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In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K.
The following are the properties and relations pertaining to air:
Specific heat at constant pressure, cp = 1.005 kJ/kg K;
Specific heat at constant volume, cv = 0.718 kJ/kg K;
Characteristic gas constant, R = 0.287 kJ/kg K.
Enthalpy, h = cpT
Internal energy, u = cvT
Q.
If the pressure at station Q is 50 kPa, the change in entropy in kJ/kg K is
- a)– 0.155
- b)0
- c)0.160
- d)0.355
Correct answer is option 'C'. Can you explain this answer?
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Given data:
- Pressure at station P (P1) = 150 kPa
- Temperature at station P (T1) = 350 K
- Temperature at station Q (T2) = 300 K
- Pressure at station Q (P2) = 50 kPa
- Specific heat at constant pressure (cp) = 1.005 kJ/kg K
- Specific heat at constant volume (cv) = 0.718 kJ/kg K
- Characteristic gas constant (R) = 0.287 kJ/kg K
Approach:
To find the change in entropy, we can use the entropy change equation for an adiabatic process:
ΔS = cp * ln(T2/T1) - R * ln(P2/P1)
Solution:
1. Calculate the logarithmic terms:
- ln(T2/T1) = ln(300/350) = ln(0.857) ≈ -0.1505
- ln(P2/P1) = ln(50/150) = ln(0.333) ≈ -1.0986
2. Substitute the values into the entropy change equation:
ΔS = 1.005 * (-0.1505) - 0.287 * (-1.0986)
ΔS ≈ -0.1512 + 0.3171
ΔS ≈ 0.1659 kJ/kg K
3. Round the value to three decimal places:
ΔS ≈ 0.160 kJ/kg K
Therefore, the change in entropy is approximately 0.160 kJ/kg K. Hence, the correct answer is option 'C'.
- Pressure at station P (P1) = 150 kPa
- Temperature at station P (T1) = 350 K
- Temperature at station Q (T2) = 300 K
- Pressure at station Q (P2) = 50 kPa
- Specific heat at constant pressure (cp) = 1.005 kJ/kg K
- Specific heat at constant volume (cv) = 0.718 kJ/kg K
- Characteristic gas constant (R) = 0.287 kJ/kg K
Approach:
To find the change in entropy, we can use the entropy change equation for an adiabatic process:
ΔS = cp * ln(T2/T1) - R * ln(P2/P1)
Solution:
1. Calculate the logarithmic terms:
- ln(T2/T1) = ln(300/350) = ln(0.857) ≈ -0.1505
- ln(P2/P1) = ln(50/150) = ln(0.333) ≈ -1.0986
2. Substitute the values into the entropy change equation:
ΔS = 1.005 * (-0.1505) - 0.287 * (-1.0986)
ΔS ≈ -0.1512 + 0.3171
ΔS ≈ 0.1659 kJ/kg K
3. Round the value to three decimal places:
ΔS ≈ 0.160 kJ/kg K
Therefore, the change in entropy is approximately 0.160 kJ/kg K. Hence, the correct answer is option 'C'.
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In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K.The following are the properties and relations pertaining to air:Specific heat at constant pressure, cp = 1.005 kJ/kg K;Specific heat at constant volume, cv = 0.718 kJ/kg K;Characteristic gas constant, R = 0.287 kJ/kg K.Enthalpy, h = cpTInternal energy, u = cvTQ.If the pressure at station Q is 50 kPa, the change in entropy in kJ/kg K isa)– 0.155b)0c)0.160d)0.355Correct answer is option 'C'. Can you explain this answer?
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In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K.The following are the properties and relations pertaining to air:Specific heat at constant pressure, cp = 1.005 kJ/kg K;Specific heat at constant volume, cv = 0.718 kJ/kg K;Characteristic gas constant, R = 0.287 kJ/kg K.Enthalpy, h = cpTInternal energy, u = cvTQ.If the pressure at station Q is 50 kPa, the change in entropy in kJ/kg K isa)– 0.155b)0c)0.160d)0.355Correct answer is option 'C'. 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 In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K.The following are the properties and relations pertaining to air:Specific heat at constant pressure, cp = 1.005 kJ/kg K;Specific heat at constant volume, cv = 0.718 kJ/kg K;Characteristic gas constant, R = 0.287 kJ/kg K.Enthalpy, h = cpTInternal energy, u = cvTQ.If the pressure at station Q is 50 kPa, the change in entropy in kJ/kg K isa)– 0.155b)0c)0.160d)0.355Correct answer is option 'C'. 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 In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K.The following are the properties and relations pertaining to air:Specific heat at constant pressure, cp = 1.005 kJ/kg K;Specific heat at constant volume, cv = 0.718 kJ/kg K;Characteristic gas constant, R = 0.287 kJ/kg K.Enthalpy, h = cpTInternal energy, u = cvTQ.If the pressure at station Q is 50 kPa, the change in entropy in kJ/kg K isa)– 0.155b)0c)0.160d)0.355Correct answer is option 'C'. Can you explain this answer?.
