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------------------------------------------------ A simple ideal refrigeration cycle operates between 140 kPa and 900 kPa. The cycle operates steadily and uses refrigerant R134a. Determine the rate of cooling of the refrigerated space per kg R134a (kJ/kg). Indicate the correct answer from the list provided. If none of the choices are within 5% of the correct answer, or if the question is unanswerable, indicate that choice instead.

A. 107 kJ/kg
B. The question is unanswerable: it is missing information
C. 0.652 kJ/kg
D. 38.8 kJ/kg
E. 0.561 kJ/kg
F. 0 kJ/kg
G. 176 kJ/kg
H. 138 kJ/kg
I. None of these are within 5% of the correct solution
J. 17.2 kJ/kg

Question

College

Answer :

AngelinaMatthews AngelinaMatthews · Answer 1 · 2023-12-31T02:02:59-05:00

The rate of cooling of the refrigerated space per kg R134a (kJ/kg) is given by:

Q = h1 - h4 = 320.65 - 355.23= -34.58 kJ/kg ≈ -34.6 kJ/kg

Answer: d. -34.6 kJ/kg

A simple ideal refrigeration cycle operates between 140 kPa and 900 kPa.

The cycle operates steadily and uses refrigerant R134

a. The specific volume of R134a at 140 kPa and 10°C is 0.103 m3/kg and that at 900 kPa is 0.0328 m3/kg.

Determine the rate of cooling of the refrigerated space per kg R134a (kJ/kg).

Now, the rate of cooling of the refrigerated space per kg R134a (kJ/kg) is given by:

Q = h1 - h4 (refrigeration effect per unit mass of refrigerant)

Where h1 is the enthalpy of the refrigerant at state 1, and h4 is the enthalpy of the refrigerant at state 4.

Here, state 1: P1 = 140 kPa, T1 = -10°C (from saturation table), and state 4:

P4 = 900 kPa, T4 = 30°C (from saturation table)The refrigeration effect per unit mass of refrigerant is given by

Q = h1 - h4 = hfg - h2

Now, hfg can be obtained from the saturation table at 140 kPa or 900 kPa and h2 can be obtained from the saturation table at 900 kPa

.The specific enthalpies of R134a at the given states are as follows:

h1 = 320.65 kJ/kgh4 = 355.23 kJ/kg

Thus, the rate of cooling of the refrigerated space per kg R134a (kJ/kg) is given by:

Q = h1 - h4

= 320.65 - 355.23

= -34.58 kJ/kg

≈ -34.6 kJ/kg

Answer: d. -34.6 kJ/kg

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