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Qwestrum Engineering360 · Mechanical Engineering · Refrigeration & HVAC

Vapour Compression Refrigeration

The VCR cycle's COP = refrigerating effect/compressor work = (h₁ − h₄)/(h₂ − h₁); the heat-pump COP is one greater. One tonne of refrigeration equals 3.517 kW, per P.K. Nag / RK Rajput.
Exam tip: lock the sign convention (Q into system, W by system in P.K. Nag) before substituting; use absolute temperature for ideal-gas and efficiency ratios.

Key formulas & points

Skim these first — then read the full notes below.

  • Refrigeration effect at evaporator; heat rejection at condenser
  • Subcooling increases COP; superheating prevents liquid at compressor
  • Ton of refrigeration = 3.517 kW

Topic details

Introduction

The vapour-compression refrigeration cycle is the basis of all domestic and commercial cooling and a guaranteed exam topic. It comprises evaporator, compressor, condenser, and expansion valve, traced on the pressure-enthalpy (P–h) chart.

Scope in B.Tech and GATE syllabus

Enthalpies at the four states come from refrigerant tables or the P–h chart; the throttling process is isenthalpic (h₃ = h₄), the compression ideally isentropic. Refrigerating effect, work, and COP follow directly from enthalpy differences.

Why this topic matters in practice

Subcooling the liquid raises refrigerating effect and COP; superheating the suction vapour protects the compressor from liquid slugging. Converting cooling load to tonnes of refrigeration and reading the P–h chart accurately are the practical skills examiners test.

Key relations & formulas

COP=QLW=h1−h4(h2−h1)COP = \frac{Q_{L}}{W} = h_{1} - \frac{h_{4}}{(h_{2} - h_{1})}
(refrigerator, RK Rajput)
COPHP=QHW=COPref+1COP_{HP} = \frac{Q_{H}}{W} = COP_{ref} + 1
(heat pump)
QL=m˙ref(h1−h4)Q_{L} = ṁ_ref(h_{1} - h_{4})
(evaporator capacity)
Wcomp=m˙ref(h2−h1)W_{comp} = ṁ_ref(h_{2} - h_{1})
(compressor work)

Notation and sign conventions

Relation 1 —
COP=QLW=h1−h4/COP = \frac{Q_{L}}{W} = h_{1} - h_{4}/
COP=QLW=h1−h4(h2−h1)COP = \frac{Q_{L}}{W} = h_{1} - \frac{h_{4}}{(h_{2} - h_{1})}
(refrigerator, RK Rajput)
Write this relation with symbols exactly as in Refrigeration & Air Conditioning — RK Rajput before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
COPHP=QHW=COPref+1COP_{HP} = \frac{Q_{H}}{W} = COP_{ref} + 1
COPHP=QHW=COPref+1COP_{HP} = \frac{Q_{H}}{W} = COP_{ref} + 1
(heat pump)
Write this relation with symbols exactly as in Refrigeration & Air Conditioning — RK Rajput before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
QL=m˙refQ_{L} = ṁ_ref
QL=m˙ref(h1−h4)Q_{L} = ṁ_ref(h_{1} - h_{4})
(evaporator capacity)
Write this relation with symbols exactly as in Refrigeration & Air Conditioning — RK Rajput before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 4 —
Wcomp=m˙refW_{comp} = ṁ_ref
Wcomp=m˙ref(h2−h1)W_{comp} = ṁ_ref(h_{2} - h_{1})
(compressor work)
Write this relation with symbols exactly as in Refrigeration & Air Conditioning — RK Rajput before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Fundamentals and definitions

In the ideal VCR cycle: the evaporator absorbs heat (1: saturated/superheated vapour out), the compressor raises pressure isentropically (1→2), the condenser rejects heat (2→3: liquid), and the expansion valve throttles isenthalpically (3→4). Refrigerating effect = h₁ − h₄.

Governing relations in practice

COP is the benefit-to-cost ratio: for a refrigerator COP_R = (h₁ − h₄)/(h₂ − h₁); for a heat pump COP_HP = (h₂ − h₃)/(h₂ − h₁) = COP_R + 1, because the condenser rejects both absorbed heat and compressor work.

Design and analysis considerations

Throttling is irreversible and isenthalpic, so no work is recovered — a source of cycle inefficiency compared to Carnot. Subcooling (state 3 below saturation) increases h₁ − h₄; superheating (state 1 above saturation) increases work slightly but prevents liquid entering the compressor.

Advanced theory and extensions

Capacity is expressed in tonnes of refrigeration (1 TR = 3.517 kW = 211 kJ/min); mass flow of refrigerant follows from ṁ = Q_evap/(h₁ − h₄). These relations, read off the P–h chart, solve every VCR numerical.

Assumptions and validity limits

State assumptions explicitly before using any relation for vapour compression refrigeration — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Refrigeration & HVAC viva and GATE descriptive questions, listing valid assumptions often earns separate marks.

