ENGRD 221 – Prof. N. Zabaras
10/05/07
SOLUTIONS TO HOMEWORK 6
Problem 1
Known: The schematic and steady state operating data are provided for a vapor power
plant.
Find:
(a) Sketch the cycle on T-s coordinates.
(b) Determine the thermal efficiency and compare with the thermal efficiency of a Carnot
cycle operating between the same maximum and minimum temperatures.
Schematic and Given Data:
Assumptions:
1. The overall system shown in the figure and thus each component is at steady
state.
2. The only significant heat transfers occur with the two reservoirs.
3. Kinetic and potential energy effects can be ignored.
4. The maximum and minimum temperatures correspond respectively, to the
saturation temperatures at 10 bars and 0.2 bar.
Analysis:
(b) The thermal efficiency is
η=
•
•
Wnet
QC
•
QH
= 1−
•
QH
The net transfer rates are found by reducing mass and energy balances for the boiler and
condenser.
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ENGRD 221 – Prof. N. Zabaras
•
•
•
•
10/05/07
QH = m(h1 − h4 )
QC = m(h2 − h3 )
Thus,
⎛ h2 − h3 ⎞
⎟⎟
⎝ h1 − h4 ⎠
η = 1 − ⎜⎜
From Table A-3, h1-h4 = 2015.3 kJ/kg.
Also, h2 = hf + x2 (hg-hf) and h3 = hf + x3 (hg-hf), therefore
h2 – h3 = (x2 – x3) (hg - hf)
= (0.88 – 0.18)(2358.3) = 1650.8kJ/kg
Substituting it back to the efficiency equation,
⎛ 1650.8 ⎞
⎟ = 0.181
⎝ 2015.3 ⎠
η = 1− ⎜
For a Carnot Cycle,
η max = 1 −
TC
⎛ 60.06 + 273.15 ⎞
= 1− ⎜
⎟ = 0.265
TH
⎝ 179.9 + 273.15 ⎠
Problem 2
Known: A inventor claims that the electricity-generating unit shown in the Fig. receives a
heat transfer at the rate of 250 Btu/s at a temperature of 500°R, a second heat transfer at
the rate of 350 Btu/s at 700°R, and a third at the rate of 500 Btu/s at 1000°R.
Find: For operation at steady state, evaluate this claim.
Schematic and Given Data:
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ENGRD 221 – Prof. N. Zabaras
10/05/07
Assumptions:
1. The system is in steady state.
2. Heat transfer rates are each positive in the direction of the accompanying arrow.
Analysis:
Applying Eq. 6.28 (steady state)
•
•
•
dS Q1 Q2 Q3 •
=
+
+
+σ = 0
dt T1 T2 T3
From above equation we get
•
•
⎛ •
⎞
⎜ Q1 Q2 Q3 ⎟
+
σ = −⎜ +
⎜ T1 T2 T3 ⎟⎟
⎝
⎠
•
Substituting all the known values, we get
•
Btu
⎛ 250 Btu / s 350 Btu / s 500 Btu / s ⎞
+
+
⎟ = −1.5 D
D
D
D
700 R
1000 R ⎠
s R
⎝ 500 R
σ = −⎜
Since the entropy production has to be positive, the claim cannot be valid.
Problem 3
Known: An isolated system of total mass m is formed by mixing two equal masses of the
same liquid initially at the temperature T1 and T2.
Find:
(a) Show that the amount of entropy produced is
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ENGRD 221 – Prof. N. Zabaras
10/05/07
⎡ T1 + T2 ⎤
1/ 2 ⎥
⎣ 2(T1T2 ) ⎦
σ = mc ln ⎢
(b) Demonstrate that σ must be positive.
