17 [ ] = 1.289 Answer
kmol fuel 142 kg fuel kmol fuel kg fuel

15-67
15-62 In an oil-fired furnace, the fuel is burned with 30% excess air. The fuel, which is composed of 83%
carbon, 14.2% hydrogen, and 2.8% inert material on a weight basis, has a flow rate of 9500 kg/hr.
Determine the volumetric flow rate of the products at 101.3 kPa, 400 °C (in m3/hr).

Approach:
Treating the gases and mixture as ideal gases, the volume
flow rate of the products can be determined once the air fuel
ratio is known, since the fuel flow rate is given. The
stoichiometric balance on the chemical reaction equation is
determined first, and then a second balance is performed
with excess air.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion products
only CO2, H2O, O2, and N2.

Solution:
mP RTP
Assuming the products of combustion behave as an ideal gas mixture, using the ideal gas law: VP =
Pp M P
The total flow of products is obtained from the mass air-fuel ratio:
mP = ma + mF = mF ⎡⎣( ma mP ) + 1⎤⎦ = mF ( AFR + 1)
The air fuel ratio is obtained by solving the chemical balance equation. First, for the stoichiometric conditions
assuming air is composed of oxygen and nitrogen in the usual proportions (note that the inert solids do no affect
the balance) and the mass percentages are put in moles:
0.83 0.142
C+ H 2 + B (O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 )
12.01 2.018
Balances on the elements give:
C: 0.83 12.01 = E
H2: 0.142 2.018 = D
N2: 3.76 B = C
O: 2 B = D + 2 E → B = 0.104 and C = 0.392
For the excess air, the new reaction equation is:
0.83 0.142
(C) + (H 2 ) + (0.30 + 1) B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
12.01 2.018
Using B=0.104 from the stoichiometric balance; these new elemental balances are:
C: 0.83 12.01 = E → E = 0.0691
H2: 0.142 2.018 = D → D = 0.0704
N2: (0.3 + 1) B (3.76) = C → C = 0.508
O: (0.3 + 1) B (2) = D + 2 E + 2 F → F = 0.0309
The air fuel ratio on a mass basis is: AFR = ( 0.1352(32.0) + 0.1352(3.76)(28.01) ) 1 = 18.57 kg air kg fuel
To calculate the mixture molecular weight, M P = ∑ Yi M i , we need the total moles in the products.
nP = 0.508 + 0.0704 + 0.0691 + 0.0309 = 0.6784 moles
M P = (1 0.6784 ) [0.508(28.01) + 0.0704(18.02) + 0.0691(44.01) + 0.0309(32)] = 28.78 kg kmol
mP = ( 9500 kg hr ) (18.57 + 1) = 185,870 kg hr
(185,870 kg hr)(8.314 kJ kmol K)(400 + 273)K m3
VP = = 356, 730 Answer
(101.3kN m 2 )(28.78 kg kmol) hr

15-68
15-63 Methane (CH4) is burned completely with 200% theoretical air, which is at 103 kPa, 15 °C, and 45%
relative humidity. Determine:
a. the balanced reaction equation
b. the dew-point temperature of the combustion products at 103 kPa (in °C).

Approach:
The chemical reaction equation is balanced with dry air.
Water vapor in the moist combustion air is determined with
psychrometric equations and is added directly to the water
vapor in the products is used to evaluate the water’s partial
pressure and dewpoint temperature.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion products
only CO2, H2O, O2, and N2.

Solution:
a) Assuming air is composed of oxygen and nitrogen in the usual proportions, the chemical balance equation for
200% theoretical air for 1 mole of fuel is:
A(CH 4 ) + (1 + 1) B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As noted in example 15-10, F = B × (excess air) = B × 1 = B
Balance on each element gives:
C: A = E
H: 4 A = 2 D
N2: 2 B (3.76) = C
O: 2 B (2) = D + 2 E + 2 F = D + 2 E + 2 B
Arbitrarily setting A=1, and solving the balance equations:
E = 1, D = 2, B = 2, C = 15.04, F = 2
So the balance equation is:
CH 4 + 4(O 2 + 3.76N 2 ) ⇒ 15.04(N 2 ) + 2(H 2 O) + CO 2 + 2(O 2 )
For the number of moles of water vapor in the combustion air:
nH 2 O
PH 2O = YH 2 O Ptot = (103kPa)
4(1 + 3.76) + nH 2 O
PH 2O = φ Pg (15°C) = (0.45)(1.705kPa) = 0.767kPa
Solving for nH 2 O = 0.143mole
This is added directly to both sides of the equation:
CH 4 + 4(O 2 + 3.76N 2 ) + 0.143(H 2 O) ⇒ 15.04(N 2 ) + 2.143(H 2 O) + CO 2 + 2(O 2 ) Answer
b) For the dewpoint temperature in the products of combustion:
nH O 2.143
PH 2O = 2 Ptot = (103kPa) = 10.9kPa
ntot 15.04 + 2.143 + 1 + 2
From Appendix A-11
TDP = TSAT (10.9kPa) 47.4°C Answer

15-69
15-64 Ethane (C2H6) burns with stoichiometric air; both gases enter the combustion chamber at 14.7 lbf/in.2, 77
°F. The products of combustion are cooled to 14.7 lbf/in.2, 77 °F. Determine the heat transfer rate (in
Btu/lbmol fuel).

Approach:
For a steady system with no work, the chemical balance
equation is solved first. Using the stoichiometric
coefficients, the energy equation is then solved for the
heat transfer rate.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, and N2.
5. The system is steady with no work or potential or
kinetic energy effects.

Solutions:
For the stoichiometric conditions and assuming air is composed only of oxygen and nitrogen in the usual
proportions, the chemical balance equation is:
A(C 2 H 6 ) + B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 )
Balances on each element gives:
C: 2A = E
H: 6 A = 2 D
N2: 3.76 B = C
O: 2 B = D + 2 E
Arbitrarily setting A=1, we solve the above equations:
E = 2, D = 3, B = 3.5, C = 13.16
The complete balance equation is:
C2 H 6 + 3.5(O 2 + 3.76N 2 ) ⇒ 13.16(N 2 ) + 3(H 2 O) + 2(CO 2 )
Because the products of combustion are cooled to a low temperature, we must determine if any of the water vapor
formed during combustion condenses. If water condenses, then the remaining gas would be saturated
(φ = 100%) . Therefore, the maximum moles of water vapor in the products would be:
nH 2 O
PH 2O = YH 2O Ptot = Ptot and PH 2 O = φ Pg (77°F)
nN2 + nCO2 + nH2 O
From Appendix B-10, PH 2O (77°F) = 0.4598lbf in.2 . Solving for nH 2 O
PH 2O (nN2 + nCO2 ) (0.4598)(13.16 + 2)
nH 2 O = = = 0.489 moles
Ptot − PH 2O 14.7 − 0.4598
Because the number of moles of water vapor in the chemical balance equation is greater than the maximum
possible, then water vapor has condensed. The amount condensed is 3 - 0.489 = 2.511moles. This must be taken
into account in the energy balance.

