# Question Deriving Mass Flow Rates for an Engine

I am trying to follow the calculations made in this blog post for a $$50$$ lbf. IPA/Nitrous Oxide engine, but cannot seem to get the values the author made for their propellant flow rates. Since the mixture ratio isn't mentioned, I figured I'd derive the net flow rate through the injector and check if the fuel and oxidizer flow rates sum to the net flow rate.

Since the flow rates for fuel and oxidizer are given in volume per unit time, I converted them to mass per second using the densities as

\begin{align*} \rho_{\text{IPA}} & =\frac {786\text{ kg}}{\text{m}^3}=\frac {6.56\text{ lbm}}{\text{gal}}\\ \rho_{\text{N}_2\text{O}} & =\frac {1220\text{ kg}}{\text{m}^3}=\frac {10.181\text{ lbm}}{\text {gal}} \end{align*}

My first attempt was using specific impulse. Since $$I_{sp}=2224.3$$ sec, then

$$\dot{w}=0.02248\text{ lbf/s}\qquad\implies\qquad\dot{m}=6.98\times10^{-4}\text{ lbm/s}\tag{1}$$

Where I have divided by $$g=32.2\text{ ft}^2\text{/s}$$ to convert from units of force to units of mass. This value is supposedly for net flow rate into the engine and the value is so small. For our propellants, using the values mentioned in the blog post, then

\begin{align*} \dot{m}_{\text{IPA}} & =\frac {12.24\text{ gal}}{\text{hr}}\frac {\text{6.56}\text{ lbm}}{\text{gal}}=0.0221\text{ lbm/s}\tag2\\ \dot{m}_{\text{N}_2\text{O}} & =\frac {75.74\text{ gal}}{\text{hr}}\frac {10.81\text{ lbm}}{\text{gal}}=0.2142\text{ lbm/s}\tag3 \end{align*}

By the way, this would imply that the mixture ratio is

$$r=\frac {0.2142}{0.0221}=9.69$$

Which is oxidizer-rich than the supposed stoichiometric mixture ratio of $$6.591$$ (I obtained this value by balancing the chemical reaction). Does this sound right?

Adding the two m-dots obviously does not give the m-dot found in $$(1)$$. Am I using the specific impulse wrong? I double checked all my calculations and am sure that I did not make a mistake.

My second attempt was to use the $$\dot{m}$$ formula found in Sutton.

$$\dot{m}=A_t(p_c)_{\text{ns}}\sqrt{\frac {\gamma}{R(T_c)_{\text{ns}}}\left(\frac 2{\gamma+1}\right)^{(\gamma+1)/(\gamma-1)}}$$

I correctly calculated the area of the throat as $$A_t=0.124\text{ in}^2$$ and referencing the CEA output in the link above: $$(T_c)_{\text{ns}}=6035.958^{\circ}\text{R}$$, $$\gamma=1.2592$$, $$(p_c)_{\text{ns}}=294\text{ psia}$$, and

$$R=\frac {R_u}{\mathfrak{M}}=53.326\text{ lbf ft/lbm}^{\circ}\text{R}$$

Then

$$\dot{m}=0.0424\text{ lbm/s}$$

This value is also wrong again because the m-dot for $$(3)$$ is greater than the supposed total m-dot from the equation. I've spent a couple of hours on this and have no idea what I'm doing wrong, can anyone help?

• No chemical rocket has that high of a specific impulse in seconds. Perhaps that number is the ideal exhaust velocity in m/sec. Also weight flow rate is not in lbf/sec. It looks like you have the units wrong at the very beginning, recheck and redo. Sep 3, 2022 at 22:31
• @OrganicMarble Wait, wait, hold on a sec, units of weight flow rate is not lbf/sec? I thought weight flow rate was simply $\dot{m}g_0$ and units of $\dot{m}$ are lbm/sec, so multiplying by $32.2$ ft/s${}^2$ gives lbm ft/s${}^2$ which is lbf, is this wrong? Sep 3, 2022 at 23:09

Looking at the page you linked - and what a dreadful mishmash of units that page is - the "Specific Impulse" is given as a velocity in m/sec.

To calculate what I normally think of as Isp, divide that number by g0. Sticking with metric units, that gives an Isp of ~226 sec which is a reasonable number for a bad chemical rocket engine.

Again sticking with metric units, the "50 lbf" rocket has a thrust of ~222 N. Using these values for Isp, thrust, and g0 in the specific impulse equation and calculating mass flowrate, I get ~0.1 kg/sec.

In English units, this is ~0.22 lbm/sec, which is roughly equivalent to the sum of the weight flow rates you calculated in your equations (2) and (3).

Image cropped and annotated from https://www.grc.nasa.gov/www/k-12/airplane/specimp.html

• This is starting to make more sense to me. If I stick with the English Engineering units for calculating $\dot{m}$, then I get$$\dot{m}=\frac {50\text{ lbf}}{226\text{ s}\cdot g}$$But to convert from lbf to lbm, I have to multiply by $g$ since that's the definition of lbf. The $g$ cancels out so$$\dot{m}=\frac {50}{226}=0.22\text{ lbm/s}$$And if I use the $\dot{m}$ formula, I have to multiply the result by $\sqrt g$ when converting from lbf to lbm and I get roughly$$\dot{m}=A_tp\sqrt{\frac {\gamma g}{RT}\left(\frac 2{\gamma+1}\right)^{(\gamma+1)/(\gamma-1)}}=0.241$$Which is close enough? Sep 5, 2022 at 0:29
• @FrankW Looks good! I have to admit, I did not run the numbers on the Sutton m-dot formula, because you had a lot of mixed units there (inches, feet, lbm, etc) that I didn't feel like dealing with. As you play with this stuff, you'll start to get a "feel for the numbers", such that an Isp of 2000 + secs for a chemical engine will automatically set off the bs alarm. Sep 5, 2022 at 1:26
• @FrankW Also if you want to use English units, I seriously advise always using mass in slugs and flowrate in slugs/sec. That really is the mass unit. Use pressure in lbf/ft^2. You can always convert back to the more friendly units at the end, and you will save yourself a world of trouble along the way. See space.stackexchange.com/a/34713/6944 Sep 5, 2022 at 1:33
• Yeah, I should've realized that $2226$ is an insanely high Isp for an engine, and I thought the m/sec units were just a mistake. I prefer slugs but when I began reading Modern Engineering for Design of Liquid Propellant Rocket Engines, the authors never use slugs as a unit of mass, only lbs, which is confusing since lbf and lbm have two different meanings Sep 5, 2022 at 3:49