Question

Air enters the compressor of a gas turbine power plant at 96 kPa and 25 °C. It comes out at 1 MPa, 327 °C and a velocity of 120 m/s. Velocity at the entrance is small and can be ignored. The compressor is cooled at a rate of 1500 kJ/min. The compressor input power is 300 kW. Calculate the air mass flow rate and cross-sectional area of the outlet. (R = 0.287 kJ/kg K, Cp = 1.005 kJ/kg K)

Name: air enters the compressor of a gas turbine power plant at 96 kpa and 25 c it came out at mpa 327 c and velocity of 120 ms velocity at the entrance is small and can be ignored the compressor 96998
Uploaded: 2022-03-20T07:09:57-08:00
Duration: 4 min 17 s
Channel: Deepanshu Kumar
Description: air enters the compressor of a gas turbine power plant at 96 kpa and 25 c it came out at mpa 327 c and velocity of 120 ms velocity at the entrance is small and can be ignored the compressor 96998

          Air enters the compressor of a gas turbine power plant at 96 kPa and 25 °C. It comes out at 1 MPa, 327 °C and a velocity of 120 m/s. Velocity at the entrance is small and can be ignored. The compressor is cooled at a rate of 1500 kJ/min. The compressor input power is 300 kW. Calculate the air mass flow rate and cross-sectional area of the outlet. (R = 0.287 kJ/kg K, Cp = 1.005 kJ/kg K)

Added by Luisa I.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Deepanshu Kumar

03/20/2022

Step 1

- Convert cooling rate from kJ/min to kW: \(1500 \text{ kJ/min} \times \frac{1 \text{ min}}{60 \text{ s}} = 25 \text{ kW}\) - Note: The compressor input power is already given as 300 kW. Show more…

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Air enters the compressor of a gas turbine power plant at 96 kPa and 25 °C. It comes out at 1 MPa, 327 °C and a velocity of 120 m/s. Velocity at the entrance is small and can be ignored. The compressor is cooled at a rate of 1500 kJ/min. The compressor input power is 300 kW. Calculate the air mass flow rate and cross-sectional area of the outlet. (R = 0.287 kJ/kg K, Cp = 1.005 kJ/kg K)