Using conservation of energy, and assuming steady, no work, negligible potential and kinetic energy effects, and
the gases behave as ideal gases and mixtures, then Eq. 15-60 is applicable. Rearranging the equation to group
inflow and outflow terms (as was done in Example 15-11):
Q
= ∑ ni (∆h f° + ∆h )i − ∑ ni (∆h f° + ∆h )i
nF out in

Expanding each term:

15-70
 Q
= nN2 , P (∆h f° + hP − href ) N2 , P + nH 2 O , P , g (∆h f° + hP − href ) H 2O , P , g + nH 2 O , P , (∆h f° + hP − href ) H 2 O , P ,
nF
+ nCO2 , P (∆h f° + hP − href )CO2 , P + nC2 H 6 (∆h f° + hP − href )C2 H 6
−nO2 , R (∆h f° + hP − href )O2 , R − nN2 , R (∆h f° + hP − href ) N2 , R
Because products and reactants are at 77oF, the enthalpy difference, cancel in every term because products and
reactants are at the same temperature. The ∆h f° terms for O2 and N2 are zero because those compounds exist.
Therefore, the equation is:
Q
= nH 2 O , P , g (h f° ) H 2 O , P , g + nH 2O , P , (h f° ) H 2O , P , + nCO2 , P (h f° )CO2 , P − nC2 H 6 (h f° )C2 H 6
nF
Using data from Appendices B-18 and B-19 and the stoichiometric coefficients:
Q
= 0.489(−104, 040) + 2.511( −122,970) + 2(−169,300) − 1(−36, 420)
nF
Btu
= −661,833 Answer
lbmol fuel

15-71
15-65 Methane (CH4) at 101.3 kPa, 34 °C with a volumetric flow rate of 0.98 m3/min is burned with 20% excess
air. If the products are to be cooled from 600 °C to 200 °C, determine the heat transfer rate (in kW).

Approach:
An energy balance is used on control volume II to
determine the heat transfer rate. The chemical
reaction equation must be balanced, so that the gas
composition entering control volume II at 600oC is
known.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for
each mole of O2.
2. The products of combustion behave as an ideal
gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, O2, and N2.
5. Control volume II is steady with no work or
potential or kinetic energy effects.

Solution:
An energy and mass balance on control volume II, is performed. Assume steady, no work, negligible potential
and kinetic energy and ideal gas mixture behavior:
Q
Q = ∑ mi hi −∑ mi hi → = nN2 (h2 − h1 ) N + nH 2O (h2 − h1 ) H 2O + nCO2 (h2 − h1 )CO2 + nO2 (h2 − h1 )O2
out in nf
The number of moles in the products is determined by balancing the chemical reaction equations. Assuming air is
composed of oxygen and nitrogen in its usual proportion, the equation for 20% excess is:
A(CH 4 ) + (0.20 + 1) B (O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As noted in Example 15-10, F = B × (excess air) = 0.20 B
Balances on elements:
C: A= E
H: 4 A = 2 D
N2: (0.20 + 1) B(3.76) = C
O: (0.20 + 1) B(2) = D + 2 E + 2 F = D + 2 E + 2(0.20 B )
Arbitrarily setting A=1, and solving the balance equations:
E = 1, D = 2, B = 2.0, C = 9.024, F = 0.4
Using data from Appendix A-18 at the appropriate temperatures and interpolating
Q
= (9.024)(13782 − 26024) + 2(15882 − 30754) + 1(16484 − 35979) + 0.40(13935 − 27002)
nf
kJ
= −164940
kmol fuel
For the ideal gas mixture:
PV (101.3kN m 2 )(0.98 m3 min) kmol
n fuel = = = 0.03989
RT (8.314 kJ kmol K)(34 + 273)K min
kmol kJ kJ
Q = (0.0389 )(−164940 ) = −6414 = −106.9 kW Answer
min kmol min

15-72
15-66 A heat transfer rate of 10 MW is to be obtained by burning benzene (C6H6) at 101.3 kPa, 25 °C with
stoichiometric air at the same conditions. The products of combustion exit at 477 °C. Determine the
required fuel mass flow rate (in kg/s).

Approach:
For a steady system with no work, the chemical balance
equation is solved first. Then, using the stoichiometric
coefficients from the chemical equation, the conservation
of energy is solved for the required fuel flow rate.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, and N2.
5. The system is steady with no work or potential or
kinetic energy effects.

Solutions:
For the stoichiometric conditions, and assuming air is composed only of oxygen and nitrogen in the usual
proportions, the chemical balance equation is:
A(C6 H 6 ) + B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 )
Balances on each element gives:
C: 6A = E
H: 6 A = 2 D
N2: 3.76 B = C
O: 2 B = D + 2 E
Arbitrarily setting A=1, the balance equations are solved:
E = 6, D = 3, B = 7.5, C = 28.20
Using conservation of energy, and assuming steady, no work, negligible potential and kinetic energy effects, and
the gases and mixture behave as ideal gases, then Eq. 15-60 is applicable. Rearranging the equation to group
inflow and out flow terms:
Q
= ∑ ni (∆h f° + ∆h )i − ∑ ni (∆h f° + ∆h )i
nf out in

Expanding the terms, with “P” representing products and “R” reactants:
Q
= nN2 , P (∆h f° + hp − href ) N2 , P + nH 2 O , P (∆h f° + hp − href ) H 2O , P + nCO2 , P (∆h f° + hp − href )CO2 , P
nf
+ nC6 H 6 (∆h f° + hp − href )C6 H 6 − nO2 , R (∆h f° + hp − href )O2 , R − nN2 , R (∆h f° + hp − href ) N2 , R
Using data from the Appendices A-18 and A-19 and the stoichiometric coefficients from the balance equations:
Q
= (28.20)(0 + 20604 − 8669) + (3)(−241820 + 24088 − 9904) + (6)(−393520 + 27125 − 9364)
nF
kJ
−(1)(82930) − (7.5)(0 + 8682 − 8682) − (6)(0 + 9364 − 9364) = −2, 684, 000
kmol fuel
With Q = −10, 000kW
−10, 000 kW kmol fuel
nF = = 0.00373
−2684000 kJ kmol fuel s
With M C6 H 6 = 78.11
kmol kg kg
mF = (0.00373 )(78.11 ) = 0.291 Answer
s kmol s

15-73
15-67 The boiler in a simple Rankine cycle uses methane (CH4) at 101.3 kPa, 25 ºC which combusts with 110%
theoretical air at 101.3 kPa, 25 ºC. The products of combustion exit the smoke stack at 101.3 kPa, 150 ºC.
The cycle thermal efficiency is 40%. Determine the required fuel mass flow rate per MW of net power
output (in kg/hr MW).

Approach:
The required input heat transfer rate is determined from the
definition of cycle efficiency. The required methane flow
rate is determined, first by balancing the chemical reaction
equation, and then by applying conservation of energy to
the gas side of the boiler.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion products
only CO2, H2O, O2, and N2.
5. The system is steady with no work or potential or
kinetic energy effects.

Solution:
The cycle thermal efficiency is defined as ηcycle = Wnet Qin
Inverting this expression, we obtain the required input heat transfer per MW of net power output.
Qin 1 MW heat transfer input
= = 2.5
Wnet 0.4 MW of net power output
To obtain the required methane flow rate, start with the chemical balance equation. Assuming air is composed of
oxygen and nitrogen in its usual proportions, and taking account excess air:
A(CH 4 ) + (1.1) B (O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As noted in Example 15-10, F = B × (excess air)=0.1B
Balances on each element:
C: A = E
H: 4 A = 2 D
N2: (1.1) B(3.76) = C
O: (1.1) B (2) = D + 2 E + 2 F
Arbitrarily setting A=1 and solving the balance equations:
E = 1, D = 2, B = 2, C = 8.272, F = 0.2
CH 4 + 2.2(O 2 + 3.76N 2 ) ⇒ 8.272(N 2 ) + 2(H 2 O) + CO 2 + 0.2(O 2 )
Using conservation of energy on the reaction, and assuming shady, no work, negligible potential and kinetic
energy effects, and the gases and mixtures behave as ideal gas mixtures, then Eq. 15-60 is applicable. Grouping
in flow and out flow terms, and expanding the terms with “P” representing products and “R” reactants:
Q
= nN2 , P (∆h fo + hP − href ) N2 , P + nH 2 O , P (∆h fo + hP − href ) H 2 O , P + nCO2 , P (∆h fo + hP − href )CO2 , P + nO2 , P (∆h fo + hP − href )O2 , P
nF
−nCH 4 (∆h fo + hP − href )CH 4 − nO2 , R (∆h fo + hP − href )O2 , R − nN2 , R (∆h fo + hP − href ) N2 , R

Using data from Appendices A-18 and A-19 and the stoichrometric coefficients from the chemical balance
equation:

15-74
Q
= (8.272)(0 + 12313 − 8669) + (2)(−241820 + 14147 − 9904) + (1)(−393520 + 14333 − 9364) + 0.2(0 + 12405 − 8682)
nP
−(1)(−74850) − (2.2)(0 + 8682 − 8682) − 8.272(0 + 8669 − 8669)
kJ
= −757970
kmol fuel
MW heat transfer input kW
−(2.5 )(1000 )
MW net power output MW kmol
nF = = 0.00330
kJ s ⋅ MW net power
−757970
kmol fuel
kmol kg kg kg
mF = (0.0033 )(16.04 ) = 0.0529 = 190.5 Answer
s kmol s MW hr MW

15-75
15-68 Gaseous pentane (C5H12) at 25ºC is burned with 22% excess air that is at 37 ºC. The products of
combustion are cooled to 400 °C. Determine the heat transfer per kg mole of fuel (in kJ/kgmol).

Approach:
An energy balance is used on the control volume to
determine the heat transfer rate. The chemical balance
equation must be balanced, so that the number of moles
or reactants and products can be used in the energy
equation.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, O2, and N2.
5. The system is steady with no work and negligible
potential and kinetic energy effects.

Solution:
Assuming air is composed of only oxygen and nitrogen in its usual proportions, the chemical balance equation for
22% excess air is:
A(C5 H12 ) + (0.22 + 1) B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As noted in Example 15-10, F = B × (excess air) = 0.22 B
Balances on each element give:
C: 5A = E
H: 12 A = 2 D
N2: (0.22 + 1) B(3.76) = C
O: (0.22 + 1) B(2) = D + 2 E + 2 F = D + 2 E + 2(0.22 B )
Arbitrarily setting A=1, and solving the balance equations:
E = 5, D = 6, B = 8, C = 36.70, F = 1.76
Using conservation of energy, and assuming steady, no work, negligible potential and kinetic energy effects, and
the gases and mixture behave as ideal gases, then Eq. 15-60 is applicable. Rearranging the equation to group
inflow and outflow terms:
Q
= ∑ ni (∆h f° + ∆h )i −∑ ni (∆h f° + ∆h )i
nf out in

Expanding the terms and using “P” to represent products and “R” reactants

Q
= nN2 , P (∆h f° + hp − href ) N2 , P + nH 2O , P (∆h f° + hp − href ) H 2 O , P + nCO2 , P (∆h f° + hp − href )CO2 , P + nO2 , P (∆h f° + hp − href ) N2 , P
nF
− nC5 H12 (∆h f° + hp − href )C5 H12 − nO2 , R (∆h f° + hp − href )O2 , R + nN2 , R (∆h f° + hp − href ) N2 , R
Using Appendices A-18 and A-19 and the stoichiometric coefficients from the chemical balance equation:
Q
= (36.70)(0 + 20197 − 8669) + (6)(−24182 + 22970 − 9904) + (5)(−393520 + 25648 − 9364)
nF
+(1.761)(0 + 20197 − 8682) − (1)(−146440 + h298 − h298 ) − (9.76)(0 + 9030 − 8682) − (36.70)(0 + 9014 − 8669)
kJ
= 2.685 × 106 Answer
kmol fuel

15-76
15-69 Propane (C3H8) and pure oxygen (O2) in stoichiometric proportions are combusted in a water-cooled
burner. The reactants both enter at 90 °F and 2 atm and exit at 850 °F. The products (CO2 and H2O only)
have a flow rate of 250 lbm/hr. The specific heat of the propane is 17.83 Btu/lbmolR. Determine the heat
transfer rate (in Btu/hr).

Approach:
The chemical balance equation is solved first. Then,
using the stoichiometric coefficients from the chemical
equation, conservation of energy is solved for the heat
transfer rate.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2 and H2O.
5. The system is steady with no work and negligible
potential and kinetic energy effects.

Solution:
For stoichiometric conditions, the chemical balance equation is:
A(C3 H8 ) + B(O 2 ) ⇒ C (H 2 O) + D(CO 2 )
Balance on each element gives:
C: 3A = D
H: 8 A = 2C
N2: 2 B = C + 2 D
Arbitrarily setting A=1 → D = 3, C = 4, B = 5
C3 H8 + 5(O 2 ) ⇒ 4(H 2 O) + 3(CO 2 )
Using conservation of energy, and assuming steady, no work, negligible potential and kinetic energy effects, and
the gases and mixture behave as ideal gases, then Eq. 15-60 is applicable. Rearranging the equation to group
inflow and outflow terms, and expanding the terms:
Q
= nH 2 O (∆h f° + hp − href ) H 2 O + nCO2 (∆h f° + hp − href )CO2 − nC3 H8 (∆h f° + hp − href )C3 H8 − nO2 (∆h f° + hp − href )O2
nF
Using Appendices B-18 and B-19 and the stoichiometric coefficients from the chemical balance equations:
Btu Btu
∆hC3 H8 = (17.83 )(90 − 77)R = 231
lbmolR lbmol
Q
= (4)(−104040 + 10805 − 4258) + 3(−169300 + 12257 − 3880)
nF
Btu
− (1)(−44680 + 231) − 5(0 + 9492 − 3725) = −857,127
lbmol fuel
To obtain the molar fuel flow rate:
mP
mP = mO2 + mF = mF (mO2 mF + 1) and mF = → nF = mF M F
mO2 mF + 1
For propane M F = 44.09 lbm lbmol. From the chemical balance equation:
mO2 nO2 M O2 5 32.0 250 lbm hr lbm
= = ( )( ) = 3.629 → MF = = 54.0
mF nF M F 1 44.09 3.629 + 1 hr
54.0 lbmol
nF = = 1.225
44.09 hr
Btu lbmol Btu
Q = (−857,127 )(1.225 ) = −1, 049,940 Answer
lbmol hr hr

15-77
15-70 Overall plant efficiency in a Rankine cycle is improved by using the exhaust gases from the boiler to
preheat the incoming combustion air. Consider a system in which methane (CH4) at 101.3 kPa, 25 °C is
burned in the boiler with 150% theoretical air. After the products of combustion are used to boil the water
in the boiler, the gases leave the boiler at 727 °C, enter a heat exchanger to preheat the combustion air, and
leave the heat exchanger at 587 °C. The air enters the heat exchanger at 25 °C. Determine the combustion
air temperature as it leaves the heat exchanger and enters the boiler (in °C).

Approach:
Conservation of energy is applied to the control volume
around the air preheater. The relative flow of the air and
products of the combustion are determined by balancing
the chemical reaction equation.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion products only CO2, H2O, O2, and N2.

