Sample Exam Papers and Solutions

Sample Exam Papers and Solutions
The midterm and final exam paper of the 2103241 (thermodynamics I) course is composed of two
parts as follows:
PART A This is an analytical part. Students must demonstrate the step-by-step procedure for analyzing
and solving thermodynamic problems such as choosing the system under consideration,
choosing a right thermodynamic model, drawing a diagram to help understanding the process,
writing appropriate thermodynamic laws, etc. In order to obtain a full credit, students must
show all details of calculation and provide sufficiently accurate results.
PART B This is a conceptual part. The exam question is focused on the basic concept and fundamentals
of thermodynamics. Students need not to show any of the calculations but can draw diagrams or
equations as a supplement. However, students must provide the theoretical supports to their
answers in order to obtain a full credit.
PART A: จงแสดงวิธีทําอยางละเอียด
1) A cylindrical tank with 0.5-m diameter and 0.8-m height contains CO2 at 5.9 MPa, 92oC. Estimate the
mass of CO2 in the tank. (10 points)
Given:
Find:
Solution:
D, H, P, T
m
Model:
The compressibility-factor generalized chart for CO2
Properties of CO2, R = 0.1889 kJ/kg-K, Pc = 7.38 MPa, Tc = 304.1 K
We notice that the value of P is relatively comparable to Pc. Therefore, we will choose the
compressibility-factor generalized chart as a model to find the property of CO2
5.9 MPa
P
Pr =
=
= 0.80
Pc 7.38 MPa
T 365.15 K
Tr =
=
= 1.20
Tc
304.1 K
Go to the compressibility-factor generalized chart shown below
Z = 0.83
From the compressibility-factor generalized chart, we have
Z = 0.83
Find the volume of the tank which is equal to that of CO2
π
π
V = D 2 H = (0.52 m 2 )(0.8 m )
4
4
3
V = 0.15708 m
State of CO2: P = 5.9 MPa, T = 92oC = 365.15 K, V = 0.15708 m3
P V = Zm R T
(5,900 kPa )(0.15708 m )
3
kJ ⎞
⎛
= ( 0.83) ( m ) ⎜ 0.1889
⎟ (365.15 K )
kg − K ⎠
⎝
m = 16.188 kg
Answer
Note: If the ideal gas law is used to find the mass of CO2, its value will deviate from the one obtained by
the answer above by 17 percent.
2) Oxygen is compressed in a piston/cylinder device under the polytropic process with n = 1.4. The
initial pressure and temperature of the oxygen is 100 kPa and 30oC, respectively. The specific work
done on this process is 200 kJ/kg. Determine the final temperature and pressure. (10 points)
Given:
Find:
Solution:
n, P1, T1, 1w2
T2, P2
CM
O2
Control mass: Oxygen in the piston/cylinder device
Process:
Polytropic with n=1.4
Model:
Ideal gas for oxygen
Properties of oxygen at 25oC: R = 0.2598 kJ/kg-K
State 1: P1 = 100 kPa, T1 = 30oC = 303.15 K
Ideal gas law
P1 v 1 = R T2
(1)
State 2:
Ideal gas law
(2)
P2 v 2 = R T2
In case of a polytropic relation: PVn = constant, we can substitute this relation into the moving boundary
work equation and obtain
2
1W2 =
∫
P dV =
1
P2 V2 − P1 V1
1− n
P
2
Polytropic
n=1.4
Rewriting the equation by dividing by m gives
1
w2
P v −Pv
= 2 2 1 1
1− n
1
100 kPa
V2
We can substitute the ideal gas law, i.e., equations (1) and (2) into the above equation.
1 w2 =
RT2 − RT1
R (T2 − T1 )
=
1− n
1− n
W2
1
V1
V
We substitute all known variables into the previous equation to determine T2. Remind that the value of the
specific work is negative due to the compression process (dV < 0).
− 200
kJ (T2 − 303.15 K )
kJ
= 0.2598
kg 1 − 1.4
kg
T2 = 611.08 K
Answer
From the polytropic relation (per unit mass)
P1 v 1 n = P2 v 2 n
(3)
Dividing equation (2) by equation (1) gives
Pv
P1 v 1
= 2 2
T2
T1
We raise the above equation to the nth power.
