5.3 Gas Power Cycles (Brayton, Otto, Diesel, Dual & Combined Cycle)

Q: An air-standard Brayton cycle operates with a compressor inlet temperature of 300 K and a pressure ratio of rp = 11.0. Assuming cold-air-standard conditions with k = 1.40, what is the ideal thermal efficiency of this gas turbine?

49.6%. The thermal efficiency of an ideal air-standard Brayton cycle depends exclusively on the pressure ratio: \eta_{th} = 1 - 1 / r_p^{(k-1)/k}. Calculating the exponent: (k-1)/k = (1.4 - 1)/1.4 = 0.28571. Then r_p^{0.28571} = 11.0^{0.28571} = 1.9839. Thus \eta_{th} = 1 - (1 / 1.9839) = 1 - 0.50406 = 0.49594, or 49.6%.

Q: A combined cycle power plant consists of a topping gas turbine with an individual thermal efficiency of 38% and a bottoming steam turbine with an individual thermal efficiency of 32%. Assuming all exhaust heat from the gas turbine is captured by the heat recovery steam generator (HRSG), what is the overall combined cycle thermal efficiency?

57.8%. The overall combined cycle efficiency is given by \eta_{cc} = \eta_{gas} + \eta_{steam} - \eta_{gas} \eta_{steam}. Substituting the values: \eta_{cc} = 0.38 + 0.32 - (0.38 \times 0.32) = 0.70 - 0.1216 = 0.5784, or 57.8%. Simply adding the efficiencies to get 70% is physically invalid because the steam turbine only operates on the heat rejected by the gas turbine.

Q: In a 4-stroke single-cylinder reciprocating engine with a bore of 100 mm and a stroke of 120 mm, the net work produced per complete thermodynamic cycle is measured to be 650 J. What is the Mean Effective Pressure (MEP) of this engine?

690 kPa. First, compute the piston displacement volume: V_d = (\pi / 4) \times D^2 \times L = (\pi / 4) \times (0.100\text{ m})^2 \times 0.120\text{ m} = 0.00094248\text{ m}^3. The Mean Effective Pressure is the net work divided by displacement volume: MEP = W_{net} / V_d = 650\text{ J} / 0.00094248\text{ m}^3 = 689,670\text{ N/m}^2 = 689.7\text{ kPa} \approx 690\text{ kPa}.

Q: Why does a stationary gas turbine power plant incorporate intercooling between compressor stages in a multi-stage compression system?

It reduces the specific volume of the gas, thereby minimizing total compressor work input.. Steady-flow compression work is given by w = \int v dP. Because cooling air between stages reduces its temperature and specific volume v, less work is required to compress the denser fluid across subsequent high-pressure stages. This decreases the overall compressor work input and elevates net cycle work output.

Key Takeaways

The ideal air-standard Brayton cycle consists of isentropic compression, constant-pressure combustion, isentropic expansion, and constant-pressure heat rejection, with efficiency dictated entirely by the pressure ratio: $\eta_{th} = 1 - 1 / r_p^{(k-1)/k}$ .
Gas turbine cycles exhibit very high Back Work Ratios ( $BWR = w_c / w_t \approx 40-60\%$ ) because compressing a low-density compressible gas requires substantial mechanical work compared to liquid pumping.
Brayton modifications—multi-stage compression with intercooling, multi-stage expansion with reheat, and exhaust recuperation/regeneration—greatly increase net work output and thermal efficiency.
Combined Cycle Gas Turbine (CCGT) power plants combine a topping Brayton turbine with a bottoming Rankine steam generator (HRSG) to achieve world-class thermal efficiencies of 55–65%.
Internal combustion reciprocating engine cycles (Otto for spark-ignition, Diesel for compression-ignition, and Dual) are characterized by compression ratio $r$ , cutoff ratio $r_c$ , and Mean Effective Pressure ( $MEP$ ).

Last updated: August 2026

Gas Power Cycles (Brayton, Otto, Diesel, Dual & Combined Cycle)

Gas power cycles utilize a gaseous working fluid throughout the entire thermodynamic cycle without undergoing phase change. Gas turbine engines operating on the Brayton cycle power commercial aviation, peaking utility power generation, and baseload Combined Cycle Gas Turbine (CCGT) power plants. Reciprocating internal combustion engines operating on Otto, Diesel, and Dual cycles drive transportation and distributed power generation.

