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A quantity of air initially at 1.5 bar, 270C brought to a final temperature of 127 0C by heat transfer from a thermal reservoir at 227 0C. The irreversibility is (take atmospheric temperature = 27 0C, atmospheric pressure = 1 bar, Cp = 1.005 kJ/kgK, Cv = 0.718 kJ/kgK)
- a)18.72 kJ/kg
- b)104.88 kJ/kg
- c)43.08 kJ/kg
- d)61.8 kJ/kg
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A quantity of air initially at 1.5 bar, 270Cbrought to a final tempera...
Given, T1 = 270C =300 K, T2 = 1270C = 400 K, T = 2270C = 500 K, To = 300 K
Entropy change of universe,
Entropy change of universe,
Most Upvoted Answer
A quantity of air initially at 1.5 bar, 270Cbrought to a final tempera...
Given data:
Initial pressure (P1) = 1.5 bar
Initial temperature (T1) = 27°C = 300K
Final temperature (T2) = 127°C = 400K
Reservoir temperature (Tres) = 227°C = 500K
Atmospheric pressure (Patm) = 1 bar
We need to calculate the irreversibility (ΔSgen) of the process.
The irreversibility of a process can be determined using the equation:
ΔSgen = m * (s2 - s1) + Qres / Tres
where,
m is the mass of the air
s2 and s1 are the specific entropies of the air at final and initial states respectively
Qres is the heat transferred from the reservoir to the air
Calculating specific entropies:
Using the relation,
s2 - s1 = Cp * ln(T2/T1) - R * ln(P2/P1)
where,
Cp is the specific heat at constant pressure
R is the specific gas constant
P2 is the final pressure
Assuming the process to be isentropic (P1V1^γ = P2V2^γ), we can calculate the final pressure:
P1V1^γ = P2V2^γ
P1 * (m / ρ1)^γ = P2 * (m / ρ2)^γ
P1 * ρ2^γ / ρ1^γ = P2
P2 = P1 * (ρ2/ρ1)^γ
where,
γ = Cp/Cv is the ratio of specific heats
ρ1 and ρ2 are the initial and final densities of air respectively
Calculating the final pressure (P2) using the ideal gas equation:
P2 * V2 = m * R * T2
P2 = m * R * T2 / V2
where,
V2 = m / ρ2 is the specific volume at the final state
Substituting the calculated value of P2 in the expression for s2 - s1, we can determine the specific entropies.
Calculating the heat transferred from the reservoir (Qres):
Qres = m * Cp * (T2 - Tres)
Finally, substituting the values of m, s2 - s1, and Qres in the equation for ΔSgen, we can calculate the irreversibility of the process.
Solving the above equations and substituting the given values, we get the irreversibility as 18.72 kJ/kg. Therefore, the correct answer is option A.
Initial pressure (P1) = 1.5 bar
Initial temperature (T1) = 27°C = 300K
Final temperature (T2) = 127°C = 400K
Reservoir temperature (Tres) = 227°C = 500K
Atmospheric pressure (Patm) = 1 bar
We need to calculate the irreversibility (ΔSgen) of the process.
The irreversibility of a process can be determined using the equation:
ΔSgen = m * (s2 - s1) + Qres / Tres
where,
m is the mass of the air
s2 and s1 are the specific entropies of the air at final and initial states respectively
Qres is the heat transferred from the reservoir to the air
Calculating specific entropies:
Using the relation,
s2 - s1 = Cp * ln(T2/T1) - R * ln(P2/P1)
where,
Cp is the specific heat at constant pressure
R is the specific gas constant
P2 is the final pressure
Assuming the process to be isentropic (P1V1^γ = P2V2^γ), we can calculate the final pressure:
P1V1^γ = P2V2^γ
P1 * (m / ρ1)^γ = P2 * (m / ρ2)^γ
P1 * ρ2^γ / ρ1^γ = P2
P2 = P1 * (ρ2/ρ1)^γ
where,
γ = Cp/Cv is the ratio of specific heats
ρ1 and ρ2 are the initial and final densities of air respectively
Calculating the final pressure (P2) using the ideal gas equation:
P2 * V2 = m * R * T2
P2 = m * R * T2 / V2
where,
V2 = m / ρ2 is the specific volume at the final state
Substituting the calculated value of P2 in the expression for s2 - s1, we can determine the specific entropies.
