7.1 Standard Vapor-Compression Refrigeration Cycle on P-h & T-s Diagrams

7.1 Standard Vapor-Compression Refrigeration Cycle on P-h & T-s Diagrams

The vapor-compression refrigeration cycle is the foundational thermodynamic mechanism for modern air conditioning, commercial refrigeration, thermal storage, and heat pump systems. By circulating a phase-changing working fluid (refrigerant) between two distinct pressure levels, thermal energy is extracted from a low-temperature conditioned space (evaporator) and rejected to a higher-temperature ambient sink (condenser). On the PE Mechanical: HVAC and Refrigeration examination, mastery of state-point identification, enthalpy tracking across Pressure-Enthalpy ($P$-$h$) and Temperature-Entropy ($T$-$s$) diagrams, isentropic compressor efficiencies, and volumetric displacement rates is essential for solving core thermal-fluid problems.

---

1. Thermodynamic Architecture of the Four Core Components

A standard single-stage vapor-compression system consists of four primary components connected in a closed continuous loop:

+-----------------------------------------------------------------------------------------+
| STANDARD SINGLE-STAGE VAPOR-COMPRESSION REFRIGERATION LOOP                              |
+-----------------------------------------------------------------------------------------+
|                                 CONDENSER (High Pressure: P_cond)                       |
|                         +-----------------------------------------------+               |
|                         | Desuperheating -> Condensing -> Subcooling     |               |
|                         +-----------------------------------------------+               |
|                         ^ State 2 (Superheated Vapor)   | State 3 (Saturated Liquid)    |
|                         |                               v                               |
|                 +---------------+               +---------------+                       |
|                 |  COMPRESSOR   |               | EXPANSION DEV |                       |
|                 |  (Work In: W) |               | (Throttling)  |                       |
|                 +---------------+               +---------------+                       |
|                         ^                               |                               |
|                         | State 1 (Saturated Vapor)     v State 4 (Two-Phase Liquid/Vap)|
|                         +-----------------------------------------------+               |
|                         | Evaporation -> Latent Heat Absorption (Q_evap)|               |
|                         +-----------------------------------------------+               |
|                                 EVAPORATOR (Low Pressure: P_evap)                       |
+-----------------------------------------------------------------------------------------+

Detailed State-Point Sequence

1. State 1 $\rightarrow$ State 2 (Compression Process):

   • Ideal (Isentropic): Reversible, adiabatic compression of saturated (or slightly superheated) vapor from low evaporating pressure ($P_{\text{evap}}$) to high condensing pressure ($P_{\text{cond}}$) along a line of constant entropy ($s_1 = s_{2s}$).
   • Actual (Non-Isentropic): Internal friction, valve pressure drops, and irreversible turbulence increase refrigerant entropy ($s_2 > s_1$), resulting in a higher discharge enthalpy ($h_2 > h_{2s}$) and elevated discharge temperature ($T_2 > T_{2s}$).
   • Compressor Work Input: $w_{\text{in}} = h_2 - h_1 = \frac{h_{2s} - h_1}{\eta_s}$

2. State 2 $\rightarrow$ State 3 (Isobaric Heat Rejection Process):

   • High-pressure superheated vapor enters the condenser, desuperheats to saturation temperature ($T_{\text{cond}}$), undergoes isothermal latent heat condensation to saturated liquid, and leaves as saturated (or subcooled) liquid at state 3 ($h_3 = h_f$ at $P_{\text{cond}}$).
   • Condenser Heat Rejection Rate: $q_{\text{cond}} = h_2 - h_3$

3. State 3 $\rightarrow$ State 4 (Isenthalpic Throttling Process):

   • Saturated high-pressure liquid passes through an adiabatic expansion device (thermostatic expansion valve, electronic expansion valve, capillary tube, or fixed orifice). No shaft work is performed and heat transfer is negligible, rendering the process strictly isenthalpic ($h_4 = h_3$).
   • The sudden pressure drop causes a portion of the liquid to boil instantly into flash gas, yielding a low-temperature two-phase mixture at state 4 with vapor quality $x_4$: $x_4 = \frac{h_4 - h_{f,\text{evap}}}{h_{fg,\text{evap}}} = \frac{h_3 - h_{f,\text{evap}}}{h_{g,\text{evap}} - h_{f,\text{evap}}}$

4. State 4 $\rightarrow$ State 1 (Isobaric/Isothermal Heat Absorption Process):

   • The low-pressure two-phase mixture enters the evaporator tubes, absorbing latent heat from the refrigerated space or chilled water stream at constant evaporating temperature ($T_{\text{evap}}$) and constant pressure ($P_{\text{evap}}$) until fully vaporized to saturated vapor at state 1 ($h_1 = h_g$ at $P_{\text{evap}}$).
   • Refrigeration Effect: $q_{\text{evap}} = h_1 - h_4 = h_1 - h_3$

---

2. Thermodynamic Property Mapping: P-h & T-s Diagrams

Understanding the geometry of refrigerant saturation domes on $P$-$h$ and $T$-$s$ coordinates allows rapid verification of cycle performance and identifying irreversibilities.

