--- title: "Fuel-air ratio in a gas turbine combustor" subject: "Turbo Machines" chapter: 3 key: "semester_7/turbo_machines:combustion-chamber-fuel-air-ratio-numerical" source: https://www.bctengineer.com/q/semester_7/turbo_machines/combustion-chamber-fuel-air-ratio-numerical site: BCT Engineering --- # Fuel-air ratio in a gas turbine combustor Turbo Machines · Chapter 3: Gas Turbine - Asked in: year not given, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/turbo_machines/combustion-chamber-fuel-air-ratio-numerical - Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-3 ## Question A combustion chamber receives 20 kg/s of air at 520 K and delivers products at 1250 K. The lower calorific value of the fuel is 43 000 kJ/kg and the combustion efficiency is 98%. Taking mean specific heats of 1.005 kJ/kg K for air and 1.148 kJ/kg K for the products and a common datum of 0 K for enthalpy, find the fuel flow rate, the fuel-air ratio and the heat released. If the pressure at inlet is 800 kPa and the pressure loss is 4%, find the outlet pressure. ## Short answer Using an energy balance on the combustion chamber, fuel mass flow = 0.4483 kg/s (1614 kg/h), fuel-air ratio = 0.0224, heat released = 18891 kW, outlet pressure = 768 kPa. ## Answer **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which energy balance is used to find the fuel flow in the combustion chamber? A. Enthalpy of air in + useful heat from fuel (efficiency x LHV) = enthalpy of products out B. Enthalpy of products out = LHV of fuel only, ignoring the air C. Enthalpy of air in = enthalpy of products out D. Heat released = pressure loss x air flow Answer: A. The balance is m_a cp_a T2 + eta_b m_f LHV = (m_a + m_f) cp_g T3, so the entering air enthalpy plus the effective fuel heat equals the products' enthalpy. (Key point: Energy balance) 2. In the fuel flow formula, what is the denominator (eta_b LHV - cp_g T3) in physical terms? A. Heat released per kg air B. Net energy per kg of fuel after the fuel's own share of products is heated to T3 C. The pressure drop per kg of fuel D. Total heat lost to the walls Answer: B. Each kg of fuel adds its effective heat 0.98 x 43000 but also adds mass that must be heated to T3, so the net is 42140 - 1435 = 40705 kJ/kg fuel. (Key point: Fuel flow formula) 3. The result f = 0.0224 is far below the stoichiometric value of about 0.067. What does this show? A. Combustion is incomplete B. The fuel is of poor quality C. Most of the air is used for dilution and cooling of the products D. The pressure loss is too high Answer: C. A very lean overall mixture means only part of the air burns; the rest dilutes and cools the gas to the turbine-limited temperature. (Key point: Fuel-air ratio meaning) 4. With inlet pressure 800 kPa and a 4% pressure loss, the outlet pressure is A. 832 kPa B. 800 kPa C. 760 kPa D. 768 kPa Answer: D. p3 = p2 (1 - 0.04) = 800 x 0.96 = 768 kPa. 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