# Calculate Thrust of Rocket Nozzle - Off Design

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.

1. 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)

2. 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

• I dont care to find sources so Im writing a comment instead of an answer: yes, with a frozen nozzle geometry, you can use that equation (the rocket equation) to compute off-design thrust. Commented Mar 15, 2023 at 3:49