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Rec 10 solutions

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MAE 241 - Summer 2022
Recitation 10 - Administered 16-June, 2022
Instructions
- You should attempt problems 1 and 2 to receive credit. We just want to see your attempt,
regardless correct or incorrect.
- Provide as much details as you can in your solutions
- Both problems are good practice problems for Exam 2.
- At the end of the recitation please upload your work to GradeScope; individual submission.
Problem 1– Four stroke heat engine modeled as a Carnot Cycle
A particular engine is modeled as heat engine that operates as an air-standard Carnot cycle is executed
in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after
the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ,
please answer the following.
TH= 1200 K and TL = 350 K
a. Sketch the cycle on a T-s diagram. Include the isobars for 350 kPa and 300 kPa and the heat
interactions between the cycle and the surroundings according to the processes that exchange
heat.
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MAE 241 – Summer 2022
Recitation 10
b. Using variable specific heats, determine the maximum pressure in the cycle. Hint: if you want
to visualize where the maximum pressure occurs and the T-s diagram is not helping you,
represent the cycle on the P-v diagram.
The maximum pressure in the cycle will be after isentropic compression process (4-1). To
visualize it better, P-v diagram of Carnot cycle is shown below,
Process 4-1 is an isentropic compression process. So, based on variable specific heats,
𝑃4
π‘ƒπ‘Ÿ4
π‘ƒπ‘Ÿ1
( )
=
=> 𝑃1 = ( ) 𝑃4
𝑃1 𝑠=π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ π‘ƒπ‘Ÿ1
π‘ƒπ‘Ÿ4
From Table A-17
At 350 K , π‘ƒπ‘Ÿ4 = 2.379 and At 1200 K π‘ƒπ‘Ÿ1 = 238.0
238.0
𝑃1 = (
) 300 π‘˜π‘ƒπ‘Ž = 30,012.61 π‘˜π‘ƒπ‘Ž
2.379
c. Compute the heat transfer to the air in the cycle. Hint: look at your cycle diagram on part (a)
Conservation of the Energy Equation for process 1-2 as a closed system is given by
𝐸𝑖𝑛 − πΈπ‘œπ‘’π‘‘ = βˆ†πΈπ‘ π‘¦π‘ π‘‘π‘’π‘š
(𝑄𝑖𝑛 + π‘Šπ‘–π‘› + πΈπ‘šπ‘Žπ‘ π‘  𝑖𝑛 ) − (π‘„π‘œπ‘’π‘‘ + π‘Šπ‘œπ‘’π‘‘ + πΈπ‘šπ‘Žπ‘ π‘  π‘œπ‘’π‘‘ ) = βˆ†π‘ˆ + βˆ†πΎπΈ + βˆ†π‘ƒπΈ
Let’s make some assumptions that will simplify the equation:
1- The system is stationary, βˆ†πΎπΈ = βˆ†π‘ƒπΈ = 0
2- Since the piston is a closed system then, there is no mass enter or leave the system,
πΈπ‘šπ‘Žπ‘ π‘  𝑖𝑛 = πΈπ‘šπ‘Žπ‘ π‘  π‘œπ‘’π‘‘ = 0
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MAE 241 – Summer 2022
Recitation 10
3- No heat leaves the piston-cylinder assembly, π‘„π‘œπ‘’π‘‘ = 0
4- Air behaves as an ideal gas
Then, we will end up with the following equation,
𝑄𝑖𝑛 + π‘Šπ‘–π‘› − π‘Šπ‘œπ‘’π‘‘ = βˆ†π‘ˆ
Or in terms of specific quantities (since mass inside the system is constant always)
π‘žπ‘–π‘› + 𝑀𝑖𝑛 − π‘€π‘œπ‘’π‘‘ = βˆ†π‘’ (1)
Process 1-2 is an expansion process. So, 𝑀𝑖𝑛 = 0. The boundary work for an isothermal
expansion is given by
𝑃
π‘€π‘œπ‘’π‘‘,12 = 𝑅𝑇1 ln 𝑃1 (2)
2
Finding π‘·πŸ
From ideal gas law, 𝑣4 =
𝑅𝑇4
𝑃4
0.2867
=
π‘˜π½
.
π‘˜π‘”−𝐾
350𝐾
300 π‘˜π‘ƒπ‘Ž
π‘š3
= 0.3344 π‘˜π‘”
Process 4-1 is an isentropic process. So from variable specific heats,
𝑣4 π‘£π‘Ÿ4
π‘£π‘Ÿ1
=
=> 𝑣1 = ( ) 𝑣4
𝑣1 π‘£π‘Ÿ1
π‘£π‘Ÿ4
From Table A-17
At 350 K , π‘£π‘Ÿ4 = 422.2 and At 1200 K π‘£π‘Ÿ1 = 14.470
π‘£π‘Ÿ1
14.470
π‘š3
π‘š3
𝑣1 = ( ) 𝑣4 = (
) 0.3344
= 0.01146
π‘£π‘Ÿ4
422.2
π‘˜π‘”
π‘˜π‘”
From ideal gas law, 𝑣3 =
𝑅𝑇3
𝑃3
=
0.2867
π‘˜π½
.
