Vapor Compression Refrigeration Cycle Analysis

The vapor compression cycle forms the foundation of modern refrigeration and air conditioning. Understanding cycle thermodynamics enables equipment selection, performance prediction, and efficiency optimization.

Ideal Vapor Compression Cycle

graph LR
    A[1: Compressor Inlet<br/>Saturated Vapor] -->|Isentropic Compression| B[2: Compressor Discharge<br/>Superheated Vapor]
    B -->|Constant Pressure<br/>Desuperheating + Condensing| C[3: Condenser Outlet<br/>Saturated Liquid]
    C -->|Isenthalpic Expansion| D[4: Evaporator Inlet<br/>Two-Phase]
    D -->|Constant Pressure<br/>Evaporation| A

Process descriptions:

1→2: Isentropic compression (ideal)
2→3: Constant pressure heat rejection
3→4: Isenthalpic throttling
4→1: Constant pressure heat absorption

Pressure-Enthalpy (P-h) Diagram

The P-h diagram visualizes refrigeration cycles:

Key regions:

Subcooled liquid: Left of saturated liquid line
Two-phase mixture: Between sat. liquid and sat. vapor lines
Superheated vapor: Right of saturated vapor line

State points:

State 1: Compressor suction (saturated or superheated vapor)
State 2: Compressor discharge (superheated vapor)
State 3: Condenser outlet (subcooled liquid)
State 4: Evaporator inlet (two-phase mixture)

Thermodynamic Analysis

Refrigeration effect (cooling capacity):

$$Q_e = \dot{m}_r (h_1 - h_4)$$

Where:

$Q_e$ = evaporator capacity (Btu/hr or tons)
$\dot{m}_r$ = refrigerant mass flow rate (lb/hr)
$h_1, h_4$ = enthalpies at states 1 and 4 (Btu/lb)

Compressor work (ideal):

$$W_c = \dot{m}_r (h_2 - h_1)$$

Heat rejection:

$$Q_c = \dot{m}_r (h_2 - h_3)$$

Energy balance:

$$Q_c = Q_e + W_c$$

Coefficient of Performance (COP)

Refrigeration COP:

$$COP_R = \frac{Q_e}{W_c} = \frac{h_1 - h_4}{h_2 - h_1}$$

Heat pump COP:

$$COP_{HP} = \frac{Q_c}{W_c} = \frac{h_2 - h_3}{h_2 - h_1}$$

Relationship:

$$COP_{HP} = COP_R + 1$$

Carnot COP (theoretical maximum):

$$COP_{Carnot} = \frac{T_e}{T_c - T_e}$$

Where temperatures in absolute units (°R or K)

Worked Example 1: Basic Refrigeration Cycle Analysis

Given:

Refrigerant: R-134a
Evaporator temperature: 40°F
Condenser temperature: 100°F
Compressor isentropic efficiency: 75%
Refrigeration capacity: 10 tons

Find: COP, compressor power, mass flow rate

Solution:

Step 1: Determine state properties

State 1 (saturated vapor at 40°F):

$P_1 = 51.7$ psia
$h_1 = 107.5$ Btu/lb
$s_1 = 0.2228$ Btu/(lb·°R)

State 2s (isentropic compression to 100°F, $s_2s = s_1$):

$P_2 = 138.9$ psia
$h_{2s} = 116.5$ Btu/lb

Actual compressor discharge (with 75% efficiency): $$h_2 = h_1 + \frac{h_{2s} - h_1}{\eta_c} = 107.5 + \frac{116.5 - 107.5}{0.75} = 119.5 \text{ Btu/lb}$$

State 3 (saturated liquid at 100°F):

$h_3 = 38.8$ Btu/lb

State 4 (isenthalpic expansion):

$h_4 = h_3 = 38.8$ Btu/lb

Step 2: Calculate COP

$$COP = \frac{h_1 - h_4}{h_2 - h_1} = \frac{107.5 - 38.8}{119.5 - 107.5} = \frac{68.7}{12.0} = 5.73$$

Step 3: Calculate compressor power

$$Q_e = 10 \times 12,000 = 120,000 \text{ Btu/hr}$$

$$W_c = \frac{Q_e}{COP} = \frac{120,000}{5.73} = 20,941 \text{ Btu/hr} = 6.14 \text{ hp}$$

Step 4: Calculate mass flow rate

$$\dot{m}_r = \frac{Q_e}{h_1 - h_4} = \frac{120,000}{68.7} = 1,747 \text{ lb/hr}$$

Answers:

COP: 5.73
Compressor power: 6.14 hp (4.58 kW)
Mass flow rate: 1,747 lb/hr
Carnot COP (for comparison): $\frac{500}{560 - 500} = 8.33$
Carnot efficiency: $5.73 / 8.33 = 68.8%$

Superheat and Subcooling

Superheat: Temperature above saturation at compressor inlet

$$\Delta T_{superheat} = T_1 - T_{sat,evap}$$

Typical values: 5-15°F

Benefits:

Ensures no liquid enters compressor
Increases refrigeration effect slightly
Prevents compressor damage

Penalties:

Increases compressor discharge temperature
Slightly increases compression work

Subcooling: Temperature below saturation at condenser outlet

$$\Delta T_{subcool} = T_{sat,cond} - T_3$$

Typical values: 5-20°F

Benefits:

Increases refrigeration effect
Prevents flash gas in liquid line
Improves COP

No penalty

Worked Example 2: Effect of Subcooling on COP

Given: Same cycle as Example 1, but with 10°F subcooling

Solution:

State 3 (subcooled liquid at 100°F - 10°F = 90°F):

$h_3 = 38.8 - (c_p \times \Delta T) \approx 38.8 - (0.32 \times 10) = 35.6$ Btu/lb

Refrigeration effect: $$Q_e = h_1 - h_4 = 107.5 - 35.6 = 71.9 \text{ Btu/lb}$$

Increase: $(71.9 - 68.7) / 68.7 = 4.7%$

New COP: $$COP = \frac{71.9}{12.0} = 5.99$$

Improvement: $(5.99 - 5.73) / 5.73 = 4.5%$

Answer: 10°F subcooling improves COP by ~4.5%

Real Cycle Deviations from Ideal

Compressor inefficiency: Isentropic efficiency 70-85%
Pressure drops: Evaporator and condenser internal friction
Heat transfer: Superheat and subcooling
Non-ideal expansion: Real expansion devices have losses

Multi-Stage Compression

When required:

Large pressure ratios (> 8:1)
Low evaporator temperatures (< 0°F)
High condenser temperatures (> 120°F)

Benefits:

Reduced discharge temperature
Improved volumetric efficiency
Better COP

Configuration: Two-stage with intercooling

$$P_{intermediate} = \sqrt{P_e \times P_c}$$

Performance Optimization Strategies

Maximize subcooling: Use liquid subcoolers or suction-liquid heat exchangers
Minimize superheat: 5-10°F optimal for most systems
Reduce pressure drops: Proper pipe sizing, minimize bends
Floating head pressure: Lower condenser pressure when possible
Variable speed compressors: Match capacity to load

Refrigeration Tons and Capacity

1 ton of refrigeration = 12,000 Btu/hr

Origin: Heat removal to freeze 1 ton of water in 24 hours

Capacity calculation:

$$Tons = \frac{Q_e}{12,000}$$

Related Technical Guides:

References:

ASHRAE Handbook of Refrigeration, Chapter 1: Thermodynamics and Refrigeration Cycles
Stoecker, W.F. Industrial Refrigeration Handbook
AHRI Standard 540: Performance Rating of Positive Displacement Refrigerant Compressors