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Looking at the SSME Wikipedia RS-25 Space Shuttle Engines, the data shown that the thrust at sea level ($F_{sl}$) is 1.816*10$^6$ N, while the thrust in vacuum ($F_{vac}$) is 2.278*10$^6$ N. Knowing only these two pieces of information, I need to find out the exit plane area of the Nozzle.

I understand $\dot{m}_{sl} = \dot{m}_{vac}$, but I cannot find the correct relationship to include the formula of Thrust $F_{*} = \rho*A_e*v^2_{e}$.

I found this question in the book I am currently studying. It is problem 1.2 of "Astronautics. The Physics of Space Flight", Second Edition, Ulrich Walter.

Organic Marble
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Julio Gonzalez-Saenz
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1 Answers1

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You can use the method I explained at the start of this answer: Falcon 9 Merlin 1d thrust calculated through every moment of flight

Quoting from that, you use the thrust equation:

$ \ \ \ F = \dot{m}_\mathrm e V_\mathrm e + (p_\mathrm e - p_0) A_\mathrm e$

You actually know four things, not two. You know the sea level thrust, the sea level ambient pressure, the vacuum thrust, and the vacuum ambient pressure.

Fill in those values and subtract the two equations from each other. The things you don't know (m-dot, Ve, and pe) drop out.

That gives you 462000 N = 101325 N/m^2 * Ae which resolves to

Ae = 4.55 m^2

which is close to the true area. (This fact sheet gives the exit diameter as 2.3 m)

Organic Marble
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    My mistake was to assume that the exit velocity at sea level was different from the one in the vacuum. – Julio Gonzalez-Saenz Sep 25 '22 at 20:14
  • @JulioGonzalez-Saenz: I too would have assumed a difference but the reason I think there's a difference results in a difference too small to matter. – Joshua Sep 26 '22 at 17:51
  • @Joshua the exhaust velocity only depends on the combustion chamber properties and the nozzle geometry. See the first equation here https://space.stackexchange.com/q/43918/6944 Note Pe in that equation is the exit plane pressure, not the ambient pressure – Organic Marble Sep 26 '22 at 17:57