π‘˜π‘”−𝐾
350𝐾
150 π‘˜π‘ƒπ‘Ž
π‘š3
= 0.6689 π‘˜π‘”
Similarly 2-3 is an isentropic expansion process. So, from variable specific heats,
𝑣2 π‘£π‘Ÿ2
π‘£π‘Ÿ2
=
=> 𝑣2 = ( ) 𝑣3
𝑣3 π‘£π‘Ÿ3
π‘£π‘Ÿ3
From Table A-17
At 350 K , π‘£π‘Ÿ3 = 422.2 and At 1200 K π‘£π‘Ÿ2 = 14.470
π‘£π‘Ÿ2
14.470
π‘š3
) 𝑣3 = (
) 𝑣3 = 0.02292
π‘£π‘Ÿ3
422.2
π‘˜π‘”
Process 2-3 is an isothermal process. So,
𝑃1 𝑣1 = 𝑃2 𝑣2
𝑣2 = (
𝑃2 = (
𝑣1
)𝑃 =
𝑣2 1
0.01146
π‘š3
π‘˜π‘”
π‘š3
0.02292
π‘˜π‘” )
(
Alternatively, 𝑃2 can be found as follows,
30,012.61 π‘˜π‘ƒπ‘Ž = 15,001.46 π‘˜π‘ƒπ‘Ž
Similarly 2-3 is an isentropic expansion process. So, from variable specific heats,
𝑃2 π‘ƒπ‘Ÿ2
π‘ƒπ‘Ÿ2
=
=> 𝑃2 = ( ) 𝑃3
𝑃3 π‘ƒπ‘Ÿ3
π‘ƒπ‘Ÿ3
From Table A-17
At 350 K , π‘ƒπ‘Ÿ3 = 2.379 and At 1200 K π‘ƒπ‘Ÿ2 = 238.0
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MAE 241 – Summer 2022
Recitation 10
𝑃2 = (
π‘ƒπ‘Ÿ2
238.0
) 𝑃3 = (
) 150 π‘˜π‘ƒπ‘Ž = 15,006.30 π‘˜π‘ƒπ‘Ž
π‘ƒπ‘Ÿ3
2.379
So, from (2)
π‘€π‘œπ‘’π‘‘ = 𝑅𝑇1 ln
𝑃1
π‘˜π½
30,012.61 π‘˜π‘ƒπ‘Ž
π‘˜π½
= 0.2867
. 1200 𝐾 ln
= 238.47
𝑃2
π‘˜π‘” − 𝐾
15,006.30 π‘˜π‘ƒπ‘Ž
π‘˜π‘”
From (1)
π‘žπ‘–π‘› = π‘€π‘œπ‘’π‘‘ + βˆ†π‘’ = π‘€π‘œπ‘’π‘‘,12 + 𝑒2 − 𝑒1
𝑒2 = 𝑒1 since both states are at same temperature.
π‘˜π½
π‘žπ‘–π‘› = π‘€π‘œπ‘’π‘‘,12 = 238.47
π‘˜π‘”
d. Compute the entropy change per unit mass, in kJ/kg-K, for the high temperature process.
Entropy balance for a closed system and the process 1-2 is given by,
𝑄
𝑄
βˆ†π‘†12 = ∑ − ∑ + 𝑆𝑔𝑒𝑛
𝑇
𝑇
𝑖𝑛
π‘œπ‘’π‘‘
In a Carnot cycle, all the processes are reversible. So, 𝑆𝑔𝑒𝑛 = 0. In the high temperature process,
there is only heat addition. So, π‘„π‘œπ‘’π‘‘ = 0.
βˆ†π‘†12 =
𝑄𝑖𝑛
𝑇𝐻
In terms of specific quantities,
π‘˜π½
π‘žπ‘–π‘› 238.47 π‘˜π‘”
π‘˜π½
βˆ†π‘ 12 =
=
= 0.1987
𝑇𝐻
1200 𝐾
π‘˜π‘” − 𝐾
e. Determine the mass of air in the cycle, in kg, with three significant digits.
