Question 16.3: During a ground test for a turbojet engine at ambient condit......

The ambient conditions are the same as the compressor inlet conditions as the engine is during ground run.

T_{01}=288 \text{ K}, \quad P_{01}=1 \text{ bar}

At 90% of design speed

\Delta T_{012}=\frac{T_{01}}{\eta_{\text{c}}} \left[\left(\frac{P_{02}}{P_{01}} \right)^{(\gamma_{\text{c}}-1)/\gamma_{\text{c}}} -1\right] =148.5 \text{ K} \\ W_{\text{c}}=W_{\text{t}} \\ \Delta T_{034}= \Delta T_{012} \times \frac{Cp_{\text{a}}}{Cp_{\text{g}}} =130 \text{ K} \\ \left(\frac{P_{04}}{P_{03}} \right) =\left[1-\frac{\Delta T_{034}}{\eta_{\text{t}}T_{03}} \right] ^{\gamma_{\text{h}}/(\gamma_{\text{h}}-1)}=0.535 \\ \frac{P_{03}}{P_{04}} =1.866 \\ T_{04}=T_{03}-\Delta T_{034}=870 \text{ K} \\ \frac{T_{03}}{T_{04}} =1.149

During acceleration, the pressure and temperature ratio across the turbine is constant for a choked condition. Moreover, the rotational speed is proportional to the square root of the turbine inlet temperature thus,

\frac{P_{04}}{P_{03}} =\text{const.}, \quad \frac{T_{04}}{T_{03}} =\text{const.}, \quad \text{and } \quad N \alpha \sqrt{T_{03}}

At 95% of design speed

\sqrt{\frac{T_{03}}{1000} }=\frac{0.95}{0.9} \\ \therefore T_{03}=1114 \text{ K}

At 100% design speed

\sqrt{\frac{T_{03}}{1000} }=\frac{1.0}{0.9} \\ \therefore T_{03}=1234.6 \text{ K}

The specific thrust at ground run and choked nozzle is given by the relation:

\frac{T}{\dot{m}} =V_{\text{c}}+\frac{A_{\text{e}}}{\dot{m}} (P_{\text{c}}-P_{\text{a}})

Here, the governing equations for \dot{m},V_{\text{c}},P are given:

\frac{P_{\text{c}}}{P_{04}} =\left(1-\frac{1}{\eta_{\text{n}}}\frac{\gamma-1}{\gamma+1} \right) ^{\gamma/(\gamma-1)} \\ P_{\text{C}}=\frac{P_{\text{C}}}{P_{04}} \frac{P_{04}}{P_{03}} \frac{P_{03}}{P_{02}} \frac{P_{02}}{P_{01}} P_{01} \\ T_{04}=T_{03} \times \frac{T_{04}}{T_{03}} =\frac{T_{03}}{1.149}

The exhaust speed is calculated from Equation 16.19a

V_5= \sqrt{\gamma RT_5}= \sqrt{\frac{2\gamma RT_{04}}{\gamma+1} } \quad \quad \quad (16.19\text{a}) \\ \dot{m}=\left(\frac{\dot{m}\sqrt{T_{01}}}{P_{01}} \right) (P_{01})\left(\frac{1}{\sqrt{T_{01}}} \right) =\left(\frac{\dot{m}\sqrt{T_{01}}}{P_{01}} \right) \times \frac{1}{\sqrt{288}} =\frac{\left(\dot{m}\sqrt{T_{01}}/P_0\right) }{16.97} \\ \frac{A_{\text{e}}}{\dot{m}} =\frac{0.5}{\dot{m}}

Calculations are arranged in the following table. Plots for the temperature at the inlet and outlet of the turbine as well as at the nozzle outlet against the rotational speed are shown in Figure 16.15.
Plots for the exhaust (here critical) speed and the specific thrust against the rotational speed ratio are given in Figure 16.16.