1. Air-Standard Assumptions & The Ideal Brayton Cycle

To analyze complex gas power cycles analytically, engineers employ the standard Air-Standard Assumptions:

The working fluid is a fixed mass of air that behaves throughout as an ideal gas.
All processes are internally reversible.
The combustion process is modeled as an external constant-pressure or constant-volume heat addition process from an external source.
The exhaust stroke/blowdown is modeled as a constant-pressure or constant-volume heat rejection process to an external sink.
Under Cold-Air-Standard Assumptions, specific heats are evaluated at ambient room temperature ( $300\text{ K} / 520{}^\circ\text{R}$ , $k = 1.40, c_p = 1.005\text{ kJ/(kg}\cdot\text{K)}$ ).

+-----------------------------------------------------------------------------------------+
|                                 THE IDEAL BRAYTON CYCLE                                 |
|                                                                                         |
|                               Combustion Chamber (q_in)                                 |
|                            +------------------------------+                             |
|                            |     P_2 = P_3 = constant     |                             |
|                            +------------------------------+                             |
|                              ^                          |                               |
|                 State 2      | Compressed Air           | State 3: Hot Combustion Gas   |
|                 (High-P, T2) |                          v (High-P, T3 = T_TIT)          |
|             +------------+   |                     +------------+                       |
|  w_c,in --> | COMPRESSOR |---+                     |  TURBINE   | ----> w_t,out         |
|             +------------+                         +------------+                       |
|                   ^                                      |                              |
|          State 1  | Ambient Air                          | State 4: Hot Exhaust Gas     |
|          (P1, T1) |                                      v (Low-P, T4)                  |
|                            +------------------------------+                             |
|                            |     P_4 = P_1 = constant     |                             |
|                            +------------------------------+                             |
|                                 Exhaust Cooler (q_out)                                  |
+-----------------------------------------------------------------------------------------+

       P-v Diagram: Brayton Cycle                    T-s Diagram: Brayton Cycle
   P ^                                           T ^
     |        Combustion (2->3, P=const)           |                 3 (TIT: Max Temp)
     |         2 *-----------* 3                   |                /|
     |          /             \                    |               / |
     |   Comp. /               \ Turb.             |    2         /  | Expansion (3->4s)
     |  (1->2)/                 \ (3->4)           |   /|________/   |
     |       *-------------------*                 |  / |  q_in      | 
     |       1  Heat Rejection   4                 | 1--+------------+-- 4 (Exhaust)
     |          (4->1, P=const)                    | |  |  q_out     | 
     +-----------------------------------> v       +------------------------> s

Mathematical Formulation of the Ideal Brayton Cycle

Let the Pressure Ratio be defined as $r_p = P_2 / P_1 = P_3 / P_4$ .

For isentropic compression ( $1 \to 2$ ) and expansion ( $3 \to 4$ ):

\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} = r_p^{\frac{k-1}{k}} \implies T_2 = T_1 \cdot r_p^{\frac{k-1}{k}}

\frac{T_3}{T_4} = \left(\frac{P_3}{P_4}\right)^{\frac{k-1}{k}} = r_p^{\frac{k-1}{k}} \implies T_4 = \frac{T_3}{r_p^{\frac{k-1}{k}}}

w_{c,in} = c_p (T_2 - T_1), \quad w_{t,out} = c_p (T_3 - T_4)

q_{in} = c_p (T_3 - T_2), \quad q_{out} = c_p (T_4 - T_1)

Thermal Efficiency & Back Work Ratio

\eta_{th,Brayton} = \frac{w_{net}}{q_{in}} = 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}}}

BWR = \frac{w_{c,in}}{w_{t,out}} = \frac{T_2 - T_1}{T_3 - T_4} \approx 0.40 \text{ to } 0.60

