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xxMikexx [17]
2 years ago
3

A 0.12-m3 rigid tank contains saturated refrigerant-134A at 800 kPa. Initially, 25 percent of the volume is occupied by liquid a

nd 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 A valve at the bottom of the tank is now opened, and liquid is withdrawn from the tank.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.
Engineering
1 answer:
poizon [28]2 years ago
6 0

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

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