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

[Adi6677]

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?

 

[TSny]

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.

 

[Adi6677]

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 ??

 

[TSny]

K thanks but the area of the opening of cone isn't given can it be found out in terms of A ??

Yes.

 

[Adi6677]

Yes.

Okay I tried but I don't know what's the relationship between them is it something to do with similarity ??

 

[TSny]

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 $A$.

 

[Delta2]

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...

 

[Adi6677]

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

 

[Adi6677]

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 $$ V' = √(6gh)/√(1-16a^2/A^2)$$

 

[Adi6677]

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 $$ V' = √(6gh)/√(1-16a^2/A^2)$$

 

[TSny]

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 $A$?

 

[Adi6677]

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 $4A$

 

[TSny]

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.

 

[Adi6677]

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

 

[TSny]

 

[Adi6677]

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?

 

[TSny]

I believe you had this correct in your first post.

 

[Adi6677]

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

 

[TSny]

Good work.