Level 1, Unit 3, Section 5 – 5.2 Plotting the Vapour Compression Cycle

5.2.1 — The Four State Points

A standard single-stage vapour compression cycle is defined by four state points, labelled 1 through 4. Each point sits in a specific location on the P–h diagram and corresponds to a specific location in the physical system.

Point	Location in System	Region on P–h Diagram	Refrigerant Condition
1	Compressor inlet (suction)	Superheated vapour — right of dome, low pressure	Low-pressure superheated vapour. Temperature slightly above T_sat (evaporator SH)
2	Compressor outlet (discharge)	Superheated vapour — right of dome, high pressure	High-pressure, high-temperature superheated vapour. Hottest point in the system.
3	Condenser outlet / liquid line	Subcooled liquid — left of dome, high pressure	High-pressure subcooled liquid. Temperature below T_sat at condenser pressure.
4	Expansion device outlet / evaporator inlet	Two-phase mixture — inside dome, low pressure	Low-pressure liquid–vapour mixture. Temperature = T_sat at low-side pressure.

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Point 4 has the same enthalpy as Point 3

The expansion device (TXV, EEV, or capillary tube) throttles the refrigerant without doing any work and without exchanging heat with the surroundings. This means the process is isenthalpic — enthalpy does not change. On the P–h diagram, the 3→4 line is purely vertical: same h, lower P. The enthalpy doesn’t change, but the refrigerant crosses the saturation boundary and becomes a two-phase mixture.

5.2.2 — The Four Processes

The four processes connecting the state points each have a characteristic shape on the P–h diagram. Recognising these shapes is how a technician quickly reads a cycle plotted from real field data.

Low → High P

1 → 2 Compression

Near-vertical line rising steeply in the superheated region. Ideal: perfectly vertical (isentropic). Real: tilts right (irreversibilities add heat, increasing discharge enthalpy).

→

High P

2 → 3 Condensation

Horizontal line moving left at high pressure. Crosses from superheated region through the dome (condensation) into the subcooled region. Total length = heat rejected per kg.

→

High → Low P

3 → 4 Expansion

Perfectly vertical line dropping from high to low pressure. No horizontal movement — enthalpy is constant. The refrigerant enters the dome at Point 4 as a two-phase mixture.

→

Low P

4 → 1 Evaporation

Horizontal line moving right at low pressure. Starts inside the dome (boiling), crosses the saturated vapour line, and ends in the superheated region. Total length = refrigeration effect per kg.

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The Two Horizontal Lines — Why They Are the Key to the Diagram

The two horizontal lines (evaporation at low pressure, condensation at high pressure) are where all the heat transfer occurs. Their horizontal lengths are proportional to the heat exchanged per kilogram of refrigerant — no calculation needed, just measurement on the diagram.

Long evaporation line = large refrigeration effect per kg
Long condensation line = large heat rejection per kg
A short, steep compression line = efficient compressor (low work input)
A cycle where the condensation line is much longer than the evaporation line has a high heat of compression — wasteful

5.2.3 — Energy Calculations from the Diagram

Because the horizontal axis is specific enthalpy, any horizontal distance directly represents an energy quantity per unit mass. Three key energy values define the cycle:

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Refrigeration Effect q_evap

Heat absorbed per kg in the evaporator. The horizontal distance from Point 4 to Point 1.

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Heat of Compression w_comp

Compressor work added per kg. The horizontal distance from Point 1 to Point 2.

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Heat Rejected q_cond

Heat released per kg in the condenser. The horizontal distance from Point 2 to Point 3. Always equals q_evap + w_comp.

q_evap = h₁ − h₄ (kJ/kg or BTU/lb)

w_comp = h₂ − h₁ (kJ/kg or BTU/lb)

q_cond = h₂ − h₃ (kJ/kg or BTU/lb)

Energy balance check: q_cond = q_evap + w_comp

To convert from specific values (per kg) to total system capacity (kW or BTU/hr), multiply by the refrigerant mass flow rate ˙m:

Q_evap = ˙m × (h₁ − h₄) [kW or BTU/hr]

W_comp = ˙m × (h₂ − h₁) [kW]

Q_cond = ˙m × (h₂ − h₃) [kW or BTU/hr]

5.2.4 — Coefficient of Performance (COP)

COP is the ratio of useful output to work input. On the P–h diagram it is simply the ratio of two horizontal distances — the evaporation line length divided by the compression line length.

