Siphon in a Juice Bar?

Mastering Physics frustrates me in that I finish all the problems except for one or two and those just sit there as I try until my tries run out.

I looked at this problem and thought Bernouli’s equation. A siphon’s velocity is based on the difference between the water level of the glass the water is flowing from and either (i) the water level of the glass the fluid is flowing to, or (ii) if the fluid exits above water level, the height the fluid exits. But I kept reading and re-reading and nothing made sense.

Then I realized your problem was ambiguous. The following is the correct way the second to last sentence should be constructed:

“The beginning of the straw is below the surface of the cranberry juice. The end of the straw is a point above the surface of the orange juice. Let h_o be the height differential between the surface of cranberry juice and the end point of the straw located above the orange juice glass.”

Also Read :   What’s a subdermal neurophone?

Anyhow, with this reading, it will be v = sqrt (2*g*h_o)

The easy way this is derived is that the Kinetic Energy at exit is equal to the potential energy of the drop in elevation. Basically 1/2 mv^2=mgh or v=sqrt (2gh)

For details see http://en.wikipedia.org/wiki/Siphon and scroll down to Bernouli’s equation.

Siphon Bar

Leave a Comment