P1n v 1n
P2n v 2n
= n
T1n
T2
(4)
Dividing equation (4) by equation (3) to eliminate v1 and v2 and rearranging the equation yields
n -1
⎛ P2 ⎞
⎛T ⎞
⎜⎜ ⎟⎟ = ⎜⎜ 2 ⎟⎟
⎝ P1 ⎠
⎝ T1 ⎠
n
⎛T ⎞
P2
= ⎜⎜ 2 ⎟⎟
P1
⎝ T1 ⎠
n/(n -1)
Since P1, T1, and T2 are known, we can solve for P2 as follows:
611.08 ⎞1.4/0.4
P2
⎛
= ⎜
⎟
100 kPa
⎝ 303.15 ⎠
P2 = 1,162.9 kPa
Answer
3) N2 initially at 110 K is heated in a 30-L rigid tank until it reaches the critical point. Determine the heat
transfer of this process. (10 points)
Given:
Find:
Solution:
T1, V, P2, T2
1Q2
CM
N2
Control mass: N2 in the rigid tank
Process:
Isochoric
Model:
Table of thermodynamics for N2
Since the process is isochoric from an initial state to the critical point, it will be very useful to draw the PV diagram of this process as follows:
P
Psat@110 K
2 (critical point)
1
v1 = v2
v
It can be seen that in order to heat N2 to the critical point by an isochoric process, the initial state must be a
two-phase mixture. First, we find the properties at the critical point which is the final state.
State 2 (final state):
P2 = Pc = 3.3978 MPa
T2 = Tc = 126.2 K
fixed state: critical point
v2 = vc = 0.003194 m3/kg
u2 = uc = 18.94 kJ/kg
We can find the mass inside the tank from here. Note that V = constant = 30 L =0.03 m3
0.03 m 3
V
m= =
= 9.3926 kg
v 1 0.003194 m 3 / kg
State 1 (initial state):
T1 = 110 K
v2 = v1 = 0.003194 m3/kg
v 1 = (1 − x 1 ) v f + x 1 v g
fixed state: two-phase mixture
3
v 1 − v f 0.003194 − 0.001610 m / kg
x1 =
=
v g − v f 0.01595 − 0.001610 m 3 / kg
x 1 = 0.11046
kJ ⎞
kJ ⎞
⎛
⎛
u 1 = (1 − x 1 ) u f + x 1 u g = ( 0.88954 )⎜ − 50.81 ⎟ + ( 0.11046 )⎜ 62.31 ⎟
kg ⎠
kg ⎠
⎝
⎝
kJ
u 1 = − 38.3147
kg
The moving boundary work in this problem is zero since the process is isochoric or dV = 0.
2
1W2
=
∫ P dV
= 0
1
From the first law of thermodynamics for a control mass, we can write
1
Q 2 − 1W2 = ∆E = ∆U + ∆KE + ∆PE
Since there is no change in the kinetic and potential energy, ∆KE = 0 and ∆PE = 0. In addition, 1W2 = 0
from the above equation. The first law becomes
1 Q2
= ∆U = m ( u 2 − u 1 )
kJ ⎞
⎛
Q 2 = 9.3926 kg ⎜ 18.94 − ( − 38.3184 ) ⎟
kg ⎠
⎝
1 Q 2 = 537.8 kJ
1
Answer
Note: In this problem, we use table of thermodynamics for nitrogen table to find the properties. However,
the common mistake arises here because most students always think about applying the ideal gas equation
to substances appearing in gaseous phase under the atmospheric condition such as air, N2, O2 etc. without
paying attention on the P-V diagram.
4) Steam flows into a steam turbine at a rate
of 2 kg/s, and the heat transfer from the
turbine to the surrounding is 11 kW. The
following data in the table on the right are
known for the steam entering and leaving
the turbine. Determine the power output
of the turbine in kW. (10 points)
Given:
Find:
Solution:
Pressure
Temperature
Quality
Velocity
Elevation above
a reference plane
Inlet
Conditions
2.0 MPa
350oC
Exit
Conditions
0.1 MPa
100 %
100 m/s
3m
50 m/s
6m
steam
P1, T1, V1, Z1, P2, x2, V2, Z2, m& , Q&
W&
CV
1
Control volume: Steam turbine
Process:
Steady-state
Model:
Table of thermodynamics for water
W&
Q&
2
From the given data, we can find the properties of state 1 and state 2.
State 1:
State 2:
P1 = 2 MPa
T1 = 350oC
h1 = 3,136.96 kJ/kg
fixed state: superheat vapor
P2 = 1 MPa
x2 = 1
h2 = 2,778.08 kJ/kg
fixed state: saturated vapor
Since there is only one inlet and one exit for the turbine, we can write
Conservation of mass
m& 1 = m& 2 = m& = 2 kg / s
For the first law of thermodynamics for a control volume, we have information related to the kinetic and
potential energy. The heat loss from the turbine is also included. Thus, we can write
The first law of thermodynamics
0 = Q& − W& +
∑
V2
m& i ⎛⎜ h i + i + gZ i ⎞⎟ −
2
⎠
⎝
∑
V2
m& e ⎛⎜ h e + e + gZ e ⎞⎟
2
⎝
⎠
V1 2
V2 2
⎛
⎛
⎞
W& = Q& + m& ⎜ h 1 + + gZ1 ⎟ − m& ⎜ h 2 + + gZ 2 ⎞⎟
2
2
⎠
⎝
⎠
⎝
V1 2 − V2 2 ⎞
⎡
⎤
⎛
W& = Q& + m& ⎢(h 1 − h 2 ) + ⎜
⎟ + g (Z1 − Z 2 )⎥
⎝ 2 ⎠
⎣
⎦
kJ ⎞
kg ⎡⎛
−
3
,
136
.