COP_cooling = q_evap ÷ w_comp = (h₁ − h₄) ÷ (h₂ − h₁)

COP_heating = q_cond ÷ w_comp = (h₂ − h₃) ÷ (h₂ − h₁)

Note: COP_heating = COP_cooling + 1 (heat pump identity)

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High COP (efficient cycle)

Long evaporation line, short compression line. Achieved by: low condensing temperature, high evaporating temperature, adequate subcooling, and low compressor discharge superheat.

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Low COP (inefficient cycle)

Short evaporation line, long compression line. Caused by: high condensing temperature (dirty condenser, hot ambient), low evaporating temperature (restricted airflow, frosted coil), or low isentropic efficiency (worn compressor).

5.2.5 — Flash Gas Quality at the Expansion Device Outlet

When subcooled liquid crosses the saturation boundary at Point 4, a portion of the liquid immediately vaporises to satisfy the new, lower saturation condition. This vapour is flash gas. It does not absorb useful heat in the evaporator because it already consumed part of the available enthalpy during flashing.

x₄ = (h₄ − h_f) ÷ (h_g − h_f)

h_f and h_g are the saturated liquid and vapour enthalpies at the low-side (evaporating) pressure. x = 0 means pure liquid; x = 1 means pure vapour.

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Worked Example — Flash Gas Quality (R-410A)

At the condenser outlet, liquid R-410A is subcooled to 115°F (46.1°C) at high-side pressure (condensing at 130°F). The enthalpy at this state: h₃ = h₄ ≈ 50.5 BTU/lb (117.4 kJ/kg).

After expansion to the low side (40°F saturation, 118.2 psig):

h_f at 40°F = 42.3 BTU/lb (98.3 kJ/kg)
h_g at 40°F = 122.5 BTU/lb (285.0 kJ/kg)
h₄ = 50.5 BTU/lb (same as h₃ — isenthalpic)

x₄ = (50.5 − 42.3) ÷ (122.5 − 42.3) = 8.2 ÷ 80.2 ≈ 0.10

Approximately 10% of the mass is flash vapour at the evaporator inlet. The remaining 90% (still liquid) absorbs useful latent heat in the evaporator. This highlights why subcooling improves capacity — even a few extra degrees of subcooling reduce x₄, sending more liquid into the evaporator.

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Subcooling reduces flash gas — every degree counts

If the condenser outlet were 5°F warmer (120°F instead of 115°F), h₃ would increase to approximately 52.4 BTU/lb, pushing x₄ to about 0.13 (13%). That 3% more flash gas means 3% less refrigerant doing useful cooling in the evaporator — a direct capacity loss. Maintaining adequate subcooling through proper charge level and condenser maintenance directly protects system capacity.

5.2.6 — Complete Worked Example — R-410A Residential A/C

The following example uses real R-410A thermodynamic property data to calculate all cycle performance values from a set of field measurements.

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Given Conditions

Refrigerant: R-410A
Low-side (suction) pressure: 118.2 psig → T_sat = 40°F (4.4°C)
Evaporator superheat: 10°F → suction line temp = 50°F (10°C)
High-side (discharge/liquid) pressure: 411 psig → T_sat = 130°F (54.4°C)
Condenser subcooling: 15°F → liquid line temp = 115°F (46.1°C)
Compressor isentropic efficiency: 73% (typical scroll)

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Step 1 — Locate and Record the State Point Enthalpies

Point 1 — Compressor Inlet (40°F sat + 10°F SH):

h₁ = 124.5 BTU/lb (289.7 kJ/kg)

