Answer
a) $T_2=85.8\ °C$
b) $w_{in}=184.1\ kJ/kg$
c) $s_{gen}=0.0068\ kJ/kg.K$
Work Step by Step
a) From the entropy change relation:
$\Delta s_{air}=c_p\ln{\dfrac{T_2}{T_1}}-R\ln{\dfrac{P_2}{P_1}}$
Given: $\Delta s_{air}=-0.40\ kJ/kg.K,\ c_p=1.005\ kJ/kg.K,\ R=0.287\ kJ/kg.K,\ T_1=295\ K=T_{surr}$,
$P_2=800\ kPa,\ P_1=100\ kPa$:
We can solve for $T_2=358.8\ K=85.8\ °C$
b) From the specific energy balance on the compressor:
$w_{in}=c_p(T_2-T_1)+q_{out}$ with $q_{out}=120\ kJ/kg$
$w_{in}=184.1\ kJ/kg$
c) From the definition:
$\Delta s_{surr}=\dfrac{q_{out}}{T_{surr}}=0.4068\ kJ/kg.K$
$s_{gen}=\Delta s_{total}=\Delta s_{surr}+\Delta s_{air}$
$s_{gen}=0.0068\ kJ/kg.K$