Energy balance for the entire cycle
Conservation of the Energy Equation is given by
𝐸𝑖𝑛 − πΈπ‘œπ‘’π‘‘ = βˆ†πΈπ‘ π‘¦π‘ π‘‘π‘’π‘š
(𝑄𝑖𝑛 + π‘Šπ‘–π‘› + πΈπ‘šπ‘Žπ‘ π‘  𝑖𝑛 ) − (π‘„π‘œπ‘’π‘‘ + π‘Šπ‘œπ‘’π‘‘ + πΈπ‘šπ‘Žπ‘ π‘  π‘œπ‘’π‘‘ ) = βˆ†π‘ˆ + βˆ†πΎπΈ + βˆ†π‘ƒπΈ
Let’s make some assumptions that will simplify the equation:
1- The system is stationary, βˆ†πΎπΈ = βˆ†π‘ƒπΈ = 0
2- Since the piston is a closed system then, there is no mass enter or leave the system,
πΈπ‘šπ‘Žπ‘ π‘  𝑖𝑛 = πΈπ‘šπ‘Žπ‘ π‘  π‘œπ‘’π‘‘ = 0
For a cycle, βˆ†π‘ˆ = 0 we will end up with the following equation,
π‘Šπ‘œπ‘’π‘‘ − π‘Šπ‘–π‘› = π‘Šπ‘›π‘’π‘‘ = 𝑄𝑖𝑛 − π‘„π‘œπ‘’π‘‘ = π‘š(π‘žπ‘–π‘› − π‘žπ‘œπ‘’π‘‘ )
π‘š=π‘ž
π‘Šπ‘›π‘’π‘‘
𝑖𝑛−π‘žπ‘œπ‘’π‘‘
(3)
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MAE 241 – Summer 2022
Recitation 10
Conservation of the Energy Equation for process 3-4 is given by
𝐸𝑖𝑛 − πΈπ‘œπ‘’π‘‘ = βˆ†πΈπ‘ π‘¦π‘ π‘‘π‘’π‘š
(𝑄𝑖𝑛 + π‘Šπ‘–π‘› + πΈπ‘šπ‘Žπ‘ π‘  𝑖𝑛 ) − (π‘„π‘œπ‘’π‘‘ + π‘Šπ‘œπ‘’π‘‘ + πΈπ‘šπ‘Žπ‘ π‘  π‘œπ‘’π‘‘ ) = βˆ†π‘ˆ + βˆ†πΎπΈ + βˆ†π‘ƒπΈ
Let’s make some assumptions that will simplify the equation:
1- The system is stationary, βˆ†πΎπΈ = βˆ†π‘ƒπΈ = 0
2- Since the piston is a closed system then, there is no mass enter or leave the system,
πΈπ‘šπ‘Žπ‘ π‘  𝑖𝑛 = πΈπ‘šπ‘Žπ‘ π‘  π‘œπ‘’π‘‘ = 0
3- No heat is added to the piston, π‘„π‘œπ‘’π‘‘ = 0
Then, we will end up with the following equation,
π‘„π‘œπ‘’π‘‘ = π‘Šπ‘–π‘› − π‘Šπ‘œπ‘’π‘‘ + βˆ†π‘ˆ
Or in terms of specific quantities (since mass inside the system is constant always)
π‘žπ‘œπ‘’π‘‘ = 𝑀𝑖𝑛 − π‘€π‘œπ‘’π‘‘ + βˆ†π‘’ (4)
The term 𝑀𝑖𝑛 − π‘€π‘œπ‘’π‘‘ will be the work done during isothermal process. Calculating π‘€π‘œπ‘’π‘‘
and identifying its sign will help us identify if the work is transferred out or into the
system
𝑃
π‘˜π½
150
π‘˜π½
π‘€π‘œπ‘’π‘‘ = 𝑅𝑇 ln 𝑃3 = 0.2867 π‘˜π‘”−𝐾 350 𝐾 ln 300 = −69.55 π‘˜π‘”
4
π‘˜π½
π‘˜π‘”
From (4) after enforcing βˆ†π‘’ = 0, since both 2 and 3 are at same temperatures
π‘˜π½
π‘žπ‘œπ‘’π‘‘ = 69.55
π‘˜π‘”
From (3)
0.5 π‘˜π½
π‘š=
= 2.960 × 10−3 π‘˜π‘”
π‘˜π½
π‘˜π½
238.47 π‘˜π‘” − 69.55 π‘˜π‘”
• π‘žπ‘–π‘› − π‘žπ‘œπ‘’π‘‘ is the area of rectangle in the T-s diagram which will give the net
work.
π‘†π‘œ, 𝑀𝑖𝑛 = 69.55
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MAE 241 – Summer 2022
Recitation 10
Problem 2 (Problem 9.52 ideal Diesel cycle)
An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the
compression process, air is at 95 kPa and 27°C. Accounting for the variation of specific heats with
temperature, determine
(a) the temperature after the heat-addition process,
(b) the thermal efficiency, and (c) the mean effective pressure.
Air-standard assumptions for diesel cycle
From textbook Page 480 (heat addition at constant pressure for diesel)
State 1:
𝑃1 = 95 π‘˜π‘ƒπ‘Ž, 𝑇1 = 27 ℃
From ideal gas law, we can calculate 𝑣1 =
𝑅𝑇1
𝑃1
=
(0.2867
π‘˜π½
)(273+27)𝐾
π‘˜π‘”−𝐾
95 π‘˜π‘ƒπ‘Ž
π‘š3
= 0.9063 π‘˜π‘”
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MAE 241 – Summer 2022
Recitation 10
State 2:
Compression ratio π‘Ÿ =
𝑣
So, 𝑣2 = 161 =
π‘š3
0.9063
π‘˜π‘”
16
π‘£π‘šπ‘Žπ‘₯
π‘£π‘šπ‘–π‘›
𝑣
= 𝑣1 = 16
2
= 0.0566
π‘š3
π‘˜π‘”
Also, Process 1-2 is an isentropic compression process. Assuming variable specific heats,
𝑣2
π‘£π‘Ÿ2
( )
=
= 16
𝑣1 𝑠=π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ π‘£π‘Ÿ1
From Table A-17 at 300 K we get π‘£π‘Ÿ1 = 621.2. So, π‘£π‘Ÿ2 =
π‘£π‘Ÿ1
16
=
(621.2)
16
= 38.825.
