Considering that pumps create flow and not pressure, if a pump runs water through a pipe into the air (high velocity water), is it true to say that: the pressure that pump shows - atm pressure = pressure drop in the system?
I assume you have an open system (not a closed circuit). And I assume by "the pressure that pump shows" you mean the pressure difference before-after the pump.
Then you'll have atmospheric pressure on both ends of the system. Consequently, it does not appear in the pump pressure. Except, if there is a significant difference in atmospheric pressure on both ends, the difference has to be included in the pump pressure.
@ingo Riess thanks ingo for the answer, when you are running the system with a specific volumetric flow rate, you can also read the pressure in pump display. the water is running in the air at the end, so atmospheric pressure at the end of the pipe makes sense but I did not understand why we have atm press at pump side? my problem is I calculate pressure loss using darcy formula and it is much smaller than: what pump shows - atm pressure
The source of the water is not shown in your drawing. It must come from somewhere. And there is - most likely - atmospheric pressure. Well, it could also be some pressurised water supply system...
And I don't know what is shown on the display of your pump. It could be all sorts of pressure (static, dynamic, total, total difference etc.).
@ ingo Riess sorry for that yes source of water is from outside. our pump only provides flow so for each flow rate and depending on my config, the pump displays a pressure (I assume that the pressure shown on the pump is resistance to flow). I calculated pressure drop in my system but it is not equal to pressure the pump displays - even 2 atm.
- Badges
- Science topic
- Similar topics
- Thermodynamics
- Pressure