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2 months ago

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Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid–vapor mixture at a pressure

of 140 kPa. The refrigerant absorbs 180 kJ of heat from the cooled space, which is maintained at -10oC, and leaves as saturated vapor at the same pressure. Determine (a) the entropy change of the refrigerant, (b) the entropy change of the cooled space, and (c) the total entropy change for this process.

1 answer:

grin007 [323]2 months ago

4 0

Response:

(a). The entropy change of the refrigerant is 0.7077 $\frac{KJ}{K}$

(b). The entropy change of the cooled area is $dS_{space} = - 0.6844$ $\frac{KJ}{K}$

(c). The overall entropy change, dS, is 0.0232 $\frac{KJ}{K}$

Clarification:

Provided parameters

Saturation pressure = 140 K pa

Saturation temperature referenced from property table

$T_{sat}$ = - 18.77 °c = - 18.77 + 273 = 254.23 K

(a). The change in entropy for the refrigerant is calculated by

$dS_{ref} = \frac{Q}{T_{sat}}$

Given that the heat absorbed by the refrigerant Q = 180 KJ

$dS = \frac{180}{254.23}$

dS = 0.7077 $\frac{KJ}{K}$

(b). The entropy change of the cooled space is

$dS_{space} = - \frac{Q}{T_{space}}$

$T_{space}$ = - 10 °c = 263 K

$dS_{space} = - \frac{180}{263}$

$dS_{space} = - 0.6844$ $\frac{KJ}{K}$

(c). The total change in entropy is given by

$dS = dS_{ref} + dS_{space}$

dS = 0.7077 - 0.6844

dS = 0.0232 $\frac{KJ}{K}$

This represents the overall entropy change value.

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