Gas Turbine Power Plant

The thermal efficiency of a simple open-cycle gas turbine is low (about 20–30 %) because the compressor absorbs a large part of the turbine work (60–70 %) and the exhaust leaves at a high temperature. Its efficiency and output are improved by the following methods.

Methods of efficiency improvement

1. Regeneration (heat exchanger): hot turbine exhaust preheats the compressed air before the combustion chamber, so less fuel is needed for the same turbine inlet temperature.
2. Intercooling: compression is done in stages with cooling of air between stages; compressor work falls.
3. Reheating: expansion is done in stages; gas is reheated between turbines, raising turbine work.
4. Higher turbine inlet temperature using better blade materials and blade cooling.
5. Higher (optimum) pressure ratio with high component (isentropic) efficiencies.
6. Combined cycle / waste-heat recovery: exhaust heat raises steam for a steam turbine.
7. Water or steam injection, and reducing pressure losses in ducts and combustor.

Intercooling

In intercooling the air is compressed in a low-pressure (LP) compressor, cooled in an intercooler at nearly constant pressure back to (ideally) the inlet temperature, and then compressed in a high-pressure (HP) compressor.

 Air  +-----+   +-------+   +-----+   +----+
 -1-->| LPC |-2>| Inter |-3>| HPC |-4>| CC |
      +-----+   | cooler|   +-----+   +----+
         |      +-------+      |        |5
         +========= shaft =====+====+---+--+
                                    |  T   |-6-> exhaust
                         Gen <======+------+

 T
 |                   5
 |                  /|
 |                 / |
 |         4     /   |
 |         |---'     |
 |         |   2     6
 |         |  /|    /
 |         | / |   /
 |         3   1--'
 +----------------------- s
 1-2 LP comp, 2-3 intercool,
 3-4 HP comp, 4-5 heating,
 5-6 turbine, 6-1 exhaust

How it improves the plant:

• Work of compression is proportional to the absolute temperature of the air ($W_c = c_p T_{in}[r_p^{(\gamma-1)/\gamma} - 1]$). Cooling the air to $T_3 \approx T_1$ before the HP stage reduces total compressor work. With perfect intercooling and equal stage pressure ratios ($r_{p1} = r_{p2} = \sqrt{r_p}$), compressor work is minimum.
• Turbine work is unchanged, so net work output and work ratio increase considerably.
• However, the air leaves the HP compressor colder ($T_4 < T_2'$ of single-stage), so more fuel must be burnt from 4 to 5. Alone, intercooling may give only a small increase (or even a decrease) in efficiency.
• When intercooling is used with a regenerator, the lower compressor delivery temperature allows more exhaust heat to be recovered, and thermal efficiency increases appreciably.

Thus intercooling mainly raises specific output and, combined with regeneration, raises thermal efficiency.

Cycle: 1–2 compressor, 2–x regenerator (air side), x–3 combustion chamber, 3–4 turbine, 4–y regenerator (gas side) to exhaust.

Data: $p_1 = 100$ kPa, $T_1 = 288$ K, $\dot m = 6$ kg/s, $r_p = 6$, $T_3 = 900$ K, $\varepsilon = 0.8$, $\eta_c = \eta_t = 0.85$, $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$, so $\frac{\gamma-1}{\gamma} = 0.2857$ and $r_p^{0.2857} = 6^{0.2857} = 1.6685$.

     +----------- Regenerator ---------+
     |  x (hot air)        y (exhaust) |
 1   |2                     4          |
 -> [C] ------> [CC] -3-> [T] ---------+
      \=========shaft=====/  ==> Gen

Compressor

$$\begin{aligned} T_{2s} &= T_1 r_p^{0.2857} = 288 \times 1.6685 = 480.53\ \text{K} \\ T_2 &= T_1 + \frac{T_{2s}-T_1}{\eta_c} = 288 + \frac{192.53}{0.85} = 514.51\ \text{K} \\ W_c &= c_p (T_2 - T_1) = 1.005 \times 226.51 = 227.64\ \text{kJ/kg} \end{aligned}$$

Turbine

$$\begin{aligned} T_{4s} &= \frac{T_3}{r_p^{0.2857}} = \frac{900}{1.6685} = 539.40\ \text{K} \\ T_4 &= T_3 - \eta_t (T_3 - T_{4s}) = 900 - 0.85 \times 360.60 = 593.49\ \text{K} \\ W_t &= c_p (T_3 - T_4) = 1.005 \times 306.51 = 308.04\ \text{kJ/kg} \end{aligned}$$

Regenerator and heat supplied

Since $T_4 > T_2$, regeneration is possible:

$$\begin{aligned} T_x &= T_2 + \varepsilon (T_4 - T_2) = 514.51 + 0.8(593.49 - 514.51) = 577.70\ \text{K} \\ Q_s &= c_p (T_3 - T_x) = 1.005(900 - 577.70) = 323.92\ \text{kJ/kg} \end{aligned}$$

Efficiency and power

$$\begin{aligned} W_{net} &= W_t - W_c = 308.04 - 227.64 = 80.40\ \text{kJ/kg} \\ \eta_{th} &= \frac{W_{net}}{Q_s} = \frac{80.40}{323.92} = 0.2482 = 24.82\% \\ P_{net} &= \dot m W_{net} = 6 \times 80.40 = 482.4\ \text{kW} \end{aligned}$$

For comparison, without the regenerator $Q_s = 1.005(900 - 514.51) = 387.42$ kJ/kg and $\eta = 20.75\%$, so the regenerator raises efficiency by about 4.1 percentage points.

Answer: Plant thermal efficiency = 24.82 %, net power developed = 482.4 kW (with the given data the plant cannot actually deliver its rated 800 kW at 6 kg/s).

Arrangement: compressor (C) driven by the HP turbine; the LP (power) turbine drives the generator on a separate shaft.

