Question

Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a velocity of 5 m/s. At the exit, the refrigerant is a saturated vapor at -16oC. The evaporator flow channel has constant diameter of 1.7 cm. Determine the mass flow rate of the refrigerant, in kg/s, and the velocity at the exit, in m/s.

Answer 1

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Step-by-step explanation:

From the information given:

Inlet:

Temperature
`T_1 = -16^0\ C`

Quality
`x_1 = 0.2`

Outlet:

Temperature
`T_2 = -16^0 C`

Quality
`x_2 = 1`

The following data were obtained at saturation properties of R134a at the temperature of -16° C

`v_f= 0.7428 * 10^(-3) \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg`

`v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 * 10^(-3) + 0.2 (0.1247 -(0.7428 * 10^(-3))) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg`

`m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = (1)/(0.0255) * (\pi)/(4)* (1.7 * 10^(-2))^2* 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}`

`\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies (1)/(0.0255) *6 = (1)/(0.1247)* (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}`