Again, from Table A-17, we get 𝑇2 = 862.35 𝐾 after linear interpolation
So, from ideal gas law,
π‘˜π½
𝑅𝑇2 (0.2867 π‘˜π‘” − 𝐾 ) (862.35 )𝐾
𝑃2 =
=
= 4368.12 π‘˜π‘ƒπ‘Ž
π‘š3
𝑣2
0.0566 π‘˜π‘”
State 3:
𝑃2 = 𝑃3 = 4368.12 π‘˜π‘ƒπ‘Ž
𝑣
Cut off ratio π‘Ÿπ‘ = 𝑣3 = 2. So, 𝑣3 = 2. 𝑣2 = 2 (0.0566
2
𝑇3 =
𝑃3 𝑣3
=
𝑅
π‘š3
π‘š3
) = 0.1132 π‘˜π‘”
π‘˜π‘”
(4368.12 π‘˜π‘ƒπ‘Ž) (0.1132
π‘˜π½
0.2867 π‘˜π‘” − 𝐾
π‘š3
π‘˜π‘” )
= 1724.69 𝐾
State 4:
π‘š3
𝑣4 = 𝑣1 = 0.9063
π‘˜π‘”
Process 3-4 is isentropic expansion. So, assuming variable specific heats
𝑣4 π‘£π‘Ÿ4
=
𝑣3 π‘£π‘Ÿ3
𝑣4
π‘£π‘Ÿ4 = ( ) π‘£π‘Ÿ3
𝑣3
From Table A-17, at 𝑇3 = 1724.69 𝐾, we get π‘£π‘Ÿ3 = 4.5380
π‘š3
0.9063
𝑣4
π‘˜π‘”
π‘£π‘Ÿ4 = ( ) π‘£π‘Ÿ3 = π‘£π‘Ÿ4 = (
) 4.5380 = 36.332
π‘š3
𝑣3
0.1132 π‘˜π‘”
From Table A-17, we get 𝑇4 = 882.41 𝐾
From ideal gas law
Page 7 of 15
MAE 241 – Summer 2022
Recitation 10
π‘˜π½
𝑅𝑇4 (0.2867 π‘˜π‘” − 𝐾 ) 882.41 𝐾
𝑃4 =
=
= 279.14 π‘˜π‘ƒπ‘Ž
π‘š3
𝑣4
0.9063 π‘˜π‘”
a. the temperature after the heat-addition process,
π‘»πŸ‘ = πŸπŸ•πŸπŸ’. πŸ”πŸ— 𝑲
b. the thermal efficiency,
The thermal efficiency of the cycle is given by,
πœ‚π‘‘β„Ž = 1 − (
π‘žπ‘œπ‘’π‘‘
)
π‘žπ‘–π‘›
The heat addition (Process 2-3):
Noting that the Diesel cycle is executed in a piston–cylinder device, which forms a closed system.
Conservation of the Energy Equation is given by
𝐸𝑖𝑛 − πΈπ‘œπ‘’π‘‘ = βˆ†πΈπ‘ π‘¦π‘ π‘‘π‘’π‘š
(𝑄𝑖𝑛 + π‘Šπ‘–π‘› + πΈπ‘šπ‘Žπ‘ π‘  𝑖𝑛 ) − (π‘„π‘œπ‘’π‘‘ + π‘Šπ‘œπ‘’π‘‘ + πΈπ‘šπ‘Žπ‘ π‘  π‘œπ‘’π‘‘ ) = βˆ†π‘ˆ + βˆ†πΎπΈ + βˆ†π‘ƒπΈ
Let’s make some assumptions that will simplify the equation:
1. The system is stationary, βˆ†πΎπΈ = βˆ†π‘ƒπΈ = 0
2. Since the piston is a closed system then, there is no mass enter or leave the system,
πΈπ‘šπ‘Žπ‘ π‘  𝑖𝑛 = πΈπ‘šπ‘Žπ‘ π‘  π‘œπ‘’π‘‘ = 0
3. No heat leaves the piston, π‘„π‘œπ‘’π‘‘ = 0
4. Only boundary work is present and is assumed to be transferred to the surrounding at a
constant pressure so, π‘Šπ‘–π‘› = 0
Then, we will end up with the following from Eq3:
𝑄𝑖𝑛 − π‘Šπ‘œπ‘’π‘‘ = βˆ†π‘ˆ
𝑄𝑖𝑛 = π‘š(𝑃2 (𝑣3 − 𝑣2 ) + (𝑒3 − 𝑒2 )) = π‘š(β„Ž3 − β„Ž2 )
π‘œπ‘Ÿ π‘žπ‘–π‘› =
𝑄𝑖𝑛
π‘š
= (β„Ž3 − β„Ž2 ) (1)
The heat rejection (Process 1-4)
Conservation of the Energy Equation is given by
𝐸𝑖𝑛 − πΈπ‘œπ‘’π‘‘ = βˆ†πΈπ‘ π‘¦π‘ π‘‘π‘’π‘š
(𝑄𝑖𝑛 + π‘Šπ‘–π‘› + πΈπ‘šπ‘Žπ‘ π‘  𝑖𝑛 ) − (π‘„π‘œπ‘’π‘‘ + π‘Šπ‘œπ‘’π‘‘ + πΈπ‘šπ‘Žπ‘ π‘  π‘œπ‘’π‘‘ ) = βˆ†π‘ˆ + βˆ†πΎπΈ + βˆ†π‘ƒπΈ
Let’s make some assumptions that will simplify the equation:
1- The system is stationary , βˆ†πΎπΈ = βˆ†π‘ƒπΈ = 0
Page 8 of 15
