Fuel-air ratio in a gas turbine combustor

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.