A 0.06-m3 rigid tank initially contains refrigerant- 134a at 0.8 MPa and 100 percent quality. The tank is connected by a valve to a supply line that carries refrigerant- 134a at 1.2 MPa and 36°C. Now the valve is opened, and the refrigerant is allowed to enter the tank. The valve is closed when it is observed that the tank contains saturated liquid at 1.2 MPa. Determine (a) the mass of the refrigerant that has entered the tank and (b) the amount of heat transfer.

a) 0.50613

b) 22.639 kJ

Explanation:

From table A-11 , we will make use of the properties of Refrigerant R-13a at 24°C

first step : calculate the  volume of R-13a  ( values gotten from table A-11 )

V = m1 * v1 = 5 * 0.0008261 = 0.00413 m^3

next : calculate final specific volume ( v2 )

v2 = V / m2 = 0.00413 / 0.25 ≈ 0.01652 m^3/kg

a) Calculate the mass of refrigerant that entered the tank

v2 = Vf + x2 * Vfg

v2 = Vf +  [ x2 * ( Vg - Vf ) ] ( 1 )

where:  Vf = 0.0008261 m^3/kg, V2 = 0.01652 m^3/kg , Vg = 0.031834 m^3/kg  ( insert values into equation 1 above )

x2 = ( 0.01652 - 0.0008261 ) / 0.031834

     = 0.50613 ( mass of refrigerant that entered tank )

b) Calculate the amount of heat transfer

Final specific internal energy = u2 = Uf + ( x2 + Ufg ) ( 2 )

uf = 84.44 kj/kg , x2 = 0.50613 , Ufg = 158.65 Kj/kg

therefore U2 = 164.737 Kj/kg

The mass balance  ( me ) = m1 - m2 --- ( 3 )

energy balance( Qin ) = ( m2 * u2 ) - ( m1 * u1 ) + ( m1 - m2 ) * he

therefore Qin = 41.184 - 422.2 + 403.655  = 22.639 kJ

twist boundary

explanation:

answer:

look in settings

explanation: