## Chapter 4

## Q. 4.8

In an afterburning jet engine, the gases enter the afterburner at a temperature T_0 = 900 K whether the afterburner is operating or not. Then the gases’ temperature rises to 2000 K when the burner is operating and the stagnation pressure is about 5% less than its value without burning.

If the average specific heat in the nozzle rises from 1.1 to 1.315 kJ/kg · K because of burning, how much relative change in exhaust nozzle area is required as the burner is ignited? It may be assumed that the nozzle is choked in both cases and that the mass flow is unchanged.

## Step-by-Step

## Verified Solution

Since R = 287 J/kg · K, Cp=\frac{\gamma R}{\gamma -1} ,then

1. For (C_{\text{p}})_{\text{inoperative ab}} = 1100 \text{ J/kg} \cdot \text{K,then }\gamma_1=1.353

2. For (C_{\text{p}})_{\text{operative ab}} = 1315 \text{ J/kg} \cdot \text{K,then }\gamma_2=1.279

The process is plotted in Figure 4.22 for operative and inoperative afterburner.

Since the mass flow rate \dot{m}=P \times A \times M \sqrt{ \gamma/RT} , assuming no losses in the nozzle, then the mass flow for a choked nozzle is

\dot{m}_{\text{choked}}=P^{\ast} \times A^{\ast}\sqrt{\frac{\gamma}{RT^\ast} }

This equation can be expressed in terms of the total conditions between states (6) and (7) as

\dot{m}_{\text{choked}}=\frac{A^\ast P_{06}}{\sqrt{RT_{06}}} \sqrt{\gamma}\left(\frac{2}{\gamma+1} \right) ^{(\gamma+1)/2(\gamma-1)}

where A^\ast is the throat area, which is here the outlet area of the convergent nozzle.

Since in both cases of operative and inoperative afterburner, the nozzle is choked and the mass flow is constant

\therefore \frac{A^\ast_1 P_0}{\sqrt{R \times 900}} \sqrt{\gamma_1}\left(\frac{2}{\gamma_1+1} \right) ^{(\gamma_1+1)/2(\gamma_1-1)}=\frac{A^\ast_2 \times 0.95 P_0}{\sqrt{R \times 2000}} \sqrt{\gamma_2}\left(\frac{2}{\gamma_2+1} \right) ^{(\gamma_2+1)/2(\gamma_2-1)} \\ \therefore \frac{A^\ast_2}{A^\ast_1} =\sqrt{\frac{2000}{900} }\frac{1}{0.95} \sqrt{\frac{1.353}{1.279} }\frac{(0.85)^{3.3329}}{(0.8776)^{4.084}}

The area ratio is

A^\ast_2 / A^\ast_1=1.4907 \times 1.0526 \times 1.0285 \times \frac{0.5817}{0.5867} =1.6.