Question

Refrigerant-134a is throttled from the saturated liquid state at 700 kPa to a pressure of 160 kPa. Determine the temperature drop during this process and the final specific volume of the refrigerant.

Answer 1

Refrigerant-134a is throttled by a valve. The temperature drop of the refrigerant and specific volume after expansion are to be determined.

Assumptions:

1. This is a steady-flow process since there is no change with time.
2. Kinetic and potential energy changes are negligible.
3. Heat transfer to or from the fluid is negligible.
4. There are no work interactions involved.

Properties: The inlet enthalpy of R-134a is, from the refrigerant tables (Tables A-11 through 13).

Analysis: There is only one inlet and one exit, and thus \(\overset •m_1 = \overset •m_2 = \overset •m\). We take the throttling valve as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Obviously h_f < h₂ < h_g, thus the refrigerant exists as a saturated mixture at the exit state and thus T₂ = T_sat = -15.60°C. Then the temperature drop becomes

\(\Delta\)T = T₂ - T₁

= -15.60 - 26.69

= - 42.3°C

The quality at this state is determined from

\(x_2 = \frac{h_2 - h_f}{h_{fg}}\)

\(= \frac{88.82 - 31.21}{209.90}\)

= 0.2745

Thus, 

v₂ = v_f + x₂v_fg

= 0.0007437 + 0.2745 x (0.12348 - 0.0007437)

= 0.0344 m³/kg