Question

5. A compressor takes air from the atmosphere, which is at 100 kPa and 17 °C, and compresses it to 500 kPa. The compressor is cooled at a rate of 100 kJ / kg of air throughput and the heat rejected to the surrounding air. Irreversibilities in the compressor increase the entropy of the air flowing through it by 0.2 kJ / kg. Calculate the work done and the entropy generated per unit mass of air.

Answer 1

First writing given data mentioned in a problem. Initial condition of air before compression T = 170 = 290k P = 100 kPa Final pressure of air after compression is - 500 kPa The air is taken from atmosphere with Its temperature eguals to 17°C Tatm = 170 = 17+2 73 = 290k at a rate of The compressor és cooled 9 = 100 *3kg irreve. Increase in entropy of air due to rsibilities associated with compressor. Asair - 0.2 kJ/kg gas. We know the values that for air as an édeal of R & Cp are - R = 287 1/4g.k p = 1004.6 k5/4g.k

The increase in entropy of air es given R en Asair - Epen ( 72 ) ...0.2 x 1000 = 1004.6 in () - 28750 (9 T = 560.46 K ... Final temperaf 2.90 ture of air The flow specific work is given by using steady energy equation. 9-W = Cp (T2-T,) W = cp (T2 -T.) - 9 -N = 1004.6 (560.46 - 290)- +100x10²) ... – the term g is negative as compress. or rejecting its heat to surrounding. .-W = 371704. 116 5/ W =-371.704 kJ / Negative ségn indicates that compressor es consuming work from surrounding W- 371.704 kJ/

Increase in entropy of an atmosphere Aaton - Tatm - 100 Tatm 2go Asam = 0.3448 48 / 2.k Total entropy (sgen) is, gen - Asair & Asam = 0.2+ 2538 Saen = 0.5448 kJ/g. 4