# Flow through a pipe question

1. Jan 10, 2014

### O2F

I have a flow through a pipe that changes in diameter to become narrower. I can measure the pressure P1 at the wider end where it is applied and I want to calculate the pressure P2 at the narrow end.

I can't directly apply Bernoulli's equation here because I have no way of measuring the flow rate at either end. I do however know all of the relative dimensions of the pipe (i.e. I know the cross-secional area through which the water flows at P1 and the smaller area at P2), and assuming the water is incompressible the volume flow is constant, I can substitute v2 = (A1/A2)v1. But simplfying leaves P2 in terms of v1: -

P2 = P1 + 0.5*rho*( V1^2 - ( (A1/A2)*V1 )^2 )

Is the pressure at the narrow end simply scaled linearly by the ratio of the cross sections? I have a feeling I have solved this problem before and found this to be the case, I just can't see it at the moment.

2. Jan 10, 2014

### Staff: Mentor

That equation is definitely not linear.

3. Jan 10, 2014

### O2F

Yes, thank you, I can see this.

I should have kept this simpler: look at the attached drawing, the diameters d1 and d2 and the pressure P1 are all known quantities.... how can I calculate P2?

Any ideas?

#### Attached Files:

• ###### drawing1.png
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4. Jan 10, 2014

### Andy Resnick

The value of P2 depends on the detailed geometry ('head loss')- the shape of the contraction and how far downstream you are measuring P2- there is a region directly past the vena contracta where the velocity head is reconverted back into a pressure head and flow is unsteady.

Using the continuity equation, the head loss 'h' due to the contraction is:

h = (1/C -1)^2* V2^2/2g, where V2 is the exit velocity well downstream of the contraction, g is gravitational acceleration, and C the 'contraction coefficient'.

5. Jan 10, 2014

### Staff: Mentor

I don't understand: you provided the correct equation in your first post. Have you tried to use it yet?

6. Jan 13, 2014

### O2F

Thank you both for your help.

I'm not sure I completely understand Andy's reply.
I probably should have said at the start that I am just trying to get an estimate of the pressure at the constriction (the tightest part of the cone), I drew the tube afterwards to simplify things (in reality the cone sits inside a larger reservoir which is at atmospheric pressure).

I know there is a simpler way to solve this starting with Bernoulli's expression, we are definitely overcomplicating this, I have done it before a long time ago I just can't remember how.

Thank you for the reply Russ, I can't use the equation in the first post because I cannot measure the flow velocity. I know the input pressure and the diameters at each end of the cone.

7. Jan 13, 2014

### sophiecentaur

The pressure difference depends upon the flow rate so you can't ignore it. With no flow, the pressures will all be the same static value (for a level section of pipe).