2.2 Thermodynamic Cycles: Carnot, Rankine, Brayton & Reversed Carnot Limit

2.2 Thermodynamic Cycles: Carnot, Rankine, Brayton & Reversed Carnot Limit

Thermodynamic cycles describe the continuous sequence of processes that convert thermal energy into mechanical power (heat engines) or utilize mechanical power to transfer heat from low to high temperature reservoirs (refrigerators and heat pumps). On the PE Mechanical HVAC exam, mastery of cycle benchmarks allows engineers to evaluate plant performance, verify equipment claims, and size combined heat and power (CHP) systems.

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1. The Carnot Cycle & Theoretical Thermal Limits

Proposed by Nicolas Léonard Sadi Carnot in 1824, the Carnot Cycle is a totally reversible ideal cycle operating between a high-temperature source at $T_H$ and a low-temperature sink at $T_L$. It consists of four internally and externally reversible processes:

1. 1 to 2: Reversible Isothermal Heat Addition at $T_H$ ($q_{in} = T_H (s_2 - s_1)$).
2. 2 to 3: Isentropic (Reversible Adiabatic) Expansion from $T_H$ to $T_L$ ($s_2 = s_3$).
3. 3 to 4: Reversible Isothermal Heat Rejection at $T_L$ ($q_{out} = T_L (s_3 - s_4) = T_L (s_2 - s_1)$).
4. 4 to 1: Isentropic (Reversible Adiabatic) Compression from $T_L$ to $T_H$ ($s_4 = s_1$).

Temperature (T)
      ^
   TH +--------1==============2-------- (Isothermal Heat Addition, q_in = TH*Delta_s)
      |        |              |
      |        | (Isentropic) | (Isentropic Expansion)
      |        |              |
   TL +--------4==============3-------- (Isothermal Heat Rejection, q_out = TL*Delta_s)
      |        |              |
      +--------+--------------+--------> Entropy (s)
               s1=s4          s2=s3

Carnot Heat Engine Efficiency

$\eta_{th} = \frac{W_{net,out}}{Q_{in}} = \frac{Q_H - Q_L}{Q_H} = 1 - \frac{Q_L}{Q_H}$

Because the cycle is totally reversible, $\frac{Q_L}{Q_H} = \frac{T_L}{T_H}$ (where $T_L$ and $T_H$ are in absolute temperature: $^\circ\text{R} = ^\circ\text{F} + 459.67$ or $\text{K} = ^\circ\text{C} + 273.15$):

$\eta_{th, Carnot} = 1 - \frac{T_L}{T_H}$

Carnot Principles:

1. The efficiency of an irreversible heat engine is always less than the efficiency of a reversible heat engine operating between the same two reservoirs ($\eta_{actual} < \eta_{Carnot}$).
2. The efficiencies of all reversible heat engines operating between the same two reservoirs are identical, regardless of the working fluid.

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2. The Reversed Carnot Cycle: Refrigeration & Heat Pump Limits

Running the Carnot cycle in reverse (counter-clockwise on $P-v$ and $T-s$ diagrams) produces the theoretical benchmark for all vapor-compression and absorption refrigeration systems.

Coefficients of Performance (COP)

$\text{Refrigeration Mode: } \text{COP}_{R, Carnot} = \frac{\text{Desired Output}}{\text{Required Work Input}} = \frac{Q_L}{W_{net,in}} = \frac{Q_L}{Q_H - Q_L} = \frac{T_L}{T_H - T_L}$

$\text{Heat Pump Mode: } \text{COP}_{HP, Carnot} = \frac{\text{Desired Output}}{\text{Required Work Input}} = \frac{Q_H}{W_{net,in}} = \frac{Q_H}{Q_H - Q_L} = \frac{T_H}{T_H - T_L}$

Notice the fundamental thermodynamic relationship between heat pump and refrigeration COPs:

$\text{COP}_{HP} = \text{COP}_R + 1$

Converting Between HVAC Performance Metrics

On the PE exam, equipment ratings are expressed in diverse imperial and metric units:

$\text{kW/ton} = \frac{12 \text{ kBtu/ton} \times 0.293071 \text{ kW/kBtu}}{\text{COP}_R \times 1.0} = \frac{3.51685}{\text{COP}_R}$

$\text{EER (Btu/W}\cdot\text{hr)} = 3.41214 \times \text{COP}_R = \frac{12}{\text{kW/ton}}$

Second Law (Exergy) Efficiency

The Second Law efficiency ($\eta_{II}$) measures how close an actual refrigeration machine approaches the theoretical Carnot ceiling under identical reservoir temperatures:

$\eta_{II} = \frac{\text{COP}_{actual}}{\text{COP}_{Carnot}} = \frac{\text{kW/ton}_{Carnot}}{\text{kW/ton}_{actual}}$

Modern high-efficiency water-cooled centrifugal chillers typically exhibit $\eta_{II} \approx 0.60 - 0.75$, while small direct-expansion (DX) rooftop split systems achieve $\eta_{II} \approx 0.35 - 0.50$.

