Question

Refrigerant-134a enters the expansion valve of a refrigeration system at 120 psia as a saturated liquid and leaves at 20 psia. Determine the temperature and internal energy changes across the valve. Use data from the steam tables.

Answer 1

`$P_1 = 120 \ psia$`

`$P_2 = 20 \ psia$`

Using the data table for refrigerant-134a at P = 120 psia

`$h_1=h_f=40.8365 \ Btu/lbm$`

`$u_1=u_f=40.5485 \ Btu/lbm$`

`$T_(sat)=87.745^\circ F$`

∴
`$h_2=h_1=40.8365 \ Btu/lbm$`

For pressure, P = 20 psia

`$h_(2f) = 11.445 \ Btu/lbm$`

`$h_(2g) = 102.73 \ Btu/lbm$`

`$u_(2f) = 11.401 \ Btu/lbm$`

`$u_(2g) = 94.3 \ Btu/lbm$`

`$T_2=T_(sat)=-2.43^\circ F$`

Change in temperature,
`$\Delta T = T_2-T_1$`

`$\Delta T = -2.43-87.745$`

`$\Delta T=-90.175^\circ F$`

Now we find the quality,

`$h_2=h_f+x_2(h_g-h_f)$`

`$40.8365=11.445+x_2(91.282)$`

`$x_2=0.32198$`

The final energy,

`$u_2=u_f+x_2.u_(fg)$`

`$=11.401+0.32198(82.898)$`

`$=38.09297 \ Btu/lbm$`

Change in internal energy

`$\Delta u= u_2-u_1$`

= 38.09297-40.5485

= -2.4556