# Calculate the discharge between two pipes

1. Dec 30, 2008

### amjid1709

Hi,

I have a question, it is related to flows between two resorvoirs where I need to calculate the discharge between two pipes, with different pipe diameters.

The problem I have come across is where there is an expansion in the pipe size from 250mm diameter to 300mm, the formula states that I need to use is (V1 - V2) squared does any one know what this means.

I am certain its velocity in pipe 1 minus pipe two but how to I calculate this?

Anyone?

2. Dec 30, 2008

### stewartcs

Re: Hydraulics

Hi amjid1709,

Can you provide a drawing or sketch of your setup?

CS

3. Dec 31, 2008

### amjid1709

Re: Hydraulics

Two reservoirs have a difference of 25 metres in water levels, a pipe line which of 125m long overall joins them. The upper 75m of the pipeline is 250mm diameter and the lower 50m is 300mm diameter. Local losses at entry and exit are of the form kv2/2g and K has the following values: at entry, 0.5; at exit, 1.0. At the sudden expansion v1 minus V2 squared where v1 is the velocity in the upper pipe and v2 is the velocity in the lower pipe. The friction factor is 0.028 for the upper pipe and 0.018 for the lower pipe.

Determine the rate of discharge from one reservoir to the other. Determine also the head losses along the pipeline and sketch the hydraulic gradient and total energy line.

To get the velocities in the pipes

V1 = 300/250 squared which = 1.44

if you square 1.44 you get V2 = 2.0376

and then to calculate the losess this is the method that was followed:

Entry pipe = 0.5 * 2.0376 = 1.0368

Upper pipe 0.028 * 75/ 0.025 * 2.0376 = 17.4182

Expansion V1 - V2 squared the answer I have been given is 0.44 squared which =0.1936
but I dont know how to get the answer to 0.44, I can do the rest of the calculations just this has snookered me, I bet its something easy but I just cant see it.

Any help would be appreciated.

4. Jan 4, 2009

### Chunmao Jia

Re: Hydraulics

i dont think you get the correct V1 and V2.

“V1 = 300/250 squared which = 1.44
if you square 1.44 you get V2 = 2.0376 “ (300/250)^2 you get the ratio of V1 to V2, not V1. so V1-V2=1.44V2-V2=0.44V2

5. Jan 4, 2009

### amjid1709

Re: Hydraulics

I still dont understand I have tried your workings but still cant figure it out can you elaborate please.

6. Jan 4, 2009

### nvn

amjid1709: You would need to write the Bernoulli energy balance equation, including the head losses, then solve for v1 or v2. The head loss due to a sudden expansion is he = (0.5/g)(v1 - v2)^2.

Last edited: Jan 5, 2009
7. Jan 4, 2009

### amjid1709

Re: Hydraulics

I am able to do that at the end when all head losses have been calculated the problem I'm having is where there is the sudden expansion in the pipe how do I calculate the losses at this point.

The formula given is V1 - V2 divde by 2 times gravity which is 9.81 x 2, I dont know how to do this part.

Can you help.

8. Jan 4, 2009

### nvn

You can't compute the head losses until you solve for the velocity. The head losses are a function of velocity. Include expressions for the head losses in the Bernoulli equation, then solve for v1 or v2. Is this a school assignment?

9. Jan 4, 2009

### amjid1709

Re: Hydraulics

To get the velocities in the pipes I did as follows

V1 = 300/250 squared which = 1.44

if you square 1.44 you get V2 = 2.0376

and then to calculate the losess this is the method that was followed:

Entry pipe = 0.5 * 2.0376 = 1.0368

Upper pipe 0.028 * 75/ 0.025 * 2.0376 = 17.4182

I know that I need to divde these by 2 x gravity at the end but the next step is to calculate the loss at the expansion.

Sorry but I really dont know what to do with the expansion in the pipe

10. Jan 4, 2009

### nvn

amjid1709: v1 = 1.44*v2. Therefore, if you square v1, you get 2.0736*v2^2, not 2.0376*v2^2.

The head loss due to a sudden expansion is he = (0.5/g)(v1 - v2)^2. In your case, v1 = 1.44*v2; therefore, he = (0.5/g)(1.44*v2 - v2)^2 = 0.1936(v2^2)/(2*g).

Write your Bernoulli equation from reservoir 1 to reservoir 2. The kinetic energy (velocity head) in each reservoir is zero; therefore, we can omit those two terms from the Bernoulli equation. Assuming the pipeline entrance and exit are at the same depth below each reservoir surface, the pressure head is equal in reservoirs 1 and 2, and thus cancels out of the Bernoulli equation. This leaves you with the following Bernoulli equation, where h1 = reservoir 1 elevation head relative to reservoir 2, hL = major and minor head losses, ki = inlet head loss coefficient, ko = outlet head loss coefficient, f1 and f2 are your friction factors given in post 3, and the head losses (multiplied by 2*g) are listed sequentially from beginning to end of the pipeline. Solve for v2.

h1 = hL,
where (2*g)*hL = ki*v1^2 + f1*L1*(v1^2)/D1 + 0.1936*v2^2 + f2*L2*(v2^2)/D2 + ko*v2^2.

Last edited: Jan 5, 2009
11. Jan 5, 2009

### Chunmao Jia

Re: Hydraulics

no need for me to elaberate, nuv has done it