MAE 241 – Summer 2022
Recitation 10
2- Since the piston is a closed system then, there is no mass enter or leave the system,
πΈπ‘šπ‘Žπ‘ π‘  𝑖𝑛 = πΈπ‘šπ‘Žπ‘ π‘  π‘œπ‘’π‘‘ = 0
3- No heat enters the piston, 𝑄𝑖𝑛 = 0
4- No work interactions , π‘Šπ‘–π‘› = 0 and π‘Šπ‘œπ‘’π‘‘ = 0
Then, we will end up with the following from Eq3:
−π‘„π‘œπ‘’π‘‘ = βˆ†π‘ˆ
π‘„π‘œπ‘’π‘‘ = π‘š(𝑒4 − 𝑒1 )
π‘žπ‘œπ‘’π‘‘ =
π‘„π‘œπ‘’π‘‘
π‘š
or
= 𝑒4 − 𝑒1 (2)
So,
πœ‚π‘‘β„Ž = 1 − (
π‘žπ‘œπ‘’π‘‘
𝑒4 − 𝑒1
)=1−(
)
π‘žπ‘–π‘›
β„Ž3 − β„Ž2
From Table A-17,
State 1: at 300 K, 𝑒1 = 214.07
π‘˜π½
π‘˜π‘”
π‘˜π½
State 2: at 862.35 𝐾, β„Ž2 = 890.89 π‘˜π‘”
π‘˜π½
State 3: at 1724.6 𝐾 β„Ž3 = 1910.36 π‘˜π‘”
State 4: at 882.41 𝐾 𝑒4 = 659.95
π‘˜π½
π‘˜π‘”
π‘˜π½
π‘˜π½
659.95 π‘˜π‘” − 214.07 π‘˜π‘”
𝑒4 − 𝑒1
πœ‚π‘‘β„Ž = 1 − (
)= 1−(
) = 56.26 %
π‘˜π½
π‘˜π½
β„Ž3 − β„Ž2
1910.36 π‘˜π‘” − 890.89 π‘˜π‘”
c. Mean effective pressure
𝑀𝑛𝑒𝑑
π‘žπ‘–π‘› − π‘žπ‘œπ‘’π‘‘ (β„Ž3 − β„Ž2 ) − (𝑒4 − 𝑒1 )
=
=
π‘£π‘šπ‘Žπ‘₯ − π‘£π‘šπ‘–π‘›
𝑣1 − 𝑣2
𝑣1 − 𝑣2
π‘˜π½
π‘˜π½
π‘˜π½
π‘˜π½
(1910.36 π‘˜π‘” − 890.89 π‘˜π‘”) − (659.95 π‘˜π‘” − 214.07 π‘˜π‘”)
=
= 675.05 π‘˜π‘ƒπ‘Ž
π‘š3
π‘š3
0.9063 π‘˜π‘” − 0.0566 π‘˜π‘”
π‘€π‘’π‘Žπ‘› 𝑒𝑓𝑓𝑒𝑐𝑑𝑖𝑣𝑒 π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ =
Answers in textbook: (a) 1725 K, (b) 56.3 percent, (c) 675.9 kPa
Page 9 of 15
MAE 241 – Summer 2022
Recitation 10
Problem 3 – Ideal steam power cycle (10-95)
Steam enters the turbine of a steam power plant that operates on a simple ideal Rankine cycle at a
pressure of 6 MPa, and it leaves as a saturated vapor at 7.5 kPa. Heat is transferred to the steam in the
boiler at a rate of 40,000 kJ/s. Steam is cooled in the condenser by the cooling water from a nearby
river, which enters the condenser at 15°C. Show the cycle on a T-s diagram with respect to saturation
lines, and determine (a) the turbine inlet temperature, (b) the net power output and thermal efficiency,
and (c) the minimum mass flow rate of the cooling water required.
a) the turbine inlet temperature (T3)
π‘˜π½
State 4: β„Ž4 = β„Žπ‘”@7.5π‘˜π‘ƒπ‘Ž = 2574.0 π‘˜π‘” (A-5)
π‘˜π½
Process 3-4 is an isentropic process. So, 𝑠3 = 𝑠4 = 𝑠𝑔@7.5π‘˜π‘ƒπ‘Ž = 8.2501 π‘˜π‘”−𝐾 (A-5)
At 6 MPa, value of 𝑠3 > 𝑠𝑔@6π‘€π‘ƒπ‘Ž . So, state 3 should be in superheated state. (A-5)
From Table A-6, under 6 MPa, 𝑇3 = 1089.18 𝐾 after linear interpolation.
b) the net power output and thermal efficiency
• To calculate the net power output, mass flow rate is needed. It can be calculated from
heat input to the boiler which is already given.
