Homework Help: Bernoulli's equation does not always work?

1. Feb 26, 2013

theBEAST

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

3. The attempt at a solution
Alright so first I found a relationship between z1 (depth of inflow) and z2 (depth of outflow) using mass conservation. I found that 5 * z1 = z2 and I know this is correct because the answer key has the same relationship.

Next I decided to use bernoulli's equation to find another equation to relate z1 and z2. So using the streamline on the free surface:

P1 + 0.5 * rho * V1^2 + rho * g * z1 = P2 + 0.5 * rho * V2^2 + rho * g * z2

Since the streamline is on the free surface, P1 = P2 = Patm, so they pressures will cancel out and I am left with:

0.5 * rho * V1^2 + rho * g * z1 = 0.5 * rho * V2^2 + rho * g * z2

Plugging the numbers in, cancelling the rhos and substituting z2 = 5 * z1:

0.5 * (5)^2 + (9.81) * z1 = 0.5 * (1)^2 + 9.81 * (5 * z1)

Solving this yields z1 = 0.3058m...

HOWEVER this is not the same answer as the one given in the solution manual... The solution manual uses conservation of momentum. Why is it that bernoulli's does not work in this case?

2. Feb 27, 2013

voko

Hydraulic jumps dissipate energy. Bernoulli's equation is about conservation of energy. The two are not compatible.

3. Feb 27, 2013

theBEAST

Why and how does it dissipate energy?

4. Feb 27, 2013

voko

The laminar flow is turned into turbulent; the velocity of 1 m/s is the average velocity of the flow, not the real (local) velocity anywhere of the fluid, which is much greater.

See http://en.wikipedia.org/wiki/Hydraulic_jump