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Air at 100 KPa and 280 KPa is compressed steadily to 600 KPa and 400 KPa. The mass flow rate of air is 0.03 Kg/s and a heat loss of 12 kJ/Kg occurs during the process. The power input to compressor will be_____kW
Given h(280k) = 280.13 kJ/Kg, h(400k) = 400.98 kJ/K
(Important - Enter only the numerical value in the answer)
Correct answer is between '3.9,4.1'. Can you explain this answer?
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Air at 100 KPa and 280 KPa is compressed steadily to 600 KPa and 400 K...
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Air at 100 KPa and 280 KPa is compressed steadily to 600 KPa and 400 K...
Given data:
- Initial pressure (P1) = 100 kPa
- Final pressure (P2) = 600 kPa
- Initial pressure (P3) = 280 kPa
- Final pressure (P4) = 400 kPa
- Mass flow rate of air (m) = 0.03 kg/s
- Heat loss during the process (Q) = 12 kJ/kg
- Specific enthalpy at 280 K (h3) = 280.13 kJ/kg
- Specific enthalpy at 400 K (h4) = 400.98 kJ/kg
Assumptions:
- The air behaves as an ideal gas.
- The specific heat capacity at constant pressure (Cp) is constant.
Calculations:
1. Calculation of work done by the compressor:
The work done by the compressor can be calculated using the equation:
W = m * (h2 - h1)
Where:
m = mass flow rate of air
h2 = specific enthalpy at 600 kPa (final pressure)
h1 = specific enthalpy at 100 kPa (initial pressure)
Since the specific heat capacity at constant pressure (Cp) is constant for air, we can use the equation:
Cp = (h2 - h1) / (T2 - T1)
Where:
T2 = final temperature
T1 = initial temperature
Now, rearranging the equation, we can find (h2 - h1):
(h2 - h1) = Cp * (T2 - T1)
Substituting the values:
Cp = (h4 - h3) / (T4 - T3)
Where:
T3 = initial temperature
T4 = final temperature
Now, rearranging the equation, we can find (h4 - h3):
(h4 - h3) = Cp * (T4 - T3)
Since the heat loss during the process is given by Q = m * (h3 - h4), we can substitute the value of (h4 - h3) in the equation:
Q = m * Cp * (T4 - T3)
Solving for (T4 - T3), we get:
(T4 - T3) = Q / (m * Cp)
Substituting the given values:
(T4 - T3) = 12 kJ/kg / (0.03 kg/s * Cp)
We know that Cp = (h4 - h3) / (T4 - T3), so substituting the value of (h4 - h3) and (T4 - T3), we get:
Cp = 12 kJ/kg / (0.03 kg/s * (T4 - T3))
Now, rearranging the equation, we can find (T4 - T3):
(T4 - T3) = 12 kJ/kg / (0.03 kg/s * Cp)
Substituting the value of Cp = (h4 - h3) / (T4 - T3), we get:
(T4 - T3) = 12 kJ/kg / (0.03 kg/s * ((h4 - h3) / (T4 - T3)))
Simplifying the equation, we get:
(T4 -
- Initial pressure (P1) = 100 kPa
- Final pressure (P2) = 600 kPa
- Initial pressure (P3) = 280 kPa
- Final pressure (P4) = 400 kPa
- Mass flow rate of air (m) = 0.03 kg/s
- Heat loss during the process (Q) = 12 kJ/kg
- Specific enthalpy at 280 K (h3) = 280.13 kJ/kg
- Specific enthalpy at 400 K (h4) = 400.98 kJ/kg
Assumptions:
- The air behaves as an ideal gas.
- The specific heat capacity at constant pressure (Cp) is constant.
Calculations:
1. Calculation of work done by the compressor:
The work done by the compressor can be calculated using the equation:
W = m * (h2 - h1)
Where:
m = mass flow rate of air
h2 = specific enthalpy at 600 kPa (final pressure)
h1 = specific enthalpy at 100 kPa (initial pressure)
Since the specific heat capacity at constant pressure (Cp) is constant for air, we can use the equation:
Cp = (h2 - h1) / (T2 - T1)
Where:
T2 = final temperature
T1 = initial temperature
Now, rearranging the equation, we can find (h2 - h1):
(h2 - h1) = Cp * (T2 - T1)
Substituting the values:
Cp = (h4 - h3) / (T4 - T3)
Where:
T3 = initial temperature
T4 = final temperature
Now, rearranging the equation, we can find (h4 - h3):
(h4 - h3) = Cp * (T4 - T3)
Since the heat loss during the process is given by Q = m * (h3 - h4), we can substitute the value of (h4 - h3) in the equation:
Q = m * Cp * (T4 - T3)
Solving for (T4 - T3), we get:
(T4 - T3) = Q / (m * Cp)
Substituting the given values:
(T4 - T3) = 12 kJ/kg / (0.03 kg/s * Cp)
We know that Cp = (h4 - h3) / (T4 - T3), so substituting the value of (h4 - h3) and (T4 - T3), we get:
Cp = 12 kJ/kg / (0.03 kg/s * (T4 - T3))
Now, rearranging the equation, we can find (T4 - T3):
(T4 - T3) = 12 kJ/kg / (0.03 kg/s * Cp)
Substituting the value of Cp = (h4 - h3) / (T4 - T3), we get:
(T4 - T3) = 12 kJ/kg / (0.03 kg/s * ((h4 - h3) / (T4 - T3)))
Simplifying the equation, we get:
(T4 -
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Air at 100 KPa and 280 KPa is compressed steadily to 600 KPa and 400 KPa. The mass flow rate of air is 0.03 Kg/s and a heat loss of 12 kJ/Kg occurs during the process. The power input to compressor will be_____kWGiven h(280k) = 280.13 kJ/Kg, h(400k) = 400.98 kJ/K(Important - Enter only the numerical value in the answer)Correct answer is between '3.9,4.1'. Can you explain this answer?