Solution:
Apply conservation of mass to the control volume around the preheater. Assume steady, adiabatic, no work,
negligible potential and kinetic energy effects, and each gas and mixtures of gases behave as ideal gases. Using
Eq. 15-58, dividing by the molar flow rate of fuel, and rearranging terms:
0 = ∑ ni hi − ∑ ni hi
out in

The number of moles of each substance is obtain by balancing the chemical reaction equations. Assuming air is
composed of oxygen and nitrogen in their usual proportions:
A(CH 4 ) + (1.5) B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As noted in Example15-10, F = B × (excess air) = 0.5B
Balances on each element gives:
C: A = E
H: 4 A = 2 D
N2: (1.5) B (3.76) = C
O: (1.5) B(2) = D + 2 E + 2 F
Arbitrarily setting A=1 and solving the balance equations:
E = 1, D = 2, B = 2, C = 11.28, F = 1
CH 4 + 3(O 2 + 3.76N 2 ) = 11.28(N 2 ) + 2(H 2 O) + CO 2 + O 2
Now applying the conservation of energy equation:
0 = [3hO2 + 11.28hN2 ]2 + [11.28hN2 + 2hH 2O + hCO2 + hO2 ]4 − [3hO2 + 11.28hN2 ]1 − [11.28hN2 + 2hH 2O + hCO2 + hO2 ]3
Using data from Appendix A-18 and combining terms:
0 = 3[hO2 ,2 − 8682] + 11.28[ hN2 ,2 − 8669] + 11.28[25610 − 30129] + 2[30240 − 35882] + [35296 − 42769] + [26559 − 31389]
198,394 = 3hO2 ,2 + 11.28hN2 ,2
An iterative solution is required. For a first estimate, assume air is all nitrogen:
198,394 ≈ (3 + 11.28)hN2 ,2 → hN2 ,2 ≈ 13,893kJ kmol → T2 ≈ 470 K

Guess T2 (K) hN2 ( kJ kmol ) hO2 ( kJ kmol ) RHS of Equation

470 13693 13842 195,983
480 13988 14151 200,238
By interpolation, T2 ≈ 476K ≈ 203°C Answer

15-78
15-71 In the boiler of a Rankine cycle, methane gas (CH4) at 101.3 kPa, 25 °C is burned with 40% excess air at
101.3 kPa, 127 °C. The products of combustion exit at 101.3 kPa, 427 °C. The heat transfer is used to
vaporize and superheat water, which enters as a saturated liquid at 8 MPa and leaves at 480 °C; the water
flow rate is 100 kg/s. Determine the required volumetric flow rate of the methane (in m3/s).

Approach:
The required heat transfer rate is determined by applying
conservation of energy to the water. The methane flow
rate is determined, first by balancing the chemical
reaction equation, and then by applying conservation of
energy to the gas side of the boiler.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, O2, and N2.
5. The system is steady with no work or potential or
kinetic energy effects

Solution:
Define a control volume around the water in the boiler. Assume steady, no work, negligible potential and kinetic
energy effects, no pressure drop, and apply conservation of mass and energy: Q = mW (h1 − h2 )
From Appendices A-11 and A-12 h1 = h f ( P1 ) = 1316.64 kJ kg h2 ( P2 , T2 ) = 3347.8 kJ kg
Q = (100 kg s)(1316.64 − 3347.8) kJ kg = −203,100 kW
For the chemical balance equation, air is composed only of O2 and N2 in its usual proportion. For 40% excess air:
A(CH 4 ) + (1 + 0.4) B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As noted in Example 15-10, F = B × (excess air) = 0.4 B . Performing a balance on each element:
C: A = E
H: 4 A = 2 D
N2: 1.4 B (3.76) = C
O: 1.4 B(2) = D + 2 E + 2 F
Arbitrarily setting A = 1 , and solving the balance equations:
E = 1, D = 2, B = 2, C = 10.53, F = 0.8
CH 4 + 2.8(O 2 + 3.76N 2 ) ⇒ 10.53(N 2 ) + 2(H 2 O) + CO 2 + 0.8(O 2 )
Using conservation of energy on the gas side, with the same assumptions as before and that the gases and mixture
behave as ideal gases, then Eq. 15-60 is applicable. Rearranging the equation to group inflow and outflow terms
and expanding the terms with “P” represents products and “R” reactants.
Q
= nN2 , P (∆h f° + hp − href ) N2 , P + nH 2O , P (∆h f° + hp − href ) H 2 O , P + nCO2 , P (∆h f° + hp − href )CO2 , P
nf
+ nO2 , P (∆h f° + hp − href )O2 , P − nCH 4 (∆h f° + hp − href )CH 4 − nO2 , R (∆h f° + hp − href )O2 , R − nN2 , R (∆h f° + hp − href ) N2 , R
Using Appendices A-18 and A-19 and the stoichiometric coefficients from the chemical balance equation:
Q nP = (10.53)(0 + 20604 − 8669) + (2)(−241820 + 24088 − 9904) + (1)(−393520 + 27125 − 9364)
kJ
+(0.8)(0 + 21184 − 8682) − (1)(−74850) − (2.8)(0 + 11711 − 8682) − (10.53)(0 + 11640 − 8669) = −660, 270
kmol fuel
Using Q → nF = −203,100 kW −660, 270 kJ kmol = 0.308 kmol s
nRT (0.308 kmol s)(8.314 kJ kmolK)(298K) m3
Using the ideal gas law: V= = 2
= 7.52 Answer
P 101.3kN m s

15-79
15-72 Determine the adiabatic flame temperature for hydrogen (H2) burning with 200% theoretical air. The
hydrogen and air both enter at 101.3 kPa, 25 °C (in °C).

Approach:
First, the chemical reaction equation is balanced, and then
conservation of energy is applied assuming no heat transfer.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion products
only CO2, H2O, O2, and N2.

Solution:
The chemical balance equation, assuming air is composed of oxygen and nitrogen in their usual proportions for
stoichiometric conditions is:
A(H 2 ) + B(O 2 + 3.65N 2 ) = C (N 2 ) + D(H 2 O)
Balances on each element give:
H2: A = D
O: 2 B = D
N2: 3.76 B = C
Arbitrarily setting A=1, and solving the balance equations: D = 1, B = 0.5, C = 1.88
For 200% theoretical air:
A(H 2 ) + (2) B (O 2 + 3.76N 2 ) = C (N 2 ) + D(H 2 O) + E (O 2 )
Balances on each element give:
H2: A = D
O: (2) B (2) = D + 2 E
N2: (2) B(3.76) = C
Using A = 1 and B = 0.5 from the stoichiometric conditions: D = 1, C = 3.76, E = 0.5
H 2 + (O 2 + 3.76N 2 ) = 3.76(N 2 ) + (H 2 O) + 0.5(O 2 )
Now applying conservation of energy, with the assumptions, used to develop Eq. 15-60, expanding each term, and
letting “R” represents reactants and “P” products,
nH 2 (∆h f° + hp − href ) H 2 + nO2 , R (∆h f° + hp − href )O2 , R + nN2 , R (∆h f° + hp − href ) N2 , R
= nN2 , P (∆h f° + hp − href ) N2 , P + nH 2 O (∆h f° + hp − href ) H 2 O + nO2 , P (∆h f° + hp − href )O2 , P
Evaluating the properties, using Appendices A-18 and A-19
(1)(0 + hp − href ) H 2 + (1)(0 + hp − href )O2 + (3.76)(0 + hp − href ) N2
= 3.76(0 + hp − 8669) N2 + (1)(−241820 + hp − 9904) H 2 O + 0.5(0 + hp − 8682)O2
288, 660 = 3.76hP , N2 + hP , H 2 O + 0.5hP ,O2
Using data from the tables, an iterative solution is obtained. First, assume all products are nitrogen to get an
initial estimate of the temperature:
288, 660 ≈ (3.76 + 1 + 0.5)hP , N2 → hP , N2 ≈ 54,983kJ kmol
From the nitrogen table, TP ≈ 1720 K
RHS of
Guess TP (K) hN2 ( kJ kmol ) hH 2O ( kJ kmol ) hO2 ( kJ kmol ) equation
1720 54807 68567 57394 303338
1660 52686 65643 55172 291328
1640 51980 64675 54434 287337
By interpolation TP 1647K Answer

15-80
15-73 A hot gas at 1140 ºF is needed for a process. Gaseous propane (C3H8) is to be burned with air in an
adiabatic steady flow combustion chamber; both gases enter at 77 ºF. Determine the required air-fuel mass
ratio and the percent excess air.

