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An air standard Otto cycle has thermal efficiency of 0.5 and the mean effective pressure of the cycle is 1000 kPa. For air, assume specific heat ratio γ = 1.4 and specific gas constant R = 0.287 kJ/kg.K, If the pressure and temperature at the beginning of the compression stroke are 100 kPa and 300 K, respectively, then the specific net work output of the cycle is ___________kJ/kg (round off to two decimal places).
Correct answer is '708.8'. Can you explain this answer?
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An air standard Otto cycle has thermal efficiency of 0.5 and the mean ...
Given data
P1 = 100kPa, T1 = 300K, mep = 1000 kPa, y=1.4, efficiency(η)=0.5, R = 0.287 kJ/kg-K
Using P1V1 = RT1
Efficiency(η)
where, r is the compression ratio = V1/V2
Or, r = 5.657
V1/V2 = 5.657
⟹V2=0.1522 m3/kg
Swept volume = V1 – V2 = 0.861 – 0.1522 = 0.7088 m3/kg
Went = Swept volume ×mep = 0.7088 × 1000 = 708.8 kJ/kg
Most Upvoted Answer
An air standard Otto cycle has thermal efficiency of 0.5 and the mean ...
We need to find the compression ratio (r) of the air standard Otto cycle.
From the given data, we know the thermal efficiency (η) and the mean effective pressure (MEP) of the cycle:
η = 0.5
MEP = 1000 kPa
We also know that the air in the cycle follows the specific heat ratio (γ) of 1.4. This means:
γ = cp/cv = 1.4
where cp is the specific heat at constant pressure and cv is the specific heat at constant volume.
Using the air standard assumptions, we can write the expressions for the thermal efficiency and MEP of the Otto cycle in terms of the compression ratio r:
η = 1 - (1/r)^(γ-1)
MEP = (1/γ) * P1 * [(r^(γ) - 1)/(r - 1)]
where P1 is the pressure at the start of the compression stroke.
Substituting the given values, we get:
0.5 = 1 - (1/r)^(1.4-1)
1000 = (1/1.4) * P1 * [(r^(1.4) - 1)/(r - 1)]
Solving these two equations simultaneously, we get:
r = 8.71
Therefore, the compression ratio of the air standard Otto cycle is approximately 8.71.
From the given data, we know the thermal efficiency (η) and the mean effective pressure (MEP) of the cycle:
η = 0.5
MEP = 1000 kPa
We also know that the air in the cycle follows the specific heat ratio (γ) of 1.4. This means:
γ = cp/cv = 1.4
where cp is the specific heat at constant pressure and cv is the specific heat at constant volume.
Using the air standard assumptions, we can write the expressions for the thermal efficiency and MEP of the Otto cycle in terms of the compression ratio r:
η = 1 - (1/r)^(γ-1)
MEP = (1/γ) * P1 * [(r^(γ) - 1)/(r - 1)]
where P1 is the pressure at the start of the compression stroke.
Substituting the given values, we get:
0.5 = 1 - (1/r)^(1.4-1)
1000 = (1/1.4) * P1 * [(r^(1.4) - 1)/(r - 1)]
Solving these two equations simultaneously, we get:
r = 8.71
Therefore, the compression ratio of the air standard Otto cycle is approximately 8.71.
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An air standard Otto cycle has thermal efficiency of 0.5 and the mean effective pressure of the cycle is 1000 kPa. For air, assume specific heat ratio γ = 1.4 and specific gas constant R = 0.287 kJ/kg.K, If the pressure and temperature at the beginning of the compression stroke are 100 kPa and 300 K, respectively, then the specific net work output of the cycle is ___________kJ/kg (round off to two decimal places).Correct answer is '708.8'. Can you explain this answer?
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An air standard Otto cycle has thermal efficiency of 0.5 and the mean effective pressure of the cycle is 1000 kPa. For air, assume specific heat ratio γ = 1.4 and specific gas constant R = 0.287 kJ/kg.K, If the pressure and temperature at the beginning of the compression stroke are 100 kPa and 300 K, respectively, then the specific net work output of the cycle is ___________kJ/kg (round off to two decimal places).Correct answer is '708.8'. 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 An air standard Otto cycle has thermal efficiency of 0.5 and the mean effective pressure of the cycle is 1000 kPa. For air, assume specific heat ratio γ = 1.4 and specific gas constant R = 0.287 kJ/kg.K, If the pressure and temperature at the beginning of the compression stroke are 100 kPa and 300 K, respectively, then the specific net work output of the cycle is ___________kJ/kg (round off to two decimal places).Correct answer is '708.8'. 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 An air standard Otto cycle has thermal efficiency of 0.5 and the mean effective pressure of the cycle is 1000 kPa. For air, assume specific heat ratio γ = 1.4 and specific gas constant R = 0.287 kJ/kg.K, If the pressure and temperature at the beginning of the compression stroke are 100 kPa and 300 K, respectively, then the specific net work output of the cycle is ___________kJ/kg (round off to two decimal places).Correct answer is '708.8'. Can you explain this answer?.
An air standard Otto cycle has thermal efficiency of 0.5 and the mean effective pressure of the cycle is 1000 kPa. For air, assume specific heat ratio γ = 1.4 and specific gas constant R = 0.287 kJ/kg.K, If the pressure and temperature at the beginning of the compression stroke are 100 kPa and 300 K, respectively, then the specific net work output of the cycle is ___________kJ/kg (round off to two decimal places).Correct answer is '708.8'. 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 An air standard Otto cycle has thermal efficiency of 0.5 and the mean effective pressure of the cycle is 1000 kPa. For air, assume specific heat ratio γ = 1.4 and specific gas constant R = 0.287 kJ/kg.K, If the pressure and temperature at the beginning of the compression stroke are 100 kPa and 300 K, respectively, then the specific net work output of the cycle is ___________kJ/kg (round off to two decimal places).Correct answer is '708.8'. 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 An air standard Otto cycle has thermal efficiency of 0.5 and the mean effective pressure of the cycle is 1000 kPa. For air, assume specific heat ratio γ = 1.4 and specific gas constant R = 0.287 kJ/kg.K, If the pressure and temperature at the beginning of the compression stroke are 100 kPa and 300 K, respectively, then the specific net work output of the cycle is ___________kJ/kg (round off to two decimal places).Correct answer is '708.8'. Can you explain this answer?.
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