# Assignment6_solutions

```ENSC 388
Assignment #6
Assignment date: Wednesday Oct. 21, 2009
Due date: Wednesday Oct. 28, 2009
Problem 1
A turbine operating at steady state receives air at a pressure of π1 = 3.0 πππ and a
temperature of π1 = 390πΎ. Air exits the turbine at a pressure of π2 = 1.0 πππ.
The work developed is measured as 74 ππ½ per ππ of air flowing through the
turbine. The turbine operates adiabatically, and changes in kinetic and potential
energy between inlet and exit can be neglected. Using the ideal gas model for air,
determine the turbine efficiency.
πΜπΆπ
= 1.0 πππ
πΜ
2
π2 = 1.0 πππ
π1 = 390πΎ
1
π2 = 1.0 πππ
TURBINE
Problem 2
Components of a heat pump for supplying heated air to a dwelling are shown in the
schematic below. At steady state, Refrigerant 134a enters the compressor at 6 β,
3.2 πππ and is compressed adiabatically to 75 β, 14 πππ. From the compressor,
the refrigerant passes through the condenser, where it condenses to liquid at 28 β,
14 πππ. The refrigerant then expands through a throttling valve to 3.2 πππ. The
states of the refrigerant are shown on the accompanying T-s diagram. Return air
M. Bahrami
ENSC 388
Assignment # 6
1
from the dwelling enters the condenser at 20 β, 1 πππ with a volumetric flow rate
of 0.42 π3 /π  and exits at 50 β with a negligible change in pressure. Using the
ideal gas model for the air and neglecting kinetic and potential energy effects, (a)
determine the rates of entropy production, in ππ/πΎ, for control volumes enclosing
the condenser, compressor, and expansion valve, respectively. (b) Discuss the
sources of irreversibility in the components considered in part (a).
Indoor return air
π5 = 20β
π5 = 1 πππ
(π΄π)5 = 0.42 π3 /π
Condenser
π3 = 28β 3
π3 = 14 πππ
Expansion
Valve
π4 = 3.2 πππ
4
Supply air
π6 = 50β
π6 = 1 πππ
π2 = 75β
π2 = 14 πππ
2
3
14 πππ
28β
Turbine
3.2 πππ
1
Evaporator
M. Bahrami
75β
2
T
π1 = 6 β
π1 = 3.2 πππ
1 6β
4
s
Outdoor air
ENSC 388
Assignment # 6
2
Problem 1:
Known:
ππ‘ππ‘ππ  ππππππππ‘πππ πππ π‘βπ ππππππ π ππ .
Find:
- Turbine efficiency.
π
3 πππ
1
π1 = 390πΎ
Actual
expansion
Isentropic
expansion
2s
2
1 πππ
π
Assumptions:
- The expansion is adiabatic and the changes in kinetic and potential energy
between inlet and exit can be neglected.
- The air is modeled as an ideal gas.
Analysis:
Isentropic turbine efficiency can be obtained from
πΜπΆπ
πΜ
ππ‘ =
πΜ
( ππΆπ
Μ )π
The numerator of the efficiency is known. The denominator is evaluated as
follows.
M. Bahrami
ENSC 388
Assignment # 6
3
The work developed in an isentropic expansion from the given inlet state to the
specified exit pressure is
πΜπΆπ
(
) = β1 − β2π
πΜ π
From Table A-17, at 390πΎ, β1 = 390.88 ππ½/ππ. To determine β2π  , applying an
π2
ππ (π2π  ) = ( ) ππ (π1 )
π1
With π1 = 3 πππ, π2 = 1 πππ and ππ (π1 ) = 3.481 form Table A-17 at 390πΎ
1
ππ (π2π  ) = ( ) &times; 3.481 = 1.1603
3
Interpolating in Table A-17 gives β2π  = 285.27 ππ½/ππ. Thus
πΜπΆπ
ππ½
ππ½
ππ½
(
) = 390.88 [ ] − 285.27 [ ] = 105.6 [ ]
πΜ π
ππ
ππ
ππ
Hence isentropic turbine efficiency will be
ππ½
74 [ ]
ππ
ππ‘ =
= 0.7 ππ 70%
ππ½
105.6 [ ]
ππ
M. Bahrami
ENSC 388
Assignment # 6
4
Problem 2:
Known:
ππ‘ππ‘ππ  ππππππππ‘πππ πππ π‘βπ ππππππ π ππ .
Find:
- The entropy production rates for control volumes enclosing the condenser,
compressor, and expansion valve, respectively and discuss the sources of
irreversibility in these components.
Assumptions:
- Each component is analyzed as a control volume at steady state.
- The compressor operates adiabatically, and the expansion across the valve is
a throttling process.
- For the control volume enclosing the condenser, πΜπΆπ = 0 and πΜπΆπ = 0.
- Kinetic and potential energy effects can be neglected.
- The air is modeled as an ideal gas with constant ππ = 1.005 ππ½/ππ. πΎ.
Analysis:
(a) Lets us begin by obtaining property data at each of the principal refrigerant
states located on the T-s diagram. At the inlet to the compressor, the refrigerant is a
superheated vapour at 6 β, 3.5 πππ, so from Table A-13, π 1 = 0.9415 ππ½/ππ. πΎ.
Similarly, at state 2, the refrigerant is superheated vapour at 75 β, 14 πππ, so
interpolating in Table A-13 gives π 2 = 0.9561 ππ½/ππ. πΎ and β2 = 291.3 ππ½/ππ.
