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The compression ratio of air-standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is 100 kPa, 35°C and 600 cm3. The temperature at the end of the isentropic expansion process is 800 K. Property data for air: Cp = 1.005 kJ/kg K, Cv = 0.718 kJ/kg K, R = 0.287 kJ/kg K. The highest pressure in the cycle is (in MPa) ________.
Correct answer is between '5.9,6.2'. Can you explain this answer?
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The compression ratio of air-standard Otto cycle is 9.5. Prior to the ...
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The compression ratio of air-standard Otto cycle is 9.5. Prior to the ...
°C. The maximum temperature during the cycle is 2200 K. Determine the pressures and temperatures at all the four corners of the cycle and the thermal efficiency of the cycle.
Assumptions:
- Air is an ideal gas with constant specific heats.
- The combustion process is replaced by heat addition process.
We can start by using the compression ratio to find the volumes at state 1 and state 2:
r = V1/V2
9.5 = V1/V2
V1 = 9.5V2
At state 1:
P1 = 100 kPa
T1 = 35°C + 273.15 = 308.15 K
V1 = 9.5V2
At state 2:
V1/V2 = r
V2 = V1/r
V2 = V1/9.5
V2 = 0.105V1
Using the ideal gas law, we can find the pressures at state 2 and state 3:
P1V1/T1 = P2V2/T2
At state 2:
P2 = P1(V1/V2)(T2/T1)
P2 = 100(9.5)(2200/308.15)
P2 = 1577.38 kPa
At state 3:
P3 = P2
V3 = V4
P3V3/T3 = P4V4/T4
We can use the compression ratio again to find V4:
r = V3/V4
9.5 = V3/V4
V3 = 9.5V4
Substituting and rearranging:
P3V3/T3 = P4V4/T4
P3(9.5V4)/T3 = P4V4/T4
P3/P4 = T3/T4(9.5)
P4 = P3/T3/T4(9.5)V4
Now we need to find T3 and T4. We know that the maximum temperature during the cycle is 2200 K, which occurs at state 3:
T3 = 2200 K
The heat added during the cycle is:
Qin = mCp(T3 - T2)
We don't know the mass or the specific heat, but we can use the thermal efficiency to find the ratio of Qout/Qin:
ηth = 1 - Qout/Qin
ηth = Wnet/Qin
Wnet = Qin - Qout
Since Qout = mCp(T4 - T1), we can write:
ηth = 1 - (T4 - T1)/(T3 - T2)
We can solve for T4:
T4 = T1 + (1 - ηth)(T3 - T2)
Substituting and rearranging:
P4 = P3/T3/T1 + (1 - ηth)(T3 - T2)(9.5)V4
Now we have all the pressures and temperatures at the corners of the cycle:
State 1:
P1 = 100 kPa
T1 = 35°C + 273.15 = 308.15 K
State 2:
P2 = 1577.38 kPa
V2 = V1/9.5
T2 = P1(V1
Assumptions:
- Air is an ideal gas with constant specific heats.
- The combustion process is replaced by heat addition process.
