I can't find what i'm looking for in the engineering stack exchange because it's all way over my head. need basic help answer the basics to get started. so maybe a link or video that helps me understand how to even ask the right question on motor needed. I'm not going to become an engineer or learn everything i simply want to understand how to solve one type of problem.

so here's my basic need:

I have a water pump bought at a store that needs to be driven by a drill. But i want to use a motor from another appliance to run the water pump so i can use my drill for other things and have the motor that runs the water pump just working on that. I took apart a spice/ coffee grinder and rewired it and hooked it up the water pump and its simply doesn't move from lack of power ( it works fine and spins fine not attached to the water pump). So how do i calculate the motor i need to run the water pump?

also i will need this calculation to serve other projects... I'm also going to want to make a yeast stirer and use the motor from an orange juicer to spin a magnet that will affect a stirer stick through a glass container to stir yeast for 24- 48 hours. Not a lot of power needed but will needed to run all day and night.

so I want to understand just the very basics i need to determine how to choose a motor or how to very simply adjust or reconfigure any motor set up (i.e. use cogs or some kind of added system of wheels to increase the motors power) to: 1) be able to run a water pump or spin anything that requires a lot of strength and can run continuously for an hour to two hours. 2) Secondly, I need to calculate and determine needs to run a motor that spins a light load but continuously for 48-72 hours without burning out.

I cant find answers on google or stack exchange because when i ask the question i get a ton of irrelevant or very complex answers. i'd like help in getting enough info to keep my projects going without having to study engineering for weeks. i dont need to understand every single thing but enough to get these jobs done and repeat the process for similar future projects. Thank you!

  • $\begingroup$ Let me ask the obvious question. Why not simply get a cheap secondhand drill and use that to drive your pump? $\endgroup$
    – DLS3141
    Jul 26, 2017 at 6:17
  • $\begingroup$ Also, for a stir plate, you'll want speed control, that makes it a bit more complicated. My stir plate uses a magnet glued to a PC fan with the blades removed and has a speed control circuit attached to that $\endgroup$
    – DLS3141
    Jul 26, 2017 at 6:21
  • $\begingroup$ @DLS3141 you should post your pc fan & magnet with sped control as an answer -I'd vote... $\endgroup$
    – Solar Mike
    Jul 27, 2017 at 16:00
  • $\begingroup$ @SolarMike I would, but it's not my design and I didn't even build it, I just took it apart enough to see how it worked. The guy who designs and builds them started a company and sells them for less than I could make one. stirstarters.com $\endgroup$
    – DLS3141
    Jul 27, 2017 at 16:19
  • $\begingroup$ Or , you could do it the easy, cheap way. There are submersible aquarium pumps for several dollars that will do your stirring with only a few watts. they run for years . $\endgroup$ Mar 8, 2021 at 20:20

3 Answers 3


So how do i calculate the motor i need to run the water pump?

The ideal hydraulic power required to drive a pump is simply a function of:

  • Mass flow rate
  • Fluid density
  • Total head loss of the system.

So, the ideal power required is:

$$ P_{hydraulic-kW} = (q*\rho*g*h)/(3.6*10^6)$$


$q = flow rate (\frac {m^3}{hr})$

$ \rho = density(\frac {kg}{m^3})$

$ g = gravity(9.81 \frac {m}{s^2})$

$ h = Head Loss (m)$

It's pretty much plug 'n chug to get this far, but this is just the amount of power that the pump has to impart to the fluid in order to achieve the flow rate. Since you're interested in the motor and the pump is not perfect, the efficiency of the pump in converting mechanical power fom the motor into fluid power has to be taken into account.

$$ P_{shaft-kW} = P_{hydraulic-kW}(\eta_{pump}) $$


$\eta_{pump}$ is the pump efficiency

Now, the value for $\eta_{pump}$ of your drill pump is either something you'll have to look up, estimate or measure. (I'd guess it's 0.4-ish, but that's a guess, nothing more)

  • $\begingroup$ I dont understand the math formulas above. i was hoping my question could have a simpler answer. for example: a 600 watt motor from any standard appliance can run a water pump spinning 300rpm and pump 3 litres of water per minute. if its 1200watts it will spin 600rpm and pump 6lpm. ..an answer that sounds like something along those lines. but thank you for taking the time and helping. I'm very grateful but feel i would spend hours and hours looking up each aspect of the formula above and not actually build anything and i wont nec get it. looking for a simple version of the above 4a novice. $\endgroup$
    – djalexis
    Jul 29, 2017 at 9:01
  • $\begingroup$ @djalexis that is the simple formula, and yes it includes unknown things such as efficiency. That is usually the case as you do not know the inner losses of your system. $\endgroup$
    – joojaa
    Jul 29, 2017 at 16:42
  • $\begingroup$ That's as simple as it gets. $\endgroup$
    – DLS3141
    Jul 29, 2017 at 23:42

