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A refrigeration system operating on reversed brayton cycle has a temperature of 250 k at inlet to the compressor. If the temperature at the end of constant pressure cooling is 300 k and rise in temperature of air in refrigerator is 50 k. Then the net work of compression will be _______ kJ/kg (for air cp = 1 kJ/kg K)
Correct answer is '25'. Can you explain this answer?
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Given data:
- Inlet temperature to the compressor (T1) = 250 K
- Temperature at the end of constant pressure cooling (T3) = 300 K
- Rise in temperature of air in refrigerator (ΔT) = 50 K
- Specific heat of air at constant pressure (cp) = 1 kJ/kg K
To find:
- Net work of compression (Wnet)
Solution:
1. The reversed Brayton cycle consists of four processes:
- Process 1-2: Isentropic compression
- Process 2-3: Constant pressure cooling
- Process 3-4: Isentropic expansion
- Process 4-1: Constant pressure heating
2. As per the given data, the process 2-3 is a constant pressure cooling process. Therefore, we can use the following equation to find the heat rejected (Qr) during this process:
Qr = cp * m * ΔT
Where m is the mass of air flowing through the refrigeration system.
3. The net work of compression (Wnet) is equal to the sum of work done during process 1-2 and process 3-4. Therefore, we can use the following equation to find Wnet:
Wnet = W1-2 + W3-4
4. As per the reversed Brayton cycle, the process 1-2 is an isentropic compression process. Therefore, we can use the following equation to find the work done (W1-2) during this process:
W1-2 = m * cp * (T2 - T1)
Where T2 is the temperature at the end of process 1-2.
5. As per the reversed Brayton cycle, the process 3-4 is an isentropic expansion process. Therefore, we can use the following equation to find the work done (W3-4) during this process:
W3-4 = m * cp * (T3 - T4)
Where T4 is the temperature at the end of process 3-4.
6. As per the reversed Brayton cycle, the temperatures at the end of process 2-3 and process 4-1 are equal. Therefore, we can assume that T4 is equal to T1.
7. Substituting the given values in the above equations, we get:
Qr = 1 * m * 50 = 50m
W1-2 = m * 1 * (T2 - 250)
W3-4 = m * 1 * (300 - 250) = 50m
Wnet = (m * 1 * (T2 - 250)) + (m * 1 * (300 - 250)) = m * 50 + m * (T2 - 250)
8. Simplifying the above equation, we get:
Wnet = m * (T2 - 200)
9. As per the reversed Brayton cycle, the compressor and turbine have the same mass flow rate of air. Therefore, we can assume that the mass flow rate of air is constant throughout the cycle. Therefore, we can cancel out the mass flow rate (m) from the above equation.
10. Substituting the given value of T3 in the above equation, we get:
Wnet = 50 * (T2 - 200) =
- Inlet temperature to the compressor (T1) = 250 K
- Temperature at the end of constant pressure cooling (T3) = 300 K
- Rise in temperature of air in refrigerator (ΔT) = 50 K
- Specific heat of air at constant pressure (cp) = 1 kJ/kg K
To find:
- Net work of compression (Wnet)
Solution:
1. The reversed Brayton cycle consists of four processes:
- Process 1-2: Isentropic compression
- Process 2-3: Constant pressure cooling
- Process 3-4: Isentropic expansion
- Process 4-1: Constant pressure heating
2. As per the given data, the process 2-3 is a constant pressure cooling process. Therefore, we can use the following equation to find the heat rejected (Qr) during this process:
Qr = cp * m * ΔT
Where m is the mass of air flowing through the refrigeration system.
3. The net work of compression (Wnet) is equal to the sum of work done during process 1-2 and process 3-4. Therefore, we can use the following equation to find Wnet:
Wnet = W1-2 + W3-4
4. As per the reversed Brayton cycle, the process 1-2 is an isentropic compression process. Therefore, we can use the following equation to find the work done (W1-2) during this process:
W1-2 = m * cp * (T2 - T1)
Where T2 is the temperature at the end of process 1-2.
5. As per the reversed Brayton cycle, the process 3-4 is an isentropic expansion process. Therefore, we can use the following equation to find the work done (W3-4) during this process:
W3-4 = m * cp * (T3 - T4)
Where T4 is the temperature at the end of process 3-4.
6. As per the reversed Brayton cycle, the temperatures at the end of process 2-3 and process 4-1 are equal. Therefore, we can assume that T4 is equal to T1.
7. Substituting the given values in the above equations, we get:
Qr = 1 * m * 50 = 50m
W1-2 = m * 1 * (T2 - 250)
W3-4 = m * 1 * (300 - 250) = 50m
Wnet = (m * 1 * (T2 - 250)) + (m * 1 * (300 - 250)) = m * 50 + m * (T2 - 250)
8. Simplifying the above equation, we get:
Wnet = m * (T2 - 200)
9. As per the reversed Brayton cycle, the compressor and turbine have the same mass flow rate of air. Therefore, we can assume that the mass flow rate of air is constant throughout the cycle. Therefore, we can cancel out the mass flow rate (m) from the above equation.
10. Substituting the given value of T3 in the above equation, we get:
Wnet = 50 * (T2 - 200) =
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A refrigeration system operating on reversed brayton cycle has a temperature of 250 k at inlet to the compressor. If the temperature at the end of constant pressure cooling is 300 k and rise in temperature of air in refrigerator is 50 k. Then the net work of compression will be _______ kJ/kg (for air cp = 1 kJ/kg K)Correct answer is '25'. Can you explain this answer?
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A refrigeration system operating on reversed brayton cycle has a temperature of 250 k at inlet to the compressor. If the temperature at the end of constant pressure cooling is 300 k and rise in temperature of air in refrigerator is 50 k. Then the net work of compression will be _______ kJ/kg (for air cp = 1 kJ/kg K)Correct answer is '25'. 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 A refrigeration system operating on reversed brayton cycle has a temperature of 250 k at inlet to the compressor. If the temperature at the end of constant pressure cooling is 300 k and rise in temperature of air in refrigerator is 50 k. Then the net work of compression will be _______ kJ/kg (for air cp = 1 kJ/kg K)Correct answer is '25'. 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 A refrigeration system operating on reversed brayton cycle has a temperature of 250 k at inlet to the compressor. If the temperature at the end of constant pressure cooling is 300 k and rise in temperature of air in refrigerator is 50 k. Then the net work of compression will be _______ kJ/kg (for air cp = 1 kJ/kg K)Correct answer is '25'. Can you explain this answer?.
A refrigeration system operating on reversed brayton cycle has a temperature of 250 k at inlet to the compressor. If the temperature at the end of constant pressure cooling is 300 k and rise in temperature of air in refrigerator is 50 k. Then the net work of compression will be _______ kJ/kg (for air cp = 1 kJ/kg K)Correct answer is '25'. 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 A refrigeration system operating on reversed brayton cycle has a temperature of 250 k at inlet to the compressor. If the temperature at the end of constant pressure cooling is 300 k and rise in temperature of air in refrigerator is 50 k. Then the net work of compression will be _______ kJ/kg (for air cp = 1 kJ/kg K)Correct answer is '25'. 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 A refrigeration system operating on reversed brayton cycle has a temperature of 250 k at inlet to the compressor. If the temperature at the end of constant pressure cooling is 300 k and rise in temperature of air in refrigerator is 50 k. Then the net work of compression will be _______ kJ/kg (for air cp = 1 kJ/kg K)Correct answer is '25'. Can you explain this answer?.
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