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For a small water pump like this one here, how can I roughly estimate how much thrust it would deliver considering the maximal pressure (PSI) and the flow rate (L/min). As a nozzle let's assume it is just a tube with a diameter d. I am a little bit confused because pressure and flow rate probably depend on each other as well as on the nozzle diameter.

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This would be a very rough estimation.

$$P = F v$$

Now you assume that the pumps efficiency $\eta$ is 1.

And for the power we assume 2.7 A x 12 V = 32.4 W.

For the flow rate you assume $4.3 \mathrm{\frac{L}{min}} = 7,16 \cdot 10^{-5} \mathrm{\frac{m^3}{s}}$.

This is due to the fact that flow rate and pressure are linked, as you already pointed out.

enter image description here

You see that the lower the pump head, for example the height the fluid has to overcome, the lower the flow rate. So I assumed the specified flow rate is specified for a 0 m head. Typically you would have diagrams that indicate at which pressure heads which flow rates could be achieved. In some / most cases the manufacturers of cheap pumps state $\dot{Q}$ for the lowest pressure head and $p$ for the lowest flow rate.

With the diameter of 10 mm the equations yield

$$32,4 \: W = F \cdot \frac{\dot{Q}}{A} = F \cdot \frac{7,16 \cdot 10^{-5} \mathrm{\frac{m^3}{s}}}{0,005^2 \pi \:\mathrm{m^2}}$$

$$F = 35,54 \: N$$

This would be the highest possible thrust that pump could achieve. Considering that all assumptions we made were to maximize the thrust it will be much lower in reality.

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  • $\begingroup$ doesn't the tube diameter affect the flow rate and the pressure? Otherwise I would just keep making it smaller and smaller increasing the exit speed and the produced thrust, but I guess at some point the pump will reach the maximal 35psi if I my understanding is correct. But I am still confused now, in your equation having a thinner tube/nozzle leads to a smaller thrust, does that make sense? $\endgroup$ – Mehdi Aug 23 '15 at 9:48
  • $\begingroup$ In the specs there is diameter of 10mm given. So that's what I used. If you use the continuity equation v1A1=v2A2 you see that for smaller the exit velocity increases. However since $vA=\dot{V}$ you see that the flow rate is not affected since it's constant. Your pump will reach 35psi at the lowest flow rate. Look at the diagram, you would need those to get information about the pump. Otherwise you would just assume that both values are the max. possible values as I did. $\endgroup$ – idkfa Aug 23 '15 at 9:52
  • $\begingroup$ So a smaller area (using a nozzle) should result in a higher speed, which normally results in more thrust. In your equation the opposite happens and F gets smaller. $\endgroup$ – Mehdi Aug 23 '15 at 10:03
  • $\begingroup$ If I'm not mistaken this is correct. Since your P is fixed. I should add to my answer that I assumed the pump behaves like jet engine and the exit velocity is the speed at which the pump "travels". When I'm not mobile anymore I'll add this to my answer. This is a very rough estimation though. $\endgroup$ – idkfa Aug 23 '15 at 10:32
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Firstly in the link you've provided it isn't clear whether the pressure is 351 psi or 35 psi. In the description it says 35 psi and in the images it looks like this too, but in the title it states 351 psi. I think it may be 35 psi as stated here.

The thrust force of something like a nozzle or rocket is derived from the fact that force is the change in momentum over time. The general thrust equation of a fluid stream jet is:

$F = \dot{m_e}V_e + (P_e - P_a)A_e$

where:

$\dot{m_e} = \text{mass flow rate exiting the nozzle (kg/s)} \Rightarrow 4.3 \text{ L/min} = 0.07167 \text{ kg/s} $

$\dot{V_e} = \text{velocity of stream exiting the nozzle (m/s)} = \text{??}$

$P_e = \text{pressure of fluid exiting the nozzle (N/m$^2$ or Pa)} = 35 \text{ psi}= 241.3 \text { kPa}$

$P_a = \text{atmospheric pressure (N/m$^2$ or Pa)} = 101.3 \text{ kPa}$

$A_e = \text{area of outlet (m$^2$)} = \dfrac{\pi d^2}{4}$

If you wanted to get a pretty accurate result, the only piece of information missing is the velocity of the exit stream, as well as the diameter of the nozzle, which you may be able to get from the manufacturer.

For instance say $d = 10 \text{ mm}$ and $V_e = 3 \text{ m/s}$ then:

$F = \dot{m_e}V_e + (P_e - P_a)A_e$

$F = (0.07167)(3) + (241,300 - 101,300)\dfrac{\pi(0.01^2)}{4}$

$F = 0.215 + 11$

$F = 11.215 \text{ N}$

As you can see the first term $\dot{m_e}V_e$ looks like it contributes a negligible amount to the force - so we can probably ignore $V_e$ anyways as it is mostly the pressure difference between the nozzle outlet and atmosphere contributing to the force. So provided you can find $d$ of the nozzle outlet then you could get a pretty decent approximation!

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  • $\begingroup$ That is actually my problem, the exit velocity. It can be increased by making the tube thinner and thinner but then the pressure is not constant, and that would probably affect the flow rate as well. I don't see which equation describes the relation between those. $\endgroup$ – Mehdi Aug 23 '15 at 9:43
  • $\begingroup$ @Mehdi The pump has a set flow rate, pressure and outlet area. Given this info the thrust force can be calculated. The flow rate will NOT change if the diameter is changed but rather the velocity will. $\endgroup$ – pauloz1890 Aug 23 '15 at 10:52
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This is best done by doing a momentum balance instead of trying to calculate pressure.

For a first order estimate, use the flow rate and the nozzle crossection area to compute the exit speed of the water. This is the momentum of the ejected water per unit mass. This divided by the time per ejection of each unit mass (which you get from the flow rate) is the force.

Force = momentum / time = mass * speed / time

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