96
2
,
778
.
08
⎜
⎟
kg ⎠
s ⎢⎣⎝
⎛ 50 2 − 100 2 m 2
1 kJ / kg ⎞ ⎛ m
1 kJ / kg ⎞⎤
⎜
⎟
⎜
⎟
+
×
+ 9.81 2 (6 − 3 m )×
2
2 2 ⎟ ⎜
2 2 ⎟⎥
⎜
2
s 1,000 m / s ⎠ ⎝ s
1,000 m / s ⎠⎦⎥
⎝
W& = − 11 kW + 2
W& = − 11 kW + 2
kg ⎡⎛
kJ ⎞ ⎛
kJ ⎞ ⎛
kJ ⎞⎤
358
.
88
3
.
75
0
.
02943
+
−
+
⎜
⎟
⎜
⎟
⎜
⎟
s ⎢⎣⎝
kg ⎠ ⎝
kg ⎠ ⎝
kg ⎠⎥⎦
W& = − 11 kW + 710.319 kW
W& = 699.319 kW
Answer
Note: It is clear that the change of kinetic and potential energy is much smaller than the change of
enthalpy of the steam. By assuming that the change of kinetic and potential energy is negligible, the error
relative to the above answer is approximately 1 percent. Thus, this assumption is always employed in
most engineering practice.
6) In winter, an actual heat pump supplies 120 kW of heat to a building at 300 K. The ambient air
temperature is 270 K outside the building. If the coefficient of performance of this heat pump is 60
percent of the coefficient of performance of a Carnot heat pump operating between the same
temperatures above. Find the input power of this heat pump. (10 points)
Given:
Find:
Solution:
Q& H , TH, TL, β′actual / β′ideal
W& in
We can draw the diagram representing this heat pump shown on the
right side. Since TL and TH are given, we can calculate β′ideal as
follows:
TH
β′ideal =
TH − TL
300
β′ideal =
300 − 270
β′ideal = 10
Building
TH = 300 K
Q& H =120 kW
Heat Pump
W& in
Q& L
Ambient
TL = 270 K
Next, β′actual can be determined according to the given ratio between β′actual and β′ideal .
β′actual
= 0.6
β′ideal
β′actual = 0.6 β′ideal = 0.6 ( 10 )
β′actual = 6
Finally, we use the definition of β′ to find the input power
Q&
β′actual = H
W& in
120 kW
Q&
W& in = H =
6
β′actual
W& in = 20 kW
Answer
Note: since β′actual = 6, it means that to produce 1 unit of heat, a heat pump requires 1/6 unit of the
electrical work whereas an electric heater requires 1 unit of that, resulting in more efficient work-to-heat
conversion.
7) A piston/cylinder has ammonia at 2,000 kPa, 80oC with a volume of 0.1 m3. This piston is loaded with
a linear spring, and the outside ambient is at 20oC. The ammonia now cools down to 20oC at which
point it has a quality of 10%. Find (a) the work, (b) the heat transfer, and (c) the total entropy
generation in the process. (10 points)
Given:
Find:
Solution:
P1, T1, V1, T0, T2, x2
1W2, 1Q2, Sgen total
CM
Control mass: Ammonia
Process:
Linear spring
Model:
Table of thermodynamics for ammonia
The process is under the influence of the linear spring attached on the top
of the piston. We can determine the properties of ammonia as follows:
NH3
Q
T0
State 1 (initial state):
P1 = 2 MPa
fixed state: superheat vapor
T1 = 80oC
3
v1 = 0.07595 m /kg
u1 = 1,421.6 kJ/kg
s1 = 5.0707 kJ/kg-K
We can find the mass of ammonia inside the cylinder.
