Prompt Title: ASE334 Propulsion Question

Next, we can calculate the nozzle area ratio:

$\frac{A_e}{A_t} = \frac{1}{M_e}\left(\frac{2}{y+1}\left(1 + \frac{y-1}{2}M_e^2\right)\right)^{\frac{y+1}{2(y-1)}}$

Substituting the given values, we get:

$\frac{A_e}{A_t} = \frac{1}{1.68}\left(\frac{2}{1.4+1}\left(1 + \frac{1.4-1}{2}(1.68)^2\right)\right)^{\frac{1.4+1}{2(1.4-1)}} \approx 2.28$

Now, we can use the following equation to calculate the uninstalled thrust:

$T_u = \left(\dot{m}_a + \dot{m}_f\right)V_e - \dot{m}_a(V_{\infty} - V_e)$

where $\dot{m}_a$ is the mass flow rate of air, $\dot{m}f$ is the mass flow rate of fuel, $V_e$ is the exhaust velocity and $V{\infty}$ is the free stream velocity.

The exhaust velocity can be calculated as:

$V_e = M_e\sqrt{\gamma R T_e}$

Substituting the given values, we get:

$V_e = 1.68\sqrt{1.4\times 287 \times 1200} \approx 801.3 \text{ m/s}$

The free stream velocity can be calculated as:

$V_{\infty} = M_{\infty}\sqrt{\gamma R T_{\infty}}$

Substituting the given values, we get:

$V_{\infty} = 0.8\sqrt{1.4\times 287 \times 230} \approx 222.8 \text{ m/s}$

Now, we can calculate the uninstalled thrust:

$T_u = \left(1.63\text{ kg/s} + 0.03\text{ kg/s}\right) \times 801.3 \text{ m/s} - 1.63\text{ kg/s} \times (222.8 \text{ m/s} - 801.3 \text{ m/s}) \approx 187.7 \text{ kN}$

Therefore, the uninstalled thrust of the engine is approximately 187.7 kN.