Why this topic is important for ECET?
The Vapour Compression Cycle (VCC) is the heart of refrigeration and air conditioning systems, used in refrigerators, ACs, and heat pumps. Questions are frequently asked in ECET about COP, T-s and P-h diagrams, and numerical calculations. Mastering this topic helps you score easily because the formulas are straightforward if concepts are clear.
📘 Concept Notes
🔹 What is Vapour Compression Cycle?
- It is the most widely used refrigeration cycle in practical systems (domestic refrigerators, window AC, split AC, industrial cooling).
- The working substance is a refrigerant (like R134a, R22, R600a).
- Operates on a closed cycle involving four major processes.
🔹 Main Components of VCC
- Compressor → Compresses low-pressure refrigerant vapor → high-pressure, high-temperature vapor.
- Condenser → Rejects heat to surroundings → vapor condenses into high-pressure liquid.
- Expansion Valve (Throttle) → Drops the pressure suddenly, liquid becomes low-pressure mixture.
- Evaporator → Absorbs heat from the space → refrigerant evaporates → provides cooling effect.
🔹 Processes (Ideal VCC)
- 1–2: Isentropic Compression (Compressor)
- Low-pressure vapor → compressed to high-pressure, high-temperature vapor.
- 2–3: Isobaric Heat Rejection (Condenser)
- Heat is rejected at constant pressure → vapor turns into liquid.
- 3–4: Throttling Expansion (Expansion Valve)
- Sudden pressure drop, enthalpy remains constant.
- 4–1: Isobaric Heat Absorption (Evaporator)
- Refrigerant absorbs heat → produces cooling.
🔹 Diagrams
- T–s diagram: Rectangle shape (idealized).
- P–h diagram: Commonly used, shows enthalpy changes clearly.
🔹 Advantages of VCC
- High COP compared to air cycle.
- Compact and suitable for wide applications.
- Easy to control and maintain.
🔹 Applications
- Domestic refrigerators.
- Window and split ACs.
- Industrial refrigeration systems.
- Heat pumps.
⚙️ Formulas
- Refrigerating Effect (RE):
Work of Compression:
COP (Coefficient of Performance):
Heat Rejected in Condenser:
🔟 10 MCQs
Q1. The Vapour Compression Cycle mainly works with:
a) Air
b) Refrigerant
c) Steam
d) Water
Q2. In the VCC, the throttling process is assumed to be:
a) Isothermal
b) Adiabatic
c) Constant enthalpy
d) Constant entropy
Q3. The device where low-pressure refrigerant absorbs heat is:
a) Compressor
b) Condenser
c) Evaporator
d) Expansion valve
Q4. In an ideal VCC, the compression process is:
a) Isothermal
b) Isobaric
c) Isentropic
d) Isochoric
Q5. If h1=200 kJ/kgh_1 = 200 \, kJ/kgh1=200kJ/kg, h2=250 kJ/kgh_2 = 250 \, kJ/kgh2=250kJ/kg, h3=100 kJ/kgh_3 = 100 \, kJ/kgh3=100kJ/kg, h4=100 kJ/kgh_4 = 100 \, kJ/kgh4=100kJ/kg, find COP.
Q6. Which diagram is commonly used to analyze VCC performance?
a) T–s
b) P–h
c) P–v
d) h–s
Q7. Which of the following increases COP of a refrigeration system?
a) Increasing compressor work
b) Decreasing refrigerating effect
c) Superheating of vapor
d) Decreasing compressor work
Q8. The unit of refrigerating effect is:
a) kJ/kg
b) kW
c) °C
d) bar
Q9. Which process in VCC is irreversible in practice?
a) Compression
b) Condensation
c) Throttling
d) Evaporation
Q10. For the same refrigerant and operating conditions, VCC has COP ______ air refrigeration cycle.
a) Lower than
b) Higher than
c) Equal to
d) Independent of
✅ Answer Key
| Q | Ans |
|---|---|
| 1 | b |
| 2 | c |
| 3 | c |
| 4 | c |
| 5 | 2.0 |
| 6 | b |
| 7 | d |
| 8 | a |
| 9 | c |
| 10 | b |
🧠 Explanations
- Q1: Refrigerant is the working fluid in VCC → (b).
- Q2: Throttling is an isenthalpic process → (c).
- Q3: Evaporator absorbs heat → (c).
- Q4: Compression is ideally isentropic → (c).
- Q5: COP = (h1–h4)/(h2–h1) = (200–100)/(250–200) = 100/50 = 2 → (2.0).
- Q6: P–h diagram is widely used for refrigeration cycles → (b).
- Q7: Lower work input increases COP → (d).
- Q8: Refrigerating effect = enthalpy difference, unit kJ/kg → (a).
- Q9: Throttling is irreversible in practice → (c).
- Q10: VCC has higher COP than air cycle → (b).
🎯 Motivation / Why Practice Matters
ECET frequently tests VCC numericals (COP, enthalpy differences) and conceptual diagrams.
If you practice regularly:
- You can solve COP problems in <1 minute.
- You’ll recognize tricky conceptual traps (isenthalpic throttling, isentropic compression).
- This gives you a competitive edge in time management and boosts overall rank.
📲 CTA
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