Net power output:
Analyzing the boiler as a control volume:
Starting with the general mass balance:
Page 10 of 15
MAE 241 – Summer 2022
Recitation 10
∑𝑖𝑛 π‘šΜ‡ − ∑π‘œπ‘’π‘‘ π‘šΜ‡ =
π‘‘π‘šπΆπ‘‰
Eq1
𝑑𝑑
The energy balance equation:
𝐸̇𝑖𝑛 − πΈΜ‡π‘œπ‘’π‘‘ = π‘‘πΈπ‘ π‘¦π‘ π‘‘π‘’π‘š /𝑑𝑑
We can expand it knowing that the boiler is under steady state condition as follows:
𝑄̇𝑖𝑛 + π‘ŠΜ‡π‘–π‘› + ∑ π‘šΜ‡ πœƒ = π‘„Μ‡π‘œπ‘’π‘‘ + π‘ŠΜ‡π‘œπ‘’π‘‘ + ∑ π‘šΜ‡ πœƒ
𝑖𝑛
π‘œπ‘’π‘‘
Then it can be written as:
2
2
𝑉
𝑉
𝑄̇𝑖𝑛 + π‘ŠΜ‡π‘–π‘› + ∑𝑖𝑛 π‘šΜ‡ (β„Ž + + 𝑔𝑧) = π‘„Μ‡π‘œπ‘’π‘‘ + π‘ŠΜ‡π‘œπ‘’π‘‘ + ∑π‘œπ‘’π‘‘ π‘šΜ‡ (β„Ž + + 𝑔𝑧)
2
2
Eq2
The assumptions:
1- This is a steady-flow process since there is no change with time,
π‘‘π‘šπΆπ‘‰
𝑑𝑑
= 0.
2- Potential energy and kinetic energy changes are negligible, so small compared with (βˆ†β„Ž ≫
βˆ†πΎπΈ & βˆ†β„Ž ≫ βˆ†π‘ƒπΈ) so, βˆ†πΎπΈ = 0, & βˆ†π‘ƒπΈ = 0.
3- There is one inlet and one outlet.
4- There is no heat out from the boiler, π‘„Μ‡π‘œπ‘’π‘‘ = 0.
5- No work interactions between the surroundings and the system, π‘ŠΜ‡π‘–π‘› = 0, π‘ŠΜ‡π‘–π‘› = 0
Using the assumptions, Eq10 become as follows to represent the total mass of the boiler (π‘šΜ‡ ):
π‘šΜ‡π‘–π‘› = π‘šΜ‡π‘œπ‘’π‘‘ = π‘šΜ‡
Eq3
Using the assumptions, Eq11 becomes as follows:
𝑄̇𝑖𝑛 + π‘ŠΜ‡π‘–π‘› + ∑ π‘šΜ‡ (β„Ž +
𝑖𝑛
𝑉2
𝑉2
Μ‡
Μ‡
+ 𝑔𝑧) = π‘„π‘œπ‘’π‘‘ + π‘Šπ‘œπ‘’π‘‘ + ∑ π‘šΜ‡ (β„Ž +
+ 𝑔𝑧)
2
2
π‘œπ‘’π‘‘
Which will lead to:
𝑄̇𝑖𝑛 = π‘šΜ‡ [β„Ž3 − β„Ž2 ]
Eq4
Let’s analyze the pump as control volume as follows:
Starting with the general mass balance:
∑𝑖𝑛 π‘šΜ‡ − ∑π‘œπ‘’π‘‘ π‘šΜ‡ =
π‘‘π‘šπΆπ‘‰
𝑑𝑑
Eq5
The energy balance equation:
𝐸̇𝑖𝑛 − πΈΜ‡π‘œπ‘’π‘‘ = π‘‘πΈπ‘ π‘¦π‘ π‘‘π‘’π‘š /𝑑𝑑
We can expand it knowing that the pump is under steady state condition as follows:
𝑄̇𝑖𝑛 + π‘ŠΜ‡π‘–π‘› + ∑ π‘šΜ‡ πœƒ = π‘„Μ‡π‘œπ‘’π‘‘ + π‘ŠΜ‡π‘œπ‘’π‘‘ + ∑ π‘šΜ‡ πœƒ
𝑖𝑛
π‘œπ‘’π‘‘
Page 11 of 15
MAE 241 – Summer 2022
Recitation 10
Then it can be written as:
2
2
𝑉
𝑉
𝑄̇𝑖𝑛 + π‘ŠΜ‡π‘π‘’π‘šπ‘,𝑖𝑛 + ∑𝑖𝑛 π‘šΜ‡ (β„Ž + + 𝑔𝑧) = π‘„Μ‡π‘œπ‘’π‘‘ + π‘ŠΜ‡π‘œπ‘’π‘‘ + ∑π‘œπ‘’π‘‘ π‘šΜ‡ (β„Ž + + 𝑔𝑧)
2
2
π‘€π‘π‘’π‘šπ‘,𝑖𝑛 =
π‘ŠΜ‡π‘π‘’π‘šπ‘,𝑖𝑛
π‘šΜ‡
Eq6
= β„Ž2 − β„Ž1 Eq7
Since this is an ideal Rankine cycle, the pump is adiabatic and reversible, i.e. isentropic. For open
flow devices, the reversible specific work input is
2
𝑀rev,in = ∫ 𝑣 𝑑𝑃 + Δπ‘˜π‘’ + Δ𝑝𝑒
1
2
𝑀rev,in = ∫ 𝑣 𝑑𝑃
1
where we have utilized the above assumptions that Δπ‘˜π‘’ and Δ𝑝𝑒 are both small compared to any other
relevant energy terms. The inlet to the pump (state 1) is a saturated liquid, and the outlet is a compressed
liquid, so we can treat the water as an incompressible (constant 𝑣) fluid as it flows through the pump. The
specific work in the pump then becomes
2
π‘€π‘π‘’π‘šπ‘,𝑖𝑛 = 𝑀rev,in = ∫ 𝑣 𝑑𝑃 = 𝑣1 (𝑃2 − 𝑃1 ).
