Bernoulli's equation

Gold Member
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...

Homework Statement

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).

In the solution

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!

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Spinnor
Gold Member
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.

Gold Member
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?

Spinnor
Gold Member
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?
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!

I like Serena
Homework Helper
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?

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

I like Serena
Homework Helper
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.

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

I like Serena
Homework Helper
Oh, they are.

Kinematic pressure is V^2/2g

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

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

Static pressure is just static pressure.
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?

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Gold Member
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)....