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I have been launching water bottle rockets with my kids and we are in the process of creating a simulation of the launch using numerical methods. I am a mechanical engineer, but fluid dynamics is not my strong suit.

Can you give me an intuitive explanation why the two methods below give a factor of 2.0 difference for thrust? Which method is correct?


First Method

I found these formulas for calculating thrust at the following website. https://www.ohio.edu/mechanical/programming/rocket/analysis1.html

Thrust equals mass flow rate times the exhaust velocity.

$F=\dot{m}v $

Mass flow rate is found using the water density, nozzle area, and exhaust velocity.

$\dot{m}=\rho A v$

Combining we get

$F=\rho A v^2$

The square of the exhaust velocity is equal to 2 times the internal gauge pressure divided by the water density.

$v^2=2P/\rho$

Combining we see that the thrust is equal to 2 times the nozzle area times the pressure.

$F=2AP$


Second Method

In the image below, it seems that the thrust on the rocket should be equal to the force imbalance as shown by the arrows. This force imbalance is the internal gage pressure times the nozzle area.

$F=AP$

enter image description here


----- UPDATE -----

Based on all the great answers and comments, I think the following figure is a better representation of the thrust force. Internal pressure (red arrows) drops off near the nozzle opening because the water flow speed is significant. Also, the water at the nozzle opening is not at zero pressure as I had originally assumed. Instead, there is dynamic pressure (green arrows) that is inversely proportional to the flow velocity. The force imbalance equals double the air pressure times the nozzle area, $F=2AP$.

enter image description here

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    $\begingroup$ The square of the exhaust velocity is equal to 2 times the internal gauge pressure divided by the water density. Where did you get this? $\endgroup$ – Rishabh Jain Oct 1 '19 at 12:39
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    $\begingroup$ @RishabhJain: That is equation #3 in the link I provided above. Thank you. $\endgroup$ – James Oct 1 '19 at 12:41
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    $\begingroup$ @James, your calculation assumes that only water is exiting the rocket nozzle to provide thrust. I have seen data from a device that measured thrust, and this data clearly indicated that the flow out of the bottle is chaotic, with a mixture of liquid and air bubbles leaving the bottle at the same time. This means that the thrust is far from a nice and smooth function of time or pressure as the bottle accelerates. $\endgroup$ – David White Oct 1 '19 at 15:28
  • $\begingroup$ @DavidWhite to add to that, you can trivially prove that air under pressure will also exit and provide thrust - you launch an "empty" (air only) bottle. I suspect that a vertical bottle will show less "chaos" than a horizontal one, or one at an angle. $\endgroup$ – Baldrickk Oct 2 '19 at 14:02
  • $\begingroup$ @RishabhJain it's just bernoulli's equation with no gravitational term, and assuming zero velocity at the other end. $\endgroup$ – Rick Oct 2 '19 at 15:48
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The first method is correct. In the second you have assumed that the pressure at the nozzle is still $P$ despite the water exiting with some velocity. i.e. You have neglected the dynamic pressure.

You need to use Bernoulli's principle $$ P + \frac{\rho v^2}{2} + \rho h g = {\rm constant}$$

Your first method assumes that the top surface of the water is hardly moving (because it's surface area is much bigger than the nozzle area). Applying the same idea to the second method, then we can calculate the constant both in the water and immediately below the nozzle as $$P = P_A + \frac{\rho v^2}{2},$$ where $P_A$ is atmospheric pressure and we neglect the small $\rho h g$ term which increases pressure due to to the column of liquid above the nozzle on the LHS. If we further assume that $P \gg P_A$ then $P = \rho v^2/2$ and the rate of change of momentum of liquid from the nozzle is $$ F = \rho A v^2 = 2PA$$

