# Bernoulli's equation

1. Jan 11, 2012

### Femme_physics

Again-- I'm not entirely sure if this goes to physics or engineering, but the word thing that comes in mind when I hear Bernoulli equation is PHYSICS, so I decided to post it here...

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

So I have this hydraulic system

http://img689.imageshack.us/img689/3596/clipboard01rvt.jpg [Broken]

With the figures

H1 = Height of suctioning = 3m
H2 = Height of delivery = 25m
D1 = diameter of the suctioning pipe = 150mm
D2 = diameter of the delivery pipe = 100mm
Q = 0.02 [m^3/s]
y1 = pressure lost at the suctioning line
y2 = pressure lost at the delivery line
Energy conversation efficency of the pump = 75%

Find the pressure at the pump's entrance (p1).

2. Relevant equations
In the solution

3. The attempt at a solution

So I build the equation:

http://img196.imageshack.us/img196/2145/ber1j.jpg [Broken]

http://img21.imageshack.us/img21/9740/ber2h.jpg [Broken]

HOWEVER, the result in my manual is different. It's p1 = -0.05064x10^5 [Pa]

I don't get how could we have gotten such hugely different results!

Last edited by a moderator: May 5, 2017
2. Jan 11, 2012

### Spinnor

What is the density of the fluid that is pumped? For water I get a pressure reduction of about 30,000 Pa from potential energy term and a pressure reduction of about 500 Pa from the velocity term.

3. Jan 12, 2012

### Femme_physics

The fluid is water, the density of water is 1 g/cu.cm. We work in Newton/Meter so it's 10 N/m.

How does it change Bernoulli's equation?

4. Jan 12, 2012

### Spinnor

From

http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html

Bernoulli's equation has the density in it and I don't think you mentioned water?

Good luck!

5. Jan 13, 2012

### I like Serena

Hi Fp!

I can't match your relevant equation with Bernoulli's equation.
Perhaps you can explain what the symbols in your equation stand for?

6. Jan 13, 2012

### Femme_physics

Hsa = static pressure at a
Hva = kinematic pressure at a
Za = height difference from zero point

Hsb = static pressure at b
Hvb = kinematic pressure at b
Zb = height different from zero point
Sigma Ya-b = pressure losses throughout point a to b

7. Jan 13, 2012

### I like Serena

Do you have formulas for static pressure Hs and also for kinematic pressure Hv then?

And while we're at it, also for Z and Sigma Y?

I ask, because they are not simply pressure, speed, and height.

8. Jan 13, 2012

### Femme_physics

Oh, they are.

Static pressure is just static pressure.

Kinematic pressure is V^2/2g

The height is just height, in our case a difference of 3.

9. Jan 13, 2012

### I like Serena

Okay...

But that kinematic pressure is not what you calculated in your scan...

To match with your kinematic pressure and your height, static pressure has to be:
$$H_s = {p \over \rho g}$$
where $p$ is just pressure, $\rho$ is the density of water (1000 kg/m3), and $g$ is the regular acceleration of gravity.

Where does the pressure loss come in?
In your scan you mention that it is 2, but where does that come from?

Last edited: Jan 13, 2012
10. Jan 21, 2012

### Femme_physics

ILS -- I had consulted classmates and solve it and I will post the solution later.

Our static pressure is just a value, we don't put it in a fraction. Pressure lost comes in due to the friction in pipe, length of pipe, diameter of pipe, and non-uniformity in the pipe (90 degrees change, valves in between... etc)....