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3. The Rankine Cycle (Steam Power & District Energy Cogeneration)

The Rankine cycle is the fundamental operating cycle for steam-based power plants and large campus district energy facilities producing steam for district heating, absorption chillers, and turbine drives.

   [Boiler / Steam Generator]  <==== (Qin)
       ^                   |
(4)    | Liquid            | Superheated Steam (1)
   [Feed Pump]             V
  (Win)|               [Steam Turbine] ====> (Wout)
       | Liquid            |
       |                   | Wet / Low-Pressure Vapor (2)
   [Condenser] <===========+
       =====> (Qout to Cooling Tower)
       State 3: Saturated Liquid

Ideal Rankine Cycle State Points & Energy Equations

1. 1 to 2: Isentropic Expansion in Turbine: Steam expands from boiler pressure ($P_{high}$) to condenser pressure ($P_{low}$). Shaft work output is: $w_{t} = h_1 - h_2 = \eta_t (h_1 - h_{2s})$
2. 2 to 3: Constant-Pressure Heat Rejection in Condenser: Steam condenses to saturated liquid at $P_{low}$ ($T_3 = T_{sat}(P_{low})$, $h_3 = h_{f@P_{low}}$): $q_{out} = h_2 - h_3$
3. 3 to 4: Isentropic Pumping of Liquid Feedwater: Saturated liquid is pumped from $P_{low}$ to $P_{high}$. Because liquid is virtually incompressible ($v \approx v_{f@P_{low}} = \text{constant}$): $w_{p,in} = h_4 - h_3 = \int_3^4 v\,dP \approx v_3 (P_4 - P_3) = v_{f@P_3} (P_{boiler} - P_{condenser})$ $h_4 = h_3 + w_{p,in} = h_3 + \frac{v_3 (P_{boiler} - P_{condenser})}{\eta_p}$
4. 4 to 1: Constant-Pressure Heat Addition in Boiler: Water enters as subcooled liquid, is heated to saturation, evaporated, and superheated to $T_1$: $q_{in} = h_1 - h_4$

Thermal Efficiency of the Rankine Cycle

$\eta_{th, Rankine} = \frac{w_{net,out}}{q_{in}} = \frac{w_t - w_p}{q_{in}} = \frac{(h_1 - h_2) - (h_4 - h_3)}{h_1 - h_4}$

Combined Heat & Power (CHP) / Cogeneration Utilization Factor

In district energy plants, high-pressure steam drives a backpressure extraction turbine, and low-pressure exhaust steam is delivered to building heating coils and domestic water heat exchangers:

$\epsilon_{CHP} = \frac{\dot{W}_{net} + \dot{Q}_{process}}{\dot{Q}_{in}}$

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4. The Brayton Cycle (Gas Turbines in CCHP Plants)

The Brayton Cycle is the ideal air-standard cycle for gas turbine power plants, frequently deployed in industrial HVAC plants for Combined Cooling, Heating, and Power (CCHP) with waste-heat-driven absorption chillers.

Ideal Air-Standard Brayton Cycle Processes

1. 1 to 2: Isentropic Compression in a compressor ($P_1 \to P_2$).
2. 2 to 3: Constant-Pressure Combustion / Heat Addition ($q_{in} = c_p (T_3 - T_2)$).
3. 3 to 4: Isentropic Expansion through a gas turbine ($P_2 \to P_1$, $w_t = c_p (T_3 - T_4)$).
4. 4 to 1: Constant-Pressure Heat Rejection ($q_{out} = c_p (T_4 - T_1)$).

