# Calculating mass expelled from cold gas thrusters

Consider a cold gas thruster rated at 1 lbf per firing. The thruster is on-off: there is no throttling, so when it fires it's at a constant full thrust. Assume the gas is nitrogen, and that it has a specific impulse of 72 s.

The thrust force per firing ($$F_\texttt{th}$$) is the product of mass flow rate ($$\dot{m}$$), specific impulse ($$i_\texttt{sp}$$), and standard gravity ($$g_\texttt{0}$$):

$$F_{\texttt{th}} = \dot{m} \cdot i_{\texttt{sp}} \cdot g_{\texttt{0}},$$

...so it seems I can get the mass flow rate as

$$\dot{m} = \frac{F_\texttt{th}}{i_\texttt{sp}\cdot g_\texttt{0}},$$

which I can integrate (with $$F_\texttt{th}$$ and therefore $$\dot{m}$$ as constant) to get the cumulative mass expelled over time:

$$m = \dot{m} \int_{t_\texttt{0}}^{t_\texttt{1}} dt.$$

Firing for 1 s would then expel...

$$m = \frac{1 \hspace{2pt}\texttt{lbf}\cdot\texttt{s}}{73 \hspace{2pt} \texttt{s} \cdot 32.2 \hspace{2pt}\texttt{ft/s}^2} = 4.25\texttt{e-}4 \hspace{3pt} \texttt{lbm}.$$

This would give you 2,350 1-lbf 1-s thrust firings per lbm of nitrogen. That has to be nonsense, right? What stupid thing have I done?

Update: Corrected units for g0.

## 2 Answers

The concept and formula are correct for a coherent system of units. You have chosen to use US customary units and these are not a coherent system, i.e. in the situation in the OP one is bound to run into trouble.

Looking on the bright side, your sense of what range the right answer could be in definitely helped you out here!

There are several ways of looking at it which are numerically identical in outcome, the most straight-forward are either:

• to use lbf and lbm in that relationship you have to use a different formula which is to multiply the last line by g (where that g is a dimensionless constant)
• to use the formula as it stands you have to accept that either the units of mass are the slug, not the lbm, (a slug is a much larger unit of mass, 1 slug = lbm x g) or that the unit of force on the top line is the poundal, not the lbf (the poundal is a much smaller unit of force than the lbf).

What is a coherent system of units? I have no idea what the usual practice is in the US, I grew up, professionally, on metric, but here it is:

A coherent system of units is where you can try to move 1 mass unit with 1 force unit and it results in 1 acceleration unit.

The lbm-poundal-ft/s^2 approach, the lbf-slug-ft/s^2 and metric (kg-N-m/s^2) are all consistent units. The US customary units, 1 lbf trying to move a 1lbm, would result in an acceleration of 1 g, not 1 ft/s^2.

There are dozens of confusing resources around the web, you can try https://en.wikipedia.org/wiki/Pound_(force) but I find, largely, the explanations are very poor.

• In my last equation, if you multiply the right side by g0, the units no longer match up with the left side. The units are consistent right now. And in fact, if you move g0 to the left side, you'll see you have lbm*g0, which is the definition of lbf, and which matches with the units on the right side (lbf once you cancel the s, since g0 is now on the right side). – user39728 Mar 14 at 19:05
• So the calculation results seem unreasonably small, but the units seem right, because they are consistent. US units are confusing but they still have to match up between left and right sides of an equation... – user39728 Mar 14 at 19:08
• You know, I don't know why on Earth we still use US units. They are the dumbest units anyone could use. Is there any other system less efficient and more confusing currently in use? No. It defies reason that anyone would choose US units given a choice like metric. – user39728 Mar 14 at 19:09
• In fact, from now on, I'll refuse to present calculations in US units. I'll work with them to the extent that my data is in US units, but I'll no longer communicate in US units. If someone shows me lbf, hit them with a slug. – user39728 Mar 14 at 19:11
• Many thanks, Puffin, for the thoughtful answer though. You raise some good points. Thanks for correcting my units typo too. – user39728 Mar 14 at 19:12

Puffin's answer is correct; this is just a supplemental that was a bit too involved for a comment.

A physical interpretation of specific impulse measured in seconds is "the length of time for which one mass unit of propellant can provide the force needed to balance the mass against Earth gravity". Therefore 1 lbm of propellant with 73 sec Isp should provide 73 1-second 1-lbf firings.

The result you got was 73 * 32.2 = 2350 -- for Isp given in seconds, and lbf and lbm units, $$g_0$$ is already factored in.

• Thanks, Russel! OK, two votes for g0 being out of place in my equations. I got it, I got it, should have used metric. Ha ha. This helps because I do need to get correct numbers on this. Thanks. – user39728 Mar 14 at 19:59
• Thanks for making me look at my math again. Embarrassing! – Organic Marble Mar 14 at 21:22