Schematic and Given Data:
Analysis:
(a) The final temperature Tf can be evaluated from an energy balance:
ΔU = Q − W = 0 − 0 = 0
m
⎡m
⎤
⇒ mu (T f ) − ⎢ u (T1 ) − u (T2 )⎥ = 0
2
⎣2
⎦
m
m
⇒ u (T f ) − u (T1 ) + u (T f ) − u (T2 ) = 0
2
2
[
]
[
]
Since each mass is incompressible with constant specific heat c, Δu = cΔT
[
]
[
]
m
m
c T f − T1 + c T f − T2 = 0
2
2
T + T2
⇒ Tf = 1
2
An entropy balance gives
⎛ δQ ⎞
ΔS = ∫ ⎜
⎟ + σ = 0 + σ ⇒ ΔS = σ
T ⎠b
1⎝
2
This gives
σ = ms f − ⎢ s1 + s 2 ⎥ = [(s f − s1 ) + (s f − s 2 )]
2 ⎦ 2
⎣2
⎡m
m
⎤
m
Using Eq. 6.13,
σ=
Tf
m ⎡ Tf
+ ln
c ⎢ln
2 ⎣ T1
T2
⎡ Tf ⎤
⎡ Tf 2 ⎤
⎤ m
=
c
ln
⎥
⎢
⎥ = mc ln ⎢
⎥
1
⎢⎣ (T1T2 ) 2 ⎥⎦
⎢⎣ T1T2 ⎥⎦
⎦ 2
⎡ T +T ⎤
2
⇒ σ = mc ln ⎢ 1
⎥
1
⎢⎣ 2(T1T2 ) 2 ⎥⎦
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ENGRD 221 – Prof. N. Zabaras
(b) σ ≥ 0 when
T1 + T2
2(T1T2 )
1
2
10/05/07
⎡ T +T ⎤
2
ln ⎢ 1
⎥≥0
1
⎢⎣ 2(T1T2 ) 2 ⎥⎦
≥ 1 ⇒ T1 + T2 ≥ 2(T1T2 )
1
2
Squaring both sides,
(T1 + T2 )2 ≥ 4(T1T2 )
T1 + 2T1T2 + T2 ≥ 4(T1T2 )
2
2
2
2
T1 − 2T1T2 + T2 ≥ 0
⇒ (T1 − T2 ) ≥ 0
2
The inequality is satisfied for either T1 > T2 or T2 > T1. The equality applies only when T1
= T2.
Problem 4
Known: An insulated mixing chamber at steady state receives two liquid streams of the
same substance at temperatures T1 and T2 and mass flow rates of m 1 and m 2 ,
respectively. A single stream exits at T3 and m 3 .
Find:
(a) T3 in terms of T1, T2, and the ratio of mass flow rates m 1 / m 3 .
(b) the rate of entropy production per unit of mass exiting the chamber in terms of c,
T1/T2, and m 1 / m 3 .
(c) For fixed values of c and T1/T2, determine the value of m 1 / m 3 for which the rate of
entropy production is a maximum.
Schematic and Given Data:
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ENGRD 221 – Prof. N. Zabaras
10/05/07
Assumptions:
1. The control volume shown in the figure is at steady state.
2. For the CV, Qcv = Wcv = 0, and all kinetic and potential energy effects are
negligible.
3. The liquid streams can be modeled as incompressible with constant specific heat c
and negligible effects of pressure.
Analysis:
For parts a and b, refer to HW5 Problem 5 solution.
(a) T3 = T2 + ( m 1 / m 3 )(T1 – T2)
(b)
⎡ m ⎛ T ⎞
⎛ m ⎛ T ⎞ m ⎞⎤
= c ⎢ 1 ln⎜⎜ 2 ⎟⎟ + ln⎜⎜1 + 1 ⎜⎜ 1 ⎟⎟ − 1 ⎟⎟⎥
m 3
⎝ m 3 ⎝ T2 ⎠ m 3 ⎠⎦
⎣ m 3 ⎝ T1 ⎠
σ
•
(c) Letting x =
m1
•
m3
Taking the derivative of
⎛ • ⎞
⎜
d σ • ⎟
⎜ m ⎟
⎝
3⎠
dx
•
σ , with respect to x while holding c and T1/T2 constant,
•
⎧
⎡⎛ T1 ⎞ ⎤ ⎫
⎪
⎢⎜ T ⎟ − 1⎥ ⎪
⎛ T1 ⎞
⎪
⎝ 2⎠ ⎦ ⎪
⎣
= c ⎨− ln ⎜ ⎟ +
⎬
T
⎡
⎤
⎛
⎞
T
⎝
⎠
2
1
⎪
1 + x ⎢⎜ ⎟ − 1⎥ ⎪
⎪⎩
⎣⎝ T2 ⎠ ⎦ ⎪⎭
m3
C ,T1 / T2
Setting this to zero for maximum point,
⎛T
− ln⎜⎜ 1
⎝ T2
⎡⎛ T
⎞⎡
⎟⎟ ⎢1 + x ⎢⎜⎜ 1
⎠ ⎢⎣
⎣⎝ T2
⎞ ⎤ ⎤ ⎡⎛ T1
⎟⎟ + 1⎥ ⎥ + ⎢⎜⎜
⎠ ⎦ ⎥⎦ ⎣⎝ T2
⎞ ⎤
⎟⎟ − 1⎥ = 0
⎠ ⎦
Solving
⎛ T ⎞ ⎡ ⎛ T ⎞⎤
ln⎜⎜ 1 ⎟⎟ + ⎢1 − ⎜⎜ 1 ⎟⎟⎥
⎝ T2 ⎠ ⎣ ⎝ T2 ⎠⎦
⇒x=
⎛ T ⎞ ⎡ ⎛ T ⎞⎤
ln⎜⎜ 1 ⎟⎟ ⎢1 − ⎜⎜ 1 ⎟⎟⎥
⎝ T2 ⎠ ⎣ ⎝ T2 ⎠⎦
1. In applying equation 3.20b, the term in the curly brackets below has been omitted
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ENGRD 221 – Prof. N. Zabaras
10/05/07
h2 − h1 = c(T2 − T1 ) + {v(P2 − P1 )}
Since the specific volumes of liquids are small, a substantial pressure difference (P1-P2)
would be required before the underlined term becomes significant. Effects of pressure are
neglected in assumption 3. Effects of pressure are neglected in assumption 3.
