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An engine working on air standard Otto cycle is supplied with air at 0.1 MPa and 35°C. The compression ratio is 8. The heat supplied is 500 kJ/kg. Property data for air: cp = 1.005 kJ/kg K, cv = 0.718 kJ/kg K, R = 0.287 kJ/kg K. The maximum temperature (in K) of the cycle is _________ (correct to one decimal place).
Correct answer is '1402.0 to 1406.0'. Can you explain this answer?
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An engine working on air standard Otto cycle is supplied with air at 0...
°C. The compression ratio of the cycle is 8 and the specific heat ratio of air is 1.4. Calculate the maximum temperature and pressure in the cycle.
To solve this problem, we can use the air standard Otto cycle equations. The maximum temperature can be found using the equation:
T3 = T2 * (compression ratio)^(specific heat ratio - 1)
where T3 is the maximum temperature and T2 is the temperature at the end of the compression stroke.
First, we need to convert the given pressure and temperature values to the SI unit system:
P1 = 0.1 MPa = 0.1 * 10^6 Pa
T1 = 35°C = 35 + 273.15 = 308.15 K
Next, we can use the ideal gas law to find the specific volume at state 1:
V1 = R_specific * T1 / P1
where R_specific is the specific gas constant for air. For air, R_specific = R_universal / molar mass of air.
Using the values R_universal = 8.314 J/(mol*K) and molar mass of air = 28.97 g/mol, we can calculate R_specific:
R_specific = 8.314 / 0.02897 = 287.05 J/(kg*K)
Now we can calculate V1:
V1 = 287.05 * 308.15 / 0.1 * 10^6 = 0.879 m^3/kg
Next, we can use the compression ratio to find the specific volume at state 2:
V2 = V1 / (compression ratio) = 0.879 / 8 = 0.1099 m^3/kg
Using the specific heat ratio, we can find the maximum temperature:
T3 = T2 * (compression ratio)^(specific heat ratio - 1)
We can rearrange this equation to find T2:
T2 = T3 / (compression ratio)^(specific heat ratio - 1)
Since the compression ratio is given as 8 and the specific heat ratio is given as 1.4, we can calculate T2:
T2 = T3 / 8^(1.4 - 1)
Now we need to find T3. The maximum temperature in the Otto cycle occurs at the end of the combustion process, which is also the end of the constant volume heat addition process (state 3). During this process, the heat transfer into the system is given by:
Q_in = C_v * (T3 - T2)
where Q_in is the heat transfer, C_v is the specific heat at constant volume, and T2 and T3 are the temperatures at the end of the compression and combustion processes, respectively.
We can rearrange this equation to solve for T3:
T3 = Q_in / C_v + T2
Since this is an air standard Otto cycle, the heat transfer into the system can be calculated using the equation:
Q_in = C_v * T2 * (compression ratio)^(specific heat ratio - 1) - C_v * T1
Substituting the given values, we can find Q_in:
Q_in = 0.718 * 308.15 * 8^(1.4 - 1) - 0.718 * 308.15 = 806.61 KJ
To solve this problem, we can use the air standard Otto cycle equations. The maximum temperature can be found using the equation:
T3 = T2 * (compression ratio)^(specific heat ratio - 1)
where T3 is the maximum temperature and T2 is the temperature at the end of the compression stroke.
First, we need to convert the given pressure and temperature values to the SI unit system:
P1 = 0.1 MPa = 0.1 * 10^6 Pa
T1 = 35°C = 35 + 273.15 = 308.15 K
Next, we can use the ideal gas law to find the specific volume at state 1:
V1 = R_specific * T1 / P1
where R_specific is the specific gas constant for air. For air, R_specific = R_universal / molar mass of air.
