Nozzle Exit Pressure Too High?

Question

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?

Answer 1

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"