Pressure just before the exit hole in a draining tank

AI Thread Summary
The discussion focuses on the pressure dynamics in a cylindrical tank draining viscous fluid through a small pipe. It explores the assumption that the fluid's velocity in the tank is zero, leading to a static pressure calculation, and contrasts this with the behavior of an ideal fluid using Bernoulli's principle. It highlights that, as fluid approaches the exit hole, pressure decreases from hydrostatic to atmospheric, and flow velocity increases due to radial convergence. The conversation also addresses the implications of viscosity on pressure calculations and the continuity of flow, emphasizing that velocity within the pipe remains constant despite the pressure changes. Overall, the complexities of fluid dynamics in both viscous and ideal scenarios are examined.
Conductivity
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Homework Statement


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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
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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?
 

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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.
 
BvU said:
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?
 
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.
 
Chestermiller said:
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.
 
Conductivity said:
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.
 
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Chestermiller said:
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 Velo
Chestermiller said:
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.
Does velocity increase as fluid from the tank enters the pipe and the keeps increasing inside the pipe? If velocity keeps increasing inside the pipe until the exit wouldn't that contradict the law of mass conservation, since the diameter of the pipe doesn't change
 
The velocity inside the pipe is constant.
 
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