Approach:
We are given the adiabatic flame temperature. To find the excess
air, teams of terms of the unknown excess air. Then conservation of
energy is applied to the reaction with no heat transfer.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion products only
CO2, H2O, O2, and N2.

Solution:
The chemical balance equation is written assuming air is composed of oxygen and nitrogen in their usual
proportions. Excess air is taken into account. Let X = excess air:
A(C3 H8 ) + (1 + X ) B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As noted in Example 15-10, F = XB
Balances on each element gives:
C: 3 A = E
H: 8 A = 2 D
N 2: (1 + X ) B(3.76) = C
O: ( HX ) B (2) = D + 2 E + 2 F
Arbitrarily setting A=1 and solving the balance equations:
E = 3, D = 4, B = 5, C = 18.8(1 + X ), F = 5 X
C3 H8 + (1 + X )5(O 2 + 3.76N 2 ) ⇒ 18.8(1 + X )(N 2 ) + 4(H 2 O) + 3(CO 2 ) + 5 X (O 2 )
Applying conservation of energy, with the assumptions used to develop Eq. 15-60, expanding each term, and
letting “R” represent reactants and “P” products:
nC3 H8 (∆h fo + hR − href )C3 H8 + nO2 , R (∆h fo + hR − href )O2 , R + nN2 , R (∆h fo + hR − href ) N2 , R
= nN2 , P (∆h fo + hR − href ) N2 , P + nH 2O (∆h fo + hR − href ) H 2O + nCO2 (∆h fo + hR − href )CO2 + nO2 , R (∆h fo + hR − href )O2 , P
Evaluating the properties using Appendices A-18 and A-19

(1)(−44680 + h537 − h537 )C3 H8 + (1 + X )5(0 + h537 − h537 )O2 + (1 + X )5(3.76)(0 + h537 − h537 ) N2
= 18.8(1 + X )(0 + 11410 − 3730) + 4(−104040 + 13494 − 4258) + 3(−169300 + 15829 − 4028) + 5 X (0 + 11833 − 3725)

Solving for X
−44680 = −707329 + X (144384 + 40540)
X = 3.583 or 358% excess air
ma nM (1 + 3.583)5[(1)(32.0) + (3.76)(28.01)]
AFR = = a a =
mF nF M F (1)(44.09)
lbm air
= 71.4 Answer
lbm fuel

15-81
15-74 Determine the enthalpy of combustion at 25 °C for ethane (C2H6) when gaseous ethane reacts with air
producing liquid water in the products (in kJ/kmol). Do the same for acetylene (C2H2).

Approach:
The enthalpy of combustion is defined in Eq. 15-61.
First, we must balance the chemical reaction
equation. Then, we use the definition to calculate the
enthalpy of combustion.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for
each mole of O2.
2. The products of combustion behave as an ideal
gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, and N2.

Solutions:
a) The chemical reaction equation for the ethane, assuming air is composed of oxygen and nitrogen in their usual
proportions.
A(C 2 H 6 ) + B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 )
Balances on each element give:
C: 2A = E
H: 6 A = 2 D
N2: 3.76 B = C
O: 2 B = D + 2 E
Arbitrarily setting A=1, and solving the balance equations:
E = 2, D = 3, B = 3.5, C = 13.16 and C2 H 6 + 3.5(O 2 + 3.76N 2 ) ⇒ 13.16(N 2 ) + 3(H 2 O) + 2(CO 2 )
The enthalpy of combustion is defined as:
hc° = ∑ ni (∆h f° )i − ∑ ni (∆h f° )i
out in

Expanding terms:
∆hc° = nN2 , P ∆h f°, N2 , P + nH 2 O , P ∆h f°, H 2 O , P + nCO2 , P ∆h f°,CO2 , P − nC2 H 6 ∆∆h f°,C2 H 6 − nO2 , R ∆h f°,O2 , R − nN2 , R ∆h f°, N2 , R
Using data from Appendix A-18 and A-19, and canceling the nitrogen terms:
∆hc° = (3)(−286830) + (2)(−393520) − (1)(−84680) − (3.5)(0)
= −1,562,850 kJ kmol
b. For acetylene
A(C2 H 2 ) + B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 )
Balances on each element:
C: 2A = E
H: 2 A = 2 D
N2: 3.76 B = C
O: 2 B = D + 2 E
With A=1, E=2, D=1, B=2.5, C=9.40 and C2 H 2 + B (O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 )
∆hc° = (1)(−286830) + (2)(−393520) − (1)(226730) + (2.5)(0)
= −1,300, 600 kJ kmol Answer

15-82
15-75 Butane (C4H10) is to be burned with sufficient excess air so that the products of combustion enter a gas
turbine at 1600 K. Fuel and air enter at 25 °C. Determine the required percent excess air.

Approach:
We are given the adiabatic flame temperature. To find the
required excess air, first, the chemical reaction equation is
balanced in terms of the unknown excess air. Then,
conservation of energy is applied to the reaction with no
heat transfer.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion products
only CO2, H2O, O2, and N2.

Solution:
The chemical balance equation is written assuming air is composed of oxygen and nitrogen in their usual
proportions. Excess air is taken into account. Let X=excess air.
A(C4 H10 ) + (1 + X ) B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As noted in Example 15-10, F = XB
Balances on each element give:
C: 4A = E
H: 10 A = 2 D
N2: (1 + X ) B (3.76) = C
O: (1 + X ) B (2) = D + 2 E + 2 F
Arbitrarily setting A=1, and solving the balance equations:
E = 4, D = 5, B = 6.5, C = 24.44(1 + X )(N 2 ) + 5(H 2 O) + 4(CO 2 ) + 6 X (O 2 )
Applying conservation of energy, with the assumptions used to develop Eq. 15-60, expanding each term, and
letting “R” represent reactants and “P” products.

nC4 H10 (∆h f° + hp − href )C4 H10 + nO2 , R (∆h f° + hp − href )O2 , R + nN2 , R (∆h f° + hp − href ) N2 , R
= nN2 , P (∆h f° + hp − href ) N2 , P + nH 2O , P (∆h f° + hp − href ) H 2O , P + nCO2 , P (∆h f° + hp − href )CO2 , P + nO2 , P (∆h f° + hp − href )O2 ,P
Evaluating the properties from Appendices A-18 and A-19
(1)(−126150 + h298 − h298 ) + (1 + X )(6.5)(0 + h298 − h298 ) + (1 + X )(6.5)(3.76)(0 + h298 − h298 )
= 24.44(1 + X )(0 + 50571 − 8669) + 5(−241820 + 62748 − 9904)
+4(−393520 + 76944 − 9364) + 6.5 X (0 + 52961 − 8682)
Solving for X
−126,150 = −1, 224,555 + X (1, 024, 084 + 287,814)
X = 0.837 or 83.7% excess air Answer

15-83
15-76 In a regenerative Brayton cycle, the maximum operating temperature of a gas turbine is not exceeded by
using excess air during the combustion of the fuel. Consider a turbine whose maximum temperature is
1600 °F. Methane (CH4) at 14.7 lbf/in.2, 77 °F is burned completely with air that is preheated in the
regenerator to 400 °F. Determine the required percent excess air.