1
m3
State 1; 𝑣1 = 𝑣𝑓@7.5π‘˜π‘ƒπ‘Ž = 0.001008 π‘˜π‘”
π‘€π‘π‘’π‘šπ‘,𝑖𝑛 = 𝑣1 (𝑃2 − 𝑃1 ) = 0.001008
m3
π‘˜π½
(6000 π‘˜π‘ƒπ‘Ž − 7.5 π‘˜π‘ƒπ‘Ž ) = 6.04
π‘˜π‘”
π‘˜π‘”
π‘€π‘π‘’π‘šπ‘,𝑖𝑛 = β„Ž2 − β„Ž1 = 6.04
State 1: β„Ž1 = β„Žπ‘“@7.5π‘˜π‘ƒπ‘Ž = 168.75
π‘˜π½
π‘˜π‘”
π‘˜π½
π‘˜π‘”
β„Ž2 = β„Ž1 + 6.04
π‘˜π½
π‘˜π½
= 174.79
π‘˜π‘”
π‘˜π‘”
So from Eq4,
𝑄̇𝑖𝑛 = π‘šΜ‡ [β„Ž3 − β„Ž2 ]
π‘˜π½
At state 3: β„Ž3 = 4852.18 π‘˜π‘” (Table A-6)
π‘˜π½
40,000
𝑄̇𝑖𝑛
π‘˜π‘”
𝑠
π‘šΜ‡ =
=
= 8.551
π‘˜π½
π‘˜π½
β„Ž3 − β„Ž2 4852.18
𝑠
π‘˜π‘” − 174.79 π‘˜π‘”
Page 12 of 15
MAE 241 – Summer 2022
Recitation 10
π‘ŠΜ‡π‘π‘’π‘šπ‘,𝑖𝑛 = π‘šΜ‡ π‘€π‘π‘’π‘šπ‘,𝑖𝑛 = 8.551
π‘˜π‘”
π‘˜π½
π‘˜π½
. 6.04
= 51.65
𝑠
π‘˜π‘”
𝑠
Now analyzing the turbine as a control volume:
Starting with the general mass balance:
∑𝑖𝑛 π‘šΜ‡ − ∑π‘œπ‘’π‘‘ π‘šΜ‡ =
π‘‘π‘šπΆπ‘‰
Eq8
𝑑𝑑
The energy balance equation:
𝐸̇𝑖𝑛 − πΈΜ‡π‘œπ‘’π‘‘ = π‘‘πΈπ‘ π‘¦π‘ π‘‘π‘’π‘š /𝑑𝑑
We can expand it knowing that the turbine is under steady state condition as follows:
𝑄̇𝑖𝑛 + π‘ŠΜ‡π‘–π‘› + ∑ π‘šΜ‡ πœƒ = π‘„Μ‡π‘œπ‘’π‘‘ + π‘ŠΜ‡π‘œπ‘’π‘‘ + ∑ π‘šΜ‡ πœƒ
𝑖𝑛
π‘œπ‘’π‘‘
Then it can be written as:
2
2
𝑉
𝑉
𝑄̇𝑖𝑛 + π‘ŠΜ‡π‘–π‘› + ∑𝑖𝑛 π‘šΜ‡ (β„Ž + + 𝑔𝑧) = π‘„Μ‡π‘œπ‘’π‘‘ + π‘ŠΜ‡π‘œπ‘’π‘‘ + ∑π‘œπ‘’π‘‘ π‘šΜ‡ (β„Ž + + 𝑔𝑧)
2
2
Eq9
The assumptions:
1- This is a steady-flow process since there is no change with time,
π‘‘π‘šπΆπ‘‰
𝑑𝑑
= 0.
2- Potential energy and kinetic energy changes are negligible, so small compared with (βˆ†β„Ž ≫
βˆ†πΎπΈ & βˆ†β„Ž ≫ βˆ†π‘ƒπΈ) so, βˆ†πΎπΈ = 0, & βˆ†π‘ƒπΈ = 0.
3- There is one inlet and one outlet.
4- The turbine is adiabatic , 𝑄̇𝑖𝑛 = 0 π‘Žπ‘›π‘‘ π‘„Μ‡π‘œπ‘’π‘‘ = 0.
