# Bernoulli equation with pressure tank

## Homework Statement

Water flows at a rate of 30ml/s from an opening in the bottom of a 4m high pressure tank (a tank with a plunger type lid). Calculate the flow rate when an extra 50 kPa of pressure is applied.

## Homework Equations

Bernoulli's equation and $$\frac{V}{t} = Av$$

## The Attempt at a Solution

My idea is to apply Bernoulli's equation to work out the area of the opening by finding the velocity of the water in the initial case and then to use Bernoulli again to find the speed of the same in the case where the extra pressure is applied. Then to use this together with the area to calculate the flow rate in the final case.

But I'm a bit uncertain of some of my assumptions. NOTE: subscript 2 refers to top of tank and subscript 1 refers to bottom.

Initial case (no extra pressure):
$$v_{2} << v_{1}$$, therefore $$v_{2} \approx 0$$
$$P_{2} = P_{1}$$
$$h_{1} = 0$$

therefore

$$v_{1} = \sqrt{2gh_{2}}$$

this means that area of opening is:

$$A = \left( \frac{V}{t} \right)_{initial} \frac{1}{\sqrt{2gh_{2}}}$$

Pressure case:

with application of additional pressure I did pretty much the same – using Bernoulli to work out the velocity of the out flowing water by considering the increased pressure of P2 and then using the flow rate equation to work out the new flow rate.

However I'm not sure. How does the changing water level affect this? I'm guessing you can ignore this but am still unconvinced. Also in the initial case I've assumed the pressure difference between P1 and P2 is zero. I'm not sure if this is correct either, seeing as the top of the container is closed.

thanks

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