Hi there I have a 19mm pipe. I will measure the flowrate of the water by filling a known volume container and timing it. Is it possible from this information to work out the pressure of the water coming out in psi or something similar? Help appreciated. Thanks
Welcome to PF! Hi dyllos! Welcome to PF! Hint: Bernoulli's equation will give you a relation between pressure and speed, so combine that with an equation relating speed and area and flow rate.
You didn't say if you wanted static or velocity pressure, but here's another hint: if the water is out in the atmosphere, the static pressure is zero.
At the moment the pipe is open to the atmosphere, but if the pressure is going to be high enough it will be connected to a gas hot water system (the pipe is coming from a tank higher up, by gravity). I need to know the pressure before having a go at connecting it to the hot water system as that is not a small job. Would the pressure be increased if the pipe size decreased as I think the gas hot water inlet is smaller than 19mm. Thanks Dylan
I timed it today and it took 9 minutes to fill a 200 litre container - I am guessing the pressure is going to be quite weak. Any help in solving Brenouli's for this situation would be appreciated, thanks
Pressure = Force/Area so yes, if the force remains the same, decreasing the diameter and thus the area will increase the pressure.
I'm still not exactly sure how to calculate the pressure (in psi or Pascals) - more or less - doesn't have to be extremely accurate. Thanks Dylan
Pressure = Force/Area so if you know the force, and you can measure the area, to quote john madden, "BOOM" you know the pressure. Assuming that the tank is open at the top so that it is at atmospheric pressure, then the force is just due to the mass of water. [tex]\frac{9.8 m/s * m kg}{Area}=pressure[/tex] where m is the mass of water.
Sorry I missed this: I'm still not clear on what you are trying to do, but I would say that you probably would want to cap the pipe and find the static pressure behind it in that condition. What is the source of this water? Assuming a system with a considerable amount of pressure loss, using smaller pipes will increase the static pressure, when the fluid is moving. The maximum static pressure is achieved when you close the pipe, which is a good thing to know.
That sounds like a pretty typical flow rate through a faucet of a domestic water system. If the pipe is just wide open, that's a pretty low flow rate, yes. I really think you're barking up the wrong tree there: there is so much pressure lost in a domestic water piping system that finding the exit velocity pressure is not very useful. To know how much pressure you'll have to deal with, you need to measure the static pressure when the system is closed (no flow).
I have a problem along the same lines and I'm hoping to get some help. I have a residential RO system and it wastes a lot of water by pumping the waste water down the drain. I'm trying to divert this water to my pool to reduce water needed to overcome evaporation. To do this, I've attached the RO waste line to a pool pump line which is only used a few hours per day. I'm connecting to the an input water valve used to top off the pool. Just above this line is a vacuum break, so the pressure is at atmosphere most of the time. However, I'm concerned that when the pump is operating, the line will become pressurized from the pump pushing the water along. I want to figure out how to calculate that or to just understand it. It may not have much pressure because the other end is just dumped into the pool. here is certainly some resistance in the line, but what would the back pressure be on my RO drain line? I am trying to determine if I will need to install a normally open solenoid valve that closes when the pump turns on to shut off the RO system. That will be pretty easy to do, but I can't for the life of me figure out if it's even necessary.
You can use reduced Bernoulli equation in SI units as P1 = P(atmosphere) + 500(v1^{2}-v2^{2}) 500 = half of water density. P1 is pressure of interest in the pipe, v1 is the velocity of water at the same point in the pipe. I don't see any problem in setting v2 = 0 if water is coming of the pipe [someone can correct me]. You can calculate v1 by timing and collecting water in a bucket and using flow rate v1 = Q/A. Q = volume of water per sec. = m^{3}/sec A = Area of pipe opening. BTW: someone here said if you lower the diameter of the pipe, pressure gets higher. This always confuse me because according to Bernoulli's equation, if velocity is higher , pressure is lower. Lowering diameter increases velocity. Can anyone explain?
Quick go at this. The pressure of the water coming out of the pipe is roughly: air pressure ( about 1000 kilo Pascals )
hi.. umm my topic is just a little different from the thread. i want to find a relation between pressure and time in pressurisation of a gas pipeline. any equations which will be used to get the expression relating pressure and time as i need to find how much time would it take to pressurise a gas pipeline starting from atmospheric pressure to the desired pressure. please reply asap! im sorry if im going a little out of topic
The increased pressure would propagate as a pressure wave through the pipe at the speed of sound. Can you give more details as to how you are increasing the pressure. The pressure in a closed pipe would increase in proportion to the incoming flow rate, and in inverse proportion to the volume of the closed pipe. Not sure about an open pipe.
No. You could have high pressure and lots of resistance, or low pressure and little resistance and still get the same flow. How high up is the tank from the outlet? Water exerts about 0.433 psi for every foot of height, or 1.422 psi per meter. That will give you the static pressure at the outlet. Is the pipe from the tank to the outlet all 19 mm? How long is it?
Basically its a closed pipe of 36inches(dia).pressurisation of the pipe need to be done by a compressor.we need to rise the pressure from atmospheric to 98kg/cm2g.its a compressible unsteady flow.basically need to find a relation between pressure and time as to how much time would it take to achieve that pressure.
Why not just calculate the "Head" pressure and use that? If you know how high above the outlet the top of the water is then using the the weight of a 1 square inch column of water that high gives you the PSI of the water with the pipe shut off from flow (static or "head" pressure). There are losses in the pipe due to friction and disturbed flow but with a gauge mouted to the pipe about a foot above the outlet you can measure the flow pressure easily - you could also measure the static pressure by shutting off the flow with the same gauge. Paul