Question

A 0.12-m3 rigid tank contains saturated refrigerant-134A at 800 kPa. Initially, 25 percent of the volume is occupied by liquid and the rest by vapor. A valve at the bottom of the tank is now opened, and liquid is withdrawn from the tank. Heat is transferred to the refrigerant such that the pressure inside the tank remains constant. The valve is closed when no liquid is left in the tank and the vapor starts to come out. Determine the total heat transfer for this process.

Answer 1

Answer:

201.381kJ

Explanation:

We have here a mixed fluid, so initially

$V_f+V_g = 0.12 m^3 \\V_f=25%*012 = 0.03 m^3 \\ V_g=75%*0.12 = 0.09 m^3$

From saturated tables, at 800kPa

$v_f = 0.0008458m^3/kg\\v_g=0.025621m^3/kg\\m_f=\frac{V_f}{v_f}=35.47kg\\m_g=\frac{V_g}{v_g}=3.513\\m_1=m_f+m_g=38.98Kg$

We know that at the end, the Volume is only the initial vapor,

$m_2=\frac{V}{v_g}=\frac{0.12}{0.025621} = 4.684kg$

From the tables at 800kPA

$u_f=94.79kJ/kg \\u_g=246.79kJ/kg$

So,

$U_1=m_fu_f+m_gu_g= (35.47*94.79)+(3.513*246.79)\\U_1=4229.2kJ\\U_2=m_2u_g=4.684*246.79=1155.96kJ$

The exiting fluid is saturated, so the equation that we have is

$m_e=m_1-m_2=38.98-4.648=34.3kG$

We search in the tables at 800kPa

$h_e=h_f=95.47kJ/kg$

We can make a energy balance,

$Q_{in}=m_eh_e+U_2-U_1\\Q_{in}=(34.3*95.47)+1155.96-4229.2=201.381kJ$