Answer:
[tex]Q=-38.15kJ[/tex]
Explanation:
From the question we are told that
Piston-cylinder initial Volume of air [tex]v_1=0.1 m^3[/tex]
Piston-cylinder initial temperature [tex]T_1=400k[/tex]
Piston-cylinder initial pressure [tex]P_1= 100kpa[/tex]
Supply line temperature[tex]T_s=400k[/tex]
Supply line pressure [tex]P_s= 500kpa[/tex]
Valve final pressure [tex]P_v=500kpa[/tex]
Piston movement pressure [tex]P_m=200kpa[/tex]
Piston-cylinder final Volume of air[tex]v_2=0.2 m^3[/tex]
Piston-cylinder final temperature [tex]T_2=440k[/tex]
Piston-cylinder final pressure [tex]P_2= 500kpa[/tex]
Generally the equation for conservation of mass is mathematically given by
[tex]Q=m_2 \mu_2-m_1 \mu_1 +W-(m_2-m_1)h[/tex]
where
Initial moment
[tex]m_1=\frac{p_1 V_1}{RT_1}[/tex]
[tex]m_1=\frac{100*0.1}{0.287*400}[/tex]
[tex]m_1=8.7*10^-^2kg[/tex]
Final moment
[tex]m_2=\frac{p_2 V_2}{RT_2}[/tex]
[tex]m_1=\frac{500*0.3}{0.287*440}[/tex]
[tex]m_1=79*10^{-2}kg[/tex]
Work done by Piston movement pressure
[tex]W=Pd[/tex]
[tex]W=P(v_2-v_1)[/tex]
[tex]W=200(0.2-0.1))[/tex]
[tex]W=20000J[/tex]
Heat function
[tex]h=cT_1[/tex]
[tex]h=1.005(400)[/tex]
[tex]h=402[/tex]
Therefore
[tex]Q=(0.792*0.718(440)-0.0871*0.718(400)+20-(0.792-0.087)*402))[/tex]
[tex]Q=-38.15kJ[/tex]
It is given mathematically that the system lost or dissipated Heat of
[tex]Q=-38.15kJ[/tex]