Question

In an airbreathing gas turbine engine, the combustor inlet temperature is \(600\ \text{K}\). The heating value of the fuel is \(43.4\times 10^{6}\ \text{J/kg}\). Assume\(c_p = 1100\ \mathrm{J\,kg^{-1}\,K^{-1}}\) for air and burned gases, burner efficiency \(=100\%\), and fuel-air ratio \(f\ll 1\). Neglect kinetic energy at inlet/exit. What fuel-air ratio is required to achieve \(1300\ \text{K}\) at the combustor exit?}

• A. 0.0177
• B. 0.0215
• C. 0.0127
• D. 0.0277

Hint:

For combustors, a handy formula is \(f=\dfrac{c_p(T_3-T_2)}{\text{LHV}-c_pT_3}\) (with \(100\%\) efficiency and \(f\ll1\)). The \(c_pT_3\) term accounts for heating the added fuel+air to \(T_3\).

Answer 1

The Correct Option is A

Step 1: Energy balance per kg of air (standard burner equation).
With \(f\) kg of fuel per kg air, and \(c_p\) same for reactants/products, \[ f\,\text{LHV} + c_p\,T_2 = (1+f)c_p\,T_3. \] Rearrange: \[ f(\text{LHV}-c_pT_3)=c_p\,(T_3-T_2)\;\Rightarrow\; f=\frac{c_p\,(T_3-T_2)}{\text{LHV}-c_pT_3}. \] Step 2: Substitute numbers.
\(T_2=600\ \text{K},\; T_3=1300\ \text{K}\Rightarrow \Delta T=700\ \text{K}\).
\[ \text{Numerator}=c_p\Delta T=1100\times 700=7.70\times 10^{5}\ \text{J/kg}. \] \[ \text{Denominator}=\text{LHV}-c_pT_3 =43.4\times10^{6}-1100\times1300 =4.197\times10^{7}\ \text{J/kg}. \] \[ f=\frac{7.70\times10^{5}}{4.197\times10^{7}}\approx 1.83\times10^{-2}\approx \boxed{0.0177}. \]