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HOT SEAT
Thermodynamics A 214
5th May 2020
Adam Venter
PhD Candidate
Solar Thermal Energy Research Group (STERG) | Stellenbosch University
QUESTION 7 (7-127)
Problem:
Air enters a compressor steadily at the ambient conditions of 100 kPa and 20 °C and leaves at
800 kPa. Heat is lost from the compressor in the amount of 120 kJ/kg and the air experiences
an entropy decrease of 0.40 kJ/kg.k. Using constant specific heats. Determine (a) the exit
temperature of the air, (b) the work input to the compressor, and (c) the entropy generation
during this process.
Key notes: Air, Compressor, Steadily, kJ/kg
Ideal gas
Control volume analysis
 Steady flow process
Per unit mass basis
QUESTION 7 (7-127)
1
Problem:
2
Air enters a compressor steadily at the ambient conditions of 100 kPa and 20 °C and leaves at
800 kPa. Heat is lost from the compressor in the amount of 120 kJ/kg and the air experiences
an entropy decrease of 0.40 kJ/kg.k. Using constant specific heats. Determine (a) the exit
temperature of the air, (b) the work input to the compressor, and (c) the entropy generation
during this process.
0
𝑒𝑒𝑖𝑖𝑖𝑖 − π‘’π‘’π‘œπ‘œπ‘œπ‘œπ‘œπ‘œ = Δ𝑒𝑒𝑠𝑠𝑠𝑠𝑠𝑠
800 kPa
π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
120 π‘˜π‘˜π‘˜π‘˜/π‘˜π‘˜π‘˜π‘˜
𝑀𝑀𝑖𝑖𝑖𝑖
Compressor
Air
100 kPa
20 °C
QUESTION 7 (7-127)
Problem:
Air enters a compressor steadily at the ambient conditions of 100 kPa and 20 °C and leaves at
800 kPa. Heat is lost from the compressor in the amount of 120 kJ/kg and the air experiences
an entropy decrease of 0.40 kJ/kg.k. Using constant specific heats. Determine (a) the exit
temperature of the air, (b) the work input to the compressor, and (c) the entropy generation
during this process.
𝑒𝑒𝑖𝑖𝑖𝑖 = π‘’π‘’π‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
800 kPa
π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
120 π‘˜π‘˜π‘˜π‘˜/π‘˜π‘˜π‘˜π‘˜
𝑀𝑀𝑖𝑖𝑖𝑖
0
Compressor
Air
100 kPa
20 °C
π‘žπ‘žπ‘–π‘–π‘–π‘– + 𝑀𝑀𝑖𝑖𝑖𝑖 + πœƒπœƒπ‘–π‘–π‘–π‘– = π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ + π‘€π‘€π‘œπ‘œπ‘œπ‘œπ‘œπ‘œ + πœƒπœƒπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
0
0
0
0
0
π‘žπ‘žπ‘–π‘–π‘–π‘– + 𝑀𝑀𝑖𝑖𝑖𝑖 + β„Žπ‘–π‘–π‘–π‘– + π‘˜π‘˜π‘’π‘’π‘–π‘–π‘–π‘– + 𝑝𝑝𝑒𝑒𝑖𝑖𝑖𝑖 = π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ + π‘€π‘€π‘œπ‘œπ‘œπ‘œπ‘œπ‘œ + β„Žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ + π‘˜π‘˜π‘’π‘’π‘œπ‘œπ‘œπ‘œπ‘œπ‘œ + π‘π‘π‘’π‘’π‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
ASSUMPTIONS
- Steady operating conditions exist
- π‘˜π‘˜π‘˜π‘˜ = 𝑝𝑝𝑝𝑝 = 0
- Ideal gas with constant specific
heats @ 300 K
- π‘€π‘€π‘œπ‘œπ‘œπ‘œπ‘œπ‘œ = π‘žπ‘žπ‘–π‘–π‘–π‘– = 0
QUESTION 7 (7-127)
Problem:
Air enters a compressor steadily at the ambient conditions of 100 kPa and 20 °C and leaves at
800 kPa. Heat is lost from the compressor in the amount of 120 kJ/kg and the air experiences
an entropy decrease of 0.40 kJ/kg.k. Using constant specific heats. Determine (a) the exit
temperature of the air, (b) the work input to the compressor, and (c) the entropy generation
during this process.
