Bernoulli's

[467167]

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Hey everyone.

So I just came across a problem in TPRH Science Workbook related to Bernoulli's equation and I'm confused about the relationship now.

I assumed that if pressure increases, then the velocity would decrease. However, this problem says that because the pressure higher as you get lower, velocity would increase out of a hole in a tank.

 

[sillyjoe]

It is very hard to understand specifically what you are asking. Can you post a picture of the problem or at least the text of the problem?

However, in problems like this the two pertinent equations are Bernoulli's equation and the continuity equation.

Continuity equation = Flow = (cross sectional area)1 * (velocity)1 = (cross sectional area)2 * (velocity)2

From this equation we can see that as the cross sectional area gets smaller the velocity increases.

You can then use Bernouilli's equation to apply the continuity equation:

K = Pagainst the walls + 1/2pv^2 + pgh

where K = a constant and p= density

From this equation we can see that as the velocity increases the pressure against the wall decreases and vice versa.

Since I don't have your specific question it is hard for me to try to apply what I just said to your question but hopefully the explanation can lead to some clarity when solving the problem.

 

[Schenker]

I'm assuming you are comparing two water tanks, each with a hole in its side, and one hole is higher than the other. (We can even say that both holes are on the same tank.)

Velocity and pressure do have an inverse relationship, as you said. However, once outside the tank, water will have the same pressure whether coming from the higher or lower hole, so that relationship won't be of predictive use here.

As water exits the tank, it's pressure will be 1 atm (atmospheric pressure), no matter the hole. Thus, the P term in Bernoulli's equation is the same for water going through both holes.

On the other hand, the higher hole (which has a greater height, h) has a greater (rho)(g)(h) term and thus must have a lower (1/2)(rho)(v^2) term. The only way for the (1/2)(rho)(v^2) term to be lower in a fluid of fixed density (rho is density) is for velocity (v) to be lower.

 

[Czarcasm]

I'm assuming you are comparing two water tanks, each with a hole in its side, and one hole is higher than the other. (We can even say that both holes are on the same tank.)

Velocity and pressure do have an inverse relationship, as you said. However, once outside the tank, water will have the same pressure whether coming from the higher or lower hole, so that relationship won't be of predictive use here.

As water exits the tank, it's pressure will be 1 atm (atmospheric pressure), no matter the hole. Thus, the P term in Bernoulli's equation is the same for water going through both holes.

On the other hand, the higher hole (which has a greater height, h) has a greater (rho)(g)(h) term and thus must have a lower (1/2)(rho)(v^2) term. The only way for the (1/2)(rho)(v^2) term to be lower in a fluid of fixed density (rho is density) is for velocity (v) to be lower.

Welcome back 😉

 

[467167]

Hi! Sorry about the delay. So the question is "Which plug should be removed for the emerging system of water to have the greatest exit speed?"

Also, @Schenker does that mean that the terms pgh and 1/2pv^2 have an inverse relationship as well if P = 1 atm?

 

[NextStepTutor_1]

In this case, it is important to remember that Bernoulli's equation has three terms, and for a given system, they are equal to a constant value.

So: P + 1/2[rho]*v^2+ [rho]*g* y = k.

So, it's true that for a given height, higher pressure means lower speed, but in this case, we are looking at varying heights. Thus, to see the largest speed, we need to look for how to minimize the other two terms (pressure, and height). In this case, pressure is constant for all three plugs because they are open to the air. So the only other thing we can affect is height, so we pick the plug with the lowest height and that will have the highest speed.