Should negative pressure-thrust increase or decrease the nozzle-exit velocity of a rocket?

Question

first post here.

I've just started on George Sutton & Oscar Biblarz's Rocket Propulsion Elements and have stumbled upon contradictory solutions between the Eighth (PDF linked) and Ninth Edition (my hard-copy version) and would like someone to tell me which is wrong. I think the latest version is erroneous.

In Chapter 2 sections 2-3 focus on Thrust and Exhaust Velocity respectively (pp.32-36 of the linked PDF version, pp.31-35 of my Ninth print edition).

In Equation 2-13 (p.32 v8, p.33 v9) it is shown that:

$$ F = ṁv_{2} + (p_{2}-p_{3})A_{2} $$

where F = thrust force, ṁ = mass flow rate, $v_{2}$ = nozzle-exit velocity, $p_{2}$ = nozzle-exit pressure, $p_{3}$ = ambient pressure and $A_{2}$ = nozzle-exit cross-sectional area.

The first term in this equation constitutes that momentum-thrust, and the second term the pressure-thrust. From this it is shown that when nozzle-exit pressure is less than the ambient pressure, the pressure-thrust should be negative and reduce the overall thrust of the rocket.

In Example 2-2 (pp.35-36 v8, pp.34-35 v9) of both editions the pressure-thrust is calculated as such:

$$ (p_{2} − p_{3})A_{2} = (0.070 − 0.1013) × 10^6 × 0.0574 = −1797N $$

However, whilst Edition 8 solution calculates the nozzle-exit velocity as:

$$ v_{2} = (62250 − 1797)/24.88 = 2430 m/sec $$

Edition 9 calculates:

$$ v_{2} = (62250 + 1797)/24.88 = 2574 m/sec $$

From what I read in the prior pages I'm concluding that the revision of the solution from v8 to v9 was incorrect. Why should the overall thrust increase if the pressure-thrust is negative?

But such a mistake would seem awfully silly - why change the original solution if it was correct? Especially where, of all parts of the book, readers will be scrutinising the content the most! (Or getting rather concerned about their (mis)understanding of the foundational concepts of rocketry). So is v8 and my conclusion wrong? Or the revised edition?

Answer 1

To summarize what I think your confusion is from the comments, you are trying to reconcile two statements/ideas:

1. "Thrust is proportional to exit velocity"
2. "Exit pressures below ambient decrease thrust"

And although they might seem contradictory, they aren't really. The first is just an approximate concept -- $F \propto v_2$. And that's true, to an approximate order of accuracy. If you begin to account for more physics, you begin to refine that expression. So when you account for exit pressure differences, you get:

$$ F = \dot{m} v_2 + (p_2 - p_3) A_2 $$

It's still true that thrust is proportional to exit velocity, to some order of accuracy. This improves the fidelity of our model to include pressure differences. But note, it's still not the most accurate it could be -- we could add in friction losses, or ablation, or heat losses more generally, or any number of additional effects that would improve the accuracy of our thrust equation.

Now to answer the question about which equation is right -- the equation in the 9th edition, and the equation I gave above, are the correct thrust models. You can also find them various places on the internet, like the NASA GRC pages. And the exercise you are listing is just taking a known thrust, known exit and ambient pressures, known exit area, and known $\dot{m}$ and re-arranging to get $v_2$. So doing the algebra confirms that the 9th edition is correct.

Now, finally, to answer the question about "seeing negative pressure thrust increasing the velocity" -- it's due to the construction of the problem. In this problem, the thrust is given to you as a constant value. If you have a fixed thrust and negative pressure thrust, then the velocity must be higher than it would be with no pressure thrust or positive thrust for a fixed value of thrust.

In other words, you're reading into the example more than should be read into it. For another system, it might be that the exit velocity is fixed and so negative pressure thrust would lead to less overall thrust.