𝑒𝑒𝑖𝑖𝑖𝑖 = π‘’π‘’π‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
800 kPa
π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
120 π‘˜π‘˜π‘˜π‘˜/π‘˜π‘˜π‘˜π‘˜
𝑀𝑀𝑖𝑖𝑖𝑖
Compressor
Air
100 kPa
20 °C
𝑀𝑀𝑖𝑖𝑖𝑖 + β„Žπ‘–π‘–π‘–π‘– = π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ + β„Žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
𝑀𝑀𝑖𝑖𝑖𝑖 = β„Žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ − β„Žπ‘–π‘–π‘–π‘– + π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
Δβ„Ž = 𝑐𝑐𝑝𝑝,π‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Ž Δ𝑇𝑇
𝑀𝑀𝑖𝑖𝑖𝑖 = 𝑐𝑐𝑝𝑝 𝑇𝑇2 − 𝑇𝑇1 + π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
QUESTION 7 (7-127)
Problem:
Air enters a compressor steadily at the ambient conditions of 100 kPa and 20 °C and leaves at
800 kPa. Heat is lost from the compressor in the amount of 120 kJ/kg and the air experiences
an entropy decrease of 0.40 kJ/kg.k. Using constant specific heats. Determine (a) the exit
temperature of the air, (b) the work input to the compressor, and (c) the entropy generation
during this process.
𝑀𝑀𝑖𝑖𝑖𝑖 = 𝑐𝑐𝑝𝑝 𝑇𝑇2 − 𝑇𝑇1 + π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
800 kPa
π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
120 π‘˜π‘˜π‘˜π‘˜/π‘˜π‘˜π‘˜π‘˜
𝑀𝑀𝑖𝑖𝑖𝑖
Δπ‘†π‘†π‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Ž = 𝑐𝑐𝑝𝑝 ln
Compressor
Air
100 kPa
20 °C
𝑑𝑑𝑑 𝑣𝑣𝑣𝑣𝑣𝑣
𝑑𝑑𝑑𝑑 =
−
𝑇𝑇
𝑇𝑇
2
2
𝑑𝑑𝑑
𝑣𝑣𝑣𝑣𝑣𝑣
Δ𝑠𝑠 = οΏ½
−οΏ½
𝑇𝑇
𝑇𝑇
1
1
𝑇𝑇2
𝑃𝑃2
− 𝑅𝑅𝑅𝑅𝑅𝑅
𝑇𝑇1
𝑃𝑃1
𝑇𝑇𝑇𝑇𝑇𝑇 = 𝑑𝑑𝑑 − 𝑣𝑣𝑣𝑣𝑣𝑣
𝑑𝑑𝑑 = 𝑐𝑐𝑝𝑝 𝑑𝑑𝑑𝑑
𝑅𝑅𝑅𝑅
𝑣𝑣 =
𝑃𝑃
2 𝑐𝑐 𝑑𝑑𝑑𝑑
𝑝𝑝
Δ𝑠𝑠 = οΏ½
1
𝑇𝑇
2
2
𝑅𝑅𝑑𝑑𝑑𝑑
𝑃𝑃
1
−οΏ½
2
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
Δ𝑠𝑠 = 𝑐𝑐𝑝𝑝 οΏ½
− 𝑅𝑅 οΏ½
𝑇𝑇
1
1 𝑃𝑃
QUESTION 7 (7-127)
Problem:
Air enters a compressor steadily at the ambient conditions of 100 kPa and 20 °C and leaves at
800 kPa. Heat is lost from the compressor in the amount of 120 kJ/kg and the air experiences
an entropy decrease of 0.40 kJ/kg.k. Using constant specific heats. Determine (a) the exit
temperature of the air, (b) the work input to the compressor, and (c) the entropy generation
during this process.