Important

Optimum Pressure Ratio for Maximum Net Work: While thermal efficiency increases monotonically with pressure ratio $r_p$ , the net specific work output ( $w_{net} = w_t - w_c$ ) peaks at an intermediate optimum pressure ratio: $r_{p,opt} = \left(\frac{T_{max}}{T_{min}}\right)^{\frac{k}{2(k-1)}} = \left(\frac{T_3}{T_1}\right)^{\frac{k}{2(k-1)}}$

Real Cycle Component Efficiencies

Accounting for compressor and turbine isentropic efficiencies:

w_{c,actual} = \frac{w_{c,s}}{\eta_c} = \frac{c_p (T_{2s} - T_1)}{\eta_c} \implies T_{2a} = T_1 + \frac{T_{2s} - T_1}{\eta_c}

w_{t,actual} = \eta_t \cdot w_{t,s} = \eta_t c_p (T_3 - T_{4s}) \implies T_{4a} = T_3 - \eta_t (T_3 - T_{4s})

2. Brayton Modifications: Intercooling, Reheat & Regeneration

To increase power capacity and thermal efficiency, large industrial gas turbines incorporate multi-stage modifications:

+-----------------------------------------------------------------------------------------+
|                          BRAYTON MODIFICATION TECHNOLOGIES                              |
|                                                                                         |
|   1. INTERCOOLING (Multi-Stage Compression):                                            |
|      - Cools gas between LP and HP compressor stages.                                   |
|      - Reduces total compressor work (since w = integral v dP; smaller v means less work)|
|      - Optimum intermediate pressure: P_i = sqrt(P_1 * P_2).                            |
|                                                                                         |
|   2. REHEAT (Multi-Stage Expansion):                                                    |
|      - Reheats gas between HP and LP turbine stages.                                    |
|      - Increases total turbine work output without exceeding maximum metallurgical TIT. |
|      - Optimum intermediate pressure: P_i = sqrt(P_3 * P_4).                            |
|                                                                                         |
|   3. REGENERATION / RECUPERATION:                                                       |
|      - Transfers waste heat from hot turbine exhaust (T4) to preheat compressor         |
|        discharge air (T2) before entering combustor.                                    |
|      - Drastically cuts fuel consumption (q_in).                                        |
+-----------------------------------------------------------------------------------------+

       T-s Diagram: Brayton Cycle with Intercooling, Reheat & Regeneration
   T ^
     |                 Combustor       Reheater
     |                 3 *-----------* 5 (TIT Max Temp)
     |                  /             \
     |    Regenerator  /               \ 
     |    T_x <-------+-- T_4 (Exhaust) +---> T_y (Preheated to Combustor)
     |       2 *     /                   \ * 6
     |        /|\   /                     /|\
     | Inter-/ | \ /                     / | \ Reheat
     | cool 1* |  * 2'                  4* |  * 5'
     |       | |                         | |
     +-------+-+-------------------------+-+--------> s

Regenerator Effectiveness ( $\epsilon$ )

The effectiveness of a regenerator is the ratio of actual heat transfer to maximum possible heat transfer:

\epsilon = \frac{h_5 - h_2}{h_4 - h_2} \approx \frac{T_5 - T_2}{T_4 - T_2}

Where $T_5$ is the air temperature exiting the regenerator (entering combustor), $T_2$ is compressor exit temperature, and $T_4$ is turbine exhaust temperature. The thermal efficiency of an ideal regenerator ( $\epsilon = 1.0$ ) cycle is:

\eta_{th,regen} = 1 - \left(\frac{T_1}{T_3}\right) r_p^{\frac{k-1}{k}}

Warning

When Regeneration Degrades Efficiency: If pressure ratio $r_p$ is very high, compressor exit temperature exceeds turbine exhaust temperature ( $T_2 > T_4$ ). Installing a regenerator would transfer heat backward from compressor air to exhaust, increasing fuel consumption and degrading efficiency.

3. Combined Cycle Gas Turbine (CCGT) & HRSG

A Combined Cycle Gas Turbine (CCGT) marries the high-temperature capabilities of the Brayton cycle with the low-temperature heat rejection advantages of the Rankine steam cycle.

