12.4 Gas Turbine Power Plants & Combined Cycles

Gas turbine power plants operate on the Brayton cycle, utilizing high-temperature gaseous combustion products directly as the working fluid to drive an expansion turbine. Gas turbines offer compact footprint, high power density, rapid start-up capability, and low capital cost, making them ideal for peaking power generation and combined cycle base-load stations.

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1. The Ideal Air-Standard Brayton Cycle

The simple open-cycle gas turbine consists of an axial/centrifugal air compressor, a combustion chamber (combustor), and a gas turbine mounted on a common shaft.

Pressure Ratio ($r_p$) & Temperature Relations

$$r_p = \frac{P_2}{P_1} = \frac{P_3}{P_4}$$ $$\frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{k-1}{k}} = r_p^{\frac{k-1}{k}} = \frac{T_3}{T_4}$$

Thermal Efficiency Formula

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

Optimal pressure ratio for maximum net work output per unit mass flow rate is given by:

$$r_{p,opt} = \left( \frac{T_{max}}{T_{min}} \right)^{\frac{k}{2(k-1)}} = \left( \frac{T_3}{T_1} \right)^{\frac{k}{2(k-1)}}$$

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2. Back Work Ratio ($BWR$)

In a gas turbine power plant, a substantial fraction of the mechanical work generated by the turbine is consumed directly to drive the air compressor on the same shaft:

$$BWR = \frac{w_{compressor}}{w_{turbine}} = \frac{h_2 - h_1}{h_3 - h_4} = \frac{T_2 - T_1}{T_3 - T_4} = \frac{T_1}{T_3} r_p^{\frac{k-1}{k}}$$

Unlike steam Rankine cycles where liquid pump work consumes less than $1\%$-$2\%$ of turbine work, gas turbines exhibit high Back Work Ratios ranging from $40\%$ to $60\%$ because compressing compressible gaseous air requires significantly greater specific work.

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3. Brayton Cycle Modifications

A. Regeneration / Recuperation

In simple gas turbine cycles, turbine exhaust gas leaves at a high temperature $T_4$ (often $500^\circ\text{C}$–$600^\circ\text{C}$), which is higher than compressor discharge temperature $T_2$. A counter-flow heat exchanger called a regenerator or recuperator uses turbine exhaust to preheat compressed air before entering the combustor.

• Regenerator Effectiveness ($\epsilon_{regen}$):

$$\epsilon_{regen} = \frac{h_{actual\ preheat} - h_2}{h_4 - h_2} \approx \frac{T_x - T_2}{T_4 - T_2}$$

Regeneration reduces fuel heat input $q_{in}$ from $(h_3 - h_2)$ down to $(h_3 - h_x)$, significantly boosting thermal efficiency at low-to-moderate pressure ratios.

B. Multistage Compression with Intercooling

Compressing air in multiple stages with an intermediate cooling heat exchanger (intercooler) reduces overall compressor work input. Intercooling shifts compressor process lines closer to isothermal compression. For a two-stage compressor, minimum work is achieved when pressure ratios are equal across stages:

$$P_i = \sqrt{P_1 P_2} \implies \frac{P_i}{P_1} = \frac{P_2}{P_i} = \sqrt{r_p}$$

C. Multistage Expansion with Reheating

Expanding gas through high-pressure and low-pressure turbine stages with an intermediate reheater combustion chamber increases total turbine work output $w_t$. Combining multistage intercooling, multistage reheating, and regeneration causes the cycle to approach the theoretical Ericsson cycle limit.

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4. Combined Cycle Gas Turbine (CCGT) Power Plants

To achieve maximum thermodynamic resource utilization, modern central-station power generation pairs a high-temperature gas turbine topping cycle with a steam turbine bottoming cycle.

Fuel + Air --> Combustor --> Gas Turbine (Topping) --> Electrical Power (W_GT)
                                   |
                                   v Hot Exhaust (550°C-600°C)
                     Heat Recovery Steam Generator (HRSG)
                                   |
                                   v Steam
                             Steam Turbine (Bottoming) --> Electrical Power (W_ST)

Heat Recovery Steam Generator (HRSG)

The HRSG is a multi-pressure counter-flow heat exchanger that extracts heat from hot gas turbine exhaust ($550^\circ\text{C}$–$600^\circ\text{C}$) to boil feedwater and superheat steam without requiring additional fuel firing.

