How does the reference height affect Bernoulli's principle calculations?

[Avalanche]

Homework Statement

Please click on the following link for the solution.

http://s1292.photobucket.com/albums...ew&current=Bernoullisequation_zps0d4485d7.jpg

Homework Equations

P₁ + .5(rho)v₁² + (rho)gy₁ = P₂ + .5(rho)v₂² + (rho)gy₂

The Attempt at a Solution

P₁ = 101300 Pa
y1 = 0
rho = 1015 kg/m^2
v1 = 0.1*10^-2 m/s
P2 = ?
y2 = 2.0 m
v2 = 0.5 m/s

Plugging numbers into bernoulli's principle

P2 = 101300 + .5(1015)(0.1*10^-2)² + 0 - .5(1015)(.5)^2 - 1015*9.8*2
P2 = 81281.66Pa

But according to the textbook, the answer is 120,800 Pa. I get that answer when y1 = 2.0 and y2 = 0. I don't understand how the reference height could be at P2 could be at 0 because if it was, shouldn't P2 be denoted as P1?

Any help would be appreciated.

 

[voko]

The height at the manometer is two meters LESS than the height at the beer level. No matter what you use as the reference height, y1 - y2 = 2 m.

 

[Avalanche]

So for this question, the reference height of 0 is already set at the manometer? Is it possible to use the reference height at the beer level in the container?

 

[voko]

As I said, the "zero" level can be anywhere, but you must have y1 - y2 = 2 m.

 

[Nickie]

The reference height affects Bernoulli's principle calculations because it determines the potential energy term in the equation. In this case, the reference height is taken as the height at P1, which is at ground level (y1 = 0). This means that the potential energy term at P1 is 0, since the height is 0. However, in your attempt at the solution, you have taken the reference height at P2, which is at a height of 2.0 m (y2 = 2.0). This means that the potential energy term at P2 is not 0, but rather -(rho)gy2.

In order to get the correct answer, you need to use the correct reference height. In this case, it should be taken as the height at P1, which is 0. So, the correct equation would be:

P1 + .5(rho)v12 + (rho)gy1 = P2 + .5(rho)v22 + (rho)gy2

P1 = 101300 Pa
y1 = 0
rho = 1015 kg/m^2
v1 = 0.1*10^-2 m/s
P2 = ?
y2 = 0
v2 = 0.5 m/s

Plugging these values into the equation, we get:

P2 = 101300 + .5(1015)(0.1*10^-2)^2 + 0 - .5(1015)(0.5)^2 - 1015*9.8*0
P2 = 120,800 Pa

This is the same answer as the one given in the textbook. So, in summary, the reference height affects the potential energy term in the equation, and it is important to use the correct reference height in order to get the correct answer.