+-----------------------------------------------------------------------------------------+
|                        COMBINED CYCLE ARCHITECTURE & ENERGY BALANCE                     |
|                                                                                         |
|       Fuel (Q_in)                                                                       |
|            |                                                                            |
|            v                                                                            |
|   +-----------------+                                                                   |
|   | TOPPING BRAYTON | ------> Gas Turbine Electric Power (W_dot_gas = eta_g * Q_in)     |
|   |   GAS TURBINE   |                                                                   |
|   +-----------------+                                                                   |
|            | Hot Turbine Exhaust (550 - 650 °C)                                         |
|            v                                                                            |
|   +-----------------+                                                                   |
|   |  HRSG (BOILER)  |                                                                   |
|   +-----------------+                                                                   |
|            | High-P Superheated Steam                                                   |
|            v                                                                            |
|   +-----------------+                                                                   |
|   | BOTTOMING STEAM | ------> Steam Turbine Electric Power (W_dot_steam)                |
|   |  RANKINE CYCLE  |                                                                   |
|   +-----------------+                                                                   |
|            |                                                                            |
|            v Condenser Waste Heat (Ambient Sink)                                        |
+-----------------------------------------------------------------------------------------+

Combined Cycle Thermal Efficiency

The overall combined cycle efficiency is derived directly from the individual cycle efficiencies:

\dot{W}_{total} = \dot{W}_{gas} + \dot{W}_{steam} = \eta_{gas} \dot{Q}_{in} + \eta_{steam} \dot{Q}_{in,HRSG}

Assuming all waste heat from the gas turbine enters the HRSG ( $\dot{Q}_{in,HRSG} = (1 - \eta_{gas})\dot{Q}_{in}$ ):

\eta_{cc} = \frac{\dot{W}_{total}}{\dot{Q}_{in}} = \eta_{gas} + \eta_{steam} (1 - \eta_{gas}) = \eta_{gas} + \eta_{steam} - \eta_{gas} \eta_{steam}

Modern heavy-duty F- and H-class CCGT power stations regularly achieve thermal efficiencies exceeding $60\% \text{ to } 64\%$ (LHV basis).

4. Internal Combustion Reciprocating Cycles: Otto, Diesel & Dual

Reciprocating engines compress and expand gas within a cylinder bounded by a reciprocating piston.

+-----------------------------------------------------------------------------------------+
|                           RECIPROCATING ENGINE KINEMATICS                               |
|                                                                                         |
|   Bore (D):                 Piston diameter                                             |
|   Stroke (L):               Linear distance between TDC and BDC                         |
|   Top Dead Center (TDC):    Minimum cylinder volume coordinate (Clearance Volume V_c)   |
|   Bottom Dead Center (BDC): Maximum cylinder volume coordinate (Total Volume V_max)     |
|   Displacement Volume:      V_d = V_max - V_c = (pi/4) * D^2 * L                        |
|   Compression Ratio:        r = V_max / V_min = (V_d + V_c) / V_c                       |
|   Mean Effective Pressure:  MEP = W_net / V_d = W_net / (V_max - V_min)                 |
+-----------------------------------------------------------------------------------------+

       P-v: Otto Cycle (Spark-Ignition)              P-v: Diesel Cycle (Compression-Ignition)
   P ^                                           P ^
     |          3 (Peak P & T)                     |         2  3 (Constant P Combustion)
     |          *                                  |          *--* (Cutoff Volume V3)
     |         /|                                  |         /    \
     |  Comb. / |                                  |  Comp. /      \ Expansion
     | (V=C) /  | Expansion                        | (1->2)/        \ (3->4)
     |    2 *   | (3->4)                           |     1*----------* 4 (Blowdown)
     |     /    * 4 (Blowdown)                     |      (BDC: V_max)
     |   1*-----+----------------> v               +------+------+-----------------> v
       (TDC:Vc) (BDC:Vmax)                             (TDC:V2) (V3)   (BDC:V1)

The Air-Standard Otto Cycle (Spark Ignition)

Combustion is modeled as instantaneous constant-volume heat addition ( $2 \to 3$ ):

\eta_{th,Otto} = 1 - \frac{1}{r^{k-1}}

Compression ratios in gasoline engines are constrained to $r \approx 8 - 11$ to prevent pre-ignition and engine knock (autoignition of the end-gas).

