# Nozzle Exit Pressure Too High?

I've been doing some simple rocket science for fun, and I wanted to calculate the specifications of the Lunar Module Descent Engine (LMDE). The LMDE uses Aerozine-50 (UDMH) and Dinitrogen Tetroxide($$N_2O_4$$). I wanted to calculate the exit pressure of the nozzle. The formula to calculate exit exhaust velocity is: $$V_e = \sqrt{\frac{TR}{M} \cdot \frac{2\gamma}{\gamma-1} \cdot \Biggl( 1- \bigg(\frac{P_e}{P}\bigg)^{(\gamma-1)/\gamma} \Bigg)}$$ where:

• $$V_e$$ is exhaust velocity. In this case it is equal to $$3225$$ $$m/s$$.

Calculated by $$F = \dot mV_e$$ or $$V_e = \frac{F}{\dot m}$$ where $$F$$ is $$11,965$$ $$N$$ and the mass flow rate is $$3.71$$ $$kg/s$$ at 25 % thrust.

• $$T$$ is the absolute temperature of the inlet gas. In this case is equal to $$294.216$$ $$K$$. Obtained from here. See it by finding "Nominal propellant temperature at injector inlet".
• $$R$$ is the universal gas constant which is $$8314.5$$ $$J/(kmol·K)$$
• $$M$$ is the molecular gas weight of the propellant. In this case it is $$20.58$$ $$kg/kmol$$. Obtained from here. Note that the LM has a mixture ratio of 1.6, therefore use the line labelled 1.6. Also, the LM chamber pressure is 120 psia (about 8 atm). This source also has the specific heat ratio.
• $$\gamma$$ is the isentropic expansion factor, also known as Specific Heat ratio. In this case it is $$1.232$$. (Obtained from the source above).
• $$P_e$$ is the exit nozzle pressure (in Pascals).
• $$P$$ is the pressure of inlet gas. In this case it is $$3,010,000$$ $$Pa$$ (or 437 psia). Obtained from here.

I'm trying to find the exit pressure ($$P_e$$) so I rearranged the formula to: $$P_e = P \cdot \left(1-\frac{V_e^2 \cdot M \cdot (\gamma-1)}{TR \cdot 2\gamma}\right)^{\frac{\gamma}{\gamma-1}}$$

Substitute all values. The numbers and units are $$3.01 \times 10^6$$ $$Pascals$$, $$3225$$ $$m/s$$, $$294.261$$ $$K$$, $$8314.5$$ $$J/kmol \cdot K$$, $$20.58$$ $$kg/kmol$$, and $$\gamma = 1.232$$ is a ratio. $$P_e = 3.01 \times 10^6 \cdot \left(1-\frac{(3225)^2(20.58)(1.232-1)}{(294.261)(8314.5)(2)(1.232)}\right)^{\frac{1.232}{1.232-1}}$$

if I subsitute all values into the equation, I get $$1.105 \times 10^{11}$$ $$Pa$$. About 16.8 million psi!

Another way to calculate exit pressure is with $$F=\dot mV_e + (P_e - P_a)A$$ or: $$P_e = \frac{F-\dot mV_e}{A} + P_a$$

$$A$$ is the area of nozzle exit ($$1.9$$ $$m^2$$) and $$P_a$$ is the ambient pressure which is $$0$$ $$Pa$$ because the LMDE is in a vacuum. The result is $$0.13158$$ $$Pa$$ or $$0.00001885$$ $$psi$$.

Question: What's going on here? For the first equation, I'm getting such an unrealistic, high exit pressure for the nozzle. But in the second equation, I'm getting such a low exit pressure (almost a vacuum). Shouldn't these values be approximately the same?

• Isn't $\dot m$ 3.71 kg/s at 25% thrust. I also used 11,965 N which is approx 25% of the LM's full power. May 5, 2020 at 23:37
• Done. My bad. It totally went out of my mind until you mentioned 15 kg/s. May 5, 2020 at 23:41
• Can you verify that $R$ is supposed to be the universal gas constant in the first equation? I think there are conflicting conventions, e.g.: $R = \frac {R_u} {\mathfrak M}$ where $R_u$ is universal gas constant and ${\mathfrak M}$ is molecular weight: May 6, 2020 at 1:11
• @RussellBorogove Yes. De Laval Nozzle, exhaust velocity. May 6, 2020 at 1:17
• One more thing before I call it a night. The P in the first equation should be the combustion chamber pressure. I think it was like 100 psi but maybe less at 25% throttle. Not sure where you're getting 437. Remember the exit of the combustion chamber is the entrance to the de Laval nozzle, and those are the properties you should use in the exit velocity equation. Chamber pressure and chamber temperature. May 6, 2020 at 3:59

Writing an answer to summarize discussion in comments and hopefully generalize a bit.

The most important takeaway is that the De Laval exit velocity equation (found on Wikipedia, also equation 3-14 in Sutton, 4th edition) requires that you use the properties of the combustion chamber as inputs into the equation. This includes the temperature rise due to combustion.

Unfortunately the value of the combustion chamber temperature is quite complicated to calculate properly. See my answer to this question How do you determine what the temperature will be in the combustion chamber of a rocket engine? This can be the most frustrating part of attempting to calculate combustion chamber properties from first principles. Sometimes you can get data for a particular engine as a function of, for example, chamber mixture ratio. This is the approach we used in the SSME model in the Shuttle Mission Simulator.

I've also heard there are various online calculators which can do this but I have no experience with them. A comment on that answer mentions "propep". A comment on this answer to this question Quantitative plots of v, T, p, vs position from chamber through nozzle to ambient for a few canonical modern engines? mentions "cpropep"

• Calculate combustion temperature and combustion pressure. That is the key. Thank you very much. I think I was calculating the pressure of the fuel going into the chamber. Calculating combustion temperature is indeed hard. Luckily, I found this source that states the combustion temperature and pressure. (See it by finding "TRW TR-201"). So it's about 703,265 Pa and 2,973 K. I'm forgetting about calculating it at 25% but instead at 100%. So that's about 47,860 N and 15 kg/s mass flow rate and $V_e$ of 3,190.7 m/s. May 6, 2020 at 15:37
• (Continuing comment above (word limit)). Using the values above, it's about 143.76 Pascals (0.021 psi). That seems like the correct answer. Would it be safe to assume that at 25% thrust, the exit pressure is just the exit pressure at full thrust divided by 4? I upvoted, but would you care to explain why the second equation (at full thrust, so 47.86 kN and 15 kg/s) is giving about 6.67 Pa (0.00096 psi). The second result is about 22 times lower. I wouldn't consider that "approximate". May 6, 2020 at 15:45
• @StarMan Without doing a lot of work (remember I'm lazy) I think those two equations have different assumptions. The Ve equation assumes isentropic flow and maybe other things, and the thrust equation doesn't make any assumptions about the flow. May 6, 2020 at 15:53
• I see. So the second thrust equation to find $P_e$ is in a way, similar to $P = F/A$ as its getting the amount and force of the matter being ejected, and dividing it by the area. May 6, 2020 at 16:00
• True. It's a difference of about 260 N. Anyways, thank you very much. May 6, 2020 at 16:11