 1        2         3        4         5
 --> [C] ---> [CC] ---> [HPT] ---> [LPT] -->
      \__________________/           |
         (HP shaft)                  +==> Gen

Data: $T_1 = 288$ K, $p_1 = 100$ kPa, $r_p = 7$, $\eta_c = 0.82$, $\eta_{HP} = \eta_{LP} = 0.85$, $T_3 = 610 + 273 = 883$ K. Air: $c_{pa} = 1.005$, $\gamma = 1.4$ ($\frac{\gamma-1}{\gamma} = 0.2857$). Gas: $c_{pg} = 1.15$, $\gamma = 1.333$ ($\frac{\gamma-1}{\gamma} = 0.2498$).

Compressor

$$\begin{aligned} T_{2s} &= 288 \times 7^{0.2857} = 502.17\ \text{K} \\ T_2 &= 288 + \frac{502.17 - 288}{0.82} = 549.18\ \text{K} \\ W_c &= 1.005(549.18 - 288) = 262.49\ \text{kJ/kg} \end{aligned}$$

(a) State at entry to the LP turbine

HP turbine work = compressor work (neglecting fuel mass and mechanical losses):

$$\begin{aligned} c_{pg}(T_3 - T_4) &= W_c \Rightarrow T_4 = 883 - \frac{262.49}{1.15} = 654.75\ \text{K} \\ T_{4s} &= T_3 - \frac{T_3 - T_4}{\eta_{HP}} = 883 - \frac{228.25}{0.85} = 614.47\ \text{K} \\ \frac{p_3}{p_4} &= \left(\frac{T_3}{T_{4s}}\right)^{1/0.2498} = \left(\frac{883}{614.47}\right)^{4.003} = 4.269 \\ p_4 &= \frac{700}{4.269} = 163.98\ \text{kPa} \end{aligned}$$

Gas enters the LP turbine at 1.64 bar and 654.75 K (381.8 °C).

(b) Net power per kg/s

LP turbine expands from 1.64 bar to 1 bar:

$$\begin{aligned} T_{5s} &= T_4\left(\frac{p_5}{p_4}\right)^{0.2498} = 654.75\left(\frac{100}{163.98}\right)^{0.2498} = 578.65\ \text{K} \\ T_5 &= T_4 - \eta_{LP}(T_4 - T_{5s}) = 654.75 - 0.85(76.10) = 590.07\ \text{K} \\ W_{net} &= W_{LP} = 1.15(654.75 - 590.07) = 74.38\ \text{kJ/kg} \end{aligned}$$

Net power = 74.38 kW per kg/s of air.

(c) Thermal efficiency

$$\begin{aligned} Q_s &= c_{pg}(T_3 - T_2) = 1.15(883 - 549.18) = 383.89\ \text{kJ/kg} \\ \eta_{th} &= \frac{74.38}{383.89} = 0.1938 = 19.38\% \end{aligned}$$

Answer: (a) $p_4$ = 1.64 bar, $T_4$ = 654.75 K; (b) 74.38 kW per kg/s; (c) $\eta$ = 19.38 %.

A simple open-cycle gas turbine has a low thermal efficiency (about 20–30 %) because (i) the compressor consumes a large part (60–70 %) of turbine work and (ii) the exhaust leaves at 400–550 °C, carrying away much heat. The following methods improve efficiency and/or output.

1. Regeneration

A heat exchanger (regenerator) transfers heat from the hot turbine exhaust to the compressed air before it enters the combustion chamber. Fuel required for the same turbine inlet temperature falls, so efficiency rises while net work is unchanged. Effective only when exhaust temperature is higher than compressor delivery temperature (low–moderate pressure ratios). Effectiveness $\varepsilon = \dfrac{T_x - T_2}{T_4 - T_2}$, usually 0.6–0.8.

2. Intercooling

Compression in two or more stages with cooling between stages reduces compressor work ($W \propto T_{inlet}$). Net work and work ratio increase. Alone it slightly increases fuel needed, but with regeneration it gives a definite gain in efficiency.

3. Reheating

Expansion in two turbines with reheating of gas between them (in a second combustor) increases turbine work, so net output rises; efficiency rises only when combined with regeneration because exhaust is hotter.

4. Combination (intercooling + reheating + regeneration)

Approaches the Ericsson cycle; gives the highest efficiency and work ratio, but plant is complex and costly.

5. Higher turbine inlet temperature

Efficiency and specific work increase with $T_3$. Achieved with nickel-based superalloys, ceramic coatings and blade cooling (air or steam).

6. Optimum pressure ratio and better components

For given $T_1, T_3$ there is a pressure ratio giving maximum work and another giving maximum efficiency. Increasing isentropic efficiencies of compressor and turbine (better blade design) and reducing pressure losses in ducts, filters and combustor all improve performance.

7. Waste-heat recovery / combined cycle

Exhaust heat raises steam in a heat recovery steam generator (HRSG) for a steam turbine (combined cycle, overall 50–60 %), or is used for process heating (cogeneration).

8. Water/steam injection and inlet air cooling

Injecting water or steam increases mass flow and output; cooling the inlet air (evaporative or chiller) increases air density and reduces compressor work, especially in hot weather.

Data: $T_1 = 15 + 273 = 288$ K, $r_p = 4$, $\eta_c = 0.82$, $T_3 = 600 + 273 = 873$ K, $\eta_t = 0.70$, $\varepsilon = 0.78$, $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$; $r_p^{(\gamma-1)/\gamma} = 4^{0.2857} = 1.4860$.