•
2. When
m1
= 0 , Eq. a gives T3 = T2, then
•
m3
⎡ • ⎛ ⎞
⎢ m1 ln⎜ T2 ⎟ + ln⎛⎜ T3
c
=
•
⎜T
⎢ • ⎜⎝ T1 ⎟⎠
⎝ 2
m3
⎣ m3
•
σ
⎞⎤⎥
⎟⎟ = 0
⎠⎥⎦
•
And when
m1
•
=1, Eq a gives T3 = T1, then
m3
•
⎡ ⎛T
= c ⎢ln⎜⎜ 2
⎣ ⎝ T1
m3
σ
•
⎛T
⎞
⎟⎟ + ln⎜⎜ 1
⎝ T2
⎠
⎞⎤
⎟⎟⎥ = 0
⎠⎦
In these cases there is no mixing, and thus no entropy production.
3. To verify that this locates a maximum, form the second derivative of the equation and
evaluate the resulting expression with the expression for the x here to show that this
second derivative is negative.
Problem 5
Known:
The Figure shows a simple vapor plant operating at steady state with water as the
working fluid. Data at key locations are given on the figure. The mass flow rate of the
water circulating through the components is 109kg/s.
Find:
(a) the mass flow rate of the cooling water, in kg/s.
(b) the thermal efficiency.
(c) the rates of entropy production, each in kW/K, for the turbine, condenser, and pump.
(d) Using the results of part (c), place the components in rank order, beginning with
component contributing most to inefficient operation of the overall system.
Schematic and Given Data:
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ENGRD 221 – Prof. N. Zabaras
10/05/07
Assumptions:
1. For each of four principal components, a control volume at steady state encloses the
component.
2. For the turbine, pump and condenser, Qcv = 0
3. Stray heat transfer and kinetic and potential energy effects can be ignored.
4. For liquid water, h ~ hf(T), s ~ sf(T).
5. Energy transfers are positive in the directions of the arrows.
Analysis:
State
h (kJ/kg)
s (kJ/kg.K)
Table
1
2
3
4
5
6
3425.1
2336.7
173.9
188.9
83.96
146.68
6.6622
7.4651
0.5926
0.6061
0.2966
0.5053
A-4
A-3
A-3
A-5
A-2
A-2
(a) Mass and energy rate balances for the condenser reduce to give
•
• ⎡h − h ⎤
⎡ 2336.7 − 173.9 ⎤
3
mcw = m ⎢ 2
⎥ = 109 ⎢
⎥ = 3758.7 kg / s
⎣ 146.68 − 83.96 ⎦
⎣ h6 − h5 ⎦
(b) The thermal efficiency is given by
Page 8 of 14
ENGRD 221 – Prof. N. Zabaras
•
η=
10/05/07
•
Wt − W p
•
Qin
The rate of work done by the turbine and pump, and heat transfer are given by
•
•
Wt = m(h1 − h2 ) = (109kg / s )(3425.1 − 2326.7 )kJ / kg = 118636kJ / s
•
•
•
•
W p = m(h4 − h3 ) = (109kg / s )(188.9 − 173.9)kJ / kg = 1635kJ / s
Qin = m(h1 − h4 ) = (109kg / s )(3425.1 − 188.9)kJ / kg = 352746kJ / s
Substituting these values into the efficiency equation,
η=
117001
= 0.332
352746
(c) Applying mass and energy balances,
•
•
σ t = m ( s2 − s1 ) = (109kg / s )( 7.4651 − 6.6632 )
•
•
•
•
kJ 1kW
= 87.52kW / K
kg ⋅ K 1kJ / s
σ p = m ( s4 − s3 ) = (109kg / s )( 0.6061 − 0.5926 ) = 1.47 kW / K
•
σ c = m ( s3 − s2 ) + mcw ( s6 − s5 )
= (109kg / s )( 0.5926 − 7.4651) + ( 3758.7 kg / s )( 0.5053 − 0.2966 ) = 35.34kW / K
From above calculations, order is {Turbine, condenser, pump}.
