Gas Turbine

Schematic of the open-cycle gas turbine

          fuel
            |
   +-----+  v  +-----------+  +-------+
 air-->| C  |--->| Combustor |->|   T   |--> exhaust
   +--+--+     +-----------+  +---+---+
      |                            |
      +----------- shaft ----------+--> load

Functions

1. Compressor (C): axial or centrifugal; raises the pressure of the air from atmosphere to $p_2$ (pressure ratio 6 to 30). It absorbs 50 to 60% of the turbine work.
2. Combustion chamber: fuel is burnt at nearly constant pressure with only part of the air (primary air); the rest dilutes and cools the gases to the turbine limit (about 1100 to 1500 K).
3. Turbine (T): gases expand to atmosphere and develop work. Part drives the compressor; the remainder is the net output.
4. Starting motor, fuel pump, lubricating system and auxiliaries.

Open and closed cycle

Open cycle is far more common because of its simplicity, light weight and quick start.

Ideal Brayton cycle

Processes: 1-2 isentropic compression; 2-3 constant-pressure heat addition; 3-4 isentropic expansion; 4-1 constant-pressure heat rejection.

 T-s diagram              p-v diagram
 T                        p
 |        3               |  2--------3
 |       /                |  |         \
 |  2   /  p2             |  |          \
 | /   /                  |  |           4
 |/   4                   |  1-----------
 1---/   p1               +--------------- v
 +------------ s

For a perfect gas with constant $c_p$, and pressure ratio $r_p = p_2/p_1 = p_3/p_4$:

$$q_{in} = c_p (T_3 - T_2), \qquad q_{out} = c_p (T_4 - T_1)$$ $$\eta = 1 - \frac{q_{out}}{q_{in}} = 1 - \frac{T_4 - T_1}{T_3 - T_2}$$

From the isentropic relations, $\dfrac{T_2}{T_1} = \dfrac{T_3}{T_4} = r_p^{(\gamma-1)/\gamma}$, hence $\dfrac{T_4}{T_1} = \dfrac{T_3}{T_2}$ and $\dfrac{T_4 - T_1}{T_3 - T_2} = \dfrac{T_1}{T_2}$:

$$\boxed{\eta = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}}}$$

The efficiency depends only on the pressure ratio and rises with it.

Pressure ratio for maximum net work

Let $x = r_p^{(\gamma-1)/\gamma}$, $T_2 = T_1 x$, $T_4 = T_3/x$:

$$w_{net} = c_p (T_3 - T_4) - c_p (T_2 - T_1) = c_p \left[T_3\left(1 - \frac1x\right) - T_1(x - 1)\right]$$ $$\frac{dw_{net}}{dx} = c_p\left[\frac{T_3}{x^2} - T_1\right] = 0 \;\Rightarrow\; x^2 = \frac{T_3}{T_1}$$ $$\boxed{r_{p,opt} = \left(\frac{T_3}{T_1}\right)^{\gamma/[2(\gamma-1)]}}$$

At this ratio $T_2 = T_4 = \sqrt{T_1 T_3}$ and the maximum work is

$$w_{net,max} = c_p\left(\sqrt{T_3} - \sqrt{T_1}\right)^2$$

The pressure ratio for maximum efficiency is higher than that for maximum work, but net work then falls to zero at $x = T_3/T_1$. Practical gas turbines are designed near the maximum work ratio.

$(\gamma-1)/\gamma = 0.2857$ and $r_p^{0.2857} = 8^{0.2857} = 1.8114$.