0.1 m 3
V1
= 1.31666 kg
m= =
v 1 0.07595 m 3 / kg
State 2 (final state):
T2 = 20oC
x2 = 0.1
P2 = Psat = 857.5 kPa
fixed state: two-phase mixture
m3
m3
v 2 = (1 − x 2 ) v f + x 2 v g = ( 1 − 0.1) 0.001638 + ( 0.1) 0.14922
kg
kg
m3
v 2 = 0.016396
kg
kJ
kJ
u 2 = (1 − x 2 ) u f + x 2 u g = ( 1 − 0.1) 272.89 + ( 0.1) 1,332.2
kg
kg
kJ
u 2 = 378.821
kg
s 2 = (1 − x 2 ) s f + x 2 s g = ( 1 − 0.1) 1.0408
s 2 = 1.44532
kJ
kg − K
kJ
kJ
+ ( 0.1) 5.0860
kg − K
kg − K
Because the area under the P-v diagram has a trapezoidal
shape as shown, it can be calculated as follows:
2
1W2 =
∫
PdV = 0.5 (P1 + P2 )(V2 − V1 )
1
W = 0.5 m (P1 + P2 )(v 2 − v 1 )
P
1
2,000 kPa
2
857.5 kPa
1 2
W = 0.5 (1.31666 kg )(2,000 + 857.5 kPa )
1 2
× (0.016396 − 0.07595 m 3 / kg )
0.016396 m3/kg
W = − 112.031 kJ
1 2
0.07595 m3/kg
Answer
From the first law of thermodynamics for a control mass
1
Q 2 − 1W2 = ∆E = ∆U + ∆KE + ∆PE
Because there is no change in the kinetic and potential energy, ∆KE = 0 and ∆PE = 0. The first law
becomes
1
Q 2 − 1W2 = ∆U = m (u 2 − u 1 )
kJ ⎞
⎛
(
)
(
)
Q
−
−
112
.
031
kJ
=
1
.
31666
kg
378
.
821
−
1
,
421
.
6
⎜
⎟
1 2
kg
⎝
⎠
1 Q 2 = − 1, 485.01 kJ
Answer
From the second law of thermodynamics for a control mass and surrounding
Q
S gen total = (S 2 − S1 ) − 1 2
T0
Q
S gen total = m (s 2 − s1 ) − 1 2
T0
kJ ⎞ − 1,485.01 kJ
⎛
S gen total = (1.31666 kg )⎜ 1.44532 − 5.0707
⎟−
kg − K ⎠ 20 + 273.15 K
⎝
kJ
Answer
S gen total = 0.2923
K
v
8) CO2 at 100 kPa, 300 K is compressed by a steady-state uninsulated compressor to 800 kPa, 450 K.
The input power is 1.7 MW. The flow rate of CO2 is measured to be 10 kg/s. If the ambient
temperature is at 300 K, determine the total rate of entropy generation. (10 points)
Given:
Find:
Solution:
P1, T1, P2, T2, W& , m& , T0
S& gen total
CV
T0
Control volume: Compressor
Process:
Steady-state
Model:
Ideal gas for carbon dioxide
2
W&
Q&
1
CO2
There is only one inlet and one exit in this compressor. Thus, we can write
Conservation of mass
m& 1 = m& 2 = m& = 10 kg / s
For the first law of thermodynamics, the kinetic and potential energy is negligible. However, the heat
transfer may occur because of the uninsulated compressor. Thus, we can write
The first law of thermodynamics
Vi 2
⎛
+ gZ i ⎞⎟ −
m& i ⎜ h i +
2
⎝
⎠
0 = Q& − W& + m& (h 1 ) − m& (h 2 )
0 = Q& − W& +
∑
∑
Ve 2
⎛
+ gZ e ⎞⎟
m& e ⎜ h e +
2
⎝
⎠
Q& = W& + m& (h 2 − h 1 )
To find the change of enthalpy of CO2, we use table of the ideal-gas properties of CO2:
State 1:
T1 = 300 K
h1 = 214.38 kJ/kg s 0T1 = 4.8631 kJ/kg-K
State 2:
T2 = 450 K
h2 = 351.70 kJ/kg s 0T 2 = 5.2325 kJ/kg-K
Then, substituting into the first law of thermodynamics gives
kg ⎛
kJ ⎞
Q& = − 1,700 kW + ⎛⎜ 10 ⎞⎟ ⎜ 351.70 − 214.38 ⎟
kg ⎠
⎝ s ⎠⎝
Q& = − 326.8 kW
For the second law of thermodynamics for a control volume and surrounding, the total rate of entropy
generation can be written as
The second law of thermodynamics
Q& cv
T0
Q&
= m& 2 (s 2 ) − m& 1 (s1 ) − cv
T0
Q&
= m& (s 2 − s1 ) − cv
T0
S& gen total =
S& gen total
S& gen total
∑
m& e s e −
∑
m& i s i −