Approach:
We are given the adiabatic flame temperature. To find
the required excess air, first, the chemical reaction
equation is balanced in terms of the unknown excess air.
Then, conservation of energy is applied to both the
reaction.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, O2, and N2.
5. The combustion process is steady and adiabatic with
no work or potential or kinetic energy effects.

Solution:
The chemical balance equation is written assuming air is composed of oxygen and nitrogen in their usual
proportions. Excess air is taken into account. Let X=excess air
A(CH 4 ) + (1 + X ) B (O 2 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As noted in Example 15-10, F=XB.
Balances on each element give:
C: A = E
H: 4 A = 2 D
N2: (1 + X ) B (3.76) = C
O: (1 + X ) B (2) = D + 2 E + 2 F
Arbitrarily setting A=1, and solving the balance equations:
E = 1, D = 2, B = 2, C = 7.52(1 + X ), F = 2 X
Applying conservation of energy, with the assumptions used to develop Eq. 15-60, expanding each term, and
letting “R” represent reactants and “P” products:
nCH 4 (∆h f° + hp − href )CH 4 + nO2 , R (∆h f° + hp − href )O2 , R + nN2 , R (∆h f° + hp − href ) N2 , R
= nN2 , P (∆h f° + hp − href ) N2 , P − nH 2 O , P (∆h f° + hp − href ) H 2 O , P
+ nCO2 , P (∆h f° + hp − href )CO2 , P + nO2 , P (∆h f° + hp − href )O2 , P
Evaluating properties from Appendices A-18 and A-19
(1)(−32210 + h537 − h537 ) + (1 + X )2(0 + 6042 − 3725) + (1 + X )(2)(3.76)(0 + 5986 − 3730)
= (1 + X )(7.52)(0 + 15013 − 3730) + (2)(−104040 + 18054 − 4258) + (1)(−169300 + 21818 − 4028)
+(2 X )(0 + 15672 − 3725)
Solving for X
−10611 + X (4634 + 16965) = −247150 + X (84848 + 23894)
X = 2.71 or 271% excess air Answer

15-84
15-77 Gaseous propane (C3H8) at 25 ºC is burned in an adiabatic chamber with 110% theoretical air at 25 ºC to
ensure complete combustion. Determine the exit temperature of the products of combustion (in ºC).

Approach:
First the chemical reaction equation is balance, and then
conservation of energy is applied assuming no heat transfer.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion products only
CO2, H2O, O2, and N2.
5. The system is steady and adiabatic with no work or potential or
kinetic energy effects.

Solution:
The chemical balance equation assuming air is composed of oxygen and nitrogen in their usual proportions for
110% theoretical air is:
A(C3 H8 ) + (1.1) B (O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As asked in Example 15-10, F=B×(excess air)=0.1B
Balance, on each element gives:
C: 3A = E
H: 8 A = 2 D
N2: 1.1B(3.76) = C
O: 1.1B(2) = D + 2 E + 2 F
Arbitrarily setting A=1 and solving the balance equations:
E = 3, D = 4, B = 5, C = 20.68, F = 0.5
C3 H8 + 5.5(O 2 + 3.76N 2 ) ⇒ 20.68(N 2 ) + 4(H 2 O) + 3(CO 2 ) + 0.5(O 2 )
Applying conservation of energy, with the assumptions used to develop Eq. 15-60, expanding each term, and
letting “R” represent reactants and “P” products:
nC3 H8 (∆h fo + hR − href )C3 H8 + nO2 , R (∆h fo + hR − href )O2 , R + nN2 (∆h fo + hR − href ) N2 , R
= nN2,P (∆h fo + hR − href ) N2 , P + nH 2 O (∆h fo + hR − href ) H 2 O + nCO2 (∆h fo + hR − href )CO2 + nO2 , P (∆h fo + hR − href )O2 , P
Evaluating the properties using Appendices A-18 and A-19

(1)(−103850 + h298 − h298 )C3 H8 + (5.5)(0 + h298 − h298 )O2 + (20.68)(0 + h298 − h298 ) N2
= (20.68)(0 + hP − 8669) N2 + (4)(−241820 + hP − 9904) H 2 O + (3)(−393520 + hP − 9364)CO2 + (0.5)(0 + hP − 8682)O2
2, 295,300 = 20.68hP , N2 + 4hP , H 2O + 3hP ,CO2 + 0.5hP ,O2
Using the data from Appendices A-18 and A-19, an iterative solution is obtained. First, assume all products are
nitrogen to obtain an initial estimate of the temperature:
2, 295,300 ≈ (20.68 + 4 + 3 + 0.5)hP , N2
hP , N2 ≈ 81, 450 kJ kmol
From the nitrogen table TP ≈ 2450K
Guess hN2 ( kJ kmol ) hH 2O ( kJ kmol ) hCO2 ( kJ kmol ) hO2 ( kJ kmol ) RHS of
T(K) equation
2300 75676 98199 199035 79316 2,354,500
2250 73856 95562 115984 77397 2,296,240
By extrapolation TP 2249K Answer

15-85
15-78 In a simple Brayton cycle power plant, air enters the adiabatic compressor (ηC = 83%) at 100 kPa, 22 °C
and leaves at 450 kPa. The high-pressure air enters the combustion chamber where it burns with methane
(CH4) that enters at 450 kPa, 25 °C with a molar flow rate of 0.15 kmol/s. The products of combustion at
450 kPa, 727 °C enter the adiabatic turbine (ηT = 90%) and expand to 100 kPa. Determine:
a. the percent theoretical air
b. the net power output (in kW).

Approach:
We are given the temperature at location 3. To find
the excess air, first, the chemical reaction equation is
balanced in terms of the unknown excess air. Then,
conservation of energy is applied to the reaction with
no heat transfer. The air temperature entering the
combustion/leaving the compressor is required for
that energy balance, and analyzing the compressor
will give the outlet temperature. Finally,
conservation of energy is applied to the turbine to
find the work output.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for
each mole of O2.
2. The products of combustion behave as an ideal
gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, O2, and N2.

Solution:
a) The chemical balance equation is written assuming air is composed of oxygen and nitrogen in their usual
proportions. Excess air is taken into account. Let X= excess air.
A(CH 4 ) + (1 + X ) B (O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As noted in Example 15-10, F = BX (excess air) = XB
Balances on each element give:
C: A = E
H: 4 A = 2 D
N2: (1 + X ) B (3.76) = C
O: (1 + X ) B (2) = D + 2 E + 2 F
Arbitrarily setting A=1, and solving the balance equations:
E = 1, D = 2, B = 2, C = 7.52(1 + X ), F = 2 X
CH 4 + (1 + X )2(O 2 + 3.76N 2 ) ⇒ 7.52(1 + X )(N 2 ) + 2(H 2 O) + CO 2 + 2 X (O 2 )
Applying conservation of energy to the reaction, with the assumptions used to develop Eq. 15-60, expanding each
term, and letting “R” represent reactants and “P” products:
nCH 4 (∆h f° + hp − href )CH 4 + nO2 , R (∆h f° + hp − href )O2 , R + nN2 , R (∆h f° + hp − href ) N2 , R
= nN2 , P (∆h f° + hp − href ) N2 , P − nH 2 O , P (∆h f° + hp − href ) H 2O , P
+ nCO2 , P (∆h f° + hp − href )CO2 , P + nO2 , P (∆h f° + hp − href )O2 , P
To solve the equation, we need the temperature at 2, T2. The definition of isentropic efficiency is used:
h −h h −h
nC = 2 s 1 → h2 = h1 + 2 s 1
h2 − h1 nC
Once the outlet temperature, T2s , is determined, the molar enthalpies of the oxygen and nitrogen can be obtained.
Using the isentropic relations for an ideal gas and properties from Appendix A-9.
P2 Pr 2 P 450
= → Pr 2 = 2 Pr1 = ( )(1.3068) = 5.881
P1 Pr1 P1 100