       +------ Regenerator ------+
       |  x                 y    |
 1     | 2                  4    |
 --> [C] --> [CC] --3--> [T] ----+

Compressor

$$\begin{aligned} T_{2s} &= 288 \times 1.4860 = 427.97\ \text{K} \\ T_2 &= 288 + \frac{427.97 - 288}{0.82} = 458.69\ \text{K} \\ W_c &= 1.005(458.69 - 288) = 171.54\ \text{kJ/kg} \end{aligned}$$

Turbine

$$\begin{aligned} T_{4s} &= \frac{873}{1.4860} = 587.49\ \text{K} \\ T_4 &= 873 - 0.70(873 - 587.49) = 673.14\ \text{K} \\ W_t &= 1.005(873 - 673.14) = 200.86\ \text{kJ/kg} \end{aligned}$$

Regenerator

$T_4 = 673.14$ K $> T_2 = 458.69$ K, so heat can be recovered:

$$T_x = T_2 + \varepsilon(T_4 - T_2) = 458.69 + 0.78(214.45) = 625.96\ \text{K}$$

Thermal efficiency

$$\begin{aligned} W_{net} &= 200.86 - 171.54 = 29.32\ \text{kJ/kg} \\ Q_s &= c_p (T_3 - T_x) = 1.005(873 - 625.96) = 248.27\ \text{kJ/kg} \\ \eta_{th} &= \frac{29.32}{248.27} = 0.1181 = 11.81\% \end{aligned}$$

(Without the regenerator: $Q_s = 1.005(873 - 458.69) = 416.38$ kJ/kg, $\eta = 7.04\%$.)

Answer: Thermal efficiency of the cycle = 11.81 %.

A simple open-cycle constant-pressure gas turbine draws atmospheric air, compresses it, burns fuel in it at constant pressure, expands the hot gas in a turbine and exhausts it to the atmosphere (Brayton/Joule cycle). With intercooling, compression is split into LP and HP stages with an air cooler between them.

Sketch of the plant

             water in/out
                 ||
 Air->[LPC]-2->[ IC ]-3->[HPC]-4->[CC]-5->[ T ]-6->
  1     ||                 ||      ^       ||  exh
        ++======shaft======++======|=======++==>Gen
                                  fuel

LPC = low-pressure compressor, IC = intercooler, HPC = high-pressure compressor, CC = combustion chamber, T = turbine.

T–s diagram

 T
 |                   5
 |                  /|
 |                 / |
 |         4     /   |
 |         |---'     |
 |         |   2     6
 |         |  /|    /
 |         | / |   /
 |         3   1--'
 +----------------------- s
 1-2 LPC, 2-3 intercooling at p_i,
 3-4 HPC, 4-5 combustion at p_2,
 5-6 turbine, 6-1 exhaust (p_1)

How intercooling improves the plant

• Compressor work per kg is $W = c_p T_{in}\left[r^{(\gamma-1)/\gamma} - 1\right]$, i.e. proportional to inlet temperature. Cooling the air from $T_2$ back to $T_3 \approx T_1$ at the intermediate pressure makes the HP stage work much smaller.
• Total compressor work is minimum when intercooling is perfect ($T_3 = T_1$) and the stage pressure ratios are equal: $p_i = \sqrt{p_1 p_2}$.
• Turbine work is unchanged, so net work and work ratio increase; a smaller plant gives the same output.
• Because air now leaves the HPC colder ($T_4$ lower), more fuel is needed in 4–5, so intercooling alone gives only a small change in efficiency. When combined with a regenerator, the larger gap between exhaust temperature $T_6$ and $T_4$ allows more heat recovery, and thermal efficiency rises noticeably.
• Other ways to raise efficiency: regeneration, reheating, higher turbine inlet temperature, optimum pressure ratio and better component efficiencies.

A regenerator is a counter-flow heat exchanger that uses the hot exhaust gas leaving the turbine to preheat the compressed air before it enters the combustion chamber. Heat that would be thrown away is returned to the cycle, so less fuel is burnt for the same turbine inlet temperature.

Sketch

          +---------- REGENERATOR ----------+
          |  air 2 -> x       y <- gas 4    |
          +---|-------^-------|-------^-----+
              |       |       v       |
 Air 1 ->[ C ]+       |    exhaust    |
           ||   x---->[ CC ]--3-->[ T ]+
           ||           ^          ||
           ++====shaft==|==========++==> Generator
                      fuel

Processes: 1–2 compression, 2–x air heated in regenerator, x–3 combustion, 3–4 expansion, 4–y gas cooled in regenerator, y–1 exhaust to atmosphere.

p–v and T–s diagrams

 p
 |  2    x    3
 |  *----*----*
 |   \          \
 |    \            \
 |     \              \
 |      *------*--------*
 |      1      y        4
 +--------------------------- v

 T
 |             3
 |            *
 |           /|
 |      x   / |
 |      *--'  |
 |     /      * 4
 |  2 *    _-'
 |    |  *'
 |    | / y
 |    *
 |    1
 +-------------------- s

Heat from 4–y (gas side) supplies 2–x (air side). Ideal: $T_x = T_4$, $T_y = T_2$.

How the regenerator increases performance

• Heat supplied from fuel becomes $Q_s = c_p(T_3 - T_x)$ instead of $c_p(T_3 - T_2)$; since $T_x > T_2$, fuel consumption falls.
• Compressor and turbine work are unchanged, so net work is the same while heat input is less, therefore

$$\eta = \frac{W_t - W_c}{c_p (T_3 - T_x)}$$

increases (often by 5–10 percentage points).

• Effectiveness $\varepsilon = \dfrac{T_x - T_2}{T_4 - T_2}$ (practical 0.6–0.8). Higher effectiveness needs a larger, costlier heat exchanger and gives more pressure loss.
• Regeneration is useful only when $T_4 > T_2$, i.e. at low and moderate pressure ratios. At high pressure ratios the compressor delivery is hotter than the exhaust and regeneration would reduce efficiency.
• For a fixed fuel supply the turbine inlet temperature can be raised, which increases the output; also exhaust is cooler, reducing waste heat.