The change of entropy of CO2 can be determined as follows:
P
s 2 − s1 = s 0T 2 − s 0T1 − R ln 2
P1
kJ ⎞ ⎛
kJ ⎞ ⎛⎜ 800 kPa ⎞⎟
⎛
s 2 − s1 = ⎜ 5.2325 − 4.8631
⎟ ln
⎟ − ⎜ 0.1889
kg − K ⎠ ⎝
kg − K ⎠ ⎝⎜ 100 kPa ⎟⎠
⎝
kJ
s 2 − s1 = − 0.023407
kg − K
Substituting into the second law of thermodynamics yields
− 326.8 kW
kg ⎛
kJ ⎞
S& gen total = ⎛⎜ 10 ⎞⎟ ⎜ − 0.023407
⎟−
⎝ s ⎠⎝
kg − K ⎠
300 K
kW
S& gen total = − 0.23407 + 1.08933
K
kW
Answer
S& gen total = 0.85527
K
Note: If we choose to find the change of enthalpy and entropy of CO2 by assuming that the value CP0 of
CO2 is a constant at an average temperature between T1 and T2, which is 375 K. By using the third degree
polynomial of CO2, we will have CP0 = 0.91822 kJ/kg-K. Then, substituting into the first and second law
of thermodynamics gives
Q& = W& + m& C P 0 (T2 − T1 ) = − 322.67 kW
S& gen total = m& (s 2 − s1 ) −
⎛
Q& cv
T
P ⎞ Q&
kW
= m& ⎜⎜ C P 0 ln 2 − R ln 2 ⎟⎟ − cv = 0.87058
T0
T1
P1 ⎠ T0
K
⎝
The error from this method is approximately less than 2% from the first one.
9) An ideal Diesel cycle with air as a working fluid has a compression ratio of 18.2. Air is at 0.1 MPa,
300 K at the beginning of the compression process. The maximum temperature is 2,000 K.
Determine (a) the heat addition (per unit mass) and (b) the thermal efficiency. (10 points)
Given:
Find:
Solution:
P
rV, T1, P1, T3
qH, ηth
6,000 kPa
2
4
Assumptions:
1) Ideal Diesel cycle
2) ∆KE and ∆PE are negligible.
3) Cold air-standard cycle
95 kPa
1
Vmin(TDC)
Properties of air at 25oC: R=0.287 kJ/kg-K, CP0=1.004 kJ/kg-K, k=1.4
State 1:
P1 = 100 kPa, T1 = 300 K
P1 v 1 = R T1
Ideal gas equation
(100 kPa ) v
kJ ⎞
⎛
= ⎜ 0.287
⎟ (300 K )
kg
K
−
⎝
⎠
3
v 1 = 0.861 m / kg
State 2:
3
1
rV = 18.2 = v1/v2
k −1
T2 ⎛ v 1 ⎞
= ⎜⎜ ⎟⎟ = (rV )k −1 = ( 18.2 )1.4 −1 = 3.1917
T1 ⎝ v 2 ⎠
T2 = 3.1917 (300 K ) = 957.52 K
k
P2 ⎛ v 1 ⎞
= ⎜⎜ ⎟⎟ = (rV )k = ( 18.2 )1.4 = 58.0898
P1 ⎝ v 2 ⎠
P2 = 58.0898 (100 kPa ) = 5,808.98 kPa
State 3:
P3 = P2 = 5,808.98 kPa, T3 = 2,000 K
Ideal gas equation
P3 v 3 = R T3
(5,808.98 kPa ) v
kJ ⎞
⎛
0
.
287
=
⎜
⎟ (2 ,000 K )
3
kg − K ⎠
⎝
v 3 = 0.098813 m 3 / kg
Vmax(BDC)
V
State 4:
v4 = v1 = 0.861 m3/kg
For the isentropic process from state 3 to state 4, we can write
1.4 −1
k −1
⎛ 0.098813 m 3 / kg ⎞
⎛ v3 ⎞
T4
⎟
=⎜ ⎟ =⎜
⎜ 0.861 m 3 / kg ⎟
T3 ⎜⎝ v 4 ⎟⎠
⎝
⎠
T4 = 0.42065 (2,000 K ) = 841.31 K
k
= 0.42065
1.4
3
⎛ v 3 ⎞ ⎜⎛ 0.098813 m / kg ⎞⎟
P4
= 0.048276
=⎜ ⎟ =
P3 ⎜⎝ v 4 ⎟⎠ ⎜⎝ 0.861 m 3 / kg ⎠⎟
P4 = 0.048276 (5,808.98 kPa ) = 280.44 kPa
Apply the first law of thermodynamics to the individual process
Process: Isobaric from state 2 to state 3
2 q3
= q H = h 3 − h 2 = C P 0 (T3 − T2 )
kJ
(2,000 − 957.52 K )
kg − K
kJ
q H = 1,046.65
kg
q H = 1.004
Answer
Process: Isochoric from state 4 to state 1
q = q L = u 1 − u 4 = C V 0 (T1 − T4 )
4 1
kJ
(300 − 841.31 K )
kg − K
kJ
q L = − 388.12
kg
To obtain the net specific work, applying the first law of thermodynamics for the entire cycle yields
q L = 0.717
w net = q H + q L = 1,046.65 + ( − 388.12 ) kJ / kg = 658.53 kJ / kg
The thermal efficiency of the cycle can be determined.