State 3 is compressed liquid at 28 β, 14 πππ. From Table A-11, π 3 ≈ π π@28 β =
0.33846 ππ½/ππ. πΎ and β3 ≈ βπ@28 β = 90.69 ππ½/ππ. The expansion through the
valve is a throttling process, thus β4 = β3 . Using data from Table A-11, the quality
at state 4 is
π₯4 =
M. Bahrami
β4 − βπ@3.2πππ
βππ@3.2πππ
ππ½
ππ½
90.69 [ ] − 55.16 [ ]
ππ
ππ
=
= 0.18
ππ½
196.71 [ ]
ππ
ENSC 388
Assignment # 6
5
And the specific entropy is
π 4 = π π@3.2πππ + π₯4 π ππ@3.2πππ = 0.21637 [
ππ½
ππ½
] + 0.18 &times; 0.71369 [
]
ππ. πΎ
ππ. πΎ
ππ½
→ π 4 = 0.3448 [
]
ππ. πΎ
Condenser:
Consider the control volume enclosing the condenser. With the first and third
assumptions, the entropy rate balance reduces to
0 = πΜπππ (π 2 − π 3 ) + πΜπππ (π 5 − π 6 ) + π Μπππ,ππππ
To evaluate π Μπππ requires the two mass flow rates, πΜπππ and πΜπππ , and the change
in specific entropy for the air. These are obtained next.
Evaluating the mass flow rate of air using the ideal gas model
πΜπππ
π5
π3
100[πππ]
ππ
= (π΄π)5
= 0.42 [ ] &times;
= 0.5 [ ]
ππ½
ππ
π
π
β5
0.286 [
] &times; 293 [πΎ]
ππ. πΎ
π
The refrigerant mass flow rate is determined using an energy balance for the
control volume enclosing the condenser together with the 1 st, 3rd and 4th
assumptions to obtain
πΜπππ =
πΜπππ (β6 − β5 )
β2 − β3
With the 5th assumption, β6 − β5 = ππ (π6 − π5 ). Inserting values
πΜπππ
ππ
ππ½
0.5 [ ] &times; 1.005 [
] &times; (323[πΎ] − 293[πΎ])
ππ
π
ππ. πΎ
=
= 0.075 [ ]
ππ½
ππ½
π
291.3 [ ] − 90.69 [ ]
ππ
ππ
The change in specific entropy of the air is
M. Bahrami
ENSC 388
Assignment # 6
6
π 6 − π 5 = ππ ππ
π6
π6
ππ½
323[πΎ]
1[πππ]
− πππ = 1.005 [
− π ππ
] &times; ππ
π5
π5
ππ. πΎ
293[πΎ]
β 1[πππ]
0
→ π 6 − π 5 = 0.098 [
ππ½
]
ππ. πΎ
Finally, solving the entropy balance for π Μπππ and inserting values
π Μπππ = πΜπππ (π 3 − π 2 ) + πΜπππ (π 6 − π 5 )
ππ
ππ½
ππ½
ππ
] &times; (0.33846 [
] − 0.9561 [
]) + 0.5 [ ]
π
ππ. πΎ
ππ. πΎ
π
ππ½
ππ
&times; 0.098 [
] = 2.677 &times; 10−3 [
]
ππ. πΎ
πΎ
π Μπππ,ππππ = 0.075 [
Compressor:
For the control volume enclosing the compressor, the entropy rate balance reduces
with the first and third assumptions to
0 = πΜπππ (π 1 − π 2 ) + π Μπππ,ππππ
or
ππ
ππ½
ππ½
π Μπππ,ππππ = πΜπππ (π 2 − π 1 ) = 0.075 [ ] &times; (0.9561 [
] − 0.9415 [
])
π
ππ. πΎ
ππ. πΎ
ππ
= 1.095 &times; 10−3 [
]
πΎ
Valve:
Finally, for the control volume enclosing the throttling valve, the entropy rate
balance reduces to
0 = πΜπππ (π 3 − π 4 ) + π Μπππ,π£ππ
Solving for π Μπππ,π£ππ and inserting values
M. Bahrami
ENSC 388
Assignment # 6
7
π Μπππ,π£ππ = πΜπππ (π 4 − π 3 ) = 0.075 [
ππ
ππ½
ππ½
] &times; (0.3448 [
] − 0.33846 [
])
π
ππ. πΎ
ππ. πΎ
ππ
= 4.755 &times; 10−4 [
]
πΎ
(b) The following table summarize, in rank order, the calculated entropy
production rates
ππΎ
]
π²
compressor 1.095 &times; 10−3
valve
4.755 &times; 10−4
condenser 2.677 &times; 10−3
Component
πΜ πππ [
Entropy production in the compressor is due to fluid friction, mechanical friction
of the moving parts, and internal heat transfer. For the valve, the irreversibility is
primarily due to fluid friction accompanying the expansion across the valve. The
principal source of irreversibility in the condenser is the temperature difference
between the air and refrigerant streams. In this problem, there are no pressure
drops for either stream passing through the condenser, but slight pressure drops
due to fluid friction would normally contribute to the irreversibility of condensers.
The evaporator lightly has not analyzed in this solution.
Note (1): Since the temperature difference is large in problem one, applying
approximate solution may leads to considerable error. Try to apply the
approximate solution and see how different the result is.
Note (2): Due to relatively small temperature change of the air, the specific heat ππ
can be taken constant at the average of the inlet and exit air temperature.
Note (3): Temperature in πΎ are used to evaluate πΜπππ , but since a temperature
difference is involved the same result would be obtained if temperature in β were
used. Temperatures in πΎ must be used when a temperature ratio is involved.
M. Bahrami
ENSC 388
Assignment # 6
8
```