The Air-Standard Diesel Cycle (Compression Ignition)

Combustion is modeled as constant-pressure heat addition ( $2 \to 3$ ) across the Cutoff Ratio ( $r_c = V_3 / V_2$ ):

\eta_{th,Diesel} = 1 - \frac{1}{r^{k-1}} \left[\frac{r_c^k - 1}{k (r_c - 1)}\right]

Because only air is compressed during the compression stroke, Diesel engines eliminate knock limitations and operate at high compression ratios ( $r \approx 14 - 22$ ), yielding higher actual thermal efficiencies than Otto engines.

The Dual (Sabathé) Cycle

Modern high-speed diesel engines operate on a Dual Cycle where heat addition occurs partly at constant volume ( $2 \to 3$ , pressure ratio $r_p = P_3 / P_2$ ) and partly at constant pressure ( $3 \to 4$ , cutoff ratio $r_c = V_4 / V_3$ ).

5. Step-by-Step Worked Problem: Regenerative Gas Turbine Analysis

Problem: A gas turbine operating on an air-standard regenerative Brayton cycle draws in ambient air at $P_1 = 100\text{ kPa}$ ( $14.5\text{ psia}$ ) and $T_1 = 300\text{ K}$ ( $80^\circ\text{F}$ ). The pressure ratio across the compressor is $r_p = 8.0$ . The turbine inlet temperature is $T_3 = 1300\text{ K}$ ( $1880^\circ\text{F}$ ). The compressor isentropic efficiency is $\eta_c = 84\%$ , the turbine isentropic efficiency is $\eta_t = 88\%$ , and the regenerator effectiveness is $\epsilon = 80\%$ . Assume cold-air-standard properties ( $c_p = 1.005\text{ kJ/(kg}\cdot\text{K)}, k = 1.40, (k-1)/k = 0.2857$ ). Calculate:

Actual compressor exit temperature ( $T_{2a}$ ) and actual turbine exhaust temperature ( $T_{4a}$ )
Air temperature entering the combustor ( $T_5$ )
Net specific work produced ( $w_{net}$ in $\text{kJ/kg}$ )
Cycle thermal efficiency ( $\eta_{th}$ )

+-----------------------------------------------------------------------------------------+
|                     REGENERATIVE BRAYTON CALCULATION STEPS                              |
|                                                                                         |
|   STEP 1: Compressor Calculations (P_2/P_1 = 8.0, T_1 = 300 K)                          |
|           T_2s = T_1 * (r_p)^((k-1)/k) = 300 * (8.0)^0.28571 = 300 * 1.8114 = 543.43 K  |
|           w_c,s = c_p * (T_2s - T_1) = 1.005 * (543.43 - 300) = 244.65 kJ/kg             |
|           w_c,a = w_c,s / eta_c = 244.65 / 0.84 = 291.25 kJ/kg                           |
|           T_2a = T_1 + (w_c,a / c_p) = 300 + (291.25 / 1.005) = 589.80 K                 |
|                                                                                         |
|   STEP 2: Turbine Calculations (T_3 = 1300 K, P_3/P_4 = 8.0)                            |
|           T_4s = T_3 / (r_p)^((k-1)/k) = 1300 / 1.8114 = 717.68 K                        |
|           w_t,s = c_p * (T_3 - T_4s) = 1.005 * (1300 - 717.68) = 585.23 kJ/kg            |
|           w_t,a = eta_t * w_t,s = 0.88 * 585.23 = 515.00 kJ/kg                           |
|           T_4a = T_3 - (w_t,a / c_p) = 1300 - (515.00 / 1.005) = 787.56 K                 |
|                                                                                         |
|   STEP 3: Regenerator Heat Transfer (epsilon = 0.80)                                    |
|           T_5 = T_2a + epsilon * (T_4a - T_2a)                                          |
|           T_5 = 589.80 + 0.80 * (787.56 - 589.80) = 589.80 + 0.80 * (197.76)           |
|           T_5 = 589.80 + 158.21 = 748.01 K                                              |
|                                                                                         |
|   STEP 4: Combustor Heat Input (q_in)                                                   |
|           q_in = c_p * (T_3 - T_5) = 1.005 * (1300 - 748.01) = 554.75 kJ/kg              |
|           (Note: Without regenerator, q_in would be c_p*(1300 - 589.80) = 713.75 kJ/kg) |
|                                                                                         |
|   STEP 5: Net Work, BWR & Thermal Efficiency                                            |
|           w_net = w_t,a - w_c,a = 515.00 - 291.25 = 223.75 kJ/kg                        |
|           BWR = w_c,a / w_t,a = 291.25 / 515.00 = 0.5655 = 56.55%                       |
|           eta_th = w_net / q_in = 223.75 / 554.75 = 0.4033 = 40.33%                     |
+-----------------------------------------------------------------------------------------+

