Velocity of fluid flowing from a hole at the bottom of a conical shape

In summary, the conversation discusses the application of Bernoulli's equation in a hypothetical case involving a cone-shaped container. The approach is to consider the pressure at cross section A first and to use the full Bernoulli equation to relate a point at the top of the cylinder to a point at area A. The resulting equation involves factors of 2 and the atmospheric pressure, and the area of the top of the cone can be expressed in terms of the area A. However, there is still some confusion and disagreement over the final answer.
  • #1
Adi6677
10
3
Homework Statement
So in the question we have to find the speed of flow of fluid from the hole of the conical shaped vessel
Relevant Equations
Bernoulli's Equation :
$$ P + dgh + dV^2/2 = P' + d'gh + d'V'^2/2 $$
Equation of continuity
$$VA = V'A' $$
My approach was that I consider the pressure of cross section A first
Pa= P + dgh

Now by writting Bernoulli's Equation between the cross-section A and the opening :
$$ Pa + 2dgh + 2dV^2/2 = Pb + 2dV'^2/2$$
Where Pb is the pressure of the opening which is equal to the atmospheric pressure

And also, equation of continuity between or A and B :
$$ VA = V'a$$
$$ => V' = VA/a $$

Now putting these values in Bernoulli :

$$ ( P.air + dgh) +2dgh + 2dV^2/2 = P.air + 2dV'^2/2$$
$$ => 3gh = V'^2- V^2$$
Now putting value of V in terms of V'
$$=> 3gh = V'^2 - V'^2(a/A)^2$$
$$ => 3gh = V'^2(1-(a/A)^2)$$
$$ => √3gh/√(1-(a/A)^2) = V'$$Now there is not a single option similar in the question can some help me and tell me where I am wrong?
 

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  • #2
Adi6677 said:
My approach was that I consider the pressure of cross section A first
Pa= P + dgh
Here you're assuming fluid statics which does not take into account the flow. You'll need to use the full Bernoulli equation to relate a point at the top of the cylinder to a point at area ##A##.

Now by writting Bernoulli's Equation between the cross-section A and the opening :
$$ Pa + 2dgh + 2dV^2/2 = Pb + 2dV'^2/2$$
Where Pb is the pressure of the opening which is equal to the atmospheric pressure
OK. looks good.
 
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  • #3
TSny said:
Here you're assuming fluid statics which does not take into account the flow. You'll need to use the full Bernoulli equation to relate a point at the top of the cylinder to a point at area ##A##.

OK. looks good.
K thanks but the area of the opening of cone isn't given can it be found out in terms of A ??
 
  • #4
Adi6677 said:
K thanks but the area of the opening of cone isn't given can it be found out in terms of A ??
Yes.
 
  • #5
TSny said:
Yes.
Okay I tried but I don't know what's the relationship between them is it something to do with similarity ??
 
  • #6
Adi6677 said:
Okay I tried but I don't know what's the relationship between them is it something to do with similarity ??
Yes. You might think about how the radius of the top compares to the radius of the area ##A##.
 
  • #7
I don't understand the whole configuration btw. How does the top fluid flow through the area A while the density is ##\rho## above A and ##2\rho## below A. Is there some special device at A that compresses the fluid? And how velocity of fluid at the opening remains constant in time...
 
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  • #8
Delta2 said:
I don't understand the whole configuration btw. How does the top fluid flow through the area A while the density is ##\rho## above A and ##2\rho## below A. Is there some special device at A that compresses the fluid? And how velocity of fluid at the opening remains constant in time...
It's just a hypothetical case to help one get used to applying Bernoulli I guess it is a sample paper of an exam conducted after high school so I don't think much dept is given in it
 
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  • #9
TSny said:
Yes. You might think about how the radius of the top compares to the radius of the area ##A##.
So I tried the question and reattempted the question 2-3 times but the answer I got still wasn't correct. after following ur advice the final answer I got was $$ V' = √(6gh)/√(1-16a^2/A^2)$$
 