In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K.The following are the properties and relations pertaining to air:Specific heat at constant pressure, cp = 1.005 kJ/kg K;Specific heat at constant volume, cv = 0.718 kJ/kg K;Characteristic gas constant, R = 0.287 kJ/kg K.Enthalpy, h = cpTInternal energy, u = cvTQ.If the pressure at station Q is 50 kPa, the change in entropy in kJ/kg K isa)– 0.155b)0c)0.160d)0.355Correct answer is option 'C'. 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 In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K.The following are the properties and relations pertaining to air:Specific heat at constant pressure, cp = 1.005 kJ/kg K;Specific heat at constant volume, cv = 0.718 kJ/kg K;Characteristic gas constant, R = 0.287 kJ/kg K.Enthalpy, h = cpTInternal energy, u = cvTQ.If the pressure at station Q is 50 kPa, the change in entropy in kJ/kg K isa)– 0.155b)0c)0.160d)0.355Correct answer is option 'C'. 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 In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K.The following are the properties and relations pertaining to air:Specific heat at constant pressure, cp = 1.005 kJ/kg K;Specific heat at constant volume, cv = 0.718 kJ/kg K;Characteristic gas constant, R = 0.287 kJ/kg K.Enthalpy, h = cpTInternal energy, u = cvTQ.If the pressure at station Q is 50 kPa, the change in entropy in kJ/kg K isa)– 0.155b)0c)0.160d)0.355Correct answer is option 'C'. Can you explain this answer?.
Solutions for In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K.The following are the properties and relations pertaining to air:Specific heat at constant pressure, cp = 1.005 kJ/kg K;Specific heat at constant volume, cv = 0.718 kJ/kg K;Characteristic gas constant, R = 0.287 kJ/kg K.Enthalpy, h = cpTInternal energy, u = cvTQ.If the pressure at station Q is 50 kPa, the change in entropy in kJ/kg K isa)– 0.155b)0c)0.160d)0.355Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K.The following are the properties and relations pertaining to air:Specific heat at constant pressure, cp = 1.005 kJ/kg K;Specific heat at constant volume, cv = 0.718 kJ/kg K;Characteristic gas constant, R = 0.287 kJ/kg K.Enthalpy, h = cpTInternal energy, u = cvTQ.If the pressure at station Q is 50 kPa, the change in entropy in kJ/kg K isa)– 0.155b)0c)0.160d)0.355Correct answer is option 'C'. Can you explain this answer?, a detailed solution for In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K.The following are the properties and relations pertaining to air:Specific heat at constant pressure, cp = 1.005 kJ/kg K;Specific heat at constant volume, cv = 0.718 kJ/kg K;Characteristic gas constant, R = 0.287 kJ/kg K.Enthalpy, h = cpTInternal energy, u = cvTQ.If the pressure at station Q is 50 kPa, the change in entropy in kJ/kg K isa)– 0.155b)0c)0.160d)0.355Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K.The following are the properties and relations pertaining to air:Specific heat at constant pressure, cp = 1.005 kJ/kg K;Specific heat at constant volume, cv = 0.718 kJ/kg K;Characteristic gas constant, R = 0.287 kJ/kg K.Enthalpy, h = cpTInternal energy, u = cvTQ.If the pressure at station Q is 50 kPa, the change in entropy in kJ/kg K isa)– 0.155b)0c)0.160d)0.355Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K.The following are the properties and relations pertaining to air:Specific heat at constant pressure, cp = 1.005 kJ/kg K;Specific heat at constant volume, cv = 0.718 kJ/kg K;Characteristic gas constant, R = 0.287 kJ/kg K.Enthalpy, h = cpTInternal energy, u = cvTQ.If the pressure at station Q is 50 kPa, the change in entropy in kJ/kg K isa)– 0.155b)0c)0.160d)0.355Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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