A vertical piston-cylinder device initially contains 0.1 m^3 of air at 400 K and 100 kPa. At this initial condition, the piston is resting on a stop. The piston-cylinder device is connected to a supply line with air at 400 K and a pressure of 500 kPa. The valve between the supply line and the piston-cylinder device is opened and is left open until the pressure in the piston-cylinder device reaches 500 kPa. The piston is observed to start moving when the pressure in the cylinder is 200 kPa and the piston continues to rise until it reaches a second stop. At the second stop, the piston-cylinder volume is 0.2 m^3 . The final temperature and pressure in the piston-cylinder device are 440 K and 500 kPa, respectively. Determine the heat transfer to or from the piston-cylinder device for the filling process Determine heat transfer (kJ) during the entire process.

Respuesta :

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]