Question

Air at 100 kPa and 27°C is compressed to 600 kPa and 77°C. Calculate the entropy change of air. Take c 1.005 kJ/kg K and HC=1.008 kJ/kg K

Answer 1

Air at 300 kPa and 27°C is compressed to 600 kPa and 77° C. The change in entropy is 0.359 KJ/K.

1) To find the change in entropy in this case, we have to use twin energy equation for pressure.

2) dS = m Cp ln(T2 / T1) + m R ln(P1 / P2)

3) The given known values are: Cp = 1.005 KJ/ kg K, R = 0.287 (for air), T1 = 300K , T2 = 350 K, P1 = 100 kPa , P2 = 600 KPa

4) Putting the above values in the twin equation, we get change in entropy as 0.359 KJ/K.

Answer 2

Air at 100 kPa and 27°C is compressed to 600 kPa and 77° C. The change in entropy is 0.359 KJ/K.

To find the change in entropy we have to use the energy equation for pressure.

The equation is as follows:

dS = m Cp ln($\frac{T2}{T1}$) + m R ln($\frac{P1}{P2}$)

Thus according to the question the given values are:

Cp = 1.005 KJ/ kg K

R = 0.287

T1 = 273+27 = 300K

T2 = 273+77 = 350K

P1 = 100 kPa

P2 = 600 KPa

Now put the values in the equation we get,

dS = 1$\times$ 1.005 ln($\frac{350}{300}$) + 1 $\times$ 0.287 ln($\frac{100}{600}$)

By solving we get dS = 0.359 KJ/K.

Therefore,Air at 100 kPa and 27°C is compressed to 600 kPa and 77° C. The change in entropy is 0.359 KJ/K.