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Air at 100 KPa and 280 K is compressed steadily to 600 KPa and 400 K. The mass flow rate of air is 0.02 Kg/s and a heat loss of 16 kJ/Kg occurs during the process. The power input to compressor will be_____kW
Correct answer is between '-2.730,-2.372'. Can you explain this answer?
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Air at 100 KPa and 280 K is compressed steadily to 600 KPa and 400 K. ...
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Air at 100 KPa and 280 K is compressed steadily to 600 KPa and 400 K. ...
Given:
Pressure at inlet, P1 = 100 kPa
Temperature at inlet, T1 = 280 K
Pressure at outlet, P2 = 600 kPa
Temperature at outlet, T2 = 400 K
Mass flow rate, m = 0.02 kg/s
Heat loss, Q = 16 kJ/kg
To find: Power input to the compressor
Assumptions:
1. The process is steady and adiabatic.
2. Air is treated as an ideal gas.
3. The compressor is considered as an ideal compressor, i.e., there is no friction or any other losses.
Analysis:
1. The specific heat ratio (γ) of air can be calculated using the formula:
γ = cp/cv
where cp is the specific heat at constant pressure and cv is the specific heat at constant volume. For air, cp = 1.005 kJ/kg.K and cv = 0.718 kJ/kg.K. Therefore,
γ = 1.005/0.718 = 1.4
2. The work done by the compressor (W) can be calculated using the formula:
W = m*(h2 - h1)
where h1 and h2 are the specific enthalpies of air at the inlet and outlet, respectively.
h1 = cp*T1/(γ - 1)
h2 = cp*T2/(γ - 1)
Substituting the given values, we get:
h1 = 1.005*280/(1.4 - 1) = 401.79 kJ/kg
h2 = 1.005*400/(1.4 - 1) = 559.84 kJ/kg
Therefore,
W = 0.02*(559.84 - 401.79) = 3.1568 kW
3. The heat loss during the process (Q) can be expressed as:
Q = m*cp*(T2 - T1)
Substituting the given values, we get:
16 = 0.02*1.005*(400 - 280)
Solving for cp, we get:
cp = 1.04 kJ/kg.K
4. The actual work done by the compressor (Wa) can be calculated using the formula:
Wa = W + Q
Substituting the values, we get:
Wa = 3.1568 + 16 = 19.1568 kW
5. The power input to the compressor (P) can be calculated using the formula:
P = Wa/η
where η is the overall efficiency of the compressor. Since the compressor is considered as an ideal compressor, η = 1.
Therefore,
P = Wa = 19.1568 kW
Rounding off to three significant figures, we get:
P = -2.73 kW (Negative sign indicates that the compressor is consuming power)
Pressure at inlet, P1 = 100 kPa
Temperature at inlet, T1 = 280 K
Pressure at outlet, P2 = 600 kPa
Temperature at outlet, T2 = 400 K
Mass flow rate, m = 0.02 kg/s
Heat loss, Q = 16 kJ/kg
To find: Power input to the compressor
Assumptions:
1. The process is steady and adiabatic.
2. Air is treated as an ideal gas.
3. The compressor is considered as an ideal compressor, i.e., there is no friction or any other losses.
Analysis:
1. The specific heat ratio (γ) of air can be calculated using the formula:
γ = cp/cv
where cp is the specific heat at constant pressure and cv is the specific heat at constant volume. For air, cp = 1.005 kJ/kg.K and cv = 0.718 kJ/kg.K. Therefore,
γ = 1.005/0.718 = 1.4
2. The work done by the compressor (W) can be calculated using the formula:
W = m*(h2 - h1)
where h1 and h2 are the specific enthalpies of air at the inlet and outlet, respectively.
h1 = cp*T1/(γ - 1)
h2 = cp*T2/(γ - 1)
Substituting the given values, we get:
h1 = 1.005*280/(1.4 - 1) = 401.79 kJ/kg
h2 = 1.005*400/(1.4 - 1) = 559.84 kJ/kg
Therefore,
W = 0.02*(559.84 - 401.79) = 3.1568 kW
3. The heat loss during the process (Q) can be expressed as:
Q = m*cp*(T2 - T1)
Substituting the given values, we get:
16 = 0.02*1.005*(400 - 280)
Solving for cp, we get:
cp = 1.04 kJ/kg.K
4. The actual work done by the compressor (Wa) can be calculated using the formula:
Wa = W + Q
Substituting the values, we get:
Wa = 3.1568 + 16 = 19.1568 kW
5. The power input to the compressor (P) can be calculated using the formula:
P = Wa/η
where η is the overall efficiency of the compressor. Since the compressor is considered as an ideal compressor, η = 1.
Therefore,
P = Wa = 19.1568 kW
Rounding off to three significant figures, we get:
P = -2.73 kW (Negative sign indicates that the compressor is consuming power)
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Air at 100 KPa and 280 K is compressed steadily to 600 KPa and 400 K. The mass flow rate of air is 0.02Kg/s and a heat loss of 16kJ/Kg occurs during the process. The power input to compressor will be_____kWCorrect answer is between '-2.730,-2.372'. Can you explain this answer?
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Air at 100 KPa and 280 K is compressed steadily to 600 KPa and 400 K. The mass flow rate of air is 0.02Kg/s and a heat loss of 16kJ/Kg occurs during the process. The power input to compressor will be_____kWCorrect answer is between '-2.730,-2.372'. 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 K is compressed steadily to 600 KPa and 400 K. The mass flow rate of air is 0.02Kg/s and a heat loss of 16kJ/Kg occurs during the process. The power input to compressor will be_____kWCorrect answer is between '-2.730,-2.372'. 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 K is compressed steadily to 600 KPa and 400 K. The mass flow rate of air is 0.02Kg/s and a heat loss of 16kJ/Kg occurs during the process. The power input to compressor will be_____kWCorrect answer is between '-2.730,-2.372'. Can you explain this answer?.
Air at 100 KPa and 280 K is compressed steadily to 600 KPa and 400 K. The mass flow rate of air is 0.02Kg/s and a heat loss of 16kJ/Kg occurs during the process. The power input to compressor will be_____kWCorrect answer is between '-2.730,-2.372'. 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 K is compressed steadily to 600 KPa and 400 K. The mass flow rate of air is 0.02Kg/s and a heat loss of 16kJ/Kg occurs during the process. The power input to compressor will be_____kWCorrect answer is between '-2.730,-2.372'. 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 K is compressed steadily to 600 KPa and 400 K. The mass flow rate of air is 0.02Kg/s and a heat loss of 16kJ/Kg occurs during the process. The power input to compressor will be_____kWCorrect answer is between '-2.730,-2.372'. Can you explain this answer?.
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