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    $\begingroup$ Thank you. Contrary to your second sentence, I intended to assume that the pressure at the nozzle was zero. I guess that the reason the second method predicts less force is that I didn't consider that there is dynamic pressure at the nozzle pushing UP on the water that is still in the bottle? $\endgroup$ – James Oct 1 '19 at 14:15
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    $\begingroup$ @James yes, there is a pressure gradient within the water, so the narrower the cross section as the water approaches the nozzle, the faster the fluid, the lower the pressure. $\endgroup$ – Rick Oct 2 '19 at 15:45
  • $\begingroup$ I edited the answer, hope my changes made it more clear for other readers. $\endgroup$ – Kevin Kostlan Oct 3 '19 at 0:39
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    $\begingroup$ @KevinKostian Please don't edit an answer like that. Especially one that has already received upvotes in double figures. You doubled it in length and removed the crucial point/reference about "dynamic pressure", inserted your own peripheral example and thus completely changed the nature of the answer. You should post your own answer. $\endgroup$ – Rob Jeffries Oct 3 '19 at 6:49
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Often these rockets are launched off a vertical section of pipe or rod that extends up inside the bottle, acting as a piston until the rocket has moved far enough to clear the end of the pipe. During this phase, your second method is correct: the thrust is simply the nozzle area times the pressure.

Why should the thrust increase when the rocket separates from the piston? Let me try to provide an intuitive justification. I won't prove that the thrust doubles, just dispel the notion that it should remain unchanged.

Let's say that somehow the piston is able to extend itself by constantly adding little cylindrical plugs to itself. These plugs are initially arranged in a tall rack just beside the rocket; as the rocket moves upward, the piston keeps grabbing a plug from the rack, somehow transporting it through the wall of the rocket, and adding it on to the end. The rocket never leaves the piston, and it's clear that the thrust remains just $PA$.

But this is essentially what is actually happening as the rocket expels water, with one exception. Each little bit of water leaving the nozzle can be thought of as a "plug", and the force acting to separate it from the rest of the rocket is still $PA$. But, unlike the plugs waiting in the rack at rest relative to the earth, each one of the water plugs is moving upward a bit faster than the previous one was--just fast enough to match the velocity of the rocket. The momentum transferred to the rocket by these moving plugs constitutes an additional force relative to the case in which the plugs have zero velocity.

Of course, this analogy of plugs being added during flight does not correspond to the actual case of a rocket that loses mass over time. But that difference does not affect the instantaneous thrust.

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The pressure drop from tank pressure to atmospheric pressure does not occur instantly at the nozzle but rather is spread out according to the area of the flow channel. This reduced pressure results in additional thrust that was not accounted for in your second solution. Here is one way to calculate this additional missing thrust:

Bernoulli's equation (your second to last equation) applies:

$$P+\frac12\rho v^2 = constant$$

we can combine this with your mass flow equation to get:

$$P+\frac12\rho \left(\frac{\dot m}{\rho A}\right)^2 = constant$$

Your original answer assumes negligible velocity/large area at the water surface1:

$$P_{tank} + 0 = constant$$

This gives us our constant:

$$P+\frac12\rho \left(\frac{\dot m}{\rho A}\right)^2 = P_{tank}$$

and we know the pressure at the exit is atmospheric/ 0 gauge pressure:

$$0 + \frac12\rho \left(\frac{\dot m}{\rho A_{exit}}\right)^2 = P_{tank}$$

We can solve for $\dot m$:

$$\dot m = A_{exit}\sqrt{2 \rho P_{tank}}$$

Plugging back in:

$$P+P_{tank}\left(\frac{A_{exit}}{A}\right)^2 = P_{tank}$$

Solving for pressure:

$$P = P_{tank} \left(1-\left(\frac{A_{exit}}{A}\right)^2\right)$$

So if we wanted to calculate the additional thrust due to the pressure ebeing lower near the nozzle we'd need to integrate the pressure by the area:

$$F = F_{up} - F_{down} = \int_{A_{exit}}^\infty \left (P_{tank} - P_{tank} \left(1-\left(\frac{A_{exit}}{A}\right)^2\right) \right) \; dA $$

$$F = P_{tank} \int_{A_{exit}}^\infty \frac{{A_{exit}}^2}{A^2} \; dA $$

$$F = P_{tank} A_{exit} $$

So there's the missing $PA$ from your second solution.