Problem 6
Known: The Figure provides the schematic of a heat pump using Refrigerant 134a as the
working fluid, together with steady-state data at key points. The mass flow rate of the
refrigerant is 7kg/min, and the power input to the compressor is 5.17kW.
Find:
(a) Determine the coefficient of performance for the heat pump.
(b) If the valve were replaced by a turbine, power could be produced, reducing thereby
the power requirement of the heat pump system. Would you recommend this powersaving measure? Explain.
Page 9 of 14
ENGRD 221 – Prof. N. Zabaras
10/05/07
Schematic and Data:
Assumptions:
1. The heat pump system shown in the figure operates at steady state.
2. Stray heat transfer and kinetic and potential energy effects can be ignored.
3. W and Q are positive in the directions of the arrows.
Analysis:
(a) Using Eq. 2.47 expressed in a rate basis,
•
γ =
Qout
•
Wc
•
•
An energy rate balance for the condenser gives, Qout = m(h2 − h3 )
7
•
γ=
m(h2 − h3 )
•
=
kg 1 min
(293.21 − 99.56) kJ
kg
min 60s
= 4.37
5.17 kJ / s
Wc
(b) To assess the merit of a power recovery turbine, consider an ideal turbine: one whose
isentropic turbine efficiency is 100%. Thus,
•
•
Wt = m ( h3 − h4 s ) , h3 = 99.56kJ / kg , s3 = 0.3656kJ / kg ⋅ K
Page 10 of 14
ENGRD 221 – Prof. N. Zabaras
10/05/07
Then since s4s = s3,
x4 s =
s4s − s f
sg − s f
⎡ 0.3656 − 00.1710 ⎤
= 0.259
= 1⎢
⎣ 0.9222 − 00.1710 ⎥⎦
⇒ h4 s = h f + x 4 s h fg = 42.95 + 0.259(201.14 ) = 95.05kJ / kg
Then
•
Wt = 7
kg 1min
(99.56 − 95.05) kJ = 0.53kW
min 60s
kg
Discussion
This is the maximum theoretical power that any power-recovery turbine would be able to
develop. At best, the turbine could offset about 10% of the compressor power
requirement. In most heat pump applications, such a turbine is not implemented due to
the turbine cost and operating difficulties related to the low-quality refrigerant that would
be expanding through such a turbine.
Problem 7
Known: As shown in the Fig., a steam turbine having an isentropic turbine efficiency of
90% drives an air compressor having an isentropic compressor efficiency of 85%.
Steady-state operating data are provided on the figure.
Find:
(a) Determine the mass flow rate of the steam entering the turbine, in kg of steam per kg
of air exiting the compressor.
(b) Repeat part (a) if ηt = ηc = 100%
Schematic and Data:
Assumptions:
1. Assume the ideal gas model for air.
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ENGRD 221 – Prof. N. Zabaras
10/05/07
2. Ignore stray heat transfer and kinetic and potential energy effects.
3. For control volume, Wcv = 0.
Analysis:
•
•
•
•
Mass balances give m1 = m2 and m3 = m4 . An energy balance gives
•
•
•
•
0 = Qcv − Wcv + m1 (h1 − h2 ) + m4 (h3 − h4 ) ⇒
•
•
•
• ⎡h − h ⎤
3
0 = 0 − 0 + m1 (h1 − h2 ) + m4 (h3 − h4 ) ⇒ m1 = m4 ⎢ 4
⎥
⎣ h1 − h2 ⎦
The respective isentropic efficiencies are, η t =
(1)
h − h3
h1 − h2
, η c = 4s
h1 − h2 s
h4 − h3
Substitute into equation 1,
•
m1
•
m4
=
1 ⎡ h4 s − h3 ⎤
η tη c ⎢⎣ h1 − h2 s ⎥⎦
(2)
Property Data: Table A-4, h1= 3335.5kJ/kg, s1= 7.2540kJ/kg.K
Equating s2s=s1, we can obtain the quality of state 2, with which we can obtain the h2s
value.
x2 s =
7.2540 − 1.3026
= 0.9826 ⇒ h2 s = 417.46 + 0.9826(2258) = 2636.2kJ / kg
7.3594 − 1.3066
The process 3Æ4s is an isentropic process. From Table A-22, we obtain h3 =
300.19kJ/kg, Pr3 = 1.3860, and for isentropic processes we know that Pr4 = Pr3 (P4/ P3) =
6.2496.