5- No work applied from the surroundings to the system, π‘ŠΜ‡π‘–π‘› = 0
Using the assumptions, Eq6 become as follows to represent the total mass of the turbine (π‘šΜ‡ ):
π‘šΜ‡π‘–π‘› = π‘šΜ‡π‘œπ‘’π‘‘ = π‘šΜ‡
Eq10
Using the assumptions, Eq7 becomes as follows:
𝑄̇𝑖𝑛 + π‘ŠΜ‡π‘–π‘› + ∑ π‘šΜ‡ (β„Ž +
𝑖𝑛
𝑉2
𝑉2
+ 𝑔𝑧) = π‘„Μ‡π‘œπ‘’π‘‘ + π‘ŠΜ‡π‘‘π‘’π‘Ÿπ‘,π‘œπ‘’π‘‘ + ∑ π‘šΜ‡ (β„Ž +
+ 𝑔𝑧)
2
2
π‘œπ‘’π‘‘
Which will lead to:
π‘˜π‘”
π‘˜π½
π‘˜π½
π‘˜π½
π‘ŠΜ‡π‘‘π‘’π‘Ÿπ‘,π‘œπ‘’π‘‘ = π‘šΜ‡ [β„Ž3 − β„Ž4 ] = 8.551 𝑠 (4852.18 π‘˜π‘” − 2574.0 π‘˜π‘”) = 19,480.71 𝑠 Eq11
π‘ŠΜ‡π‘›π‘’π‘‘ = π‘ŠΜ‡π‘‘π‘’π‘Ÿπ‘,π‘œπ‘’π‘‘ − π‘ŠΜ‡π‘π‘’π‘šπ‘,𝑖𝑛 = 19,480.71
π‘˜π½
π‘˜π½
π‘˜π½
− 51.65 = 19,429.06
𝑠
𝑠
𝑠
Page 13 of 15
MAE 241 – Summer 2022
Recitation 10
πœ‚π‘‘β„Ž
π‘˜π½
π‘ŠΜ‡π‘›π‘’π‘‘ 19,429.06 𝑠
=
=
= 48.57 %
π‘˜π½
𝑄̇𝑖𝑛
40,000
𝑠
c) the minimum mass flow rate of the cooling water required.
Assumptions
𝑑𝐸𝑐𝑣
π‘‘π‘šπ‘π‘£
•
Condenser is operating under steady flow conditions.
•
•
•
No mixing happens inside the Condenser.
Pressure drop is negligible for steam and cooling water.
Change in kinetic and potential energies are small compared to change in other energies. βˆ†πΎπΈ =
βˆ†π‘ƒπΈ ≈ 0
𝑑𝑑
= 0,
𝑑𝑑
=0
The cooling water should be raised to steam temperature so that steam completely cools down. So, the
temperature of the cooling water will be saturation temperature at 7.5 KPa. So, Tout=40.29 ℃.
Writing energy balance for the system chosen in this part and applying the assumptions listed above,
𝑑𝐸
𝐸̇𝑖𝑛 − πΈΜ‡π‘œπ‘’π‘‘ = 𝑑𝑑𝑐𝑣 Eq11
(𝑄̇𝑖𝑛 + π‘ŠΜ‡π‘–π‘› + ∑ π‘šΜ‡ (β„Ž +
𝑖𝑛
𝑉2
𝑉2
𝑑𝐸𝑐𝑣
Μ‡
Μ‡
)
(𝑄
+ 𝑔𝑧) − π‘œπ‘’π‘‘ + π‘Šπ‘œπ‘’π‘‘ + ∑ π‘šΜ‡ (β„Ž +
+ 𝑔𝑧)) =
2
2
𝑑𝑑
π‘œπ‘’π‘‘
π‘šΜ‡ 𝑠 (β„Ž4 ) + π‘šπ‘€
Μ‡ (β„Žπ‘–π‘› ) − π‘šΜ‡ 𝑠 (β„Ž1 ) − π‘šπ‘€
Μ‡ (β„Žπ‘œπ‘’π‘‘ ) = 0
π‘šΜ‡ π‘Ÿ (β„Ž4 − β„Ž1 ) = π‘šπ‘€
Μ‡ (β„Žπ‘œπ‘’π‘‘ − β„Žπ‘–π‘› )
π‘šΜ‡ (β„Ž4−β„Ž1 )
π‘œπ‘’π‘‘ −β„Žπ‘–π‘› )
π‘šΜ‡π‘€ = (β„Žπ‘ 
Eq12
kJ
At inlet of the cooling water, β„Žπ‘–π‘› = β„Žπ‘“@15℃ = 62.982 kg(A-4)
π‘˜π½
So, β„Žπ‘œπ‘’π‘‘ = β„Žπ‘“@40.29℃ = 168.74 π‘˜π‘”(A-4)
Page 14 of 15
MAE 241 – Summer 2022
Recitation 10
π‘˜π‘”
π‘˜π½
π‘˜π½
8.551 𝑠 (2574.0 π‘˜π‘” − 168.75 π‘˜π‘”)
π‘˜π‘”
π‘šΜ‡π‘€ =
= 194.47
π‘˜π½
kJ
𝑠
(168.74 π‘˜π‘” − 62.982 kg)
Page 15 of 15
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