Data: $r_p = 6$, $T_3 = 610 + 273 = 883$ K, $\eta_c = 0.80$, $\eta_t = 0.82$, $T_1 = 15 + 273 = 288$ K, $\dot m = 16$ kg/s. Compression: $c_p = 1.005$, $\gamma = 1.4$ ($\frac{\gamma-1}{\gamma} = 0.2857$). Expansion: $c_p = 1.11$, $\gamma = 1.333$ ($\frac{\gamma-1}{\gamma} = 0.2498$). Combustion: $c_p = 1.11$ kJ/kg K. Mass of fuel neglected.

 1        2          3        4
 --> [C] ---> [CC] ---> [T] ---> exhaust
      \================/  ==> Generator

Compressor

$$\begin{aligned} T_{2s} &= 288 \times 6^{0.2857} = 480.53\ \text{K} \\ T_2 &= 288 + \frac{480.53 - 288}{0.80} = 528.66\ \text{K} \\ W_c &= 1.005(528.66 - 288) = 241.87\ \text{kJ/kg} \end{aligned}$$

Turbine

$$\begin{aligned} T_{4s} &= \frac{883}{6^{0.2498}} = \frac{883}{1.5646} = 564.38\ \text{K} \\ T_4 &= 883 - 0.82(883 - 564.38) = 621.73\ \text{K} \\ W_t &= 1.11(883 - 621.73) = 290.01\ \text{kJ/kg} \end{aligned}$$

Power output

$$\begin{aligned} W_{net} &= 290.01 - 241.87 = 48.14\ \text{kJ/kg} \\ P &= \dot m W_{net} = 16 \times 48.14 = 770.3\ \text{kW} \end{aligned}$$

Thermal efficiency

$$\begin{aligned} Q_s &= c_p (T_3 - T_2) = 1.11(883 - 528.66) = 393.31\ \text{kJ/kg} \\ \eta_{th} &= \frac{48.14}{393.31} = 0.1224 = 12.24\% \end{aligned}$$

Answer: Generator power output ≈ 770.3 kW (assuming no mechanical/generator losses); thermal efficiency = 12.24 %.

Assumptions: no gas properties are given, so air-standard values $c_p = 1.005$ kJ/kg K and $\gamma = 1.4$ are used throughout ($\frac{\gamma-1}{\gamma} = 0.2857$, $5^{0.2857} = 1.5838$). Fuel mass is neglected. The mechanical efficiency 0.97 applies to the net shaft power delivered to the generator.

Data: $r_p = 5$, $T_3 = 550 + 273 = 823$ K, $T_1 = 20 + 273 = 293$ K, $\eta_c = 0.80$, $\eta_t = 0.83$, $\dot m = 15$ kg/s, $\eta_{mech} = 0.97$.

Compressor

$$\begin{aligned} T_{2s} &= 293 \times 1.5838 = 464.06\ \text{K} \\ T_2 &= 293 + \frac{464.06 - 293}{0.80} = 506.82\ \text{K} \\ W_c &= 1.005(506.82 - 293) = 214.89\ \text{kJ/kg} \end{aligned}$$

Turbine

$$\begin{aligned} T_{4s} &= \frac{823}{1.5838} = 519.63\ \text{K} \\ T_4 &= 823 - 0.83(823 - 519.63) = 571.20\ \text{K} \\ W_t &= 1.005(823 - 571.20) = 253.06\ \text{kJ/kg} \end{aligned}$$

Power output

$$\begin{aligned} W_{net} &= 253.06 - 214.89 = 38.16\ \text{kJ/kg} \\ P &= \eta_{mech}\,\dot m\,W_{net} = 0.97 \times 15 \times 38.16 = 555.3\ \text{kW} \end{aligned}$$

Thermal efficiency

$$\begin{aligned} Q_s &= c_p (T_3 - T_2) = 1.005(823 - 506.82) = 317.76\ \text{kJ/kg} \\ \eta_{th} &= \frac{W_{net}}{Q_s} = \frac{38.16}{317.76} = 0.1201 = 12.01\% \end{aligned}$$

(Overall, including the drive loss: $0.97 \times 12.01 = 11.65\%$.)

Work ratio

$$\text{Work ratio} = \frac{W_{net}}{W_t} = \frac{38.16}{253.06} = 0.151$$

Answer: Power output ≈ 555 kW, thermal efficiency ≈ 12.0 %, work ratio ≈ 0.151.

(If gas properties $c_p = 1.15$ kJ/kg K, $\gamma = 1.333$ are used for the expansion, as in some textbooks, the results become about 657 kW, 12.4 % and 0.174.)

Data: ideal (isentropic) compressor and turbine, $p_1 = 100$ kPa, $T_1 = 300$ K, $r_p = 10$, $T_3 = 1400$ K, $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$; $r_p^{0.2857} = 10^{0.2857} = 1.9307$.

Simple ideal Brayton cycle

$$\begin{aligned} T_2 &= T_1 r_p^{0.2857} = 300 \times 1.9307 = 579.21\ \text{K} \\ T_4 &= \frac{T_3}{r_p^{0.2857}} = \frac{1400}{1.9307} = 725.13\ \text{K} \\ W_c &= 1.005(579.21 - 300) = 280.61\ \text{kJ/kg} \\ W_t &= 1.005(1400 - 725.13) = 678.25\ \text{kJ/kg} \\ W_{net} &= 678.25 - 280.61 = 397.64\ \text{kJ/kg} \\ Q_s &= 1.005(1400 - 579.21) = 824.89\ \text{kJ/kg} \\ \eta &= \frac{397.64}{824.89} = 0.4821 = 48.21\% \end{aligned}$$

Check: $\eta = 1 - \dfrac{1}{r_p^{(\gamma-1)/\gamma}} = 1 - \dfrac{1}{1.9307} = 48.21\%$.

