# Fluid Mechanics and Bernoulli's principle

1. Jan 1, 2005

### apchemstudent

A pump and its horizontal intake pipe are located 12 m beneath the surface of a reservoir. The speed of te water in the intake pipe causes the pressure there to decrease, in accord with Bernoulli's principle. Assuming nonviscous flow, what is the maximum speed with which water can flow through the intake pipe?

I tried solving the problem like this:

density*g*12 = (density*v^2)/2

v^2 = 24*9.8

v = 15.336 m/s

2. Jan 1, 2005

### Staff: Mentor

Bernoulli's principle implies:

$\frac{1}{2}\rho_1 v_1^2\,+\,\rho_1 gz_1\,+\,P_1\,=\,\frac{1}{2}\rho_2 v_2^2\,+\,\rho_2 gz_2\,+\,P_2 = constant$

Now take $z_1 = 0$ as the reference, and z2= 12 m, the depth of the reservoir.

Let us assume that the density is the same at both elevations, i.e. $\rho_1=\rho_2 = \rho$

Also, the velocity of the fluid at the mouth of the intake pipe is approximately zero - the water in the reservoir is more or less at rest, so $v_2 = 0$.

Now, the maximum velocity would be achieved if there were zero pressure at the pump inlet, i.e. the pump puts a suction such the pressure $P_1 = 0$.

Then the above equation becomes:

$\frac{1}{2}\rho v_1^2\,=\,\rho\,gz_2\,+\,P_2$

Let P2 = 0.101325 MPa or 1 atm, and using the values you used before, so basically add the static pressure to the hydraulic pressure.

3. Jan 1, 2005

### apchemstudent

so

density * g * 12 + 1.01*10^5 Pa = 1/2 * density * v^2
root of (24 * 9.8 + 2*1.01*10^2) = v

v = 21m/s

thanks...

Last edited: Jan 1, 2005
4. Mar 23, 2005

### pattarkutty

diffusion through a membrane

hey,
can u suggest some simple methods of measuring fluid flow rates of a few ml per hour when a lighter/low viscous fluid is allowed to pass through a membrane a certain pressure?