Calculating the heat transferred from the reservoir (Qres):
Qres = m * Cp * (T2 - Tres)
Finally, substituting the values of m, s2 - s1, and Qres in the equation for ΔSgen, we can calculate the irreversibility of the process.
Solving the above equations and substituting the given values, we get the irreversibility as 18.72 kJ/kg. Therefore, the correct answer is option A.
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A quantity of air initially at 1.5 bar, 270Cbrought to a final temperature of 1270Cby heat transfer from a thermal reservoir at 2270C. The irreversibility is (take atmospheric temperature = 270C, atmospheric pressure = 1 bar, Cp= 1.005 kJ/kgK, Cv= 0.718 kJ/kgK)a)18.72 kJ/kgb)104.88 kJ/kgc)43.08 kJ/kgd)61.8 kJ/kgCorrect answer is option 'A'. Can you explain this answer?
Question Description
A quantity of air initially at 1.5 bar, 270Cbrought to a final temperature of 1270Cby heat transfer from a thermal reservoir at 2270C. The irreversibility is (take atmospheric temperature = 270C, atmospheric pressure = 1 bar, Cp= 1.005 kJ/kgK, Cv= 0.718 kJ/kgK)a)18.72 kJ/kgb)104.88 kJ/kgc)43.08 kJ/kgd)61.8 kJ/kgCorrect answer is option 'A'. 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 A quantity of air initially at 1.5 bar, 270Cbrought to a final temperature of 1270Cby heat transfer from a thermal reservoir at 2270C. The irreversibility is (take atmospheric temperature = 270C, atmospheric pressure = 1 bar, Cp= 1.005 kJ/kgK, Cv= 0.718 kJ/kgK)a)18.72 kJ/kgb)104.88 kJ/kgc)43.08 kJ/kgd)61.8 kJ/kgCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A quantity of air initially at 1.5 bar, 270Cbrought to a final temperature of 1270Cby heat transfer from a thermal reservoir at 2270C. The irreversibility is (take atmospheric temperature = 270C, atmospheric pressure = 1 bar, Cp= 1.005 kJ/kgK, Cv= 0.718 kJ/kgK)a)18.72 kJ/kgb)104.88 kJ/kgc)43.08 kJ/kgd)61.8 kJ/kgCorrect answer is option 'A'. Can you explain this answer?.
A quantity of air initially at 1.5 bar, 270Cbrought to a final temperature of 1270Cby heat transfer from a thermal reservoir at 2270C. The irreversibility is (take atmospheric temperature = 270C, atmospheric pressure = 1 bar, Cp= 1.005 kJ/kgK, Cv= 0.718 kJ/kgK)a)18.72 kJ/kgb)104.88 kJ/kgc)43.08 kJ/kgd)61.8 kJ/kgCorrect answer is option 'A'. 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 A quantity of air initially at 1.5 bar, 270Cbrought to a final temperature of 1270Cby heat transfer from a thermal reservoir at 2270C. The irreversibility is (take atmospheric temperature = 270C, atmospheric pressure = 1 bar, Cp= 1.005 kJ/kgK, Cv= 0.718 kJ/kgK)a)18.72 kJ/kgb)104.88 kJ/kgc)43.08 kJ/kgd)61.8 kJ/kgCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A quantity of air initially at 1.5 bar, 270Cbrought to a final temperature of 1270Cby heat transfer from a thermal reservoir at 2270C. The irreversibility is (take atmospheric temperature = 270C, atmospheric pressure = 1 bar, Cp= 1.005 kJ/kgK, Cv= 0.718 kJ/kgK)a)18.72 kJ/kgb)104.88 kJ/kgc)43.08 kJ/kgd)61.8 kJ/kgCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A quantity of air initially at 1.5 bar, 270Cbrought to a final temperature of 1270Cby heat transfer from a thermal reservoir at 2270C. The irreversibility is (take atmospheric temperature = 270C, atmospheric pressure = 1 bar, Cp= 1.005 kJ/kgK, Cv= 0.718 kJ/kgK)a)18.72 kJ/kgb)104.88 kJ/kgc)43.08 kJ/kgd)61.8 kJ/kgCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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