Using the values R_universal = 8.314 J/(mol*K) and molar mass of air = 28.97 g/mol, we can calculate R_specific:
R_specific = 8.314 / 0.02897 = 287.05 J/(kg*K)
Now we can calculate V1:
V1 = 287.05 * 308.15 / 0.1 * 10^6 = 0.879 m^3/kg
Next, we can use the compression ratio to find the specific volume at state 2:
V2 = V1 / (compression ratio) = 0.879 / 8 = 0.1099 m^3/kg
Using the specific heat ratio, we can find the maximum temperature:
T3 = T2 * (compression ratio)^(specific heat ratio - 1)
We can rearrange this equation to find T2:
T2 = T3 / (compression ratio)^(specific heat ratio - 1)
Since the compression ratio is given as 8 and the specific heat ratio is given as 1.4, we can calculate T2:
T2 = T3 / 8^(1.4 - 1)
Now we need to find T3. The maximum temperature in the Otto cycle occurs at the end of the combustion process, which is also the end of the constant volume heat addition process (state 3). During this process, the heat transfer into the system is given by:
Q_in = C_v * (T3 - T2)
where Q_in is the heat transfer, C_v is the specific heat at constant volume, and T2 and T3 are the temperatures at the end of the compression and combustion processes, respectively.
We can rearrange this equation to solve for T3:
T3 = Q_in / C_v + T2
Since this is an air standard Otto cycle, the heat transfer into the system can be calculated using the equation:
Q_in = C_v * T2 * (compression ratio)^(specific heat ratio - 1) - C_v * T1
Substituting the given values, we can find Q_in:
Q_in = 0.718 * 308.15 * 8^(1.4 - 1) - 0.718 * 308.15 = 806.61 KJ
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An engine working on air standard Otto cycle is supplied with air at 0.1 MPa and 35°C. The compression ratio is 8. The heat supplied is 500 kJ/kg. Property data for air: cp = 1.005 kJ/kg K, cv = 0.718 kJ/kg K, R = 0.287 kJ/kg K. The maximum temperature (in K) of the cycle is _________ (correct to one decimal place).Correct answer is '1402.0 to 1406.0'. Can you explain this answer?
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An engine working on air standard Otto cycle is supplied with air at 0.1 MPa and 35°C. The compression ratio is 8. The heat supplied is 500 kJ/kg. Property data for air: cp = 1.005 kJ/kg K, cv = 0.718 kJ/kg K, R = 0.287 kJ/kg K. The maximum temperature (in K) of the cycle is _________ (correct to one decimal place).Correct answer is '1402.0 to 1406.0'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An engine working on air standard Otto cycle is supplied with air at 0.1 MPa and 35°C. The compression ratio is 8. The heat supplied is 500 kJ/kg. Property data for air: cp = 1.005 kJ/kg K, cv = 0.718 kJ/kg K, R = 0.287 kJ/kg K. The maximum temperature (in K) of the cycle is _________ (correct to one decimal place).Correct answer is '1402.0 to 1406.0'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An engine working on air standard Otto cycle is supplied with air at 0.1 MPa and 35°C. The compression ratio is 8. The heat supplied is 500 kJ/kg. Property data for air: cp = 1.005 kJ/kg K, cv = 0.718 kJ/kg K, R = 0.287 kJ/kg K. The maximum temperature (in K) of the cycle is _________ (correct to one decimal place).Correct answer is '1402.0 to 1406.0'. Can you explain this answer?.
An engine working on air standard Otto cycle is supplied with air at 0.1 MPa and 35°C. The compression ratio is 8. The heat supplied is 500 kJ/kg. Property data for air: cp = 1.005 kJ/kg K, cv = 0.718 kJ/kg K, R = 0.287 kJ/kg K. The maximum temperature (in K) of the cycle is _________ (correct to one decimal place).Correct answer is '1402.0 to 1406.0'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An engine working on air standard Otto cycle is supplied with air at 0.1 MPa and 35°C. The compression ratio is 8. The heat supplied is 500 kJ/kg. Property data for air: cp = 1.005 kJ/kg K, cv = 0.718 kJ/kg K, R = 0.287 kJ/kg K. The maximum temperature (in K) of the cycle is _________ (correct to one decimal place).Correct answer is '1402.0 to 1406.0'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An engine working on air standard Otto cycle is supplied with air at 0.1 MPa and 35°C. The compression ratio is 8. The heat supplied is 500 kJ/kg. Property data for air: cp = 1.005 kJ/kg K, cv = 0.718 kJ/kg K, R = 0.287 kJ/kg K. The maximum temperature (in K) of the cycle is _________ (correct to one decimal place).Correct answer is '1402.0 to 1406.0'. Can you explain this answer?.
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