# Bernoulli Equation and flow loss

1. Oct 30, 2012

### danny_smith

1. The problem statement, all variables and given/known data

Liquid, specific density 0.8, flows with velocity 4 m/s
in a pipe that has a downward slope of 1:50. At a
certain point in the pipe, a pressure gauge shows a
pressure of 80 kPa. Determine the pressure at a
point 200 m downstream of the gauge if:

flow losses are ignored;
and,there is a flow loss equal to 10% of the total
2. Relevant equations

Bernoulli equation

3. The attempt at a solution
Solved the first part and got 111,360, the second part is where I'm going wrong.

So for the second part, first I find the total initial head
P1/pg +V1^(2)/2g +z = H
(80*10^3)/(800*9.8) + (4^2)/(2*9.8) + 0
I get 11.10, so 0.1* 11.1 = 1.11
therefore pgh = P = 8000*9.8 *1.11 = 8702.4,
then simply subtract for P2, 11360 -8702.4 = 102657.6 Pa

However this answer is wrong, what am I doing wrong?

Last edited: Oct 30, 2012
2. Oct 31, 2012

### Basic_Physics

I don't think it is that simple.
It is not necessary to include the velocity head since the pipe diameter is (assumed) to be constant. The flow loss means that
$0.9\frac{p_{upper}}{\rho g}=\frac{p_{lower}}{\rho g}-200$

Last edited: Oct 31, 2012
3. Oct 31, 2012

### danny_smith

But solving that equation for Plower results in the wrong answer also.

4. Oct 31, 2012

### danny_smith

By the way i think you got your Z2 mixed up it should be 4, not 200 since the slope is 1:50, either way you're answer is still wrong.

Last edited: Oct 31, 2012
5. Oct 31, 2012

### Basic_Physics

How did you get at 111,360 Pa?

6. Oct 31, 2012

### danny_smith

p1/pg = 80*10^3/800*9.8
v1/2g =0 (because it cancels)
z1=0

v2^2/2g = 0 (cancels)
z2=-4

so p2 = (80*10^(3))/(800*9.8 +4)*800*9.8 = P2

Last edited: Oct 31, 2012
7. Oct 31, 2012

### Basic_Physics

I think we have to assume that the velocity head is also going to be reduced, not just the pressure head.

8. Oct 31, 2012

### Basic_Physics

The equation then might be:
$\frac{p_{upper}}{\rho g}=-4+\frac{p_{lower}}{\rho g}+1.11$

9. Oct 31, 2012

### danny_smith

Hi Basic_Physics thanks for your input thus far, however your equation still results in the wrong answer.

10. Nov 1, 2012

### danny_smith

I know the steps to the solution but I don't understand why those steps are used, can anyone offer some input?

11. Nov 1, 2012

### Basic_Physics

As far as I can find out this type of problem is usually done via:
hupper - hlower = hloss
but the problem is that both the pressure and velocity head will be changed due to the loss.