Question

A system contains 0.15 m3 of air pressure of 3.8 bars and 1500 C. It is expanded adiabatically till the pressure falls to 1.0 bar. The air is then heated at a constant pressure till its enthalpy increases by 70 kJ. Sketch the process on a P-V diagram and determine the total work done.
Use cp=1.005 kJ/kg.K and cv=0.714 kJ/kg.K

Answer 1

(1) The total work done

$W_t=W_{ad}+W_{ib}$

$W_{ad}=\frac{1}{\gamma-1}(p_1V_1-p_2V_2)$

$V_2=V_1(\frac{p_1}{p_2})^{1/\gamma}, \gamma=\frac{C_p}{C_V}=\frac{c_p\cdot\mu}{c_V\cdot\mu}=\frac{1.005}{0.714}=1.41$

$V_2=V_1(\frac{p_1}{p_2})^{1/\gamma}=0.15\cdot(\frac{3.8}{1.0})^{1/1.41}=0.39m^3$

$W_{ad}=\frac{1}{\gamma-1}(p_1V_1-p_2V_2)=\frac{1}{1.41-1}(380000\cdot 0.15-100000\cdot 0.39)=$

$=43902J$

$W_{ib}=p\Delta V$

$\Delta H=U_2+pV_2-U_1-pV_1=\frac{m}{\mu}C_VT_2-\frac{m}{\mu}C_VT_1+p\Delta V=$

$=\frac{m}{\mu}C_V\frac{pV_2}{\frac{m}{\mu}R}-\frac{m}{\mu}C_V\frac{pV_2}{\frac{m}{\mu}R}+W_{ib}=$

$=\frac{C_V}{R}(p(V_2-V_1))+W_{ib}=\frac{C_V}{R}W_{ib}+W_{ib}=W_{ib}(1+\frac{C_V}{R})$

$W_{ib}=\frac{\Delta H}{1+\frac{C_V}{R}}=\frac{70000}{1+\frac{714\cdot0.02896}{8.31}}=20067J$

$W_t=43902+20067\approx64000J=64kJ$

(2)