Point 3 — Condenser Outlet (130°F sat − 15°F SC = 115°F liquid):

h₃ = 50.5 BTU/lb (117.4 kJ/kg)

Point 4 — Expansion Outlet (isenthalpic, same as h₃):

h₄ = h₃ = 50.5 BTU/lb (117.4 kJ/kg)

Point 2s — Ideal Isentropic Discharge (same entropy as Point 1, at high pressure):

h_2s ≈ 143.0 BTU/lb (332.6 kJ/kg) (from R-410A P–h chart along isentrope from Point 1 to 411 psig)

Point 2 — Actual Compressor Discharge (adjusted for isentropic efficiency):

h₂ = h₁ + (h_2s − h₁) ÷ η_s

h₂ = 124.5 + (143.0 − 124.5) ÷ 0.73 = 124.5 + 25.3 = 149.8 BTU/lb (348.4 kJ/kg)

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Step 2 — Calculate Cycle Performance Values

Refrigeration Effect:

q_evap = h₁ − h₄ = 124.5 − 50.5 = 74.0 BTU/lb (172.2 kJ/kg)

Heat of Compression:

w_comp = h₂ − h₁ = 149.8 − 124.5 = 25.3 BTU/lb (58.9 kJ/kg)

Heat Rejected at Condenser:

q_cond = h₂ − h₃ = 149.8 − 50.5 = 99.3 BTU/lb (231.0 kJ/kg)

Energy Balance Check:

q_evap + w_comp = 74.0 + 25.3 = 99.3 BTU/lb ✓ (matches q_cond)

COP (Cooling):

COP = 74.0 ÷ 25.3 = 2.92

Flash Gas Quality at Point 4:

x₄ = (50.5 − 42.3) ÷ (122.5 − 42.3) = 8.2 ÷ 80.2 = 0.10 (10% vapour by mass)

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Step 3 — Interpreting the Results

COP of 2.92 means the system delivers 2.92 kW of cooling for every 1 kW of electrical input — reasonable for a standard residential scroll-compressor A/C under moderate conditions.
10% flash gas means 90% of the refrigerant mass entering the evaporator is still liquid and available to absorb latent heat — a healthy result for a properly subcooled system. Improving subcooling from 15°F to 20°F would reduce flash gas to about 8%, recovering approximately 2% capacity.
Heat of compression (25.3 BTU/lb) is about 25% of heat rejected (99.3 BTU/lb) — meaning about 75% of the heat the condenser must reject originally came from the building. This is normal for a well-operating system.
If the discharge enthalpy were lower (better isentropic efficiency), COP would increase and discharge temperature would decrease, reducing thermal stress on the compressor.

5.2.7 — Real vs. Ideal Compression on the P–h Diagram

Ideal compression follows a constant-entropy line (isentrope) vertically upward on the P–h diagram. Real compression in any actual compressor deviates from this ideal due to internal friction, heat transfer through the cylinder walls, and valve losses.

Ideal Isentropic Compression (1 → 2s)

Follows a constant-entropy line — vertical on the log-P axis
No heat exchange with surroundings
Minimum possible compressor work for the pressure ratio
Actual compressors cannot achieve this
Used as the benchmark for isentropic efficiency

Real (Polytropic) Compression (1 → 2)

Curves slightly to the right of the isentrope
Higher discharge enthalpy than ideal
Higher discharge temperature
More compressor heat rejected into the refrigerant
Isentropic efficiency η_s = 0.68–0.78 for scroll compressors

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High discharge temperature = compressor wear and oil breakdown

Discharge gas temperature above 250°F (121°C) causes compressor oil to carbonise, valve reeds to warp, and motor insulation to degrade. High discharge temperature is almost always caused by one of: high compression ratio (dirty condenser, overcharge, non-condensables), excessive suction superheat (uninsulated suction line, flooded evaporator running dry), or low isentropic efficiency (worn compressor). The P–h diagram makes it immediately clear why each of these causes shifts Point 2 further right and upward.