Question

Air at 480Kpa is expanded adiabatically to a pressure of 94Kpa. It is then heated at constant volume until it again attains its temperature, when the pressure is found to be 150KPa. it then undergoes an isothermal compression to return the air to its original state. Using this information sketch the p-v diagram and calculate the value of γ. If the initial temperature of air is 190^oC determine the work done per kg during the adiabatic expansion and teh isotherman compression. Calculate the change in enthropy during the constant volume process.

Take R= 0.29KJ/KgK

Answer 1

      adiabatic                                                              isothermal

P1 = 480 kpa                                                      P1 = 94 k pa

P2   = 94 k pa                                                       P2   = 150 k pa

To find gamma γ

T1 = 190 degree centigrade = 273+190 = 363 K

R = 0.29kj/kg K

Let n =1

V1 = nRT/P1   = 1*0.29*1000*363/480*1000 =0.219

V2 = nRT/P2 =   1*0.29*1000*363/94*1000   = 1.1198

Now P2/P1 (V2/V1)^γ =1      Therefore   (94*1000/480*1000) ( 1.1198/.219) ^γ =1

(5.113)^γ    = 1*480/94     = 5.106

Take logs

log(5.113)^γ    = log 5.106

γ =log(5.106 - log 5.113 = -5.949*10^-4

To find work done (isothermal)

w    = -NRT lnVB/VA = -1*.29*1000*363ln(1.1198/0.219)    = 17178.3128 J

Work done in adiabatic process = - f P1V1( (v2/V1)^1-γ     -1)

                                              = -(-2878 *94*1000* ((1.475/1.1198)^1-(-5.949*10^-4    -1)

                                             = 2.70*10^8J

( f in above has been calculated here

Because   γ   = f+2/f

so -5.949*10^-4 = f+2/f  

f(1+5.949*10^-4) = -2

f = -2878

Here V1 = 1.1198

V2 = nRT/P2   = 1*.29* 1000*363/150*1000   = 1.475 )

Change in entropy at constant volume = delta S = nRT in PA/PB    = 1*0.29*1000* ln (94*10^3/150*10^3)

                                                                       = -812.78 j/k