Answer to Question #302264 in Mechanical Engineering for Rohit

Question

Air is a cylindrical is compressed reversible and isothermallay from 95 kPa , 25°C to 290 kPa. The initial volume is 0.162m³. Determine the work input, heat transfer and the entropy Change of the air being compressed

Answer 1

$P1v1 = p2v2$

$p2/p1 =r$

$290*10^5/95*10^5 = r =3.05$

$v2 = v1/r = 0.162/3.05$

$= 0.05306m^3$

$W = \int Vdp$

$= V (p2-p1)$

$= 0.053* (290-95)*10^5$

= $1.03mJ$

$R= 5.164*10^3$

$T2 = (290*10^5*0.05306)/5.164*10^3$

$T2 = 306K$ = $33 \degree c$

$\delta S = Cv \log (T2/T1) + R \log(V2/V1)$

= $0.178\log(33/25) + 5.164*10^3(0.05306/.162)$

= 5.763 KJ/Kg/K