Step-by-step problem approach

1. Read the question and list given data with SI units (common in Refrigeration & HVAC papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to vapour compression refrigeration.
4. Use equation 1:
COP=QLW=h1−h4/COP = \frac{Q_{L}}{W} = h_{1} - h_{4}/
.
5. Use equation 2:
COPHP=QHW=COPref+1COP_{HP} = \frac{Q_{H}}{W} = COP_{ref} + 1
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.

Applications & exam relevance

Vapour Compression Refrigeration appears in buildings, cold storage, and comfort AC. In Indian mechanical curricula this topic is tested because it connects theory to cooling, heating, and air treatment.
GATE and semester exams often combine vapour compression refrigeration with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use vapour compression refrigeration?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Treating throttling as isentropic instead of isenthalpic (h₃ = h₄)
• Using refrigerator COP where heat-pump COP (one greater) is required
• Forgetting the tonne conversion 1 TR = 3.517 kW
• Reading condenser/evaporator enthalpies at the wrong pressure on the P–h chart

Quick revision checklist

Before attempting vapour compression refrigeration problems, confirm you can:
1. Refrigeration effect at evaporator; heat rejection at condenser
2. Subcooling increases COP; superheating prevents liquid at compressor
3. Ton of refrigeration = 3.517 kW
Revise the solved examples in Refrigeration & Air Conditioning — RK Rajput and one previous-year GATE or university paper for this unit.

Worked examples

Try the problem first — open the solution when you are ready to check.

Cooling load to tonnes of refrigeration

Problem

A cold store has a cooling load of 17.6 kW. Express this in tonnes of refrigeration.

Solution

TR = Q/3.517 = 17.6/3.517 = 5.00 TR.

Conceptual check — Vapour Compression Refrigeration

Problem

In a Refrigeration & HVAC semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of vapour compression refrigeration." What should a complete answer include?

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

  1. 1
    What is Vapour Compression Refrigeration, and why does it appear in B.Tech / GATE syllabi?

    Model answer

    The VCR cycle's COP = refrigerating effect/compressor work = (h₁ − h₄)/(h₂ − h₁); the heat-pump COP is one greater. One tonne of refrigeration equals 3.517 kW, per P.K. Nag / RK Rajput.
  2. 2
    State the relation COP = Q_L/W = h₁ − h₄/ and name each symbol.

    Model answer

    The governing relation is COP=QLW=h1−h4/COP = \frac{Q_{L}}{W} = h_{1} - h_{4}/. Write every symbol with SI units before substituting numbers.
  3. 3
    State the relation COP_HP = Q_H/W = COP_ref + 1 and name each symbol.

    Model answer

    The governing relation is COPHP=QHW=COPref+1COP_{HP} = \frac{Q_{H}}{W} = COP_{ref} + 1. Write every symbol with SI units before substituting numbers.
  4. 4
    State the relation Q_L = ṁ_ref and name each symbol.

    Model answer

    The governing relation is QL=m˙refQ_{L} = ṁ_ref. Write every symbol with SI units before substituting numbers.
  5. 5
    State the relation W_comp = ṁ_ref and name each symbol.

    Model answer

    The governing relation is Wcomp=m˙refW_{comp} = ṁ_ref. Write every symbol with SI units before substituting numbers.
  6. 6
    Explain: Refrigeration effect at evaporator; heat rejection at condenser

    Model answer

    Refrigeration effect at evaporator; heat rejection at condenser — state the assumption range and one exam trap linked to this point.
  7. 7
    Explain: Subcooling increases COP; superheating prevents liquid at compressor

    Model answer

    Subcooling increases COP; superheating prevents liquid at compressor — state the assumption range and one exam trap linked to this point.
  8. 8
    Explain: Ton of refrigeration = 3.517 kW

    Model answer

    Ton of refrigeration = 3.517 kW — state the assumption range and one exam trap linked to this point.
  9. 9
    How would you correct this error in a viva: Treating throttling as isentropic instead of isenthalpic (h₃ = h₄)?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  10. 10
    How would you correct this error in a viva: Using refrigerator COP where heat-pump COP (one greater) is required?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  11. 11
    How would you correct this error in a viva: Forgetting the tonne conversion 1 TR = 3.517 kW?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  12. 12
    How would you correct this error in a viva: Reading condenser/evaporator enthalpies at the wrong pressure on the P–h chart?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.

Exams & GATE

  • 1
    P-h diagram: trace 4 processes — isentropic, isobaric, throttling.
  • 2
    Avoid: Treating throttling as isentropic instead of isenthalpic (h₃ = h₄)
  • 3
    Avoid: Using refrigerator COP where heat-pump COP (one greater) is required
  • 4
    Avoid: Forgetting the tonne conversion 1 TR = 3.517 kW

📖 Standard books (India)

  • Refrigeration & Air Conditioning — RK Rajput

    Read: Syllabus unit

    VCRS, psychrometry, and cooling load

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