# Bernolli's Equation: vb^2-va^2 = ?

1. Nov 3, 2012

### bcalkins

Bernolli's Equation: vb^2-va^2 = ?????

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

Use Bernoulli’s principle to obtain a relation between the speeds of water at points A and B, vA and vB, in terms of the acceleration of gravity g and the height of the water column h. Your expression should be of the form: vb^2-va^2 =
Let’s do a numerical check that your equation is correct. If the height of the water level is h = 27 cm, what is the value of vb^2-va^2, in SI units?

2. Relevant equations

3. The attempt at a solution
I really have no idea where to start, to me, the question doesn't seem really obvious. I just need this last problem solved to be permitted in to lab on Monday morning. Please help me in any way, thank you so much!
This is what I have as bernolli's equation in my notes, but how do I use any of this to solve for vb^2-va^2????
P1+1/2pv1^2+pgy = P2+1/2pv2^2+pgy
I know P=pressure
p=density
y= height

2. Nov 3, 2012

### Angry Citizen

Re: Bernolli's Equation: vb^2-va^2 = ?????

You don't know the algebraic manipulations required to move all terms with a "v" in it to one side, and all the terms without "v" in it to the other? Also, some context (read: a picture) of the problem would be appreciated.

3. Nov 3, 2012

### Staff: Mentor

Re: Bernolli's Equation: vb^2-va^2 = ?????

Assume P1 = P2, and let va = v1 and vb=v2. Incidentally, the y's on the two sides of the equation should have subscripts.

4. Nov 3, 2012

### bcalkins

Re: Bernolli's Equation: vb^2-va^2 = ?????

Thank you Chestermiller, I feel really stupid that I can't do simple algebra.
I think based on P1=P2 you could just take that out of the equation.
To move pgy you would subtract from both sides so you'd be left with:
1/2pva^2 = 1/2pvb^2
How would I get va^2 to the right side of the equation. Wouldn't I have to move it by dividing both side by it? So I wouldn't be left with vb^2 - va^2 but instead vb^2/va^2.
What am I doing wrong?

5. Nov 4, 2012

### Staff: Mentor

Re: Bernolli's Equation: vb^2-va^2 = ?????

P1+1/2pv1^2+pgy1 = P2+1/2pv2^2+pgy2

As I said in my previous response, there should be subscripts on the two y's,

and y2 - y2 should be called h.

6. Nov 4, 2012

### bcalkins

Re: Bernolli's Equation: vb^2-va^2 = ?????

Okay, so should P1 and P2 be taken out of the equation if they equal eachother?
If so, you'd be left with 1/2pv1^2+pgy1 = 1/2pv2^2+pgy2
So, to get the v2^2-v1^2 = ???? as the question asks; how do I get everything else to one side? The farthest I get to is:
pgy1-pgy2 = 1/2pv2^2 - 1/2pv1^2
to simplify wouldn't it be:
y1-y2 = 1/2pv2^2 - 1/2pv1^2
(y1-y2)/(1/2p * -1/2p) = v2^2-v1^2
Is that correct?

7. Nov 4, 2012

### Staff: Mentor

Re: Bernolli's Equation: vb^2-va^2 = ?????