With regenerator ($\varepsilon = 0.8$)

$T_4 = 725.13$ K $> T_2 = 579.21$ K, so regeneration is useful.

$$\begin{aligned} T_x &= T_2 + \varepsilon(T_4 - T_2) = 579.21 + 0.8(145.92) = 695.94\ \text{K} \\ Q_s' &= 1.005(1400 - 695.94) = 707.58\ \text{kJ/kg} \\ \eta' &= \frac{W_{net}}{Q_s'} = \frac{397.64}{707.58} = 0.5620 = 56.20\% \end{aligned}$$

Net work is unchanged; only the fuel heat is reduced.

Increment in efficiency

$$\Delta\eta = 56.20 - 48.21 = 7.99\ \text{percentage points}$$

(a relative increase of $7.99/48.21 = 16.6\%$).

Answer: Efficiency without regenerator = 48.21 %; with 80 % regenerator = 56.20 %; increase ≈ 7.99 % (points).

The Brayton (Joule) cycle is the ideal cycle for gas turbine plants. It has four processes: isentropic compression (1–2), constant-pressure heat addition (2–3), isentropic expansion (3–4) and constant-pressure heat rejection (4–1). Its ideal efficiency is $\eta = 1 - \dfrac{1}{r_p^{(\gamma-1)/\gamma}}$.

Open Brayton cycle

 Air in        Fuel
  1 --> [C] -2-> [CC] -3-> [T] -4-> Exhaust
         \==================/ ==> Gen

• Atmospheric air is drawn in, compressed, mixed with fuel and burnt directly in the combustion chamber; products expand in the turbine and are exhausted to the atmosphere.
• Heat rejection 4–1 takes place in the atmosphere, so a fresh charge is used every cycle.
• Advantages: simple, light, compact, low cost, quick start; no cooler needed. Used in aircraft engines and most power-station gas turbines.
• Disadvantages: working fluid is combustion gas, so turbine blades suffer deposits, erosion and corrosion; only clean fuels (gas, light oil) can be used; output depends on atmospheric conditions; filters needed.

Closed Brayton cycle

       Heater (external heat)
  +--> [Heat exch.] -3-> [T] -4-+
  |2                            |
 [C] <-1- [Cooler] <------------+
  \=======shaft=======/ ==> Gen

• The same working fluid (air, helium, CO₂, nitrogen) circulates continuously. Heat is supplied externally in a heat exchanger (by burning any fuel, or from a nuclear reactor) and rejected in a pre-cooler before the compressor.
• Advantages: clean working fluid, so no blade fouling or erosion; any fuel (coal, nuclear) can be used; higher pressure level gives smaller components and better part-load efficiency (output controlled by changing pressure level); gases with better properties (helium) can be used.
• Disadvantages: needs large heat exchangers and cooler, so heavier, costlier and slower to start; requires cooling water; leakage of working fluid must be prevented.

Data: $r_p = 6$, $T_3 = 610 + 273 = 883$ K, $\eta_c = 0.80$, $\eta_t = 0.82$, $T_1 = 15 + 273 = 288$ K, $\dot m = 16$ kg/s. Compression: $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$ ($\frac{\gamma-1}{\gamma} = 0.2857$). Expansion: $c_p = 1.11$ kJ/kg K, $\gamma = 1.333$ ($\frac{\gamma-1}{\gamma} = 0.2498$). Fuel mass and mechanical losses neglected.

 1        2          3        4
 --> [C] ---> [CC] ---> [T] ---> exhaust
      \================/  ==> Generator

Compressor work

$$\begin{aligned} T_{2s} &= T_1 r_p^{0.2857} = 288 \times 1.6685 = 480.53\ \text{K} \\ T_2 &= T_1 + \frac{T_{2s} - T_1}{\eta_c} = 288 + \frac{192.53}{0.80} = 528.66\ \text{K} \\ W_c &= 1.005(528.66 - 288) = 241.87\ \text{kJ/kg} \end{aligned}$$

Turbine work

$$\begin{aligned} T_{4s} &= \frac{T_3}{r_p^{0.2498}} = \frac{883}{1.5646} = 564.38\ \text{K} \\ T_4 &= T_3 - \eta_t (T_3 - T_{4s}) = 883 - 0.82 \times 318.62 = 621.73\ \text{K} \\ W_t &= 1.11(883 - 621.73) = 290.01\ \text{kJ/kg} \end{aligned}$$

Power output

$$\begin{aligned} W_{net} &= W_t - W_c = 290.01 - 241.87 = 48.14\ \text{kJ/kg} \\ P &= \dot m W_{net} = 16 \times 48.14 = 770.3\ \text{kW} \end{aligned}$$

(Turbine power = 4640.2 kW, of which the compressor absorbs 3869.9 kW.)

Answer: Power output of the generator ≈ 770 kW.

Assumptions: calorific value of the heavy oil is not given; a typical value CV = 42 000 kJ/kg is assumed. Heat released by the fuel equals the heat received by the gases in the combustion chamber.

Data: $p_2 = p_3 = 500$ kPa, $T_3 = 650 + 273 = 923$ K, $p_1 = p_4 = 100$ kPa, $T_1 = 27 + 273 = 300$ K, $\eta_c = 0.80$, $\eta_t = 0.85$, $c_p = 1$ kJ/kg K, $\gamma = 1.4$; $r_p = 5$, $5^{0.2857} = 1.5838$.