Brayton Thermal Efficiency & Pressure Ratio

Defining the pressure ratio $r_p = \frac{P_2}{P_1}$:

$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} = r_p^{\frac{k-1}{k}} = \frac{T_3}{T_4}$

$\eta_{th, Brayton} = 1 - \frac{q_{out}}{q_{in}} = 1 - \frac{c_p (T_4 - T_1)}{c_p (T_3 - T_2)} = 1 - \frac{1}{r_p^{\frac{k-1}{k}}}$

Back Work Ratio (BWR)

Gas turbines differ fundamentally from steam Rankine cycles in their Back Work Ratio (BWR). Because the compressor compresses a low-density gas rather than an incompressible liquid, compression consumes 40% to 60% of the gross turbine work:

$\text{BWR} = \frac{w_{comp,in}}{w_{turb,out}} = \frac{h_2 - h_1}{h_3 - h_4} = \frac{c_p (T_2 - T_1)}{c_p (T_3 - T_4)}$

In contrast, a Rankine cycle feed pump typically consumes less than 1% to 2% of the steam turbine output ($BWR_{Rankine} < 0.02$).

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5. Summary Comparison of Fundamental Thermodynamic Cycles

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6. Worked Example: Second Law Benchmarking of a Water-Cooled Chiller

Problem: A central plant water-cooled centrifugal chiller produces $600 \text{ tons}$ of cooling while maintaining chilled water leaving the evaporator at $42.0^\circ\text{F}$ (with refrigerant evaporating at $T_L = 38.0^\circ\text{F}$). The cooling tower supplies condenser water returning to the condenser at $85.0^\circ\text{F}$ (with refrigerant condensing at $T_H = 98.0^\circ\text{F}$). The electrical power input measured at the compressor motor terminals is $348.0 \text{ kW}$.

Find:

1. The actual COP and the actual operating efficiency in $\text{kW/ton}$.
2. The theoretical maximum Reversed Carnot COP ($COP_{R, Carnot}$) based on internal refrigerant operating temperatures.
3. The Second Law (exergy) efficiency $\eta_{II}$ of the chiller compressor cycle.

Step-by-Step Solution:

Step 1: Calculate actual cooling capacity and actual COP. $\dot{Q}_{evap} = 600 \text{ tons} \times 12,000 \frac{\text{Btu}}{\text{hr}\cdot\text{ton}} = 7,200,000 \frac{\text{Btu}}{\text{hr}}$

Convert electric power input to $\text{Btu/hr}$ ($1 \text{ kW} = 3,412.14 \text{ Btu/hr}$): $\dot{W}_{in} = 348.0 \text{ kW} \times 3,412.14 \frac{\text{Btu}}{\text{hr}\cdot\text{kW}} = 1,187,425 \frac{\text{Btu}}{\text{hr}}$

$\text{COP}_{actual} = \frac{\dot{Q}_{evap}}{\dot{W}_{in}} = \frac{7,200,000}{1,187,425} = 6.063$

$\text{Operating Efficiency} = \frac{348.0 \text{ kW}}{600 \text{ tons}} = 0.580 \frac{\text{kW}}{\text{ton}}$

Check consistency: $\text{COP} = \frac{3.51685}{0.580} = 6.063$ (matches).

Step 2: Calculate Carnot COP using absolute Rankine temperatures. $T_L = 38.0^\circ\text{F} + 459.67 = 497.67^\circ\text{R}$ $T_H = 98.0^\circ\text{F} + 459.67 = 557.67^\circ\text{R}$ $\Delta T = T_H - T_L = 557.67 - 497.67 = 60.0^\circ\text{R}$

$\text{COP}_{R, Carnot} = \frac{T_L}{T_H - T_L} = \frac{497.67^\circ\text{R}}{60.0^\circ\text{R}} = 8.2945$

Step 3: Calculate Second Law Efficiency ($\eta_{II}$). $\eta_{II} = \frac{\text{COP}_{actual}}{\text{COP}_{R, Carnot}} = \frac{6.063}{8.2945} = 0.7309 \implies 73.1\%$

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7. NCEES Reference Handbook Navigation & Exam Traps

• Section 1.3: Power & Refrigeration Cycles: Look up Carnot formulas, Rankine state point definitions, and Brayton cycle pressure ratio relations.
• Absolute Temperature Error: Using $^\circ\text{F}$ instead of $^\circ\text{R}$ (or $^\circ\text{C}$ instead of $\text{K}$) in Carnot COP calculations is the most common exam trap. For example, $\frac{38}{98 - 38} = 0.633$, which is completely erroneous.
• COP Subscript Confusion: Never confuse $\text{COP}R = \frac{T_L}{T_H - T_L}$ with $\text{COP}{HP} = \frac{T_H}{T_H - T_L} = \text{COP}_R + 1$. When asked for heat pump COP, it must always exceed 1.0.