+-----------------------------------------------------------------------------------------+
| PRESSURE-ENTHALPY (P-h) DIAGRAM          | TEMPERATURE-ENTROPY (T-s) DIAGRAM            |
+-----------------------------------------------------------------------------------------+
| Pressure (P)                             | Temperature (T)                              |
|    ^              Saturated Saturated    |    ^              Saturated Saturated        |
|    |                Liquid    Vapor      |    |                Liquid    Vapor          |
|    |       P_cond     3_________2  2s    |    |       P_cond     .-------2  2s          |
|    |                  |         / /      |    |                 / 3     / /             |
|    |                  |        / /       |    |                /  |    / /              |
|    |                  |       / /        |    |               /   |   / /               |
|    |       P_evap     4______1_/         |    |       P_evap /    4--1_/                |
|    |                 /        \          |    |             /         \                 |
|    +----------------------------->       |    +----------------------------->           |
|    0                 Enthalpy (h)        |    0                  Entropy (s)            |
+-----------------------------------------------------------------------------------------+

Key Curve Features on the P-h Diagram

• Constant Pressure Lines (Isobars): Horizontal straight lines spanning the entire chart.
• Constant Enthalpy Lines (Isenthalps): Vertical straight lines ($h = \text{constant}$). Expansion 3-4 drops vertically downward.
• Constant Temperature Lines (Isotherms): Vertical in the subcooled liquid region (left of dome), perfectly horizontal inside the two-phase dome ($P_{\text{sat}}$ dictates $T_{\text{sat}}$), and steep negative downward curves in the superheated vapor region (right of dome).
• Constant Entropy Lines (Isentropes): Slanted upward and to the right in the superheated vapor region. Ideal compression (1-2s) tracks directly along an isentropic contour line.
• Constant Specific Volume Lines ($v$): Slanted lines steeper than isentropes, crossing upward into the superheat region.

Key Curve Features on the T-s Diagram

• Constant Pressure Lines: Rise exponentially through the subcooled liquid region, run horizontal across the two-phase dome, and curve upward into the superheat vapor region.
• Isentropic Compression (1-2s): A vertical straight line ($s_1 = s_{2s}$).
• Throttling Irreversibility (3-4): Throttling is an irreversible process. Although $h_3 = h_4$, entropy increases ($\Delta s_{\text{throttle}} = s_4 - s_3 > 0$), shifting state 4 to the right on the $T$-$s$ diagram and creating a loss of available refrigeration capacity represented by the area under curve 3-4.

---

3. Governing Mass & Energy Balance Formulations

Quantitative analysis of the refrigeration cycle connects component enthalpies to cooling capacity, compressor horsepower, and heat rejection rates.

1. Refrigerant Mass Flow Rate ($\dot{m}_r$)

Given a required cooling capacity $\dot{Q}_{\text{evap}}$ (in $\text{Btu/hr}$ or $\text{kW}$):

$\dot{m}_r = \frac{\dot{Q}_{\text{evap}}}{q_{\text{evap}}} = \frac{\dot{Q}_{\text{evap}}}{h_1 - h_4} = \frac{\dot{Q}_{\text{evap}}}{h_1 - h_3}$

$\text{In IP Units: } \dot{m}_r \left[\frac{\text{lbm}}{\text{hr}}\right] = \frac{\text{Capacity [Tons]} \times 12,000\text{ Btu/(hr}\cdot\text{Ton)}}{h_1 - h_3\text{ [Btu/lbm]}}$

2. Compressor Power Input & Isentropic Efficiency ($\eta_s$)

The theoretical (isentropic) compressor power represents the minimum thermodynamic work required:

$\dot{W}_{s} = \dot{m}_r (h_{2s} - h_1)$

The isentropic compressor efficiency $\eta_s$ accounts for real internal losses:

$\eta_s = \frac{w_s}{w_{\text{actual}}} = \frac{h_{2s} - h_1}{h_2 - h_1}$

$h_2 = h_1 + \frac{h_{2s} - h_1}{\eta_s}$

Actual indicated shaft work rate:

$\dot{W}_{\text{comp}} = \dot{m}_r (h_2 - h_1) = \frac{\dot{m}_r (h_{2s} - h_1)}{\eta_s}$

$\text{Electrical Power [kW]} = \frac{\dot{W}_{\text{comp}} \text{ [Btu/hr]}}{3,412.14 \times \eta_{\text{motor}} \times \eta_{\text{mech}}}$

