Maximal thrust from water pump

Question

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.

Answer 1

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.

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.

Answer 2

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!

Answer 3

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