# Why is my Venturi pump much weaker than expected?

I'm trying to make a simple Venturi tee which will have air flow through it while sucking some water from down below (like paint sprayers do). I want the device to pull the water to a height of 1 meter.

That means that I need 0.1 atmospheres (104 Pascals). The dynamic pressure equation states that $P_{a}=\frac12\rho v^2$, meaning that $v=2P_a/\rho$. The density of air is $1.22 kg/m^3$, $P_a$ is 104 so $v=128\ \mathrm{m/s}$.

I experimented a bit with my air pump and found out that with a nozzle of ~5 mm in diameter I can fill a 10 litre bag with air in ~7 sec. That should mean that the velocity is around 60 m/s. With those results I took a hose of 16 mm in diameter, stuck my pump nozzle inside (from one side) , made a small hole juxatoposed to the nozzle from above (maybe 3-5 mm diameter) and stuck a straw in it. When I turn the apparatus on, hovewer, the straw (placed in a bottle of water) lifts the liquid for no more than several millimeters.

Why does my Venturi pump not work? • Two questions: (1) do I understand your calculations correctly: you calculated v=128 m/s to be the necessary speed but you only achieve 60? (2) Why do you increase the diameter after the nozzle? Thereby you are increasing the pressure again (because you increase the flow area) May 7 '15 at 18:49
• (1) Indeed 128m/s (according to my calculations) should lift the water 1 meter high, therefore I reasoned that the 60m/s I managed to achieve should at least have noticeable lifting, not the meager results I've got. (2) Huh. To be honest I thought that the fast air flow is still very close to the straw orifice, and I used a 16 mm hose so that the water have enough space to flow. You think I should narrow the exit pipe? May 7 '15 at 19:29
• Yepp @dactylo as a first very crude assumption: the speed and the pressure can be calculated using your equation for the dynamic pressure. Since (another assumption, neglecting the water) the mass flow needs to be constant. The pressure will drop. May 7 '15 at 19:33
• Thank you very much I will try it...However...now that I think about it, when I tested a 16-18 mm nozzle on my pump it filled the 10 litre bag in 1-2 sec, meaning that the air speed was around 35 m/s. Shouldn't that still generate a noticable pull (of close to 10cm)? If so, then shouldn't my current installation have done so as well? Or there is some sort of multiplication going on here (the smaller nozzle restricts the airflow to the larger nozzle/pipe)? This question is just out of curiosity. May 7 '15 at 20:03
• The pressure in the pipe will be lower because the mass flow through the smaller nozzle is smaller than the one with the larger nozzle. May 7 '15 at 21:39

The problem is that the airspeed drops as soon as the airflow leaves the nozzle. It is not the same as a water-jet expelling into air (in that case the density of the water is much greater than that of air so the water holds its velocity much better). The low-pressure air will suck in air from around it in the wider pipe and thus become turbulent around the outside edges of the cylindrical stream within a very short distance as it leaves the nozzle, and slow right down. The airflow will still be fast in the center of the cylindrical stream, but not at the edges where your intake is. You need to position the intake of your straw inside the nozzle, at its narrowest point. 