(a) Temperatures

$$T_2 = T_1 r_p^{0.2857} = 300 \times 1.8114 = 543.4\ \text{K}, \qquad T_4 = \frac{T_3}{r_p^{0.2857}} = \frac{1200}{1.8114} = 662.5\ \text{K}$$

(b) Compressor and turbine work

$$w_c = c_p (T_2 - T_1) = 1.005 \times 243.4 = 244.7\ \text{kJ/kg}$$ $$w_t = c_p (T_3 - T_4) = 1.005 \times 537.5 = 540.2\ \text{kJ/kg}$$

(c) Net work and power

$$w_{net} = w_t - w_c = 540.2 - 244.7 = 295.6\ \text{kJ/kg}, \qquad P = 5 \times 295.6 = 1478\ \text{kW}$$

(d) Thermal efficiency

$$q_{in} = c_p (T_3 - T_2) = 1.005 \times 656.6 = 659.8\ \text{kJ/kg}, \qquad \eta = \frac{w_{net}}{q_{in}} = 0.448$$

Check: $1 - 1/8^{0.2857} = 0.448$.

(e) Back work ratio

$$\text{BWR} = \frac{w_c}{w_t} = \frac{244.7}{540.2} = 0.453$$

Answer: $T_2 = 543.4$ K, $T_4 = 662.5$ K; $w_c = 244.7$, $w_t = 540.2$, $w_{net} = 295.6$ kJ/kg; $P = 1478$ kW; $\eta = 44.8\%$; BWR = 0.453.

Data: $T_1 = 288$ K, $r_p = 6$, $\eta_c = 0.80$, $\eta_t = 0.85$, $T_3 = 1073$ K, $\dot m = 15$ kg/s.

(a) Compressor

$$T_{2s} = 288 \times 6^{0.2857} = 480.5\ \text{K}$$ $$w_c = \frac{c_p (T_{2s} - T_1)}{\eta_c} = \frac{1.005 \times 192.5}{0.80} = 241.9\ \text{kJ/kg}, \qquad T_2 = 288 + \frac{241.9}{1.005} = 528.7\ \text{K}$$

(b) Turbine

Pressure at turbine inlet $p_3 = 0.95 \times 600 = 570$ kPa, so the expansion ratio is $570/100 = 5.70$.

$$\frac{\gamma - 1}{\gamma} = 0.2498, \quad T_{4s} = \frac{T_3}{r_t^{0.2498}} = \frac{1073}{1.5446} = 694.7\ \text{K}$$ $$w_t = \eta_t c_p (T_3 - T_{4s}) = 0.85 \times 1.148 \times 378.3 = 369.2\ \text{kJ/kg}$$ $$T_4 = T_3 - \frac{w_t}{c_p} = 1073 - \frac{369.2}{1.148} = 751.4\ \text{K}$$

(c) Net work

$$w_{net} = w_t - w_c = 369.2 - 241.9 = 127.3\ \text{kJ/kg}$$

(d) Thermal efficiency

$$q_{in} = c_{p,g} (T_3 - T_2) = 1.148 \times (1073 - 528.7) = 624.9\ \text{kJ/kg}$$ $$\eta_{th} = \frac{w_{net}}{q_{in}} = \frac{127.3}{624.9} = 0.204$$

(e) Work ratio and (f) power

$$\text{Work ratio} = \frac{w_{net}}{w_t} = \frac{127.3}{369.2} = 0.345 \quad (\text{back work ratio} = 0.655)$$ $$P = \dot m w_{net} = 15 \times 127.3 = 1910\ \text{kW}$$

Answer: $T_2 = 528.7$ K; $T_4 = 751.4$ K; $w_{net} = 127.3$ kJ/kg; $\eta_{th} = 20.4\%$; work ratio = 0.345; $P = 1910$ kW.

Requirements

• Complete combustion with high combustion efficiency over the whole operating range.
• Low total pressure loss (typically 2 to 8% of compressor delivery pressure).
• Stable flame over a wide air-fuel ratio, with no blow-out at altitude or during rapid throttle changes.
• Uniform outlet temperature (low "pattern factor") to protect turbine blades.
• Reliable ignition at ground and altitude, small size and weight, low smoke and emissions, long life.