Compressor

$$\begin{aligned} T_{2s} &= 300 \times 1.5838 = 475.15\ \text{K} \\ T_2 &= 300 + \frac{475.15 - 300}{0.80} = 518.93\ \text{K} \\ W_c &= 1 \times (518.93 - 300) = 218.93\ \text{kJ/kg} \end{aligned}$$

Turbine

$$\begin{aligned} T_{4s} &= \frac{923}{1.5838} = 582.77\ \text{K} \\ T_4 &= 923 - 0.85(923 - 582.77) = 633.80\ \text{K} \\ W_t &= 1 \times (923 - 633.80) = 289.20\ \text{kJ/kg} \end{aligned}$$

Thermal efficiency

$$\begin{aligned} W_{net} &= 289.20 - 218.93 = 70.26\ \text{kJ/kg} \\ Q_s &= c_p (T_3 - T_2) = 923 - 518.93 = 404.07\ \text{kJ/kg} \\ \eta_{th} &= \frac{70.26}{404.07} = 0.1739 = 17.39\% \end{aligned}$$

Power generating capacity

Heat supplied by fuel:

$$\dot Q = \dot m_f \times CV = 5 \times 42\,000 = 210\,000\ \text{kW}$$ $$P = \eta_{th} \times \dot Q = 0.1739 \times 210\,000 = 36\,518\ \text{kW} \approx 36.5\ \text{MW}$$

Corresponding air–fuel ratio (from $\dot m_f CV = (\dot m_a + \dot m_f)c_p(T_3 - T_2)$): $\dfrac{\dot m_a}{\dot m_f} = \dfrac{42\,000 - 404.07}{404.07} = 102.9$, i.e. about 515 kg/s of air.

Answer: Thermal efficiency = 17.39 %; power capacity ≈ 36.5 MW (for CV = 42 MJ/kg; in general $P = 0.1739 \times 5 \times CV$).

In a reheat gas turbine plant, expansion is done in two turbines (HP and LP). After partial expansion in the HP turbine, the gas is heated again in a reheat combustion chamber to nearly the maximum temperature, and then expanded in the LP turbine. Because gas turbine exhaust contains a lot of excess air (A/F ratio 60–100), extra fuel can be burnt directly in the reheater.

Sketch

       Fuel              Fuel
        |                 |
 1    2 v   3        4    v   5         6
 -->[C]-->[CC1]-->[HPT]-->[CC2]-->[LPT]--> exh
     ||             ||              ||
     ++=====shaft===++======shaft===++==> Gen

T–s diagram

 T
 |          3       5
 |         *       *
 |        /|      /|
 |       / |     / |
 |      /  *----'  * 6
 |   2 *   4       |  (6' without
 |     |          /    reheat is
 |     *--------'      lower)
 |     1
 +---------------------- s

1–2 compression, 2–3 combustion, 3–4 HP expansion, 4–5 reheating at intermediate pressure, 5–6 LP expansion, 6–1 exhaust.

Working and effect

• Turbine work per kg is proportional to inlet temperature. Raising the gas from $T_4$ back to $T_5 \approx T_3$ makes the LP turbine produce more work than a single turbine would from 4 onward. Total turbine work $W_t = c_p(T_3 - T_4) + c_p(T_5 - T_6)$ increases.
• Compressor work is unchanged, so net work output and work ratio increase. Maximum work occurs when the pressure ratios of the two turbines are equal: $p_4 = \sqrt{p_3 p_6}$.
• Heat supplied also increases and exhaust temperature $T_6$ is higher, so reheat alone gives little or no gain in efficiency; with a regenerator, the hot exhaust is used to preheat the air and efficiency also rises.

Applications / advantages of reheat over the simple open cycle

• Where high specific output is needed from a compact, light plant: peak-load and standby power stations, mobile/emergency sets.
• Aircraft jet engines (afterburning is a form of reheat to increase thrust for take-off and combat) and marine propulsion.
• Combined-cycle plants: the hotter exhaust from a reheat turbine gives more heat to the HRSG, increasing steam-cycle output (e.g. sequential-combustion turbines).
• Plants combined with regeneration (and intercooling) for high efficiency in base-load service.
• Reduces the size of plant (air flow) for a given output, so lower capital cost per kW.

Assumptions: the inlet air temperature is not given; it is taken as 32 °C (305 K), the value used in the standard textbook version of this problem. Air-standard values $c_p = 1.005$ kJ/kg K and $\gamma = 1.4$ are used throughout; fuel mass neglected.

Data: $p_1 = 1$ bar, $p_2 = 6.9$ bar ($r_p = 6.9$), $T_3 = 816 + 273 = 1089$ K, $\eta_c = 0.84$, $\eta_t = 0.85$, $\varepsilon = 0.60$; $6.9^{0.2857} = 1.7365$.

       +------ Regenerator ------+
       |  x                 y    |
 1     | 2                  4    |
 --> [C] --> [CC] --3--> [T] ----+

Compressor

$$\begin{aligned} T_{2s} &= 305 \times 1.7365 = 529.63\ \text{K} \\ T_2 &= 305 + \frac{529.63 - 305}{0.84} = 572.41\ \text{K} \\ W_c &= 1.005(572.41 - 305) = 268.75\ \text{kJ/kg} \end{aligned}$$

Turbine

$$\begin{aligned} T_{4s} &= \frac{1089}{1.7365} = 627.13\ \text{K} \\ T_4 &= 1089 - 0.85(1089 - 627.13) = 696.41\ \text{K} \\ W_t &= 1.005(1089 - 696.41) = 394.55\ \text{kJ/kg} \end{aligned}$$

Regenerator

$$T_x = T_2 + \varepsilon(T_4 - T_2) = 572.41 + 0.6(696.41 - 572.41) = 646.81\ \text{K}$$

(a) Thermal efficiency

$$\begin{aligned} W_{net} &= 394.55 - 268.75 = 125.80\ \text{kJ/kg} \\ Q_s &= 1.005(1089 - 646.81) = 444.40\ \text{kJ/kg} \\ \eta_{th} &= \frac{125.80}{444.40} = 0.2831 = 28.31\% \end{aligned}$$

(b) Air rate

Air needed per kWh of net output:

$$\text{Air rate} = \frac{3600}{W_{net}} = \frac{3600}{125.80} = 28.62\ \text{kg/kWh}$$

(c) Work ratio

$$\text{Work ratio} = \frac{W_{net}}{W_t} = \frac{125.80}{394.55} = 0.319$$

Answer: $\eta$ = 28.31 %, air rate = 28.62 kg/kWh, work ratio = 0.319.

