Maximum thrust

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OnePound

Member
Hi all,

I'm wondering about the theory behind maximising rocket thrust. The equation F = m(dot)V + (Pe - Pa)Ae seems to suggest that this is maximised when Pa=0. However, other sources (e.g. braeunig.us) say that you want Pe=Pa. Indeed, as far as I can see, Sutton contradicts himself and says both! Please, could someone help me out?!

Many thanks!
OnePound

cjl

Well-Known Member
It's true that for any given motor, the thrust is maximized with Pa = 0. However, in this situation, you could gain even more thrust by then extending the nozzle to a higher expansion ratio so that (for all practical purposes) Pe ~ Pa = 0. Basically, if the exit pressure is above ambient, thrust will be gained by increasing the expansion ratio until Pe=Pa.

jsdemar

Well-Known Member
Hi all,

I'm wondering about the theory behind maximising rocket thrust. The equation F = m(dot)V + (Pe - Pa)Ae seems to suggest that this is maximised when Pa=0. However, other sources (e.g. braeunig.us) say that you want Pe=Pa. Indeed, as far as I can see, Sutton contradicts himself and says both! Please, could someone help me out?!
Both are true. When Pa isn't =0, then Pe=Pa will give max thrust with an optimal expansion ratio. For Pa=0, the same holds in theory, but it's impossible to have an infinitely large nozzle to expand the exhaust gases to Pe=0. That is why you see rocket engines for use in the vacuum of Space with long nozzles compared to first-stage rockets.

butalane

Well-Known Member
Hi all,

I'm wondering about the theory behind maximising rocket thrust. The equation F = m(dot)V + (Pe - Pa)Ae seems to suggest that this is maximised when Pa=0. However, other sources (e.g. braeunig.us) say that you want Pe=Pa. Indeed, as far as I can see, Sutton contradicts himself and says both! Please, could someone help me out?!

Many thanks!
OnePound
That seems true because the non numerical form of the equation does not give insight into the contributions of each part. You're correct in saying that the (Pe-Po)Ae part of the equation is maximized in a vacuum (Po = 0) however that contribution is orders of magnitude (IIRC) less than that of the momentum contribution (Mdot*Ve).

Even when the (Pe-Po)Ae contribution grows through flight (assuming you're flying into a vacuum), the momentum portion is also growing (***or not, based on nozzle geometry*** Fixed thanks CJL)

The reason one wants Pe to equal Pa for a given pressure is to maximize the expansion of the hot gasses and thus accelerate them as much as possible, this maximizing Ve and the momentum portion of the equation.

If you have an ISP of 204 then your Ve= 2000m/s.

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cjl

Well-Known Member
That seems true because the non numerical form of the equation does not give insight into the contributions of each part. You're correct in saying that the (Pe-Po)Ae part of the equation is maximized in a vacuum (Po = 0) however that contribution is orders of magnitude (IIRC) less than that of the Momentum contribution (Mdot*V).

Even when the (Pe-Po)Ae contribution grows through flight (assuming you're flying into a vacuum), the momentum portion is also growing.
Interestingly enough, this isn't true. The exhaust velocity (assuming there isn't any flow separation in the nozzle) is entirely determined by the gas properties, the chamber conditions, and the expansion ratio. The momentum contribution will be exactly the same at sea level as it will in space. The change comes entirely from the pressure contribution, which can definitely be significant (~10% or more difference from sea level to orbit).

butalane

Well-Known Member
Interestingly enough, this isn't true. The exhaust velocity (assuming there isn't any flow separation in the nozzle) is entirely determined by the gas properties, the chamber conditions, and the expansion ratio. The momentum contribution will be exactly the same at sea level as it will in space. The change comes entirely from the pressure contribution, which can definitely be significant (~10% or more difference from sea level to orbit).
I should have been more specific, or rather not subtly injected a design factor into the discussion. I was imagining for some reason that the area ratio would change via either flow separation or an extendable nozzle skirt. For some reason I transposed the design I was working on into this example, that is a use at a much higher Pa.

Sorry for the confusion!!

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