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One kg of air in a closed system undergoes an irreversible process from an initial state of p1 = 1 bar (absolute) and T1 = 27°C, to a final state of p2 = 3 bar (absolute) and T2 = 127°C. If the gas constant of air is 287 J/kgK and the ratio of the specific heats γ = 1.4, then the change in the specific entropy (in J/kgK) of the air in the process is
- a)172.0
- b)–26.3
- c)indeterminate, as the process is irreversible
- d)28.4
Correct answer is option 'B'. Can you explain this answer?
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One kg of air in a closed system undergoes an irreversible process fro...
P1 = 1 bar, P2 = 1 bar, T1 = 300 K, T2 = 400 K,
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One kg of air in a closed system undergoes an irreversible process fro...
°C to a final state of p2 = 3 bar (absolute) and T2 = 127°C.
To determine the change in specific enthalpy, we can use the equation:
Δh = CpdT + Vdp
Where:
Δh = change in specific enthalpy
Cp = specific heat capacity at constant pressure
dT = change in temperature
V = specific volume
dp = change in pressure
First, let's determine the values for Cp and V at the initial state (state 1). We can use the ideal gas equation of state:
PV = RT
Where:
P = pressure
V = volume
R = specific gas constant
T = temperature
Rearranging the equation, we can solve for V:
V = RT/P
Substituting the given values:
V1 = (0.287 kJ/kg·K * 300 K) / (1 bar * 100 kPa/bar)
= 0.861 m^3/kg
Now, let's determine the values for Cp at state 1. For air, Cp is approximately 1.005 kJ/kg·K.
Using the same process, we can determine the values for V and Cp at the final state (state 2):
V2 = (0.287 kJ/kg·K * 400 K) / (3 bar * 100 kPa/bar)
= 0.382 m^3/kg
Cp = 1.005 kJ/kg·K
Now, we can calculate the change in specific enthalpy:
Δh = CpdT + Vdp
Δh = Cp(T2 - T1) + V(dp)
Δh = (1.005 kJ/kg·K * (127°C - 27°C)) + (0.382 m^3/kg - 0.861 m^3/kg) * (3 bar * 100 kPa/bar - 1 bar * 100 kPa/bar)
Δh = 100 kJ/kg + (-0.479 m^3/kg) * (200 kPa)
Δh = 100 kJ/kg - 95.8 kJ/kg
Δh = 4.2 kJ/kg
Therefore, the change in specific enthalpy for the air undergoing the irreversible process is 4.2 kJ/kg.
To determine the change in specific enthalpy, we can use the equation:
Δh = CpdT + Vdp
Where:
Δh = change in specific enthalpy
Cp = specific heat capacity at constant pressure
dT = change in temperature
V = specific volume
dp = change in pressure
First, let's determine the values for Cp and V at the initial state (state 1). We can use the ideal gas equation of state:
PV = RT
Where:
P = pressure
V = volume
R = specific gas constant
T = temperature
Rearranging the equation, we can solve for V:
V = RT/P
Substituting the given values:
V1 = (0.287 kJ/kg·K * 300 K) / (1 bar * 100 kPa/bar)
= 0.861 m^3/kg
Now, let's determine the values for Cp at state 1. For air, Cp is approximately 1.005 kJ/kg·K.
Using the same process, we can determine the values for V and Cp at the final state (state 2):
V2 = (0.287 kJ/kg·K * 400 K) / (3 bar * 100 kPa/bar)
= 0.382 m^3/kg
Cp = 1.005 kJ/kg·K
Now, we can calculate the change in specific enthalpy:
Δh = CpdT + Vdp
Δh = Cp(T2 - T1) + V(dp)
Δh = (1.005 kJ/kg·K * (127°C - 27°C)) + (0.382 m^3/kg - 0.861 m^3/kg) * (3 bar * 100 kPa/bar - 1 bar * 100 kPa/bar)
Δh = 100 kJ/kg + (-0.479 m^3/kg) * (200 kPa)
Δh = 100 kJ/kg - 95.8 kJ/kg
Δh = 4.2 kJ/kg
Therefore, the change in specific enthalpy for the air undergoing the irreversible process is 4.2 kJ/kg.
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One kg of air in a closed system undergoes an irreversible process from an initial state of p1 = 1 bar (absolute) and T1 = 27°C, to a final state of p2 = 3 bar (absolute) and T2 = 127°C. If the gas constant of air is 287 J/kgK and the ratio of the specific heats γ = 1.4, then the change in the specific entropy (in J/kgK) of the air in the process isa)172.0b)–26.3c)indeterminate, as the process is irreversibled)28.4Correct answer is option 'B'. Can you explain this answer?
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One kg of air in a closed system undergoes an irreversible process from an initial state of p1 = 1 bar (absolute) and T1 = 27°C, to a final state of p2 = 3 bar (absolute) and T2 = 127°C. If the gas constant of air is 287 J/kgK and the ratio of the specific heats γ = 1.4, then the change in the specific entropy (in J/kgK) of the air in the process isa)172.0b)–26.3c)indeterminate, as the process is irreversibled)28.4Correct answer is option 'B'. 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 One kg of air in a closed system undergoes an irreversible process from an initial state of p1 = 1 bar (absolute) and T1 = 27°C, to a final state of p2 = 3 bar (absolute) and T2 = 127°C. If the gas constant of air is 287 J/kgK and the ratio of the specific heats γ = 1.4, then the change in the specific entropy (in J/kgK) of the air in the process isa)172.0b)–26.3c)indeterminate, as the process is irreversibled)28.4Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One kg of air in a closed system undergoes an irreversible process from an initial state of p1 = 1 bar (absolute) and T1 = 27°C, to a final state of p2 = 3 bar (absolute) and T2 = 127°C. If the gas constant of air is 287 J/kgK and the ratio of the specific heats γ = 1.4, then the change in the specific entropy (in J/kgK) of the air in the process isa)172.0b)–26.3c)indeterminate, as the process is irreversibled)28.4Correct answer is option 'B'. Can you explain this answer?.
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