  • #10
TSny said:
Yes. You might think about how the radius of the top compares to the radius of the area ##A##.
So I tried the question and reattempted the question 2-3 times but the answer I got still wasn't correct. after following ur advice the final answer I got was $$ V' = √(6gh)/√(1-16a^2/A^2)$$
 
  • #11
Adi6677 said:
So I tried the question and reattempted the question 2-3 times but the answer I got still wasn't correct. after following ur advice the final answer I got was $$ V' = √(6gh)/√(1-16a^2/A^2)$$
What did you get for the area of the top in terms of the area ##A##?
 
  • #12
TSny said:
What did you get for the area of the top in terms of the area ##A##?
Area of the top came to be equal to ##4A##
 
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  • #13
Adi6677 said:
Area of the top came to be equal to ##4A##
Ok, good. I get an answer that agrees with one of the multiple choices. Can you show your Bernoulli equation that relates a point on the top to a point on the area ##A##? Your Bernoulli equation that relates a point on ##A## to a point at the bottom of the cone that you posted in post #1 looks correct. There are a lot of factors of 2 that need care when chugging through the algebra.
 
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  • #14
TSny said:
Ok, good. I get an answer that agrees with one of the multiple choices. Can you show your Bernoulli equation that relates a point on the top to a point on the area ##A##? Your Bernoulli equation that relates a point on ##A## to a point at the bottom of the cone that you posted in post #1 looks correct. There are a lot of factors of 2 that need care when chugging through the algebra.
Hi so here is a photo of the attempt to the question hope u get what I wrote its a little messy
 

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1571057787669.png
 
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  • #16
TSny said:
View attachment 251102
Ah sorry for that my bad but I reattempted that but still didn't got that ##17##
Can u please tell what I did wrong?
 

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  • #17
1571080688133.png


I believe you had this correct in your first post.
 
  • #18
TSny said:
View attachment 251121

I believe you had this correct in your first post.
Ah thanks finally got the answer it's quite embarassing to make soo many algebraic mistakes thanks a lot for ur time 😊
 
  • #19
Good work.
 

FAQ: Velocity of fluid flowing from a hole at the bottom of a conical shape

1. What factors affect the velocity of fluid flowing from a hole at the bottom of a conical shape?

The velocity of fluid flowing from a hole at the bottom of a conical shape is affected by the diameter of the hole, the height of the conical shape, the density and viscosity of the fluid, and the pressure difference between the top and bottom of the conical shape. These factors determine the rate at which the fluid can flow through the hole.

2. How does the diameter of the hole affect the velocity of fluid flow?

The diameter of the hole directly affects the velocity of fluid flow. A smaller hole will result in a higher velocity, as the same amount of fluid must pass through a smaller area, creating a higher flow rate. On the other hand, a larger hole will result in a lower velocity, as the fluid has more space to flow through and will flow at a slower rate.

3. What is the relationship between the height of the conical shape and the velocity of fluid flow?

The height of the conical shape also affects the velocity of fluid flow. A taller cone will result in a higher velocity, as the fluid has a longer distance to travel and will therefore flow faster. A shorter cone will result in a lower velocity, as the fluid has a shorter distance to travel and will flow at a slower rate.

4. How does the density and viscosity of the fluid impact the velocity of flow?

The density and viscosity of the fluid play a significant role in determining the velocity of flow. A denser or more viscous fluid will flow at a slower velocity, as it has more resistance to flow. On the other hand, a less dense or less viscous fluid will flow at a faster velocity, as it encounters less resistance and can flow more easily through the hole.

5. What effect does the pressure difference have on the velocity of fluid flow?

The pressure difference between the top and bottom of the conical shape also affects the velocity of fluid flow. A larger pressure difference will result in a higher velocity, as the fluid is pushed through the hole with more force. Conversely, a smaller pressure difference will result in a lower velocity, as the fluid has less force pushing it through the hole.

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