658.53 kJ / kg
w
ηth = net =
qH
1,046.65 kJ / kg
ηth = 0.6292 = 62.92 %
Answer
10) A vapor-compression refrigerator for industrial applications uses ammonia as a refrigerant. The
evaporating temperature is −5oC whereas the pressure after the compression process is 1.6 MPa. The
compressor has an isentropic efficiency of 80%. Determine (a) the quality of ammonia before entering
the evaporator and (b) the COP of the refrigerator. (10 points)
Given:
Find:
Solution:
T1, P2, ηcomp.
x4, β
P2 = P3 = 1.6 MPa
2
2s
T
3
Assumptions:
1) Ideal vapor compression refrigeration cycle except
the compressor
2) ∆KE and ∆PE are negligible.
3) From the conservation of mass, m& is constant throughout
the cycle.
State 1:
T1 = −5oC
x1 = 0
Fixed state: saturated vapor
h1 = hg = 1,436.7 kJ/kg
s1 = sg = 5.3977 kJ/kg-K
P1 = P4
−5oC
1
4
s
Q& H
3
2
Condenser
W& comp.
Expansion valve
Compressor
4
Evaporator
1
Due to the actual compressor, state 2s is determined as follows:
Q& L
State 2s:
P2s = P2 = 1,600 kPa
s2s = s1 = 5.3977 kJ/kg-K
Fixed state: superheat vapor
h2s = 1,656.99 kJ/kg (by linear interpolation)
By using the definition of the isentropic efficiency of the compressor, we can find the enthalpy of state 2.
State 2:
P2 = 1,600 kPa, ηcomp. = 0.8
h −h
ηcomp . = 1 2 s
h1 − h 2
1,436.7 − 1,656.99 kJ / kg
0.8 =
1,436.7 − h 2 kJ / kg
h 2 = 1,712.06 kJ / kg
State 3:
P3 = P2 = 1,600 kPa
x3 = 0
h3 = 376.37 kJ/kg
Fixed state: saturated liquid
State 4:
T4 = T1 = −15oC
h4 = h3 = 376.37 kJ/kg
Fixed state: two-phase mixture
h 4 = (1 − x 4 ) h f + x 4 h g
h − h 376.37 − 157.31 kJ / kg
x4 = 4 f =
h g − h f 1,436.7 − 157.31 kJ / kg
x 4 = 0.1712
Answer
The next step is to apply the first law of thermodynamics to each device in our interest.
Control volume: Evaporator
Process:
Isobaric
The first law of thermodynamics per unit mass flow rate will be simplified as follows:
q evap . = q L = h 1 − h 4
q L = 1,436.7 − 376.37 kJ / kg = 1,060.33 kJ / kg
Control volume: Compressor
Process:
Isentropic
The first law of thermodynamics per unit mass flow rate will be simplified as follows:
w comp . = h 1 − h 2
w comp . = 1,436.7 − 1,712.06 kJ / kg = − 275.36 kJ / kg
If we substitute the work with the negative sign and qL with the positive sign into the definition of the
coefficient of performance, we will get the negative value of the coefficient of performance which does not
make any sense. Therefore, we have to substitute only the magnitude of the work and qL into the definition
of the coefficient of performance.
1,060.33 kJ / kg
q
qL
β = L =
=
275.36 kJ / kg
w net
w comp .
β = 3.851
Answer
11) Steam enters the turbine of a power plant at 5 MPa and 400oC and exhausts to the condenser at 10 kPa.
The turbine produces a power output of 20,000 kW with an isentropic efficiency of 85%. Determine
(a) the mass flow rate of steam around the cycle and (b) the thermal efficiency. (10 points)
Given:
Find:
Solution:
P2 = P3, P1, W& net , ηturb .
m& , ηth
T
3
400oC
Assumptions:
1) Ideal Rankine cycle except the turbine
2) ∆KE and ∆PE are negligible.
3) From the conservation of mass, m& is constant
throughout the cycle.
