---
title: "Ideal vapour compression cycle for 5 TR"
subject: "Applied Thermodynamics"
chapter: 3
key: "semester_3/applied_thermodynamics:vcr-ideal-cycle-numerical"
source: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/vcr-ideal-cycle-numerical
site: BCT Engineering
---

# Ideal vapour compression cycle for 5 TR

Applied Thermodynamics · Chapter 3: Refrigeration

- Asked in: year not given, 8 marks
- Video: [Numerical on Simple Vapour Compression Refrigeration Cycle by Mona Yadav](https://www.youtube.com/watch?v=skFEWmKDEQ0) — mona yadav
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/vcr-ideal-cycle-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_thermodynamics/chapter-3

## Question

A refrigerator working on an ideal vapour compression cycle uses R-134a between 0.14 MPa and 0.8 MPa. From tables: dry saturated vapour at 0.14 MPa has $h_1 = 239.16$ kJ/kg and $v_1 = 0.1402$ m³/kg; the vapour after isentropic compression to 0.8 MPa has $h_2 = 275.39$ kJ/kg; saturated liquid at 0.8 MPa has $h_3 = 95.47$ kJ/kg. For a capacity of 5 TR (1 TR = 3.517 kW), determine (a) the refrigerating effect, (b) the refrigerant mass flow rate, (c) the compressor power, (d) the COP, (e) the heat rejected in the condenser, and (f) the volume flow rate at compressor inlet.

## Short answer

Refrigerating effect = 143.7 kJ/kg; mass flow rate = 0.1224 kg/s; compressor power = 4.43 kW; COP = 3.97; heat rejected = 22.0 kW; volume flow at inlet = 0.0172 m³/s.

## Answer

#### Cycle states
- State 1: dry saturated vapour, $h_1 = 239.16$ kJ/kg.
- State 2: after isentropic compression, $h_2 = 275.39$ kJ/kg.
- State 3: saturated liquid, $h_3 = 95.47$ kJ/kg.
- State 4: after throttling, $h_4 = h_3 = 95.47$ kJ/kg.

Capacity: $Q_L = 5 \times 3.517 = 17.585$ kW.

#### (a) Refrigerating effect
$$
q_L = h_1 - h_4 = 239.16 - 95.47 = 143.69\ \text{kJ/kg}
$$

#### (b) Mass flow rate
$$
\dot m = \frac{Q_L}{q_L} = \frac{17.585}{143.69} = 0.1224\ \text{kg/s}
$$

#### (c) Compressor power
$$
w = h_2 - h_1 = 275.39 - 239.16 = 36.23\ \text{kJ/kg}
$$

$$
P = \dot m\,w = 0.1224 \times 36.23 = 4.43\ \text{kW}
$$

#### (d) COP
$$
\text{COP} = \frac{q_L}{w} = \frac{143.69}{36.23} = 3.97
$$

#### (e) Heat rejected in condenser
$$
Q_H = \dot m (h_2 - h_3) = 0.1224 \times (275.39 - 95.47) = 22.02\ \text{kW}
$$

Check: $Q_L + P = 17.585 + 4.43 = 22.02$ kW, equal to $Q_H$.

#### (f) Volume flow rate at compressor inlet
$$
\dot V_1 = \dot m\,v_1 = 0.1224 \times 0.1402 = 0.0172\ \text{m}^3/\text{s}
$$

**Answer:** (a) 143.7 kJ/kg; (b) 0.1224 kg/s; (c) 4.43 kW; (d) 3.97; (e) 22.0 kW; (f) 0.0172 m³/s.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The refrigerating effect is
   A. h2 - h1 = 36.23 kJ/kg
   B. h2 - h3 = 179.9 kJ/kg
   C. h1 - h2 = -36.23 kJ/kg
   D. h1 - h4 = 143.69 kJ/kg
   Answer: D. h4 = h3, so q = 239.16 - 95.47. (Key point: Refrigerating effect)

2. The mass flow rate is
   A. 0.1224 kg/s = 17.585/143.69
   B. 1.224 kg/s
   C. 0.0122 kg/s
   D. 8.17 kg/s
   Answer: A. Capacity (5 x 3.517 kW) divided by refrigerating effect. (Key point: Mass flow)

3. The COP is
   A. 0.25
   B. 3.97
   C. 4.97
   D. 2.97
   Answer: B. 143.69/36.23 = 3.97. (Key point: COP)

4. The condenser heat rejection equals
   A. Q_L minus power, 13.2 kW
   B. compressor power only
   C. Q_L + compressor power, about 22.0 kW
   D. Q_L only
   Answer: C. Energy balance: 17.585 + 4.43 = 22.02 kW. (Key point: Condenser heat)

## Related in Applied Thermodynamics

- [VCR cycle with subcooling and superheating](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_thermodynamics%3Avcr-subcooling-superheating-numerical) — Applied Thermodynamics, chapter 3
- [Ideal versus actual vapour compression cycle](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_thermodynamics%3Aideal-vs-actual-vcr-cycle) — Applied Thermodynamics, chapter 3
- [Actual VCR cycle with compressor efficiency](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_thermodynamics%3Avcr-actual-compressor-efficiency-numerical) — Applied Thermodynamics, chapter 3
- [Simple vapour compression cycle](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_thermodynamics%3Asimple-vcr-cycle-ts-ph-diagrams) — Applied Thermodynamics, chapter 3

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