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3 years ago

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Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 21 m3/min and exits at 12 b

ar, 400 K. Heat transfer occurs at a rate of 3.5 kW from the compressor to its surroundings.
Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.

1 answer:

omeli [17]3 years ago

3 0

Answer:

- 46.5171kW

Explanation:

FIrst, the value given:

P1 = 1.05 bar (Initial pressure)

P2 = 12 bar (final pressure)

Heat transfer, Q = - 3.5 kW (It is negative because the compressor losses heat to the surroundings)

Mgaseous nitrogen = Mair = 28.0134 Kg/mol (constant)

Universal gas constant, Ru = 8.3143 Kj/Kgmolk

Specific gas constant, R = 0.28699 Kj/KgK

Initial temperature, T1 = 300 K

Final temperature, T2 = 400 K

Finding the volume:

P1V1 = RT1

V1 = RT1 ÷ P1

= (0.28699 Kj/KgK X 300k) ÷ 105

Note convert bar to Kj/Nm by multiply it by 100

V1 = 0.81997 m3/Kg

To get the mass flow rate:

m = volumetric flow rate / V1

= (21 m3/min x 1/60seconds) ÷ 0.81997 m3/Kg

= 0.4268Kg/s

Using tables for the enthalpy,

hT1 = 300.19 KJ/Kg

hT2 = 400.98 KJ/Kg

The enthalpy change = hT2 - hT1

= 100.79 KJ/Kg

Power, P = Q - (m X enthalpy change)

= - 3.5 - (0.4268 X 100.79)

= - 46.5171kW

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Where ${h_{2s}$ is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because $h_{2}-h_{1}$ can be obtained from the energy balance $\frac{W}{m}=h_{2}-h_{1}$

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$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3$ %

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