3. Condenser Heat Rejection Rate ($\dot{Q}_{\text{cond}}$) & Heat Rejection Ratio (HRR)

Applying the First Law of Thermodynamics to the entire refrigeration cycle:

$\dot{Q}_{\text{cond}} = \dot{m}_r (h_2 - h_3) = \dot{Q}_{\text{evap}} + \dot{W}_{\text{comp}}$

The Heat Rejection Ratio (HRR) is the ratio of total heat rejected at the condenser to total heat absorbed at the evaporator:

$\text{HRR} = \frac{\dot{Q}_{\text{cond}}}{\dot{Q}_{\text{evap}}} = \frac{\dot{Q}_{\text{evap}} + \dot{W}_{\text{comp}}}{\dot{Q}_{\text{evap}}} = 1 + \frac{1}{\text{COP}_R}$

---

4. Compressor Volumetric Flow Rate & Volumetric Efficiency

Refrigerant vapor entering the compressor occupies a specific volume $v_1$ (in $\text{ft}^3/\text{lbm}$ or $\text{m}^3/\text{kg}$) evaluated at suction temperature and pressure. The required actual volumetric flow rate ($\dot{V}_{\text{actual}}$) drawn into the compressor is:

$\dot{V}_{\text{actual}} = \dot{m}_r \cdot v_1$

Volumetric Efficiency ($\eta_v$)

Reciprocating and positive displacement compressors do not expel 100% of the vapor on each stroke due to clearance volume ($V_c$) between the piston crown and cylinder head. The clearance volumetric efficiency is mathematically expressed as:

$\eta_v = 1 - c \left[ \left(\frac{P_{\text{cond}}}{P_{\text{evap}}}\right)^{1/k} - 1 \right] = 1 - c \left( r_p^{1/k} - 1 \right)$

Where:

• $c = \text{Clearance factor } = V_{\text{clearance}} / V_{\text{swept}} \text{ (typically } 0.03\text{ to }0.07\text{)}$
• $r_p = \text{Pressure ratio } = P_{\text{cond}} / P_{\text{evap}} \text{ (evaluated in absolute units: psia or kPa abs)}$
• $k = c_p / c_v = \text{Isentropic exponent of the refrigerant vapor}$

The required compressor theoretical displacement (swept volume rate $\dot{V}_{\text{disp}}$) is:

$\dot{V}_{\text{disp}} = \frac{\dot{V}_{\text{actual}}}{\eta_v} = \frac{\dot{m}_r \cdot v_1}{\eta_v}$

Exam Warning on Pressure Units: Compression ratio $r_p = P_{\text{cond}} / P_{\text{evap}}$ MUST always be calculated using absolute pressures ($\text{psia} = \text{psig} + 14.696$). Using gauge pressures in compression ratios or volumetric efficiency formulas is a frequent failure point on the PE exam.

---

5. NCEES Reference Handbook Navigation Tactics

• Refrigerant Property Tables & P-h Diagrams: Navigate to Section 7: HVAC & Refrigeration Applications and locate tables and thermodynamic charts for R-134a, R-410A, and R-22.
• State 1 Lookup: Find the evaporating temperature row; read saturated vapor enthalpy $h_g$ and specific entropy $s_g$.
• State 2s Lookup: Move to the condensing pressure level; follow the constant entropy line ($s = s_1$) to determine isentropic discharge enthalpy $h_{2s}$.
• State 3 Lookup: Find the condensing temperature row; read saturated liquid enthalpy $h_f$.
• State 4 Equality: Set $h_4 = h_3$.

---

6. Worked Computational Examples

Example 1: Standard 25-Ton R-134a Chiller Cycle Analysis

A water-cooled liquid chiller utilizes R-134a to deliver $25\text{ Tons}$ of refrigeration capacity. The system operates on a standard vapor-compression cycle under the following design conditions:

• Evaporating temperature: $T_{\text{evap}} = 20^\circ\text{F}$ ($P_{\text{evap}} = 33.15\text{ psia}$)
• Condensing temperature: $T_{\text{cond}} = 110^\circ\text{F}$ ($P_{\text{cond}} = 161.1\text{ psia}$)
• Refrigerant enters compressor as dry saturated vapor and leaves condenser as saturated liquid.
• Compressor isentropic efficiency: $\eta_s = 0.82$
• Compressor clearance volumetric efficiency: $\eta_v = 0.88$

Thermodynamic Properties from R-134a Data:

• At $20^\circ\text{F}$: $h_1 = h_g = 105.30\text{ Btu/lbm}$, $s_1 = s_g = 0.2244\text{ Btu/(lbm}\cdot{}^\circ\text{R)}$, $v_1 = v_g = 1.397\text{ ft}^3/\text{lbm}$
• At $110^\circ\text{F}$: $h_3 = h_f = 47.90\text{ Btu/lbm}$
• At $P = 161.1\text{ psia}$ and $s = 0.2244\text{ Btu/(lbm}\cdot{}^\circ\text{R)}$: $h_{2s} = 119.80\text{ Btu/lbm}$

Calculate: (a) Refrigerant mass flow rate ($\text{lbm/hr}$), (b) Actual compressor discharge enthalpy and power input ($\text{kW}$), (c) Condenser heat rejection rate ($\text{Btu/hr}$), (d) Coefficient of Performance ($\text{COP}_R$), and (e) Required compressor displacement ($\text{CFM}$).

