A certain mass of air initially at a pressure of 480 kPa and temperature 190°C i

Question

A certain mass of air initially at a pressure of 480 kPa and temperature 190°C is expanded adiabatically to a pressure of 94 kPa. It is then heated at constant volume until it attains its initial temperature and a pressure of 150 kPa. Compression then takes place until the system returns to its original pressure and volume. State the type of compression necessary in the third process, and calculate: 1 The index of adiabatic expansion 2 The work done per kilogram of air For air, the characteristic gas constant R = 0.287 kJ/ kg.K

Explanation / Answer

[480]^1-t * (T1)^t = (94)^t * T2

T2 = [480]^1-t * (T1)^t / (94)^t

T2 /94= T1/150

T2 = 94/150 * 463 = 290.1466 K

It should be isothermal compression because pv=constant

[463/480]^t-1 = (94)^t*94/150

[463/480*94]^t =  [463*94/480*150]

(0.0102)^t= 0.60447

1.

t = log 0.60447 / log 0.0102 = 0.1097 = 0.11

2.

v1 occupied by 1 kg air = 0.287 *463/480 * 1= 0.2768 m3

v2 occupied by 1 kg air = 0.287 *290.14/480 * 1 =0.17347 m3

By convention, work is defined as the work on the system by its surroundings. If, for example, the system is compressed, then the work is positive and the internal energy of the system increases. Conversely, if the system expands, it does work on the surroundings and the internal energy of the system decreases.

w1 to 2 = 480*0.2768 - 94*0.17347 / 0.11 -1 = -130.963 kj

w 2 to 3 = 0 [constant volume process]

w 3 to 4 = -480*0.2768ln (0.11/0.2768) = 122.608 kj

work done per kilogram of air = w1 to 2 + w 2 to 3 + w 3 to 4 =-8.354 kj

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