Bernoulli's principle Question

[LuminousTruth]

So for example, there is a water tank that is not closed off at the top. There is a hole which is covered by the plug on the side towards the bottom end. If the plug is removed, the water level will lower and flow out of the tank from that hole the plug is covering.

How does the pressure compare at the plug's end vs the top?

I thought that since the the cross sectional area of the hole is less than the cross sectional area of the top, that means the velocity is greater through the hole than at the top. (A1v1 = A2v2)

Next, according to bernoulli's equation, P1 + 1/2pv^2 1 + pgy1 = P2 = 1/2pv^2 2 + pgy2, an increase in velocity would result in a decrease in pressure. So from that reasoning, the pressure at the hole when the plug is removed would be greater than the pressure at the top.

Yet the correct answer was that the pressure should be greater at the bottom to promote the driving force for water to escape out.

Can someone explain why my reasoning is not right? I thought I had applied Bernoulli's equation correctly with the continuity equation.

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[milski]

You should not neglect the pgy1 and pgy2 terms - at the plug y2 is lower and will result in higher pressure.

 

[Testosterone]

You are right that according to Bernoulli's equation that an increase in velocity would result in a decrease in pressure, but that only applies if the change in height between the two points of interest are zero, making pgy1 and pgy2 equal to zero. This is not the case for a water tank problem. If we assign the hole as zero, then there will be a nonzero height at the top that is equal to the depth.

A quicker, more intuitive way to think about this problem is that since the top is open, it is exposed to atmospheric pressure. Additionally, since the bottom is open, it too is also exposed to atmospheric pressure, however the bottom also has gauge pressure. Guage pressure +Atmospheric pressure > Atmospheric pressure alone.