Answer:
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
<u>a) Calculate the mass of refrigerant that entered the tank </u>
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 )
<u>b) Calculate the amount of heat transfer </u>
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