From this value of Pr4, we get h4s = 462.1kJ/kg.
(a) Given η t = 0.9, η c = 0.85
•
m1
•
m4
=
1
1
⎡ 462.1 − 300.19 ⎤
(0.232) = 0.303
=
(0.9)(0.85) ⎢⎣ 3335.5 − 2636.2 ⎥⎦ (0.9)(0.85)
Page 12 of 14
ENGRD 221 – Prof. N. Zabaras
10/05/07
(b) Given η t = 1, η c = 1
•
m1
•
= 0.232
m4
Note:
⎛ •
1 ⎜ m1
=
•
⎜ •
m4 η tη c ⎜⎝ m4
•
m1
⎞
⎟
⎟⎟
⎠ int rev
Important note: Note that the steam is the operating medium in the turbine and air (ideal
gas) is the operating medium in the compressor. When dealing with isentropic processes,
you need to pay CLOSE attention as to what is the operating medium. For example, for
the isentropic process 1Æ2s, you need to compute state 2s using the steam tables. For the
isentropic process, 3Æ4s, you need to compute 4s using the ideal gas equations (for
example Pr4 / Pr3 =P4/ P3) and the ideal gas tables (here air) that provide you for each T
the values of Pr(T).
Problem 8
Known: The Figure shows a power system operating at steady state consisting of three
components in series: an air compressor having an isentropic compressor efficiency of
80%, a heat exchanger, and a turbine having an isentropic turbine efficiency of 90%. Air
enters the compressor at 1 bar, 300K with a mass flow rate of 5.8kg/s and exits at a
pressure of 10 bar. Air enters turbine at 10 bar, 1400K and exits at a pressure of 1 bar. Air
can be modeled as an ideal gas. Stray heat transfer and kinetic and potential energy
effects are negligible.
Find: Determine, in kW, (a) the power required by the compressor, (b) the power
developed by the turbine, and (c) the net power output of the overall power system.
Schematic and Data:
Assumptions:
1. Control Volumes at steady state enclose the turbine and the compressor.
2. Assume the ideal gas model for air.
3. Ignore stray heat transfer and kinetic and potential energy effects.
Page 13 of 14
ENGRD 221 – Prof. N. Zabaras
10/05/07
Analysis:
Mass and energy rate balances reduce to give the rate of work of the turbine and
compressor and combining with their respective isentropic efficiencies, we obtain
•
•
Wt = m(h3 − h4 ), η t =
Turbine
h3 − h4
h3 − h4 s
h − h1
Compressor WC = m(h1 − h2 ), η c = 2 s
h2 − h1
•
•
•
•
⇒ Wt = mη t (h3 − h4 s )
•
⇒ Wc = −
(1)
•
m(h2 s − h1 )
ηc
(2)
From Table A-22, h1 = 300.19kJ/kg, h3 = 1515.4kJ/kg; to find h2s and h4s, use
⎡p ⎤
Pr (4s ) = Pr (3)⎢ 4 ⎥ = 450.5(1 / 10) = 45.05 ⇒ h4 s = 808.5kJ / kg
⎣ p3 ⎦
⎡p ⎤
Pr (2s ) = Pr (1)⎢ 2 ⎥ = 1.386(10 / 1) = 13.86 ⇒ h2 s = 579.9kJ / kg
⎣ p1 ⎦
Inserting values into Eqs.1 and 2,
•
•
Wt = mη t (h3 − h4 s ) = (5.8)(0.9 )(1515.4 − 808.5) = 3690kW
•
Wc = −
•
•
m(h2 s − h1 )
ηc
•
⎛ 579.9 − 300.19 ⎞
= −(5.8)(0.9 )⎜
⎟ = −2028kW
0.80
⎠
⎝
•
Wnet = Wt + Wc = 3690 + (− 2028) = 1662kW
Page 14 of 14
(a )
(b )
(c )