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Refrigerant 134a enters a well-insulated nozzle at 200 lbf/in.2, 200°F, with a velocity of 120 ft/s and exits at 50 lbf/in.2 with a velocity of 1500 ft/s. For steady-state operation, and neglecting potential energy effects, determine the temperature, in °F, and the quality of the refrigerant at the exit.

Respuesta :

Answer:

x = 0.75801 = 75.801%

T_2 = 72..78 degree F

Explanation:

From superheated R 134 a properties table

At 200 lb/in^2 and 200 degree F

[tex]h_1 = 138.99 Btu/lbm[/tex]

steady flow energy equation is givena s

[tex]h_1 + \frac{v_1^2}{2}  = h_2 + \frac{v_2^2}{2}[/tex]

[tex]138.99 + \frac{120^2}{2\times 25037} = h_2 + \frac{1500^2}{2 \times 25037}[/tex]

[tex]h_2 = 94.344 Btu/lbm[/tex]

At 90 lb/in2 Tsat = 72.78 degree F

[tex]h_f = 35.715 Btu/lbm[/tex]

hfg  = 77.345 Btu/lbm

h = hf + x hfg

[tex]94.344 = 35.715+ x \times 77.345[/tex]

solving for x we get

x = 0.75801 = 75.801%

[tex]T_2 = 72..78 degree F[/tex]

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