Zones of a combustor

 air in -> | primary | secondary | dilution | -> to turbine
           |  zone   |   zone    |   zone   |
 fuel -->  (burn at  (complete   (mix cool air,
            near      burning,    set outlet
            stoich)   cooling)    temperature)

1. Primary zone: about 15 to 20% of the air; burns fuel near stoichiometric ratio at high temperature; swirler produces recirculation that anchors the flame.
2. Secondary (intermediate) zone: more air through holes completes combustion and prevents dissociation losses.
3. Dilution zone: remaining air (about 50 to 60%) reduces the gas temperature to the turbine inlet limit and evens out the temperature profile. A portion also cools the liner wall.

Types

Definitions

• Combustion efficiency: $\eta_b = \dfrac{\text{actual heat release}}{\text{ideal heat release}} = \dfrac{(\dot m_a + \dot m_f) h_3 - \dot m_a h_2}{\dot m_f \cdot LHV}$; 98 to 99% at design.
• Pressure loss factor: $\dfrac{\Delta p_{0}}{p_{02}}$, the fall of total pressure across the chamber as a fraction of the inlet value, caused by friction in the liner and by the momentum rise of heating ("fundamental loss").

Compressor characteristic

The performance map plots the pressure ratio against the corrected mass flow $\dot m\sqrt{T_{01}}/p_{01}$ for lines of constant corrected speed $N/\sqrt{T_{01}}$, with islands of constant isentropic efficiency.

 pr |   surge line ....
    |      /\   N3 ___
    |  .../  \ /   \
    |  /  N2  \  eff.\
    | / N1     \      \
    +------------------- m*sqrt(T)/p
         choke at right end of each line

• Surge: at low flow, the blades stall, the delivery pressure collapses and the flow reverses; the cycle repeats with violent pulsations. The surge line bounds the left side of the map.
• Choking: at high flow, the Mach number reaches 1 at some section (inlet of the rotor or the throat), the mass flow cannot increase more, and the speed lines become vertical.
• Operating line must lie between surge and choke with high efficiency.

Turbine characteristic

The map of expansion ratio against $\dot m\sqrt{T_{01}}/p_{01}$ for different values of $N/\sqrt{T_{01}}$ shows that once the nozzle (or stage) is choked, the flow parameter becomes constant and independent of speed. The curves for different speeds therefore collapse at high expansion ratio. The efficiency is flat over a wide range, so turbines are less sensitive than compressors to off-design operation.

Centrifugal and axial compressors

Method: energy balance on the chamber (enthalpy datum 0 K, fuel sensible heat neglected):

$$\dot m_a c_{p,a} T_2 + \eta_b \dot m_f LHV = (\dot m_a + \dot m_f) c_{p,g} T_3$$

Fuel flow

$$\dot m_f = \dot m_a \frac{c_{p,g}T_3 - c_{p,a}T_2}{\eta_b LHV - c_{p,g}T_3}$$

Numerator: $1.148 \times 1250 - 1.005 \times 520 = 1435.0 - 522.6 = 912.4$ kJ/kg air.

Denominator: $0.98 \times 43000 - 1.148 \times 1250 = 42140 - 1435.0 = 40705$ kJ/kg fuel.

$$\dot m_f = 20 \times \frac{912.4}{40705} = 0.4483\ \text{kg/s} = 1614\ \text{kg/h}$$

Fuel-air ratio

$$f = \frac{\dot m_f}{\dot m_a} = 0.0224 \quad (\text{air-fuel ratio} = 44.6)$$

This is a very lean mixture (stoichiometric is about 0.067), which shows that most of the air is used for dilution and cooling.

Heat released

$$Q = \eta_b \dot m_f LHV = 0.98 \times 0.4483 \times 43000 = 18891\ \text{kW}$$

Outlet pressure

$$p_3 = p_2 (1 - 0.04) = 800 \times 0.96 = 768\ \text{kPa}$$

Answer: $\dot m_f = 0.4483$ kg/s (1614 kg/h); $f = 0.0224$; $Q = 18891$ kW; $p_3 = 768$ kPa.