I don't have enough reputation to leave a comment however here's some more info that might be helpful:

You can take the rpm and convert that to rad/s: $\frac{1 rev}{60 sec} = \frac{2pi}{1 rev}$ or simply multiply the rpm by $\frac{pi}{30}$ to get $\frac{rad}{s}$

You can then multiply the speed in rad/s by the volume you want to pump out per second to get your volumetric flowrate. Since radians is a placeholder unit, you'll essentially have it replaced with $m^3$

Mathematically: $V*\omega = \frac{m^3}{s}$

$V $- Desired volume in seconds (This is something you can eyeball or measure)

$\omega$ - Speed of the motor in $\frac{rad}{s}$

The Head of the pump $h$ is really just a lookup and that's as easy as it gets.

Below is an image that shows how the head of a pump relates to the pump flow rate (Look up one with SI units as this is in the Imperial System. i.e The power should be Watts not Horsepower and the flowrate should be liters per min or $\frac{m^3}{s}$ instead of gpm (gallons per min) etc).

Some guidance with this is that you can have the volume you want per sec ($V$) a fixed value and then perform the same calculations ($V*\omega$) with increasing \omega to generate your own curve as well. At the point your curve intersects with another curve is where you want to be at. Use the head at that point and the equation provided above to calculate the needed power.

$$ P_{hydraulic - Watts}= η_{pump}*(q∗ρ∗g∗h)$$

$ q - flowrate (\frac{m^3}{s})$

$ ρ - density $ (Usually 1000 $\frac{kg}{m^3}$ for water)

$ g - gravity (9.81 \frac{m}{s}) $

$ h - Head$

$ η_{pump} - Efficiency_{pump}$

Note: The efficiency is also on the same curve! Note: I have modified the equation so the calculated power will be in Watts not in KW as above

Compare the calculated power to your current motor to see if it is less or more and make decisions from there.

Pump Efficiency and Head Diagram

Image Source: https://www.enggcyclopedia.com/2011/09/pump-performance-curves/

If I have further confused you, ask for clarification where needed. I'll try to clarify where needed

  • 1
    $\begingroup$ It's brake horsepower, not break... ie horsepower working against a brake not that it is broken. $\endgroup$
    – Solar Mike
    Mar 8, 2021 at 19:28
  • $\begingroup$ Yes, that's an important clarification! $\endgroup$
    – Pelumi
    Mar 8, 2021 at 19:51
  • 1
    $\begingroup$ And just to note, why did you change from metric (used in all of your answer) to imperial units for the diagram? Or did you just find the diagram somewhere?? $\endgroup$
    – Solar Mike
    Mar 8, 2021 at 19:54
  • $\begingroup$ I just found the diagram online, I tried pointing that out too in my answer so it isn't so confusing $\endgroup$
    – Pelumi
    Mar 8, 2021 at 20:06
  • 1
    $\begingroup$ Please add the source of the diagram to your answer to give credit to whoever created this diagram. $\endgroup$ Mar 8, 2021 at 21:38

It looks like you're looking for an ultra simple way to interpret the required horsepower for your pump.

The answers above are all correct about calculating flows, energy requirements etc, but likely not what you're looking for. As simplified as possible, you need to know two things

  1. How fast does the pump need to spin (RPM)
  2. How much power (HP, kW or watts) does it need

you SHOULD be able to approximate this by two methods.

  1. using the manufactures pump curve (by googling the brand/make/model/serial number etc and finding technical documents OR
  2. by approximating what your drill can do

I have a feeling I know the style of water pump you're talking about (I've purchased them myself before), and I will use this drill as a basis

That drill runs off of 120VAC, pulls 3.4 amps and has a max speed of 3000 RPM. 3.4 amps on 120VAC is ~ 0.44 HP, or 330 watts. So tl;dr - you probably need a 1/2 HP motor, and it probably needs to spin a few thousand RPM. Is this perfect? Nope. But it will get you close enough for what you're trying to do. If you want to be safe, get a 3/4 HP.


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