I have designed a nozzle for the following input parameters
- $p_{combustion}=p_c=200*10e5 Pa$
- $T_{combustion}=T_c=2800 K$
- $F=70 kN$
- $\kappa=1,2$
- $R = 330 J/kgK$
- I assume that all my gas properties stay constant for this calculation.
The nozzle shall be working ideal at
- $p_{exit}=p_{ambient}=54048 Pa$
For having no losses this gives me $$v_e=\sqrt{\frac{2\kappa}{\kappa-1} R T_c (1-\frac{p_{exit}}{p_{c}})^{\frac{\kappa-1}{\kappa} } } = 2636 \frac{m}{s}$$
Since I know my thrust for ideal expansion this gives me the mass flow $$\dot{m}=F/v_e=26.55 \frac{kg}{s}$$
For the throat temperature I use: $$T_t=\frac{2T_{c} }{\kappa+1}=2545 K$$ And throat pressure: $$p_t=p_c*(2/(\kappa+1))^{\frac{\kappa}{\kappa-1}}=1.1289*10^7 Pa$$
And density: $$\rho_t=\frac{p_t}{R T_t} = 13.44 kg/m^3$$
So my throat area will be $$A_t=\frac{\dot{m}}{\rho_t \sqrt{\kappa R T_t}}=0.00196777 m^2$$
I hope I haven't done any mistake. Now I freeze my nozzle's geometry for Off-Design calculation.
When I use now my nozzle @SeaLevel Pressure , does my exit velocity $v_e$ change? To my mind not, because my backpressure of $p_{ambient}=101325 Pa$ is still so much lower than my combustion pressure (critical pressure ratio p_ambient/p_c=0.005< 0.53 is still valid)
As long as my nozzle geometry is fixed and the flow choked, then $v_e$ $A_e$ and $p_e$ don't change. Can I just use $$ F = \dot{m} v_e + A_e (p_e - p_{ambient} )$$
In this case, my thrust will be ruduced since $p_{ambient} > p_e $.
Is this correct?
Thank you very much for your help guys!
Best regards Lucas