State 1:
P1 = 10 kPa
x1 = 0
Fixed state: saturated liquid
v1 = vf = 0.001010 m3/kg
h1 = hf = 191.81 kJ/kg
State 2:
P2 = P3 = 5 MPa
P1 = P4
= 10 kPa
2
1
4s 4
s
3
Q& H
W& turbine
Turbine
Boiler
4
2
P2 = P3 = 3 MPa
s 2 = s1
Fixed state: compressed liquid
Pump
W& pump
Condenser
Q& L
1
However, since most thermodynamic tables do not include the compressed liquid table, we need to find h2
by using the thermodynamic property relation: T ds = dh − v dP . Under the isentropic relation (ds = 0)
and incompressible substance (v ≈ constant), we take the definite integration as follows:
2
0 =
2
∫ dh − ∫ v dP = (h − h ) − v (P − P )
2
1
1
1
2
1
1
h 2 = h 1 + v 1 (P2 − P1 )
kJ ⎛
m3 ⎞
h 2 = 191.81 + ⎜ 0.001010 ⎟ (5,000 − 10 kPa )
kg ⎝
kg ⎠
h 2 = 196.85 kJ / kg
State 3:
P3 = 5 MPa
T3 = 400oC
h3 = 3,195.64 kJ/kg
s3 = 6.6458 kJ/kg-K
Fixed state: superheat vapor
Due to the actual turbine, state 4s can be determined.
State 4s:
P4s = P4 = P1 = 10 kPa
s4s = s3 = 6.6458 kJ/kg-K
Fixed state: two-phase mixture
s − s 6.6458 − 0.6492 kJ / kg − K
= 0.7995
x 4s = 4s f =
s g − s f 8.1501 − 0.6492 kJ / kg − K
kJ
kJ
h 4 s = (1 − x 4 s ) h f + x 4 s h g = ( 1 − 0.7995) 191.81 + ( 0.7995) 2 ,584.63
kg
kg
h4s = 2,104.75 kJ/kg
By using the definition of the isentropic efficiency of the turbine, we can find the enthalpy of state 4.
State 4:
P4 = P1 = 10 kPa, ηturb . = 0.85
h −h
η turb . = 3 4
h 3 − h 4s
3,195.64 − h 4 kJ / kg
0.85 =
3,195.64 − 2,104.75 kJ / kg
h 4 = 2,268.38 kJ / kg
The next step is to apply the first law of thermodynamics to each device in our interest.
Control volume: Turbine
Process:
Adiabatic
By applying the first law of thermodynamics, we can substitute Q& H to find m& as follows:
W& turb . = m& (h 3 − h 4 )
20,000 kW = ( m& ) (3,195.64 − 2 ,268.38 kJ / kg )
m& = 21.569 kg / s
Answer
Control volume: Pump
Process:
Isentropic
The first law of thermodynamics can be written as follows:
W& pump = m& (h 1 − h 2 )
W& pump = (21.569 kg / s )(191.81 − 196.85 kJ / kg )
W& pump = − 108.71 kW
Control volume: Boiler
Process:
Isobaric
The first law of thermodynamics can be written as follows:
Q& boiler = Q& H = ( m& ) (h 3 − h 2 )
Q& H = (21.569 kg / s )(3,195.64 − 196.85 kJ / kg )
Q& H = 64 ,681.02 kW
The net specific work can be determined by combine the power from the turbine and the pump.
W& net = W& turb . + W& pump = 20,000 + ( − 108.71) kW = 19,891.29 kW
Finally, the thermal efficiency of the cycle is
20,000 kW
W&
ηth = net =
Q& H
64 ,681.02 kW
ηth = 0.3075 = 30.75 %
Answer
PART B: ตอบคําตอบโดยไมตองมีการคํานวณลงในพื้นที่ที่กําหนดใหเทานั้น การตอบนอกพื้นที่ที่
กําหนดใหหรือเขียนตัวอักษรเล็กเกินไปจนอานไมได จะไมตรวจคําตอบในขอนั้นเชนกัน
1) จงอธิบายและใหเหตุผลโดยสังเขปวาทําไมเราจึงไมเห็นคารบอนไดออกไซดปรากฏอยูในรูปของของเหลวในสภาวะ
