# Venturi effect

1. Oct 9, 2006

### laminar

A Venturi device has a diameter of 4mm at one end and a diameter of 2cm at the other. Air enters at 1200cm^3/m. Mercury is in the botom of the device. Assuming mercury's density to be 13700kg/m^3, and air's density to be 1.2kg/m^3, find how high the mercury rises. Assume air to be an ideal fluid.

I got ridiculously large numbers for this.

Q=Av

12m^3/s=pi(0.002^2)v

v=954925m/s

This is outrageously fast, and I did the same calculation for the other end and got 38197m/s, using a 1cm radius.

I don't think this is quite right. Are these velocities correct? The next tep is to plug the velocities into Bernoulli's equation:

(Mercury's density)(g)(h)=(0.5)(density of air)(v2^2-v1^2)

And solve for v, but I got around 400000m for the answer. I know it is incorrect.

Am I doing something wrong here?

2. Oct 9, 2006

### laminar

I found out that I was converting the cm^3 to m^3 wrong. I calculated the height of the mercury in the tube to be 4.1, and it is wrong according to MasteringPhysics.