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Calculating convertable energy 
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#1
Feb2113, 06:52 PM

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Hello All,
Not sure if this is the correct forum for this question but it seemed the best fit. I am trying to calculate the potential kinetic energy of water flowing through a pipe. Just your standard house pipe of 3/4 “ copper @ 40 psi and a flow rate of 5 gpm. There will be a reduction of the pipe to a .305 ID at the last 4” of pipe. It is through this section of the pipe that I am looking to calculate the energy of the water to do work. I have looked and I haven’t figured out how I would calculate this. I get a velocity of 6.6 meters/s but how do I calculate the energy potential of the water there. In useful terms like watts/HP. I am trying to design something and I need to know if there is enough energy there to make it work. My intuition says it will but that doesn’t make is so and so far I am not smart enough to figure out which equation to use or how to covert the numbers I know to get the answer. Thanks Tom 


#2
Feb2113, 11:49 PM

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Kinetic energy is ##K=\frac{1}{2}mv^2## and power would be ##P=dK/dt##
For a constant speed flow, then, ##P=\frac{1}{2}v^2dm/dt## i.e. you know v, you need to know how much mass of water flows past the point in the pipe each second. That would be the density of water times the volume flow rate. ##dm/dt = \rho dV/dt## the volume flow rate is related to the speed of the water and the crossectional area of the pipe. Enjoy. 


#3
Feb2213, 12:29 AM

P: 4

Thanks for the help but I think you just passed me in both lanes at the same time.
I can't figure out how to put the density of water in a useable number since it is 1000kg/p cubic meter. Do I need to first figure out how much water is going past per second? Then use that volume to calculate the mass as a % of 1000kg/p cubic meter. And I am not sure what each symbol in the equation represents if that is the case. dm/dt=ρdV/dt d=Density? m=mass? t=time? p=power? V=Volume? 


#4
Feb2213, 01:26 AM

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Calculating convertable energy
The "d" refers to the infinitesimal in claculus ... dm/dt means differentiate m with respect to time. ##\rho## is density. This is why I put the relation in words as well as symbols.
The formula was $$\frac{dm}{dt}=\rho\frac{dV}{dt}$$ ##\frac{dm}{dt}## is the instantaneous rate that mass is flowing through a section of pipe ... in SI units that would be kg/s An engineer would write ##\dot{m}## instead, so the expression becomes: $$\dot{m}=\rho\dot{V}$$ Putting everything in SI units will work  you'll get Watts out the other end in the power equation and Joules in the energy equation. So ##\rho=1000\text{ kg.m}^{3}## is perfectly useable as long as you put volume in cubic meters. Volume will automatically come out in cubic meters is area is square meters and speed is meters per second. You seem to be trying to do this from very little knowledge of the quantities involved. The definition of density is ##\rho=m/V## therefore ##m=\rho V## where V is the volume. The approach is to derive the equation for what you want, and then worry about the actual values. 


#5
Feb2213, 09:20 AM

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You seem to be trying to do this from very little knowledge of the quantities involved. I know the PSI of the system, I know the GPM. I know the diameter of the pipe. So I got the velocity of the water there. I will go all metric with my numbers. I don't know calculus, took years of algebra many many years ago. I will do the math with what you have provided me thanks, to see if I get anything that even seems reasonable. Thanks for taking the time, Tom 


#6
Feb2213, 10:12 AM

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It's OK  when I answer a question I have to take a guess at the background of the person asking. It could be anyone from grade school to college professors and I can normally tell from the problem being attempted. I just guessed a different kind of background from what you had that's all.
You don't have to go metric if you are unfamiliar with it  you just need to be consistent with the units. If you have area in square inches and speed in inches per second, then you want density in pounds per cubic inch ... I'm not sure how power would come out but you specified Watts so I did everything in metric. Since you got the water speed from the PSI etc then you should be able to handle the rest of the math OK ;) 


#7
Feb2213, 12:03 PM

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I am a little stuck right now, what am I not comprehending?
I did a CM conversion to try to keep it simple. With the area .45cm and the flow rate of 535.38 cm/s .45 x 535.38 = 240.73 I have a V=240 CC which equals 240 grams. So how do I get the density of 240 CC or grams of water? or is 240 that answer? I knew this would be difficult when I decided to do this before starting with making this product but I thought I was brighter than this :) 


#8
Feb2213, 08:49 PM

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Oh OK...
You want to know the power in the movement of the water given the water speed (v=6.6m/s) and the volume flow rate (dV/dt = 5gpm). Numbers off post #1. I'll walk you through the stages  the trick is to keep in mind what everything means so it's not just an abstract calculation. The formula is: ##P=\frac{1}{2}v^2\dot{m}## ##\dot{m}## is the mass flow rate and ##v## is the speed. You already have the volume flow rate (##\dot{V}##) in gpm, so we need to use ##\dot{m}=\rho\dot{V}## to get the mass flow rate. You want it in SI units. Looking up the conversion factor tells me: 1m^{3}/s = 15852gpm So what is the volume flow rate in cubicmeters per second? That is convenient units since the density of water (##\rho##) is 1000kg/m^{3} ... so what is the mass flow rate in kilograms per second? You've already done the speed calculation so v=6.6m/s So now you can find the power. 


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