อากาศปกติบนโลก
พื้นที่สําหรับคําตอบขอที่ 1
เฉลย:
หากพิจารณา phase diagram ของคารบอนไดออกไซด
จะพบวาความดันที่จุดรวมสาม (triple point pressure หรือ Ptriple)
สูงกวาความดันบรรยากาศบนโลก ดังนั้นจะเห็นจากใน phase
diagram วาถา P < Ptriple ผลที่ไดก็คือเมื่ออุณหภูมิเปลี่ยนไปตาม
เสน A-B ก็จะทําใหคารบอนไดออกไซดเปลี่ยนสถานะจาก
ของแข็งผานเสนระเหิด (sublimation line) และกลายไปเปนไอ
โดยที่ไมผานสถานะของเหลว จึงเปนเหตุใหคารบอนไดออกไซด
ไมปรากฏเปนของเหลวที่สภาวะอากาศปกติบนโลก
P
Fusion line
Solid
Vaporization line
Liquid
Triple point
Ptriple
Patm
A
Sublimation line
B
Gas
T
2) คา Z ที่ปรากฏอยูใน generalized chart มีความหมายทางกายภาพอยางไร และกาซที่มีโครงสรางโมเลกุลอยางงายทุก
ชนิดจะมีคา Z จะมีคาเขาใกล 1 ภายใตเงื่อนไขใด
พื้นที่สําหรับคําตอบขอที่ 2
เฉลย:
คา Z เปนตัวแปรที่แสดงถึงการเบี่ยงเบนของพฤติกรรมของแกสจริงที่มีตอพฤติกรรมของแกสอุดมคติ (ideal
gas) โดยที่หากคา Z = 1 ยอมหมายความวาแกสนั้นมีพฤติกรรมเปนเชนเดียวกับแกสอุดมคติ หากพิจาณาสมการของคา
Z ตามนิยามจะพบวา
Pv = ZRT
คา Z อาจเปรียบไดเปนตัวประกอบแกไข (correction factor) สําหรับสมการแกสอุดมคติก็ได
สําหรับกาซที่มีโครงสรางโมเลกุลอยางงายทุกชนิดนั้น คา Z จะเขาใกล 1 เมื่อ
1) แกสมีนั้นมีคา Pr << 1 โดยที่ไมขึ้นกับอุณหภูมิ
2) แกสมีนั้นมีคา Tr > 2 ยกเวนในกรณีที่ Pr >> 1 (ประมาณมากกวา 4)
3) มอเตอรไฟฟาเปนอุปกรณที่ใชกระแสไฟฟาเหนี่ยวนําใหเกิดสนามแมเหล็กตัดผานขดลวดทองแดงจนทําใหเกิดการ
หมุนของเพลาและสามารถสรางงานขึ้นได มอเตอรไฟฟานั้นถือวาเปน Heat Engine หรือไม อธิบาย
พื้นที่สําหรับคําตอบขอที่ 3
เฉลย:
จากแผนภาพของ heat engine ดานขาง จะเห็นไดวาองคประกอบ
ของ heat engine ก็คือจะตองมี QH ซึ่งทําหนาที่เปนความรอนซึ่งปอนเขาจาก
heat source ที่อุณหภูมิสูง TH จากนั้น QH บางสวนก็จะเปลี่ยนไปเปนงาน
หรือ Wnet ในขณะที่ความรอนสวนที่เหลือก็จะกลายเปน QL ทิ้งลงสู heat
sink ที่อุณหภูมิสูง TL ในกรณีของมอเตอรไฟฟานั้นไมมี QH ปรากฏอยู มี
แตเพียงพลังงานที่ปอนเขาในรูปของงานทางไฟฟาซึ่งจะถูกเปลี่ยนในเปน
งานทางกลเทานั้น ดังนั้นมอเตอรไฟฟาจึงไมใช heat engine
TH
QH
Heat Engine
Wnet
QL
TL
4) จงวาด component diagram ของ ideal Brayton cycle (open cycle) with a regenerator ซึ่งอุปกรณตางๆ ใน cycle มี
ประสิทธิภาพ 100% ทุกชิ้น ทั้งนี้ component diagram เปน diagram ที่แสดงใหเห็นถึงอุปกรณตางๆที่มีอยูใน cycle
พรอมทั้งวาด T-s diagram ประกอบเพื่อแสดงถึงตําแหนง state ตางๆ ที่ปรากฏอยูใน component diagram ดวย
พื้นที่สําหรับคําตอบขอที่ 4
เฉลย:
เราสามารถวาด component diagram และ T-s diagram ไดดังแผนภาพดานลาง
T
y
Regenerator
x
2
1
3
Combustion
chamber
Q& H
Compressor
P1 = Py = P4
4
x
4
3
W& net
Turbine
P2 = Px = P3
2
1
y
s
จะเห็นไดวาหาก regenerator เปนอุปกรณที่อุดมคติ เราจะไดวา Tx = T4 นั่นคืออุณหภูมิอากาศที่ถูกอุนกอนเขาสู
combustion chamber จะมีอุณหภูมิเทากับอุณหภูมิของไอเสียที่ออกจาก turbine ซึ่งจะเปนคาสูงสุดที่เปนไปได และหาก
เราใช cold air-standard assumption เราก็จะไดวา Ty = T2 เนื่องจากผลของ first law of thermodynamics ที่ regenerator