15-86
By interpolation h2 s = 454.1kJ kg
at 295K h1 = 295.17 kJ kg
(454.1 − 295.17) kJ
Therefore, h2 = 295.17 + = 486.6 → T2 = 484K
0.83 kg
Now evaluating the properties in the energy equation:
(1)(−74850) + (1 + X )2(0 + 14275 − 8662) + (1 + X )2(3.76)(0 + 14107 − 8669)
= 7.52(1 + X )(0 + 30129 − 8669) + (2)(−241820 + 35882 − 9904) + (1)(−393520 + 42769 − 9364)
+ 2 X (0 + 31389 − 8682)
−22770 + X (11186 + 40894) = −630420 + X (161379 + 45414)
X = 3.93 = 393% excess air Answer
CH 4 + 9.86(O 2 + 3.76N 2 ) ⇒ 37.07(N 2 ) + 2(H 2 O) + CO 2 + 7.86(O 2 )
b) The net power output is:
Wnet = WT − WC
Applying conservation of energy to the compressor and ignoring heat transfer and potential and kinetic energy
effects:
WC = ma (h2 − h1 ) = na M a (h2 − h1 )
Using the air fuel ratio:
AFR = na nF → na = AFR nF
(1 + 3.93)(2)(1 + 3.76) kmol air
AFR = = 46.93
1 kmol fuel
W kg kJ kJ
= AFR M a (h2 − h1 ) = (46.93)(28.97 )(486.6 − 295.17) = 260280
nF kmol kg kmol fuel
For the turbine, we need to consider isentropic expansion of an ideal gas mixture to determine the outlet
temperature: 0 = S 3 − S 4 s
P P P P
0 = 37.07[ s3° − s4° − R ln( 3 )]N2 + 2[ s3° − s4° − R ln( 3 )]H 2O + [ s3° − s4° − R ln( 3 )]CO2 + [ s3° − s4° − R ln( 3 )]O2
P4 P4 P4 P4
0 = 37.07[228.057 − s4,° N2 − 8.314 ln(4.5)] + 2[232.197 − s4,° H 2O − 8.314 ln(4.5)]
+[269.215 − s4,° CO2 − 8.314 ln(4.5)] + 7.86[243.471 − s4,° O2 − 8.314 ln(4.5)]
0 = 10503 − 37.07 s4,° N2 − 2 s4,° H 2O − s4,° CO2 − 7.86 s4,° O2
Solving this iteratively, T4 s ≈ 680K
Because nT = W Ws , using the energy equation isentropic turbine:
Ws
= 37.07( h3 − h4 ) N2 + 2( h4 − h3 ) H 2O + (h3 − h4 )CO2 + 7.86(h3 − h4 )O2
nF
= 37.07(30129 − 19991) + 2(35882 − 23342) + (42769 − 26138) + 7.86(31389 − 20524)
Ws kJ
= 502930
nF kmol fuel
W kJ
= (0.90)(502930) = 452633
nF kmol fuel
WT Wc kmol kJ
Wnet = nF ( − ) = (0.15 )[452633 − 260280]
nF nF s kmol
= 28,850 kW Answer

15-87
15-79 Methane (CH4) is burned in an insulated vessel with 250% theoretical air. Both gases initially are at 101.3
kPa, 25 °C. Determine the temperature of the products of combustion if the reaction occurs at constant
volume (in °C).

Approach:
The closed system conservation of energy equation is used to
analyze this system. We will adapt the open system equations we
have developed. The chemical reaction equation must be balanced
and then conservation of energy equation is used.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion products only
CO2, H2O, O2, and N2.
5. The system is adiabatic with no work or potential or kinetic
energy effects.

Solution:
The chemical balance equation, assuming air is composed of oxygen and nitrogen in their usual proportions and
taking into account the excess air: A(CH 4 ) + (2.5) B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As noted in Example 15-10, F = BX (excess air) = 1.5B
Balances on the elements give:
C: A = E
H: 4 A = 2 D
N2: 2.5B(3.76) = C
O: 2.5B(2) = D + 2 E + 2 F
Arbitrarily setting A=1, and solving the balance equations: E = 1, D = 2, B = 2, C = 18.80, F = 3.0
CH 4 + 5(O 2 + 3.76N 2 ) ⇒ 18.80(N 2 ) + 2(H 2 O) + CO 2 + 3(O 2 )
Applying the closed system energy equation, and assuming adiabatic, no work, no potential or kinetic energy, and
each gas and the mixture behave as ideal gases: ∆U = 0 . Also note that H = U + PV = U + nT RT , where
nT = total moles. Substituting into the energy equation: ∆H − nT R ∆T = 0
Equation 15-60 is used to describe ∆H , so expanding terms, and letting “R” represent reactants and “P” products:
nCH 4 (∆h f° + hp − href )CH 4 + nO2 , R (∆h f° + hp − href )O2 , R + nN2 , R (∆h f° + hp − href ) N2 , R − nT RTR = nN2 , P (∆h f° + hp − hhref ) N2 , P
+ nH 2O , P (∆h f° + hp − href ) H 2O , P + nCO2 , P (∆h f° + hp − href )CO2 , P + nO2 , P (∆h f° + hp − href )O2 , P − nT RTP
Using Appendices A-18 and A-19 and noting that because reactants are at 25oC, hR − href = 0 for each gas.
nT = 18.80 + 2 + 1 + 3 = 24.80 kmol
(1)(−74850) − (24.80)(8.314)(298)
= (18.80)(0 + hP − 8669) N2 + (2)(−241820 + hP − 9904) H 2 O + (1)(−393520 + hP − 9364)CO2
+ (3)(0 + hP − 8682)O2 − (24.80)(8.314)(TP )
959061 = 18.8hP , N2 + 2hP , H 2O + hP ,CO2 + 3hP ,O2 − 206.2TP
An iterative solution is required. For a first estimate, assume all products are nitrogen and TP ≈ 1620 K . Solving
for the nitrogen enthalpy, hN2 , P ≈ 51,144 kJ kmol which leads to TP = 1620 K
Guess TP hN2 hH 2O hCO2 hO2 RHS of equation
1620 51275 63709 78123 53696 996555
1600 50571 62748 76944 52961 982138
1560 49168 60838 74590 51490 953422
By interpolation TP 1568K Answer
15-88
15-80 Methane (CH4) is burned in an insulated vessel with 250% theoretical air. Both gases initially are at 101.3
kPa, 25 °C. Determine the temperature of the products of combustion if the reaction occurs at constant
pressure in a piston-cylinder assembly (in °C).

Approach:
The closed system conservation of energy equation is
used to analyze this system. The chemical reaction
equation must be balanced first, and then conservation of
energy is used.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, O2, and N2.
5. The system is adiabatic with negligible potential and
kinetic energy effects.