Combined Cycle Thermal Efficiency

Combining a gas turbine cycle of efficiency $\eta_{GT}$ and a steam bottoming cycle that converts a fraction $\eta_{ST}$ of the remaining waste heat yields an overall plant thermal efficiency:

$$\eta_{overall} = \eta_{GT} + \eta_{ST} (1 - \eta_{GT})$$

Modern heavy-duty CCGT facilities (e.g., H-class and J-class gas turbines) reach overall combined cycle thermal efficiencies of $58\%$ to $63\%$, outperforming conventional single-cycle utility power plants.

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5. Worked Step-by-Step Gas Turbine & CCGT Problem

Problem Statement: A Combined Cycle Gas Turbine (CCGT) plant features a simple Brayton gas turbine operating with a pressure ratio $r_p = 12.0$, compressor inlet air at $T_1 = 300\text{ K}$ ($27^\circ\text{C}$), and maximum turbine inlet temperature $T_3 = 1400\text{ K}$ ($1127^\circ\text{C}$). Take air properties as $k = 1.40$ and $c_p = 1.005\text{ kJ/kg}\cdot\text{K}$. Gas turbine air mass flow rate is $\dot{m}_a = 200\text{ kg/s}$. The steam bottoming cycle produces an additional net electrical output of $\dot{W}_{ST} = 40.0\text{ MW}$ from HRSG exhaust heat. Calculate:

1. Compressor discharge temperature $T_2$ and specific compressor work $w_c$
2. Turbine exhaust temperature $T_4$ and specific turbine work $w_t$
3. Gas turbine back work ratio $BWR$
4. Gas turbine net power output $\dot{W}_{GT}$ and gas turbine thermal efficiency $\eta_{GT}$
5. Total combined plant power output $\dot{W}_{overall}$ and overall efficiency $\eta_{overall}$

Step-by-Step Solution

Step 1: Calculate Temperatures T_2 and T_4

$$\frac{T_2}{T_1} = (r_p)^{\frac{k-1}{k}} = (12.0)^{\frac{1.4-1}{1.4}} = (12.0)^{0.28571} = 2.0339$$ $$T_2 = 300\text{ K} \times 2.0339 = 610.17\text{ K} \text{ (}337.2^\circ\text{C)}$$ $$T_4 = \frac{T_3}{2.0339} = \frac{1400\text{ K}}{2.0339} = 688.33\text{ K} \text{ (}415.2^\circ\text{C)}$$

Step 2: Calculate Specific Work Quantities

$$w_c = c_p (T_2 - T_1) = 1.005 \times (610.17 - 300) = 1.005 \times 310.17 = 311.72\text{ kJ/kg}$$ $$w_t = c_p (T_3 - T_4) = 1.005 \times (1400 - 688.33) = 1.005 \times 711.67 = 715.23\text{ kJ/kg}$$ $$w_{net,GT} = w_t - w_c = 715.23 - 311.72 = 403.51\text{ kJ/kg}$$

Step 3: Calculate Back Work Ratio (BWR)

$$BWR = \frac{w_c}{w_t} = \frac{311.72}{715.23} = 0.4358 = 43.58\%$$

Step 4: Calculate Gas Turbine Power and Thermal Efficiency

$$\dot{W}_{GT} = \dot{m}_a \times w_{net,GT} = 200\text{ kg/s} \times 403.51\text{ kJ/kg} = 80,702\text{ kW} = 80.70\text{ MW}$$ $$q_{in} = c_p (T_3 - T_2) = 1.005 \times (1400 - 610.17) = 1.005 \times 789.83 = 793.78\text{ kJ/kg}$$ $$\dot{Q}_{in} = \dot{m}_a \times q_{in} = 200\text{ kg/s} \times 793.78\text{ kJ/kg} = 158,756\text{ kW} = 158.76\text{ MW}$$ $$\eta_{GT} = \frac{w_{net,GT}}{q_{in}} = \frac{403.51}{793.78} = 0.5083 = 50.83\%$$

Step 5: Calculate Combined Cycle Performance

$$\dot{W}_{overall} = \dot{W}_{GT} + \dot{W}_{ST} = 80.70\text{ MW} + 40.00\text{ MW} = 120.70\text{ MW}$$ $$\eta_{overall} = \frac{\dot{W}_{overall}}{\dot{Q}_{in}} = \frac{120.70\text{ MW}}{158.76\text{ MW}} = 0.7602\text{ (ideal overall conversion)}$$ $$\text{If } \eta_{ST,eff} = 30\% \text{ of GT exhaust: } \eta_{overall} = \eta_{GT} + \eta_{ST}(1 - \eta_{GT}) = 0.5083 + 0.30(1 - 0.5083) = 0.6558 = 65.58\%$$