Question

4 m3 of air at 2 bar, 27°C is compressed up to 6 bar pressure following

pv1.2 = constant. It is subsequently expanded adiabatically to 2 bar. 

Considering the two processes to be reversible, determine the net work.

Answer 1

Solution.

$V_1=4m^3;$

$p_1=2bar;$

$T_1=300K;$

$p_2=6bar;$

$p_3=2bar;$

$pV^{1.2}=const;$

$p_1V_1=\dfrac{m}{M}RT_1\implies m=\dfrac{Mp_1V_1}{RT_1};$

$m=\dfrac{29\sdot10^{-3}\sdot2\sdot10^5\sdot4}{8.31\sdot300}=9.3kg;$

$\dfrac{T_2}{T_1}=(\dfrac{p_2}{p_1})^{\dfrac{n-1}{n}}\implies T_2=T_1(\dfrac{p_2}{p_1})^{\dfrac{n-1}{n}};$

$T_2=300\sdot(\dfrac{6}{2})^{\dfrac{1.2-1}{1.2}}=360K;$

$\dfrac{T_2}{T_3}=(\dfrac{p_2}{p_3})^{\dfrac{\gamma-1}{\gamma}}\implies T_3=T_2(\dfrac{p_3}{p_2})^{\dfrac{\gamma-1}{\gamma}};$

$T_3=300(\dfrac{2}{6})^{\dfrac{1.4-1}{1.4}}=262K;$

$W_{1-2}=\dfrac{mR(T_1-T_2)}{M(n-1)}= \dfrac{9.3\sdot8.31\sdot(300-360)}{0.029(1.2-1)}=-799479J;$

$W_{2-3}=\dfrac{mR(T_2-T_3)}{M(\gamma-1)}= \dfrac{9.3\sdot8.31\sdot(360-262)}{0.029(1.4-1)}=652908J;$

$W=-799479J+652908J=-146570J=-146.57kJ;$

Answer: $W=-146.57kJ.$