We can start by using the compression ratio to find the volumes at state 1 and state 2:
r = V1/V2
9.5 = V1/V2
V1 = 9.5V2
At state 1:
P1 = 100 kPa
T1 = 35°C + 273.15 = 308.15 K
V1 = 9.5V2
At state 2:
V1/V2 = r
V2 = V1/r
V2 = V1/9.5
V2 = 0.105V1
Using the ideal gas law, we can find the pressures at state 2 and state 3:
P1V1/T1 = P2V2/T2
At state 2:
P2 = P1(V1/V2)(T2/T1)
P2 = 100(9.5)(2200/308.15)
P2 = 1577.38 kPa
At state 3:
P3 = P2
V3 = V4
P3V3/T3 = P4V4/T4
We can use the compression ratio again to find V4:
r = V3/V4
9.5 = V3/V4
V3 = 9.5V4
Substituting and rearranging:
P3V3/T3 = P4V4/T4
P3(9.5V4)/T3 = P4V4/T4
P3/P4 = T3/T4(9.5)
P4 = P3/T3/T4(9.5)V4
Now we need to find T3 and T4. We know that the maximum temperature during the cycle is 2200 K, which occurs at state 3:
T3 = 2200 K
The heat added during the cycle is:
Qin = mCp(T3 - T2)
We don't know the mass or the specific heat, but we can use the thermal efficiency to find the ratio of Qout/Qin:
ηth = 1 - Qout/Qin
ηth = Wnet/Qin
Wnet = Qin - Qout
Since Qout = mCp(T4 - T1), we can write:
ηth = 1 - (T4 - T1)/(T3 - T2)
We can solve for T4:
T4 = T1 + (1 - ηth)(T3 - T2)
Substituting and rearranging:
P4 = P3/T3/T1 + (1 - ηth)(T3 - T2)(9.5)V4
Now we have all the pressures and temperatures at the corners of the cycle:
State 1:
P1 = 100 kPa
T1 = 35°C + 273.15 = 308.15 K
State 2:
P2 = 1577.38 kPa
V2 = V1/9.5
T2 = P1(V1
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The compression ratio of air-standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is 100 kPa, 35°C and 600 cm3. The temperature at the end of the isentropic expansion process is 800 K. Property data for air: Cp= 1.005 kJ/kg K, Cv= 0.718 kJ/kg K, R = 0.287 kJ/kg K. The highest pressure in the cycle is (in MPa) ________.Correct answer is between '5.9,6.2'. Can you explain this answer?
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The compression ratio of air-standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is 100 kPa, 35°C and 600 cm3. The temperature at the end of the isentropic expansion process is 800 K. Property data for air: Cp= 1.005 kJ/kg K, Cv= 0.718 kJ/kg K, R = 0.287 kJ/kg K. The highest pressure in the cycle is (in MPa) ________.Correct answer is between '5.9,6.2'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The compression ratio of air-standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is 100 kPa, 35°C and 600 cm3. The temperature at the end of the isentropic expansion process is 800 K. Property data for air: Cp= 1.005 kJ/kg K, Cv= 0.718 kJ/kg K, R = 0.287 kJ/kg K. The highest pressure in the cycle is (in MPa) ________.Correct answer is between '5.9,6.2'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The compression ratio of air-standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is 100 kPa, 35°C and 600 cm3. The temperature at the end of the isentropic expansion process is 800 K. Property data for air: Cp= 1.005 kJ/kg K, Cv= 0.718 kJ/kg K, R = 0.287 kJ/kg K. The highest pressure in the cycle is (in MPa) ________.Correct answer is between '5.9,6.2'. Can you explain this answer?.
The compression ratio of air-standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is 100 kPa, 35°C and 600 cm3. The temperature at the end of the isentropic expansion process is 800 K. Property data for air: Cp= 1.005 kJ/kg K, Cv= 0.718 kJ/kg K, R = 0.287 kJ/kg K. The highest pressure in the cycle is (in MPa) ________.Correct answer is between '5.9,6.2'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The compression ratio of air-standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is 100 kPa, 35°C and 600 cm3. The temperature at the end of the isentropic expansion process is 800 K. Property data for air: Cp= 1.005 kJ/kg K, Cv= 0.718 kJ/kg K, R = 0.287 kJ/kg K. The highest pressure in the cycle is (in MPa) ________.Correct answer is between '5.9,6.2'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The compression ratio of air-standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is 100 kPa, 35°C and 600 cm3. The temperature at the end of the isentropic expansion process is 800 K. Property data for air: Cp= 1.005 kJ/kg K, Cv= 0.718 kJ/kg K, R = 0.287 kJ/kg K. The highest pressure in the cycle is (in MPa) ________.Correct answer is between '5.9,6.2'. Can you explain this answer?.
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