Pressure just before the hole in a draining tank

[Conductivity]

Homework Statement

In the given picture, A cylindrical tank has a viscous fluid with density 800 kg/m^3 and viscosity of 0.8 Pa.s. A small pipe is attached to the tank with a cross sectional area of 1 cm^2. If Q = 3.12 x 10^-6 m^3/s, Find h1, h2 ,x

Homework Equations

Q = pi delta P a^4/(8 n L)
P = rho g h

The Attempt at a Solution

I was able to solve the question under the assumption that the pipe is soo small compared to the tank that velocity of water in the tank is just zero, With this assumption I can consider that the pressure in the tank is static pressure so the pressure just before the hole in a draining tank is rho g 0.4

II was wondering if we had an ideal fluid with the same setup

I figured there is a better way to find the pressure at point b using brenoulli's (better approximation).
We could consider v1 approximately 0 and then get V2 = sqrt(2g h), Using continuity get vb and then apply brenoulli's again between vb and v2 to get Pb. Isn't this a better approximation to the vale of Pb that static pressure way?

 

[BvU]

I don't see what you do with the viscosity ?
Bernoulli is indeed intended by the exercise composer. v in the bulk of the tank is 0.

 

[Conductivity]

I don't see what you do with the viscosity ?
Bernoulli is indeed intended by the exercise composer. v in the bulk of the tank is 0.

Viscosity example: we have Q = pi delta P a^4/(8 n L)
we have everything except delta P, Which we will use to find h1 and h2
For h1, Delta P = rho g (0.4) - rho g h1

Of course this is under the assumption that v of the fluid in tank is zero.

Doesnt this assumption cause a velocity discontinuity in opening of the hole?

 

[Chestermiller]

Within the tank, in the region close to the exit pipe (on the order of just a few diameters upstream of the exit hole), the flow is converging radially (in a hemi-spherical sense) toward the exit hole. As the flow converges, it is speeding up, and the pressure is decreasing. So, a few diameters upstream of the exit hole, the pressure is hydrostatic, but at the exit hole, it is atmospheric. So the pressure is not discontinuous, but it does decrease fairly rapidly in the approach to the exit hole. And, in this region, the flow velocity speeds up to the exit hole velocity.

 

[Conductivity]

Within the tank, in the region close to the exit pipe (on the order of just a few diameters upstream of the exit hole), the flow is converging radially (in a hemi-spherical sense) toward the exit hole. As the flow converges, it is speeding up, and the pressure is decreasing. So, a few diameters upstream of the exit hole, the pressure is hydrostatic, but at the exit hole, it is atmospheric. So the pressure is not discontinuous, but it does decrease fairly rapidly in the approach to the exit hole. And, in this region, the flow velocity speeds up to the exit hole velocity.

in the ideal fluid situation yes, hydrostatic pressure a bit far from the exit hole then pressure decreases as we reach the exit hole so that it becomes atmospheric.

However, In the viscous fluid problem I assumed that the pressure at the exit hole( the opening of the pipe in the tank) is hydrostatic too ( The only way this question can be solved, approximate answer). Which means the velocity of the fluid at any point in the tank is zero but in the small pipe it isnt.

 

[Chestermiller]

in the ideal fluid situation yes, hydrostatic pressure a bit far from the exit hole then pressure decreases as we reach the exit hole so that it becomes atmospheric.

However, In the viscous fluid problem I assumed that the pressure at the exit hole( the opening of the pipe in the tank) is hydrostatic too ( The only way this question can be solved, approximate answer). Which means the velocity of the fluid at any point in the tank is zero but in the small pipe it isnt.

The same thing I described for the inviscid case also happens in the viscous case (qualitatively). The way you did the problem for the viscous case is not the only way to get a good approximate answer.