6. Common Exam Traps & PE Pro-Tips

Trap 1 — Inverting Compressor vs Turbine Isentropic Efficiency: Compressor work is an input ( $w_{c,actual} = w_{c,s} / \eta_c > w_{c,s}$ ), whereas turbine work is an output ( $w_{t,actual} = \eta_t \cdot w_{t,s} < w_{t,s}$ ). Remember that irreversibility always increases work required for compression and decreases work produced during expansion.
Trap 2 — Combined Cycle Efficiency Direct Addition: Never add topping and bottoming efficiencies directly (e.g., $38\% + 32\% \ne 70\%$ ). The steam turbine operates solely on the waste thermal stream of the gas turbine, following $\eta_{cc} = \eta_{gas} + \eta_{steam} - \eta_{gas} \eta_{steam}$ .
Trap 3 — Mean Effective Pressure Volume Base: In internal combustion reciprocating calculations, $MEP = W_{net} / V_d$ . Make sure to divide by displacement volume $V_d = V_{max} - V_{min}$ , not total cylinder volume $V_{max}$ .

Test Your Knowledge

An air-standard Brayton cycle operates with a compressor inlet temperature of 300 K and a pressure ratio of rp = 11.0. Assuming cold-air-standard conditions with k = 1.40, what is the ideal thermal efficiency of this gas turbine?

38.5%

45.2%

47.8%

49.6%

Test Your Knowledge

A combined cycle power plant consists of a topping gas turbine with an individual thermal efficiency of 38% and a bottoming steam turbine with an individual thermal efficiency of 32%. Assuming all exhaust heat from the gas turbine is captured by the heat recovery steam generator (HRSG), what is the overall combined cycle thermal efficiency?

70.0%

57.8%

50.6%

44.5%

Test Your Knowledge

In a 4-stroke single-cylinder reciprocating engine with a bore of 100 mm and a stroke of 120 mm, the net work produced per complete thermodynamic cycle is measured to be 650 J. What is the Mean Effective Pressure (MEP) of this engine?

690 kPa

828 kPa

545 kPa

345 kPa

Test Your Knowledge

Why does a stationary gas turbine power plant incorporate intercooling between compressor stages in a multi-stage compression system?

It eliminates the need for air filtration before the compressor.

It raises the temperature of the air entering the combustor without consuming fuel.

It reduces the specific volume of the gas, thereby minimizing total compressor work input.

It converts the cycle into a constant-volume combustion process.

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5.3 Gas Power Cycles (Brayton, Otto, Diesel, Dual & Combined Cycle)

Key Takeaways

Gas Power Cycles (Brayton, Otto, Diesel, Dual & Combined Cycle)

1. Air-Standard Assumptions & The Ideal Brayton Cycle

Mathematical Formulation of the Ideal Brayton Cycle

Thermal Efficiency & Back Work Ratio

Real Cycle Component Efficiencies

2. Brayton Modifications: Intercooling, Reheat & Regeneration

Regenerator Effectiveness (ϵ\epsilonϵ)

3. Combined Cycle Gas Turbine (CCGT) & HRSG

Combined Cycle Thermal Efficiency

4. Internal Combustion Reciprocating Cycles: Otto, Diesel & Dual

The Air-Standard Otto Cycle (Spark Ignition)

The Air-Standard Diesel Cycle (Compression Ignition)

The Dual (Sabathé) Cycle

5. Step-by-Step Worked Problem: Regenerative Gas Turbine Analysis

6. Common Exam Traps & PE Pro-Tips

Regenerator Effectiveness ( $\epsilon$ )