𝑀𝑀𝑖𝑖𝑖𝑖 = 𝑐𝑐𝑝𝑝 𝑇𝑇2 − 𝑇𝑇1 + π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
800 kPa
π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
120 π‘˜π‘˜π‘˜π‘˜/π‘˜π‘˜π‘˜π‘˜
𝑀𝑀𝑖𝑖𝑖𝑖
Compressor
Air
100 kPa
20 °C
Δπ‘†π‘†π‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Ž = 𝑐𝑐𝑝𝑝 ln
𝑇𝑇2
𝑃𝑃2
− 𝑅𝑅𝑅𝑅𝑅𝑅
𝑇𝑇1
𝑃𝑃1
Table A-2
−0.40 π‘˜π‘˜π‘˜π‘˜⁄π‘˜π‘˜π‘˜π‘˜. 𝐾𝐾 = (1.005 π‘˜π‘˜π‘˜π‘˜⁄π‘˜π‘˜π‘˜π‘˜. 𝐾𝐾) ln
𝑇𝑇2
800 π‘˜π‘˜π‘˜π‘˜π‘˜π‘˜
− (0.287 π‘˜π‘˜π‘˜π‘˜⁄π‘˜π‘˜π‘˜π‘˜. 𝐾𝐾)𝑙𝑙𝑙𝑙
22 + 273𝐾𝐾
100 π‘˜π‘˜π‘˜π‘˜π‘˜π‘˜
𝑇𝑇2 = 358.8𝐾𝐾
QUESTION 7 (7-127)
Problem:
Air enters a compressor steadily at the ambient conditions of 100 kPa and 20 °C and leaves at
800 kPa. Heat is lost from the compressor in the amount of 120 kJ/kg and the air experiences
an entropy decrease of 0.40 kJ/kg.k. Using constant specific heats. Determine (a) the exit
temperature of the air, (b) the work input to the compressor, and (c) the entropy generation
during this process.
𝑀𝑀𝑖𝑖𝑖𝑖 = 𝑐𝑐𝑝𝑝 𝑇𝑇2 − 𝑇𝑇1 + π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
800 kPa
85.8 °C
π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
120 π‘˜π‘˜π‘˜π‘˜/π‘˜π‘˜π‘˜π‘˜
𝑀𝑀𝑖𝑖𝑖𝑖
Compressor
Air
100 kPa
20 °C
𝑀𝑀𝑖𝑖𝑖𝑖 = 1.005 π‘˜π‘˜π‘˜π‘˜⁄π‘˜π‘˜π‘˜π‘˜ °πΆπΆ 85.8 − 20 °πΆπΆ + 120 π‘˜π‘˜π‘˜π‘˜⁄π‘˜π‘˜π‘˜π‘˜
𝑀𝑀𝑖𝑖𝑖𝑖 = 184.1 π‘˜π‘˜π‘˜π‘˜/π‘˜π‘˜π‘˜π‘˜
QUESTION 7 (7-127)
Problem:
Air enters a compressor steadily at the ambient conditions of 100 kPa and 20 °C and leaves at
800 kPa. Heat is lost from the compressor in the amount of 120 kJ/kg and the air experiences
an entropy decrease of 0.40 kJ/kg.k. Using constant specific heats. Determine (a) the exit
temperature of the air, (b) the work input to the compressor, and (c) the entropy generation
during this process.
