Problem 5:10 kg of R-134a at 320 kPa fills a rigid container whose volume is 14 L. Determine the temperature and specific internal energy of the refrigerant. The container is now heated until the pressure is 600 kPa. Determine the temperature and specific internal energy of the refrigerant when the heating is completed. Find the energy transferred to the refrigerant by heat during the heating process, Q, in kJ. Neglect the changes in kinetic and potential energy.

U1 = 56.67 kJ/kg

T1 = 2.46°C

U2 = 83.80 kJ/kg

T2 = 21.55°C

Given that;

m = 10kg           }  from property table Vf = 0.0007772       Uf = 54.92

p1 = 320 kPa    }                                      Vg = 0.063604         Ufg = 176.61

now our specific volume V1 = v/m = 0.014 / 10 = 0.0014

V1 = Vf + x(Vg -Vf)

we substitute

0.0014 = 0.0007772 + x( 0.063604 - 0.0007772)

0.0006228 = 0.0628268x

x = 0.0006228 / 0.0628268

x =  0.0099129

U1 = Uf + x(Ufg)

we substitute

u1= 54.92 + 0.0099129 ( 176.61 )

u1= 56.67 kJ/kg

t1 = tset = 2.46°C

The process is with a constant volume addition

P2 = 600 kPa

V2 = V1 = 0.0014 m^3/kg

from property table

Vf = 0.0008199           Uf = 81.02

Vg = 0.034295            Ufg = 160.81

V2 = Vf + x2 (Vg - Vf)

we substitute

0.0014 = 0.0008199 + x2 ( 0.034295 - 0.0008199 )

0.0005801 = 0.0334751x2

x2 = 0.0005801 / 0.0334751

x2 = 0.017329

U2 = Uf + x2(Ufg)

U2 = 81.02 + 0.017329( 160.81 )

U2 = 83.80 kJ/kg

T2 = Tset = 21.55°C

a

explanation:

because they are cartillidge