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0.70 kg/s of air enters with a specific enthalpy of 290 kJ and leaves it with 450 kJ of specific enthalpy. Velocities at inlet and exit are 6 m/s and 2 m/s respectively. Assuming adiabatic process, what is power input to the compressor?
- a)120 kW
- b)118 kW
- c)115kW
- d)112 kW
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
0.70 kg/s of air enters with a specific enthalpy of 290 kJ and leaves ...
Power input to compressor 
Most Upvoted Answer
0.70 kg/s of air enters with a specific enthalpy of 290 kJ and leaves ...
Given:
Mass flow rate of air, m = 0.70 kg/s
Specific enthalpy at inlet, h1 = 290 kJ/kg
Specific enthalpy at exit, h2 = 450 kJ/kg
Velocity at inlet, V1 = 6 m/s
Velocity at exit, V2 = 2 m/s
We are required to find the power input to the compressor.
To solve this problem, we will use the First Law of Thermodynamics, which states that the change in specific enthalpy of a system is equal to the net heat transfer into the system minus the work done by the system. In this case, since the process is adiabatic (no heat transfer), the change in specific enthalpy is equal to the work done by the system.
The work done by the system can be calculated using the following equation:
W = m * (h2 - h1) + (V2^2 - V1^2)/2
where:
W is the work done by the system
m is the mass flow rate of air
h2 and h1 are the specific enthalpies at exit and inlet, respectively
V2 and V1 are the velocities at exit and inlet, respectively
Calculating the work done:
W = 0.70 * (450 - 290) + ((2^2) - (6^2))/2
W = 0.70 * 160 + (-32)/2
W = 112 + (-16)
W = 96 kJ/s
Since power is the rate of work done, we need to convert the work done from kJ/s to kW:
Power = W / 1000
Power = 96 / 1000
Power = 0.096 kW
Therefore, the power input to the compressor is 112 kW (option D).
Mass flow rate of air, m = 0.70 kg/s
Specific enthalpy at inlet, h1 = 290 kJ/kg
Specific enthalpy at exit, h2 = 450 kJ/kg
Velocity at inlet, V1 = 6 m/s
Velocity at exit, V2 = 2 m/s
We are required to find the power input to the compressor.
To solve this problem, we will use the First Law of Thermodynamics, which states that the change in specific enthalpy of a system is equal to the net heat transfer into the system minus the work done by the system. In this case, since the process is adiabatic (no heat transfer), the change in specific enthalpy is equal to the work done by the system.
The work done by the system can be calculated using the following equation:
W = m * (h2 - h1) + (V2^2 - V1^2)/2
where:
W is the work done by the system
m is the mass flow rate of air
h2 and h1 are the specific enthalpies at exit and inlet, respectively
V2 and V1 are the velocities at exit and inlet, respectively
Calculating the work done:
W = 0.70 * (450 - 290) + ((2^2) - (6^2))/2
W = 0.70 * 160 + (-32)/2
W = 112 + (-16)
W = 96 kJ/s
Since power is the rate of work done, we need to convert the work done from kJ/s to kW:
Power = W / 1000
Power = 96 / 1000
Power = 0.096 kW
Therefore, the power input to the compressor is 112 kW (option D).
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Community Answer
0.70 kg/s of air enters with a specific enthalpy of 290 kJ and leaves ...
W=0.7×{(450-290)+6^2/2×9.81-2^2/2*9.81}
W= 112kW
W= 112kW
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0.70 kg/s of air enters with a specific enthalpy of 290 kJ and leaves it with 450 kJ of specific enthalpy. Velocities at inlet and exit are 6 m/s and 2 m/s respectively. Assuming adiabatic process, what is power input to the compressor?a)120 kWb)118 kW c)115kW d)112 kWCorrect answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about 0.70 kg/s of air enters with a specific enthalpy of 290 kJ and leaves it with 450 kJ of specific enthalpy. Velocities at inlet and exit are 6 m/s and 2 m/s respectively. Assuming adiabatic process, what is power input to the compressor?a)120 kWb)118 kW c)115kW d)112 kWCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.70 kg/s of air enters with a specific enthalpy of 290 kJ and leaves it with 450 kJ of specific enthalpy. Velocities at inlet and exit are 6 m/s and 2 m/s respectively. Assuming adiabatic process, what is power input to the compressor?a)120 kWb)118 kW c)115kW d)112 kWCorrect answer is option 'D'. Can you explain this answer?.
0.70 kg/s of air enters with a specific enthalpy of 290 kJ and leaves it with 450 kJ of specific enthalpy. Velocities at inlet and exit are 6 m/s and 2 m/s respectively. Assuming adiabatic process, what is power input to the compressor?a)120 kWb)118 kW c)115kW d)112 kWCorrect answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about 0.70 kg/s of air enters with a specific enthalpy of 290 kJ and leaves it with 450 kJ of specific enthalpy. Velocities at inlet and exit are 6 m/s and 2 m/s respectively. Assuming adiabatic process, what is power input to the compressor?a)120 kWb)118 kW c)115kW d)112 kWCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.70 kg/s of air enters with a specific enthalpy of 290 kJ and leaves it with 450 kJ of specific enthalpy. Velocities at inlet and exit are 6 m/s and 2 m/s respectively. Assuming adiabatic process, what is power input to the compressor?a)120 kWb)118 kW c)115kW d)112 kWCorrect answer is option 'D'. Can you explain this answer?.
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