Question

Refrigerant 134a passes through a throttle valve as part of a refrigeration cycle. The fluid enters the valve as compressed liquid at a pressure of 0.8 MPa and a temperature of 25 C. It exits the valve at -20 C. Find the exit pressure and the exit specific internal energy.

Answer 1

The exit pressure will be 1.3273 bar = 0.1327 MPa and the exit specific internal energy 228.61 KJ/Kg .

1.) In graph,

1-2 = Compressor

3-4 = Throttle value

$T_{1}$ = $T_{4}$ = -20°C = 253 K

$T_{3}$ = 25°C = 298 K

$P_{3}$ = 0.8 MPa

Assuming saturated fluid is entering the value at point (3),

using table of R-134a properties

∵ $h_{3} =h_{4}$ = 234.49 KJ/Kg (at 25°C,8 bar)

at $T_{3}$ = 25° , $P_{3}$ = 0.8 MPa =8 bar

At -20°C,   $hf_{4}$ = 173.6 KJ/Kg

                 $hg_{4}$ = 386.6 KJ/Kg

$h_{4} = hf_{3} + n_{4} ( hg_{4} - hf_{4} )$ = 234.49

173.6 + $n_{4}$ [386.6 - 173.6] = 234.49

             $n_{4}$ = 0.285

$P_{4} = P_{1} = P_{sat}$ = 1.3273 bar = 0.1327 MPa

2.)

Exit specific internal energy, using table R-134a

$U_{4} = uf_{4} + n_{4} (ug_{4} - uf_{4} )$

at -20°c, $uf_{4}$ = 173.65, ufg = 192.85 KJ/ Kg

$U_{4}$ = 173.65 + 0.285 (192.85)

$U_{4}$ = 228.61 KJ/Kg

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