1: You can make your equations more accurate (especially for thin bottle rockets) by using the actual cross sectional area at the water surface instead of infinity both here and in the limit of the integral.

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As you have neglected atmospheric pressure, let's assume that you are doing the experiment in a vacuum(I know that there is a spherical cow analogy somewhere :) ). You are assuming that the pressure is exerted by air inside the ballon at the surface. The air will exert the pressure in the empty part as shown by your arrows as well as on the horizontal surface of the water. Now, this will cancel out due to a complete loop. Now, if the balloon was closed there would have been no net force as the downward force due to air pressure would have been balanced by the normal force. In your 1st method, the pressure you have taken is the air pressure present inside the ballon. In your second method, you are assuming the same pressure at the bottom of the nozzle which is incorrect.

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    $\begingroup$ I don't fully understand your answer. Are you saying that I am incorrect in the second method because the water in close proximity to the nozzle opening is at a lower pressure than the pressure of the compressed air? So the red arrows near the nozzle opening should be drawn to shorter length? Thank you. $\endgroup$ – James Oct 1 '19 at 13:34
  • $\begingroup$ The post says he uses "ga[u]ge pressure" which means that the external atmospheric pressure is already accounted for. You're on the right track here, but jumped too soon. $\endgroup$ – dmckee --- ex-moderator kitten Oct 1 '19 at 14:58
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    $\begingroup$ @James Those observations are correct (see my answer for math and details) $\endgroup$ – Rick Oct 2 '19 at 15:43
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There is a simple intuitive explanation for the factor of two difference. The thrust has two contributions, as follows.

1) The air pressure in the rocket imposes a force down the nozzle of F=PA exactly as you surmised in your second method.

2) The exhaust water at the instant it leaves the nozzle is still pressurised, and imposes a back-pressure on the rocket. The force due to the back pressure is also straightforwardly F=PA.

The combined effect is therefore 2PA.

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  • $\begingroup$ While pulse detonation propulsion is a thing, it's not really related to the thrust produced in this case. The first method is exactly calculating the thrust due to throwing mass out the back: $F=\dot m v = 2 AP$ The water doesn't expand (much) under relaxation and there is no additional thrust generated from this relaxation. $\endgroup$ – Rick Oct 2 '19 at 19:14
  • $\begingroup$ Hi Rick, many thanks. It would be really helpful if you could explain the logic for your last sentence. Best wishes. $\endgroup$ – Marco Ocram Oct 2 '19 at 19:47
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    $\begingroup$ Sure, the additional thrust generated by pulse detonation or underexpanded gas flows manifests as the pressure at the exit being above atmospheric, as the gas expands it continues to accelerate and decrease in pressure until it reaches atmospheric pressure. Since water is incompressible, it doesn't expand, which means the exit velocity is its final velocity and there is no further pressure drop. This means in the case of water (or a properly expanded gas) there is no additional thrust beyond the $\dot m v$. $\endgroup$ – Rick Oct 3 '19 at 11:15
  • $\begingroup$ Hi Rick, your second sentence is not exactly right. Water does compress, and behaves elastically within limits. It has a high bulk modulus compared to air (about 2.2GPa compared with 100kPa) so the physical change in its dimensions resulting from applied pressure is much smaller; nonetheless it releases energy when it relaxes instantaneously, which creates an impulse effect. The same effect in steel (bulk modulus 160Gpa) is what causes the central balls in a Newton's cradle to pass an impulse while appearing to remain static. $\endgroup$ – Marco Ocram Oct 3 '19 at 13:12
  • $\begingroup$ Hi Rick, I've updated the answer to include an analogous design of the rocket to illustrate the effect of ejecting a compressed elastic mass. Do you fancy calculating the thrust? $\endgroup$ – Marco Ocram Oct 3 '19 at 13:50

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