𝑠𝑠𝑔𝑔𝑔𝑔𝑔𝑔 = Δ𝑠𝑠𝑑𝑑𝑑𝑑𝑑𝑑al = Δπ‘ π‘ π‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Ž + Δ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
800 kPa
85.8 °C
π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
120 π‘˜π‘˜π‘˜π‘˜/π‘˜π‘˜π‘˜π‘˜
𝑀𝑀𝑖𝑖𝑖𝑖
184.1 π‘˜π‘˜π‘˜π‘˜/π‘˜π‘˜π‘˜π‘˜
Compressor
Air
100 kPa
20 °C
π‘žπ‘žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
120 π‘˜π‘˜π‘˜π‘˜/π‘˜π‘˜π‘˜π‘˜
Δ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
=
= 0.4068 π‘˜π‘˜π‘˜π‘˜⁄π‘˜π‘˜π‘˜π‘˜. 𝐾𝐾
𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
22 + 273 𝐾𝐾
𝑠𝑠𝑔𝑔𝑔𝑔𝑔𝑔 = −0.40 + 0.4068 = 0.0068 π‘˜π‘˜π‘˜π‘˜⁄π‘˜π‘˜π‘˜π‘˜. 𝐾𝐾
QUESTION 8 (6-100)
𝑇𝑇𝐿𝐿
Problem:
𝑇𝑇𝐻𝐻
A Carnot heat pump is to be used to heat a house and maintain an average temperature of 25°C in winter. On a day
when the average outdoor temperature remains at about 2°C, the house is estimated to lose heat at a rate of 55,000
kJ/h. if the heat pump consumes 4.8 kW of power while operating, determine
π‘Šπ‘ŠΜ‡ 𝑖𝑖𝑖𝑖 to compensate the heat lost
a. How long it will take the heat pump
b. The total heating costs, assuming an average price of 11 c/kWh for electricity;
c. The heating cost for the same day if resistance heating is used instead of a heat pump.
Key notes: Heat pump
Coefficient of performance
QUESTION 8 (6-100)
Problem:
a.
b.
c.
𝑇𝑇𝐿𝐿 = 2 °πΆπΆ
𝑇𝑇𝐻𝐻 = 25°πΆπΆ
π‘Šπ‘ŠΜ‡ 𝑖𝑖𝑖𝑖 = 4.8 π‘˜π‘˜π‘˜π‘˜
How long it will take the heat pump to compensate the heat lost
The total heating costs, assuming an average price of 11 c/kWh for electricity;
The heating cost for the same day if resistance heating is used instead of a heat pump.
𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 π‘œπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ
𝑄𝑄𝐻𝐻
𝐢𝐢𝐢𝐢𝑃𝑃𝐻𝐻𝐻𝐻 =
=
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 π‘Šπ‘Šπ‘›π‘›π‘›π‘›π‘›π‘›,𝑖𝑖𝑖𝑖
𝐢𝐢𝐢𝐢𝑃𝑃𝐻𝐻𝐻𝐻 =
𝑄𝑄𝐻𝐻
𝑄𝑄𝐻𝐻 − 𝑄𝑄𝐿𝐿
1
1
𝐢𝐢𝐢𝐢𝑃𝑃𝐻𝐻𝐻𝐻,π‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿ =
=
= 14.1
𝑇𝑇𝐿𝐿
2 + 273
1 − �𝑇𝑇
1−
25 + 273
𝐻𝐻
𝑄𝑄̇ 𝐻𝐻
𝑄𝑄̇ 𝐿𝐿
HP
π‘Šπ‘ŠΜ‡ 𝑖𝑖𝑖𝑖
QUESTION 8 (6-100)
Problem:
a.
b.
c.
𝑇𝑇𝐿𝐿 = 2 °πΆπΆ
𝑇𝑇𝐻𝐻 = 25°πΆπΆ
π‘Šπ‘ŠΜ‡ 𝑖𝑖𝑖𝑖 = 4.8 π‘˜π‘˜π‘˜π‘˜
How long it will take the heat pump to compensate the heat lost
The total heating costs, assuming an average price of 11 c/kWh for electricity;
The heating cost for the same day if resistance heating is used instead of a heat pump.