Solution:
The chemical balance equation, assuming air is composed of oxygen and nitrogen in their usual proportions and
taking into account the excess air:
A(CH 4 ) + (2.5) B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As noted in Example 15-10, F = B × (excess air) = 1.5B
Balances on each element give:
C: A = E
H: 4 A = 2 D
N2: 2.5B(3.76) = C
O: 2.5B(2) = D + 2 E + 2 F
Arbitrarily setting A=1, and solving the balance equations: E = 1, D = 2, B = 2, C = 18.80, F = 3
CH 4 + 5(O 2 + 3.76N 2 ) ⇒ 18.80(N 2 ) + 2(H 2 O) + CO 2 + 3(O 2 )
Applying the closed system energy equation, and assuming adiabatic, no potential or kinetic energy effects, and
each gas and the mixtures behave as ideal gases: ∆U = W
Assuming quasi-equilibrium, W = ∫ PdV = P∆V . Substituting this into the energy equations: ∆U = P∆V
Note that H = U + PV . Therefore, ∆H = 0
We can use Eq. 15-60. Expanding terms and letting “R” represent reactants and “P” products:
nCH 4 (∆h f° + hp − href )CH 4 + nO2 , R (∆h f° + hp − href )O2 , R + nN2 , R (∆h f° + hp − href ) N2 , R
nN2 , R (∆h f° + hp − href ) N2 , R + nH 2O (∆h f° + hp − href ) H 2 O + nCO2 (∆h f° + hp − href )CO2 + nO2 , P (∆h f° + hp − href )O2 .P
Using Appendices A-18 and A-19 and noting that because reactants are at 25oC, hR − href = 0 for each gas.
(1)(−74850) = (18.80)(0 + hP − 8669) N2 , P + (2)(−241820 + hP − 9904) H 2O , P
+ (1)(−393529 + hP − 9364)CO2 , P + (3)(0 + hP − 8682)O2 , P
1, 020,505 = 18.8hN2 , P + 2hH 2O , P + hCO2 , P + 3hO2 , P
An iterative solution is required. For a first estimate, assume all products are nitrogen.
1, 020,505 ≈ (18.8 + 2 + 1 + 3)hN2 , P → hN2 , P ≈ 41149 kJ kmol → TP ≈ 1340 K
Guess TP(K) hN2 hH 2O hCO2 hO2 RHS of equation
1340 41539 50162 61813 43475 1074400
1300 40170 48807 59522 42033 1038430
1280 39488 47912 58381 41312 1020515
TP ≈ 1280 K Answer

15-89
15-81 Consider a simple Rankine cycle power plant. Steam enters the turbine at 600 lbf/in.2, 1000 °F and
expands to 2 lbf/in.2, 96% quality. Saturated liquid leaves the condenser, and the temperature rise of the
cooling water is 15 °F. In the boiler, methane (CH4) is burned with 200% theoretical air; both gases enter
at 14.7 lbf/in.2, 77 °F, and the products of combustion leave at 460 °F. Determine:
a. the balanced reaction equation
b. the vapor mass flow rate (in lbm/lbm fuel)
c. the cooling water mass flow rate (in lbm/lbm fuel).

Approach:
Balancing the chemical reaction equation is straight
forward, as was done in examples. For the vapor flow
rate per lbm fuel, we recognize that the heat transfer from
the combustion must equal the heat transfer to the
vaporizing water; conservation of energy is applied to the
boiler to determine the vapor flow rate. Likewise,
conservation of energy applied to the condenser is used to
determine the cooling water flow rate.

Assumptions:
1. Air is composed only of 3.76 moles of N2 for each
mole of O2.
2. The products of combustion behave as an ideal gas.
3. N2 is inert.
4. Complete combustion occurs with combustion
products only CO2, H2O, O2, and N2.
5. The combustion and boiler are steady with no work
and negligible potential and kinetic energy effects.
6. The condenser is steady and adiabatic with negligible
potential and kinetic energy effects.

Solution:
a) For the chemical balance equation, assume air is composed of oxygen and nitrogen in its usual proportion, and
takes into account excess air:
A(CH 4 ) + (2.0) B(O 2 + 3.76N 2 ) ⇒ C (N 2 ) + D(H 2 O) + E (CO 2 ) + F (O 2 )
As noted in Example 15-10, F = BX (excess air) = 1.0 B
Balances on each element:
C: A = E
H: 4 A = 2 D
N2: (2) B(3.76) = C
O: (2) B (2) = D + 2 E + 2 F
Arbitrarily setting A=1 and solving the balance equations:
E = 1, D = 2, B = 2, C = 15.04, F = 2
CH 4 + 4(O 2 + 3.76N 2 ) ⇒ 15.04(N 2 ) + 2(H 2 O) + CO 2 + 2O 2 Answer
b) Using conservation of energy on the reaction and assuming steady, no work, negligible potential and kinetic
energy effects, and the gases and mixtures behave as ideal gas mixtures, then Eq. 15-60 is applicable. Grouping
inflow and outflow terms, and expanding the terms with “P” representing products and “R” reactants.
Q
= nN2 , P (∆h f° + hP − href ) N2 , P + nH 2 O , P (∆h f° + hP − href ) H 2 O , P + nCO 2 (∆h f° + hP − href )CO2
nF
+ nO2 , P (∆h f° + hP − href )O2 , P − nCH 4 (∆h f° + hP − href )CH 4 − nO2 , R (∆h f° + hP − href )O2 , R − nN2 , R (∆h f° + hP − href ) N2 , R
Using data from Appendices B-18 and B-19 and the stoichiometric coefficients from the chemical balance
equation:

15-90
 Q
= (15.04)(0 + 6409.6 − 3729.5) + (2)( −104040 + 7399.4 − 4258.0) + (1)( −169300 + 7811.4 − 4027.5)
nF
+ (2)(0 + 6486.7 − 3725.1) − (1)(−32210) − (4)(0 + href − href ) − (15.04)(0 + href − href )
Btu
= −289, 271
lbmol fuel
Conservation of mass and energy on the water flowing through the boiler using the same assumptions as above
gives:
Q = mv (h3 − h2 )
Note that the sign on this heat transfer rate is opposite of that for the combustion process.
Btu
Using the steam tables Appendix B-12 h3 ( P3 , T3 ) = 1517.8
lbm
For h2 , apply conservation of energy to the pump, assume isentropic, incompressible, no kinetic or potential
energy effects:
h2 ≈ h1 + v∆P and h1 ≈ h f ( P1 )
ft 3 lbf 144in.2 Btu Btu
h2 ≈ 94.02 + (0.01623 )(600 − 2) 2 ( )( ) = 95.8
lbm in. ft 2 778ft lbf lbm
Q Btu
= 1517.8 − 95.8 = 1422.0
mv lbm
Taking into account the signs,
Q mF 289271 lbm vapor
= = 203.4
Q mv 1422 lbmol fuel
mv 203.4 lbm vapor lbm lbm vapor
= = 12.7 Answer
mF lbmol fuel 16.04blm lbm fuel
c) Applying conservation of mass and energy to the complete condenser, and assuming steady, adiabatic, no
work, no potential or kinetic energy effects, constant liquid specific heat:
mv (h4 − h1 ) = mcw ∆hcw = mcw cρ ∆T
mcw h4 − h1
=
mv c ρ ∆T
Btu
h4 = h f + xh fg = 94.02 + (0.96)(1022.1) = 1075.2
lbm
Btu
(1075.2 − 94.02)
mcw lbm = 65.4 lbm cw
=
mv Btu lbm vapor
(1 )(15°F)
lbm°F
mcw m m lbm cooling water
= ( V )( cw ) = (12.7)(65.4) = 831 Answer
mF mF mv lbm fuel

15-91