- #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

$$ Pa + 2dgh + 2dV^2/2 = Pb + 2dV'^2/2$$

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?

`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?