𝐢𝐢𝐢𝐢𝑃𝑃𝐻𝐻𝐻𝐻,π‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿ = 14.1
𝑄𝑄𝐻𝐻 = 𝑄𝑄̇ H 1day = (55,000 π‘˜π‘˜π‘˜π‘˜⁄β„Ž) 24β„Žπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿ = 1,320 𝑀𝑀𝑀𝑀
𝑄𝑄𝐻𝐻
1,320
π‘Šπ‘Šπ‘›π‘›π‘›π‘›π‘›π‘›,𝑖𝑖𝑖𝑖 =
=
= 93,617 π‘˜π‘˜π‘˜π‘˜
𝐢𝐢𝐢𝐢𝑃𝑃𝐻𝐻𝐻𝐻
14.1
π‘Šπ‘Šπ‘›π‘›π‘›π‘›π‘›π‘›,𝑖𝑖𝑖𝑖 93,617 π‘˜π‘˜π‘˜π‘˜
Δ𝑑𝑑 =
=
= 5.42 β„Žπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿ
4.8 π‘˜π‘˜π‘˜π‘˜/𝑠𝑠
π‘Šπ‘ŠΜ‡ 𝑛𝑛𝑛𝑛𝑛𝑛,𝑖𝑖𝑖𝑖
𝑄𝑄̇ 𝐻𝐻
𝑄𝑄̇ 𝐿𝐿
HP
π‘Šπ‘ŠΜ‡ 𝑖𝑖𝑖𝑖
QUESTION 8 (6-100)
Problem:
a.
b.
c.
𝑇𝑇𝐿𝐿 = 2 °πΆπΆ
𝑇𝑇𝐻𝐻 = 25°πΆπΆ
π‘Šπ‘ŠΜ‡ 𝑖𝑖𝑖𝑖 = 4.8 π‘˜π‘˜π‘˜π‘˜
How long it will take the heat pump to compensate the heat lost
The total heating costs, assuming an average price of 11 c/kWh for electricity;
The heating cost for the same day if resistance heating is used instead of a heat pump.
Δ𝑑𝑑 = 5.42 β„Žπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿ
𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢 = π‘Šπ‘Šπ‘›π‘›π‘›π‘›π‘›π‘›,𝑖𝑖𝑖𝑖 × π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘
𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢 = (4.8 π‘˜π‘˜π‘˜π‘˜)(5.42 β„Ž)(0.11 $⁄π‘˜π‘˜π‘˜π‘˜π‘˜) = $ 2.86
πΆπΆπΆπΆπΆπΆπ‘‘π‘‘π‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿ = 𝑄𝑄𝐻𝐻 × π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = 1,320 𝑀𝑀𝑀𝑀
1 π‘˜π‘˜π‘˜π‘˜π‘˜
3600 π‘˜π‘˜π‘˜π‘˜
0.11 $⁄π‘˜π‘˜π‘˜π‘˜π‘˜ = $ 40.33
𝑄𝑄̇ 𝐻𝐻
𝑄𝑄̇ 𝐿𝐿
HP
π‘Šπ‘ŠΜ‡ 𝑖𝑖𝑖𝑖
Moving boundary work
π›Ώπ›Ώπ‘Šπ‘Šπ‘π‘ = 𝐹𝐹𝐹𝐹𝐹𝐹 = 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑃𝑃𝑃𝑃𝑃𝑃
𝐹𝐹 = 𝑃𝑃𝑃𝑃
2
𝑑𝑑𝑑𝑑 = 𝐴𝐴𝐴𝐴𝐴𝐴
π‘Šπ‘Šπ‘π‘ = ∫1 𝑃𝑃𝑃𝑃𝑃𝑃 = Area under P-V diagram
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