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.

Answer 1

The temperature and internal energy changes across the expansion valve in a refrigeration system are both zero.

Step-by-step explanation:

To determine the temperature and internal energy changes across the expansion valve in a refrigeration system, we need to apply the first law of thermodynamics. Since the refrigerant enters the expansion valve as a saturated liquid, we can assume that the process is isenthalpic, meaning there is no change in enthalpy. Therefore, the temperature change across the valve is zero. The internal energy change can be calculated using the equation ΔU = Q - W, where Q is the heat transfer and W is the work done on the system. In this case, since the process is isenthalpic, there is no heat transfer, so the internal energy change is also zero.

Answer 2

a.Internal temperature change is 90.49°F

b.Internal energy change is -2.43Btu/lbm

Step-by-step explanation:

The energy balance for this steady-flow system can be expressed as:

`\dot E_i_n-\dot E_o_u_t=\bigtriangleup \dot E_s_y_s=0\\\dot E_i_n=\dot E_o_u_t`

Therefore,

`\dot mh_1=\dot mh_2\\h_1=h_2`

#Since:
`\dot Q \approx =\dot W=\bigtriangleup ke\approx =\bigtriangleup pc\approx =0`

From tableA-1

`p_2=120pisa\\x_1=0\ \ \ \ \ \ \ \ \ \ ->h_1=41.79Btu/lbm, u_1=41.49Btm/lbm,T_1=90.49\textdegree F`

`p_2=20pisa\\\ \ \ \ \ \ \ \ \ \ \\h_2=h_1=41.79Btu/lbm, u_2=38.96Btm/lbm,T_2=-2.43\textdegree F\\\\\therefore \bigtriangleup T=T_2-T_1`

#Replacing the values in the equation:

`\bigtriangleup T=-2.43-90.49 ->|\bigtriangleup T|=92.92\textdegree F`

We have that
`\bigtriangleup u=u_2-u_1\\\bigtriangleup u=38.96-41.49=\\\bigtriangleup u=-2.43Btu/lbm`