Solving Water Flow from a Tank: Bernoulli's Theorem

[zorro]

Homework Statement

A tank is filled with water to a depth of 1m. A hole of cross section 1 cm^2 at the bottom allows the water to drain out. At what distance below the hole the cross sectional area of the stream is 0.9 times the area of the hole.

Homework Equations

The Attempt at a Solution

Applying Bernoulli's Theorem to the cross section of water stream just outside the hole to a depth 'x'
Patm + ρgh + 1/2 ρv1^2 = Patm + 1/2 ρv2^2

h=1m v1= (2gh)^0.5
v2=v1/0.9 (from equation of continuity)

On solving I got x as 0.46 m (app)
But the answer is 0.23 m

What is wrong?
Any ideas appreciated.

 

[alphysicist]

Hi Abdul Quadeer,

Homework Statement

A tank is filled with water to a depth of 1m. A hole of cross section 1 cm^2 at the bottom allows the water to drain out. At what distance below the hole the cross sectional area of the stream is 0.9 times the area of the hole.

Homework Equations

The Attempt at a Solution

Applying Bernoulli's Theorem to the cross section of water stream just outside the hole to a depth 'x'
Patm + ρgh + 1/2 ρv1^2 = Patm + 1/2 ρv2^2

h=1m v1= (2gh)^0.5
v2=v1/0.9 (from equation of continuity)

On solving I got x as 0.46 m (app)

Can you show how you got this number?

But the answer is 0.23 m

What is wrong?
Any ideas appreciated.

 

[zorro]

I'm sorry, I missed one term on LHS

[Patm + ρgh] + ρgx + 1/2 ρv1^2 = Patm + 1/2 ρv2^2
i.e. g + gx + 2g/2 = 2g/(0.9)^2
2 + x = 2/(0.9)^2
or x=0.46m

On LHS I took [Patm + ρgh] as it is the total pressure acting at the hole (due to water column above it + Patm)

 

[alphysicist]

I'm sorry, I missed one term on LHS

[Patm + ρgh] + ρgx + 1/2 ρv1^2 = Patm + 1/2 ρv2^2

No, the [Patm + ρgh] is not right here. When the fluid is touching the air, as it is when it comes out of the hole, its pressure will be air pressure.

(If there were no hole, then the pressure at the bottom of the tank would be [Patm + ρgh]. In that case the bottom of the tank would be applying that much pressure on the liquid. But we know how much pressure air applies to something--the value of the air pressure.)

i.e. g + gx + 2g/2 = 2g/(0.9)^2
2 + x = 2/(0.9)^2
or x=0.46m

On LHS I took [Patm + ρgh] as it is the total pressure acting at the hole (due to water column above it + Patm)

 

[zorro]

The fluid touches the air sideways, but it is in contact with the water in the tank vertically.
In the derivation of Bernoulli's Theorem, we take the pressure acting on the cross sections of the fluid (we do not consider any pressure acting sideways)

Refer my attachment.

 

[alphysicist]

The fluid touches the air sideways, but it is in contact with the water in the tank vertically.
In the derivation of Bernoulli's Theorem, we take the pressure acting on the cross sections of the fluid (we do not consider any pressure acting sideways)

Refer my attachment.

Your figure shows the pressure acting on the cross sections, but that is also the pressure of the cross sections (that is, the fluid element at that point is pushing out in all directions at that pressure).

Notice that you have the pressure at the lower distance to also be atmospheric pressure. That is correct, and it is also not touching the air anywhere except its sides (above it there is only water).

 

[zorro]

Thankyou!
That cleared my doubt

 

[alphysicist]

Thankyou!
That cleared my doubt

My pleasure!