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Bernoulli's equation application

  • #1
[Two circular holes, one larger than the other, are cut in the side of a large water tank whose top is open to the atmosphere. Hole 1 is the larger hole, and hole 2 is the smaller hole. The center of one of these holes is located 6.61 times as far beneath the surface of the water as the other. The volume flow rate of the water coming out of the holes is the same. (a) Decide which hole (1 or 2) is located nearest the surface of the water. (b) Calculate the ratio of the radius of the larger hole to the radius of the smaller hole, rA/rB.

I know this has something to do with bernoulli's equation but I don't know how to apply the formula to solve this problem.
 

Answers and Replies

  • #2
tiny-tim
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Hi moeraeleizhaj ! :smile:

Show us what equations you've got from Bernoulli, and where you're stuck, and then we'll know how to help. :wink:
 
  • #3
Bernoulli’s Equation

In the steady flow of a nonviscous, incompressible fluid of density (*rho*), the pressure P, the fluid speed v, and the elevation y at any two points (1 and 2) are related by :

P1+1/2[*rho*(V1)^2]+[rho*g*(y1)]=P2+1/2[*rho*(V2)^2]+[rho*g*(y2)]

^That's Bernoulli's equation.
Thanks ^^
 
  • #4
tiny-tim
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Bernoulli’s Equation

In the steady flow of a nonviscous, incompressible fluid of density (*rho*), the pressure P, the fluid speed v, and the elevation y at any two points (1 and 2) are related by :

P1+1/2[*rho*(V1)^2]+[rho*g*(y1)]=P2+1/2[*rho*(V2)^2]+[rho*g*(y2)]

^That's Bernoulli's equation.
Thanks ^^
Well, go on then …

what's the equation for the volume flow rate?
 
  • #5
P1+1/2[*rho*(V1)^2]+[rho*g*(y1)]=P2+1/2[*rho*(V2)^2]+[rho*g*(y2)]
Right, P1 and P2 are the atmospheric pressures. Since the atmospheric pressure is the same on both holes, the Ps will cancel each other out. Same with the density (*rho*), since the density is the same throughout the whole equation (water is incompressible), you can cancel it out throughout the equation.

So the equation will be:
1/2[V1)^2]+[g*(y1)]=1/2[(V2)^2]+[g*(y2)]

~> The equation for volume flowrate: Q=A*v
where Q=volume flowrate
A=cross sectional area
v=Velocity

.'. formula for v=Q/A
Since the 2 Qs are equal then we can express the v1 and v2 in the equation in terms of A and Q, i.e V1=Q/A1 and V2=Q/A2

.'. the formula will look like this:
1/2[(Q/A1)^2]+[g*(y1)]=1/2[(Q/A2)^2]+[g*(y2)]

Since "Hole 1 is the larger hole, and hole 2 is the smaller hole" .'. A1 is bigger so the V1 will be smaller cause Q is constant.

Does that mean that Y1 will be the hole in the bottom? Since we have to compensate for the smaller Q/A1 to make both sides equal.

Is my solution right?
 
  • #6
Btw, I'm kind of stuck on the second bit. The finding the ratio part of the question.
Thanks ^^
 
  • #7
tiny-tim
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1/2[V1)^2]+[g*(y1)]=1/2[(V2)^2]+[g*(y2)]

~> The equation for volume flowrate: Q=A*v

.'. formula for v=Q/A
Since the 2 Qs are equal then we can express the v1 and v2 in the equation in terms of A and Q, i.e V1=Q/A1 and V2=Q/A2

.'. the formula will look like this:
1/2[(Q/A1)^2]+[g*(y1)]=1/2[(Q/A2)^2]+[g*(y2)]
Hi moeraeleizhaj! :smile:

You see how easy it is if you're systematic? :smile:

Physics is equations … so every solution involves writing the equations, and carrying on from there. :wink:

Two points:

i] I'd have used r instead of Q, since rA and rB are given in the question: Q = πr2v

ii] Your Bernoulli equation is right, but you can actually say that both sides are zero … can you see why? :wink:

Carry on from there … the zero should make the ratio part easier! :smile:
 
  • #8
o: Ehm. sorry but why are both sides equal to zero?
 
  • #9
tiny-tim
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large tank

o: Ehm. sorry but why are both sides equal to zero?
Follow a streamline up to the top of the water …

the velocity there is zero (the question says it's a large tank, so you can assume that) …

so if you fix the height there as zero also, then the constant is zero. :smile:
 
  • #10
Ohhh, so you mean I had to solve for the depths of the holes relative to the top of the tank and not relative to each other? I had to do it separately so?

Btw, thanks very much for the help ^^
 
  • #11
tiny-tim
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Ohhh, so you mean I had to solve for the depths of the holes relative to the top of the tank and not relative to each other? I had to do it separately so?
Yup! :biggrin:

The moral is, always study the question … it often has a clue as to what you need to solve it:
… The center of one of these holes is located 6.61 times as far beneath the surface of the water as the other …
In this case, the question only gives you the ratio of the distances from the surface

so you can guess that the surface height has a particular significance! :wink:
 

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