Fluid flow rate and friction problem

In summary, the conversation discusses a problem involving finding the water flow rate in a pipe for irrigation purposes, given certain data and assumptions. The solution involves using the Bernoulli equation and the Darcy formula to estimate the velocity and head loss, and then using trial and error to find the correct velocity that balances the pressure drops and losses. The final solution is approximately 26 kg/s.
  • #1
Hello, I'm new here - so apologies in advance if this is in the wrong place, also I haven't done physics for a few years prior to the course I just started - and I've never tried to write out formula in computer text, so apologies in advance again if everything I write is completely wrong! :tongue2:

Homework Statement

A Farmer builds a water storage tank 12m above ground, as shown below:

[PLAIN]http://img408.imageshack.us/img408/6195/irrigationsystem.jpg [Broken]

The water is fed via a 200m long, 125mm diameter pipe to a field for irrigation purposes. The pipe has a friction factor of 0.008. Using the K data below, show that the water flow rate (kg/s) in the pipe is approximately 26kg/s


K Factors: Tank Exit 0.5, Bends 0.9 (x2), Valve 1.0 (Total: 3.3)
Water Density: 1000kg/m3

Homework Equations

Bernoulli's Equation: (P_1/ρ*g) + (V_1^/2*g) + z1 = (P_2/ρ*g) + (V_2^/2*g) + z2
Darcy Formula: H_f = 4*f*(L/D)*(V^/2*g)
Moody Diagram

The Attempt at a Solution

Right then, I've been asked to find the water flow rate, which I'm presuming means the mass flow rate (kg/s) - so in order to find that I need the volume flow rate, and in order to find that I need the velocity

I haven't been given the velocity in the original problem, however I think I need to use the Bernoulli Equation in order to find the velocity.

I've made a few assumptions with the Bernoulli Equation, which are that the pressure at point 1 (the water tank) is atmospheric (101325pa) - because on the picture the tank looks to be open to atmosphere, I've also assumed that the pressure at point 2 (the end of the pipe) is also atmospheric (101325pa) because the pipe is leading to a field.

As well as this I've assumed that the velocity at point 1 is approximately 0m/s (because the water is going from a large tank into a small pipe)

So at the moment, my Bernoulli equation looks like:

(101325 / (1000 x 9.81)) + (0^ / (2 x 9.81)) + 12 = (101325 / (1000 x 9.81)) + (V_2^ / (2 x 9.81)) + 0

However I understand that this equation isn't full, as I also have to take into account the friction (head loss), which would change the Bernoulli equation into:

(P_1/ρ*g) + (V_1^/2*g) + z1 = (P_2/ρ*g) + (V_2^/2*g) + z2 + h_f

So in order to accurately find the velocity using the Bernoulli equation, I have to also take into account the Head loss?

But the Darcy Formula is: H_f = 4*f*(L/D)*(V^/2*g)

So in order to use the Darcy Formula I have to know the velocity - but I don't know the velocity (as that's what I'm trying to find)? - Unless there's a way of finding the head loss without knowing the velocity?

Which somewhat puts me at a dead end.

Another route I was thinking of using to find the velocity was to use the Moody Diagram to try to find the Reynolds number, then using Re=(ρ*V*D)/μ to find the velocity (as I know all the other values.)

However the only other Moody diagram value I know is the Friction Factor of 0.008, and I'm unable to find the Relative roughness (k/D) as the problem doesn't state what sort of pipe is being used (so I can't find the roughness.)

You can probably tell by now that I'm completely stumped by this - and as far as I know I'm probably doing it all completely wrong (it seems no matter what I try to do I always end up with a missing variable), so any help would be very much appreciated. :smile:

(Updated to include picture.)
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  • #2
In a problem such as this a trial and error solution is what's generally done. You know the total head pressure in the reservoir due to its height above ground. You are given the K factors that, if I recall correctly, are multiplied by velocity squared terms to define head losses.

So what you can do is assume a flow rate. With that you can determine a velocity. Using the velocity, then compute how the pressure drops as the water moves along the pipeline recognizing that when it exits the pipeline, the pressure of the water is atmospheric. (When water comes out of a pipe, it doesn't suddenly expand in all directions. This implies atmospheric pressure.) You should see that with the supplied flow rate, the end pressure is approximately atmospheric.
  • #3
I agree with LawrenceC that the solution requires a trial and error solution. Remember that the heat loss includes the pipe AND connector losses. You need to guess a velocity so you can calculate the Darcey friction factor. I would start by neglecting the heat loss term that has the friction factor in it. That will allow you to solve for a velocity directly. That will be your first guess. Use that velocity to find Re and friction factor and solve for the pipe head loss. Your general equation will be something like 12 ft= V^2/2*g + total head loss due to friction in pipe and components. Your velocity guess can be compared to 12 ft. Choose another velocity guess, find new f plug into add terms on right side of equation and compare to 12 ft. Keep guessing a velocity compare to 12 ft. Once you have the correct velocity you know the area of the pipe and density of water you will have the mass flow rate.
  • #4
I think I understand, so V^2/2*g + the loss from pipe friction + the loss from fittings, should add up to 12m (the height of the start point) - so what I need to do is insert different velocities into the Darcy Formula until I find one that adds up to 12?

I did some trial and error with different velocities (started with 1, then 2, which were both too low - then tried 2.1, which was too high, then 2.05, which was too low, and finally 2.06, which gave me:

2.06^2 / (2 * 9.81) = 0.216

Friction: 4 * 0.008 * (200 / 0.125 ) * (2.06^2 / (2 * 9.81) = 11.074

Fittings: 3.3 * (2.06^2 / (2 * 9.81) = 0.714

Total: 0.216 + 11.074 + 0.714 = 12.004

So as 2.06 goes into the formula to make 12, that means the velocity is 2.06m/s, and such the volume flow rate is 0.0253m3/s (2.06 * 0.0123) and mass flow rate is 25.3kg/s (1000 * 0.0253) and thus the water flow rate is 25.3kg/s?

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