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

[bcalkins]

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

1. Homework Statement :

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. Homework 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 into 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
Please please please help!

 

[Angry Citizen]

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.

 

[Chestermiller]

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

 

[bcalkins]

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?

 

[Chestermiller]

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?

Your starting equation should be

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.

 

[bcalkins]

Okay, so should P1 and P2 be taken out of the equation if they equal each other?
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
Which would lead to:
(y1-y2)/(1/2p * -1/2p) = v2^2-v1^2
Is that correct?

 

[Chestermiller]

Okay, so should P1 and P2 be taken out of the equation if they equal each other?
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
Which would lead to:
(y1-y2)/(1/2p * -1/2p) = v2^2-v1^2
Is that correct?

No. Your algebra is wrong.