(If the inlet air is taken as 15 °C = 288 K instead, the same method gives $\eta$ = 30.79 %, air rate = 25.57 kg/kWh and work ratio = 0.357.)

Common methods to improve gas turbine performance

1. Regeneration – exhaust heat preheats compressed air.
2. Intercooling – multi-stage compression with cooling between stages.
3. Reheating – multi-stage expansion with reheating between turbines.
4. Combination of regeneration, intercooling and reheating.
5. Higher turbine inlet temperature (better materials, blade cooling).
6. Optimum pressure ratio and higher compressor/turbine efficiencies; low pressure losses.
7. Combined cycle / cogeneration using exhaust heat.
8. Water/steam injection and inlet air cooling.

How regeneration increases efficiency

In a simple open cycle the exhaust leaves the turbine at 400–550 °C, often hotter than the air leaving the compressor. A regenerator (counter-flow heat exchanger) passes this hot exhaust over tubes carrying the compressed air.

          +------- REGENERATOR -------+
          | air 2 --> x   y <-- gas 4 |
          +--^--------|---|--------^--+
             |        v   v        |
 Air 1 -->[ C ]    [ CC ]-3->[ T ]-+
            ||       ^ fuel  ||
            ++=====shaft=====++==> Gen
                   exhaust from y

1. Air leaves the compressor at $T_2$ and is heated in the regenerator to $T_x$ by the exhaust gas.
2. The combustion chamber only has to raise the air from $T_x$ to $T_3$, so heat supplied falls from $c_p(T_3 - T_2)$ to $c_p(T_3 - T_x)$.
3. Compressor and turbine works are not changed, so net work stays the same while fuel input decreases:

$$\eta = \frac{W_t - W_c}{c_p(T_3 - T_x)} > \frac{W_t - W_c}{c_p(T_3 - T_2)}$$

4. The exhaust leaves at the lower temperature $T_y$, so less heat is wasted.
5. Effectiveness $\varepsilon = \dfrac{T_x - T_2}{T_4 - T_2}$, practically 0.6–0.8. With an ideal regenerator ($\varepsilon = 1$), $\eta = 1 - \dfrac{T_1}{T_3}r_p^{(\gamma-1)/\gamma}$, which increases as pressure ratio decreases.

Limitation: regeneration works only when $T_4 > T_2$ (low/moderate pressure ratios). It adds cost, weight and pressure losses, so it is mainly used in stationary base-load plants.

Common methods to improve gas turbine performance

1. Reheating – expansion in stages with reheating between turbines.
2. Intercooling – compression in stages with cooling between stages.
3. Regeneration – exhaust gas preheats compressed air.
4. Combination of the above (approaches the Ericsson cycle).
5. Higher turbine inlet temperature using better materials and blade cooling.
6. Optimum pressure ratio, high component efficiencies, low pressure losses.
7. Combined cycle / cogeneration with waste-heat recovery.
8. Water/steam injection, inlet air cooling.

How reheating increases net work output

      fuel            fuel
       v               v
 1->[C]-2->[CC1]-3->[HPT]-4->[CC2]-5->[LPT]-6-> exh
     ||              ||               ||
     ++=====shaft====++=====shaft=====++==> Gen

 T
 |          3       5
 |         *       *
 |        /|      /|
 |       / |     / |
 |      /  *----'  * 6
 |   2 *   4       |
 |     |          /
 |     *--------'
 |     1
 +---------------------- s

1. Gas from the combustion chamber at $T_3$ expands in the HP turbine only to an intermediate pressure, temperature falling to $T_4$.
2. It enters a reheater (second combustion chamber) where more fuel is burnt in the excess air, raising its temperature back to $T_5 \approx T_3$.
3. It then expands in the LP turbine to atmospheric pressure ($T_6$).
4. Total turbine work:

$$W_t = c_p(T_3 - T_4) + c_p(T_5 - T_6)$$

On the T–s diagram, constant-pressure lines diverge as entropy increases, so the temperature drop $T_5 - T_6$ is larger than the drop the gas would have had from $T_4$ without reheat. Hence turbine work increases, while compressor work is unchanged, so net work output and work ratio increase. 5. Maximum work is obtained when both turbines have equal pressure ratios: $p_4 = \sqrt{p_3 p_6}$, with $T_5 = T_3$.

Effect on efficiency: extra fuel is burnt and the exhaust is hotter, so reheat alone may slightly lower efficiency; combined with regeneration, the hot exhaust preheats the air and both output and efficiency improve. Reheating is therefore used where high specific output and a compact plant are needed.

Assumptions: air-standard ideal cycle; compression in one stage (100 to 1000 kPa) and two-stage expansion are isentropic; no pressure losses; $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$, so $\frac{\gamma-1}{\gamma} = 0.2857$.

 1 ->[ C ]-2->[ Regen ]-x->[ CC ]-3->[ T1 ]-4
                 ^   |
                 |   v y (to exhaust)
 4 ->[ Reheater ]-5->[ T2 ]-6 -> hot side of Regen

States: 1 (100 kPa, 300 K) - compressor - 2 (1000 kPa) - regenerator - x - combustor - 3 (1400 K) - HP turbine - 4 (300 kPa) - reheater - 5 (1400 K) - LP turbine - 6 (100 kPa) - regenerator - y (exhaust).