Solution:

1. Refrigerant Mass Flow Rate: $\dot{Q}_{\text{evap}} = 25\text{ Tons} \times 12,000\text{ Btu/(hr}\cdot\text{Ton)} = 300,000\text{ Btu/hr}$ $q_{\text{evap}} = h_1 - h_3 = 105.30 - 47.90 = 57.40\text{ Btu/lbm}$ $\dot{m}_r = \frac{300,000\text{ Btu/hr}}{57.40\text{ Btu/lbm}} = 5,226.48\text{ lbm/hr} \quad (87.11\text{ lbm/min})$

2. Actual Compressor Discharge Enthalpy & Power Input: $\Delta h_s = h_{2s} - h_1 = 119.80 - 105.30 = 14.50\text{ Btu/lbm}$ $\Delta h_{\text{actual}} = \frac{\Delta h_s}{\eta_s} = \frac{14.50}{0.82} = 17.68\text{ Btu/lbm}$ $h_2 = h_1 + \Delta h_{\text{actual}} = 105.30 + 17.68 = 122.98\text{ Btu/lbm}$ $\dot{W}_{\text{comp}} = \dot{m}_r \times \Delta h_{\text{actual}} = 5,226.48\text{ lbm/hr} \times 17.68\text{ Btu/lbm} = 92,404\text{ Btu/hr}$ $\text{Power [kW]} = \frac{92,404\text{ Btu/hr}}{3,412.14\text{ Btu/(kW}\cdot\text{hr)}} = 27.08\text{ kW}$

3. Condenser Heat Rejection Rate: $\dot{Q}_{\text{cond}} = \dot{m}_r (h_2 - h_3) = 5,226.48 \times (122.98 - 47.90) = 5,226.48 \times 75.08 = 392,404\text{ Btu/hr}$ $\text{Check: } \dot{Q}_{\text{cond}} = \dot{Q}_{\text{evap}} + \dot{W}_{\text{comp}} = 300,000 + 92,404 = 392,404\text{ Btu/hr} \quad \checkmark$ $\text{HRR} = \frac{392,404}{300,000} = 1.308$

4. Coefficient of Performance: $\text{COP}_R = \frac{\dot{Q}_{\text{evap}}}{\dot{W}_{\text{comp}}} = \frac{300,000}{92,404} = 3.247$

5. Compressor Volumetric Displacement Rate: $\dot{V}_{\text{actual}} = \dot{m}_r \cdot v_1 = \frac{5,226.48\text{ lbm/hr} \times 1.397\text{ ft}^3/\text{lbm}}{60\text{ min/hr}} = 121.69\text{ CFM}$ $\dot{V}_{\text{disp}} = \frac{\dot{V}_{\text{actual}}}{\eta_v} = \frac{121.69\text{ CFM}}{0.88} = 138.28\text{ CFM}$

---

Example 2: Compression Ratio & Volumetric Efficiency for R-410A Heat Pump

An R-410A commercial rooftop heat pump operates with a suction pressure of $30.3\text{ psig}$ ($P_{\text{evap}} = 45.0\text{ psia}$) and a condensing pressure of $345.3\text{ psig}$ ($P_{\text{cond}} = 360.0\text{ psia}$). The compressor has a clearance factor $c = 0.045$ and the refrigerant vapor has an isentropic exponent $k = 1.18$.

Calculate: (a) The compression ratio $r_p$, and (b) The clearance volumetric efficiency $\eta_v$.

Solution:

1. Compression Ratio: $r_p = \frac{P_{\text{cond}}}{P_{\text{evap}}} = \frac{360.0\text{ psia}}{45.0\text{ psia}} = 8.00$
2. Volumetric Efficiency: $\eta_v = 1 - c \left( r_p^{1/k} - 1 \right)$ $r_p^{1/k} = 8.00^{1 / 1.18} = 8.00^{0.84746} = 5.830$ $\eta_v = 1 - 0.045 \times (5.830 - 1) = 1 - (0.045 \times 4.830) = 1 - 0.2174 = 0.7826 \quad (78.26\%$

(Note: If gauge pressures were mistakenly used, $r_{p,\text{wrong}} = 345.3 / 30.3 = 11.40$, yielding an erroneous $\eta_v = 0.645$—a massive 18% calculation error.)