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Air at 100 KPa and 280 KPa is compressed steadily to 600 KPa and 400 KPa. The mass flow rate of air is 0.03 Kg/s and a heat loss of 12 kJ/Kg occurs during the process. The power input to compressor will be_____kWGiven h(280k) = 280.13 kJ/Kg, h(400k) = 400.98 kJ/K(Important - Enter only the numerical value in the answer)Correct answer is between '3.9,4.1'. 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 Air at 100 KPa and 280 KPa is compressed steadily to 600 KPa and 400 KPa. The mass flow rate of air is 0.03 Kg/s and a heat loss of 12 kJ/Kg occurs during the process. The power input to compressor will be_____kWGiven h(280k) = 280.13 kJ/Kg, h(400k) = 400.98 kJ/K(Important - Enter only the numerical value in the answer)Correct answer is between '3.9,4.1'. 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 Air at 100 KPa and 280 KPa is compressed steadily to 600 KPa and 400 KPa. The mass flow rate of air is 0.03 Kg/s and a heat loss of 12 kJ/Kg occurs during the process. The power input to compressor will be_____kWGiven h(280k) = 280.13 kJ/Kg, h(400k) = 400.98 kJ/K(Important - Enter only the numerical value in the answer)Correct answer is between '3.9,4.1'. Can you explain this answer?.
Air at 100 KPa and 280 KPa is compressed steadily to 600 KPa and 400 KPa. The mass flow rate of air is 0.03 Kg/s and a heat loss of 12 kJ/Kg occurs during the process. The power input to compressor will be_____kWGiven h(280k) = 280.13 kJ/Kg, h(400k) = 400.98 kJ/K(Important - Enter only the numerical value in the answer)Correct answer is between '3.9,4.1'. 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 Air at 100 KPa and 280 KPa is compressed steadily to 600 KPa and 400 KPa. The mass flow rate of air is 0.03 Kg/s and a heat loss of 12 kJ/Kg occurs during the process. The power input to compressor will be_____kWGiven h(280k) = 280.13 kJ/Kg, h(400k) = 400.98 kJ/K(Important - Enter only the numerical value in the answer)Correct answer is between '3.9,4.1'. 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 Air at 100 KPa and 280 KPa is compressed steadily to 600 KPa and 400 KPa. The mass flow rate of air is 0.03 Kg/s and a heat loss of 12 kJ/Kg occurs during the process. The power input to compressor will be_____kWGiven h(280k) = 280.13 kJ/Kg, h(400k) = 400.98 kJ/K(Important - Enter only the numerical value in the answer)Correct answer is between '3.9,4.1'. Can you explain this answer?.
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Air at 100 KPa and 280 KPa is compressed steadily to 600 KPa and 400 KPa. The mass flow rate of air is 0.03 Kg/s and a heat loss of 12 kJ/Kg occurs during the process. The power input to compressor will be_____kWGiven h(280k) = 280.13 kJ/Kg, h(400k) = 400.98 kJ/K(Important - Enter only the numerical value in the answer)Correct answer is between '3.9,4.1'. Can you explain this answer?, a detailed solution for Air at 100 KPa and 280 KPa is compressed steadily to 600 KPa and 400 KPa. The mass flow rate of air is 0.03 Kg/s and a heat loss of 12 kJ/Kg occurs during the process. The power input to compressor will be_____kWGiven h(280k) = 280.13 kJ/Kg, h(400k) = 400.98 kJ/K(Important - Enter only the numerical value in the answer)Correct answer is between '3.9,4.1'. Can you explain this answer? has been provided alongside types of Air at 100 KPa and 280 KPa is compressed steadily to 600 KPa and 400 KPa. The mass flow rate of air is 0.03 Kg/s and a heat loss of 12 kJ/Kg occurs during the process. The power input to compressor will be_____kWGiven h(280k) = 280.13 kJ/Kg, h(400k) = 400.98 kJ/K(Important - Enter only the numerical value in the answer)Correct answer is between '3.9,4.1'. Can you explain this answer? theory, EduRev gives you an
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