1. Compressor exit temperature

$$\begin{aligned} T_2 &= T_1 \left(\frac{p_2}{p_1}\right)^{0.2857} = 300 \times 10^{0.2857} \\ &= 300 \times 1.9307 = 579.21\ \text{K} \end{aligned}$$

2. Turbine exit temperatures

First stage, 1000 kPa to 300 kPa:

$$T_4 = \frac{T_3}{(1000/300)^{0.2857}} = \frac{1400}{1.4106} = 992.51\ \text{K}$$

Second stage, 300 kPa to 100 kPa, after reheat to $T_5 = 1400$ K:

$$T_6 = \frac{T_5}{(300/100)^{0.2857}} = \frac{1400}{1.3687} = 1022.84\ \text{K}$$

3. Regenerator

Effectiveness $\varepsilon = \dfrac{T_x - T_2}{T_6 - T_2}$, so

$$\begin{aligned} T_x &= T_2 + 0.9(T_6 - T_2) \\ &= 579.21 + 0.9(1022.84 - 579.21) = 978.48\ \text{K} \end{aligned}$$

4. Work and heat (per kg of air)

$$\begin{aligned} W_C &= c_p (T_2 - T_1) = 1.005(579.21 - 300) = 280.61\ \text{kJ/kg} \\ W_{T1} &= c_p (T_3 - T_4) = 1.005(1400 - 992.51) = 409.53\ \text{kJ/kg} \\ W_{T2} &= c_p (T_5 - T_6) = 1.005(1400 - 1022.84) = 379.05\ \text{kJ/kg} \\ W_{net} &= 409.53 + 379.05 - 280.61 = 507.97\ \text{kJ/kg} \end{aligned}$$

Heat supplied in the combustor (x to 3) and reheater (4 to 5):

$$\begin{aligned} Q_s &= c_p (T_3 - T_x) + c_p (T_5 - T_4) \\ &= 1.005(1400 - 978.48) + 1.005(1400 - 992.51) \\ &= 423.63 + 409.53 = 833.16\ \text{kJ/kg} \end{aligned}$$

5. Thermal efficiency

$$\eta_{th} = \frac{W_{net}}{Q_s} = \frac{507.97}{833.16} = 0.6097$$

Answer: thermal efficiency $\eta_{th} \approx 61.0\%$ (0.6097).

The high value comes from the 90% regenerator: the hot turbine exhaust (1022.84 K) preheats the compressed air from 579.21 K to 978.48 K, so the fuel only has to supply the remaining heat. Reheat raises the work output, and the regenerator recovers most of the extra exhaust heat that reheat produces.

Assumptions: air-standard analysis with constant $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$, $\frac{\gamma-1}{\gamma} = 0.2857$; no pressure drops in intercooler, reheater, regenerator or combustor; steady flow; kinetic and potential energy changes neglected.

        LPC        HPC            regen          CC
 1 ->[ C1 ]-2->[IC]-3->[ C2 ]-4->[ R ]-5->[ CC ]-6
                                  ^ |
                         exhaust  | v 10 (to stack)
 6 ->[ T1 ]-7->[ RH ]-8->[ T2 ]-9-+

Pressures: 100 kPa (1), 300 kPa (2, 3, 7, 8), 1000 kPa (4, 5, 6), 100 kPa (9). Temperatures given: $T_1 = T_3 = 300$ K, $T_6 = T_8 = 1400$ K.

Compressor stages ($\eta_c = 0.8$)

Stage 1 (pressure ratio 3):

$$\begin{aligned} T_{2s} &= 300 \times 3^{0.2857} = 410.62\ \text{K} \\ T_2 &= T_1 + \frac{T_{2s} - T_1}{\eta_c} = 300 + \frac{110.62}{0.8} = 438.28\ \text{K} \end{aligned}$$

Stage 2 (pressure ratio 1000/300 = 3.333):

$$\begin{aligned} T_{4s} &= 300 \times 3.333^{0.2857} = 423.17\ \text{K} \\ T_4 &= 300 + \frac{123.17}{0.8} = 453.96\ \text{K} \end{aligned}$$

Turbine stages ($\eta_t = 0.8$)

Stage 1 (1000 to 300 kPa):

$$\begin{aligned} T_{7s} &= \frac{1400}{3.333^{0.2857}} = 992.51\ \text{K} \\ T_7 &= 1400 - 0.8(1400 - 992.51) = 1074.01\ \text{K} \end{aligned}$$

Stage 2 (300 to 100 kPa):

$$\begin{aligned} T_{9s} &= \frac{1400}{3^{0.2857}} = 1022.84\ \text{K} \\ T_9 &= 1400 - 0.8(1400 - 1022.84) = 1098.27\ \text{K} \end{aligned}$$

Regenerator ($\varepsilon = 0.8$)

$$\begin{aligned} T_5 &= T_4 + \varepsilon (T_9 - T_4) \\ &= 453.96 + 0.8(1098.27 - 453.96) = 969.41\ \text{K} \end{aligned}$$

Work and heat per kg

$$\begin{aligned} W_C &= c_p[(T_2 - T_1) + (T_4 - T_3)] \\ &= 1.005(138.28 + 153.96) = 293.70\ \text{kJ/kg} \\ W_T &= c_p[(T_6 - T_7) + (T_8 - T_9)] \\ &= 1.005(325.99 + 301.73) = 630.86\ \text{kJ/kg} \\ W_{net} &= 630.86 - 293.70 = 337.16\ \text{kJ/kg} \\ Q_s &= c_p[(T_6 - T_5) + (T_8 - T_7)] \\ &= 1.005(430.59 + 325.99) = 760.36\ \text{kJ/kg} \end{aligned}$$

(a) Thermal efficiency

$$\eta_{th} = \frac{W_{net}}{Q_s} = \frac{337.16}{760.36} = 0.4434$$

Answer (a): $\eta_{th} \approx 44.3\%$

(b) Back work ratio

$$\text{BWR} = \frac{W_C}{W_T} = \frac{293.70}{630.86} = 0.4656$$

Answer (b): back work ratio $\approx 0.466$ (46.6%)

(c) Net power

$$P = \dot m\, W_{net} = 5.807 \times 337.16 = 1957.9\ \text{kW}$$

Answer (c): net power $\approx 1958$ kW (about 1.96 MW)

Note: with 80% component efficiencies almost half of the turbine work is used by the compressor, which is why intercooling (less compressor work) and regeneration (less fuel) are both needed to get a reasonable efficiency.