# Bernoulli’s equation, static fluid, gauge pressure problem.

Hi,

I haven’t done many problems of this nature so there are a few steps in my working that i’d like to check are acceptable/agree with what the question implies.

1. The problem statement, all variables and given/known data

A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide sufficient pressure in the pipes that deliver the water to customers. The drawing (see attached) shows a reservoir that contains $5.25$x$10^{5} kg$ of water. The reservoir is vented to the atmosphere at the top. Find the gauge pressure that the water has at the faucet in house A and house B. Ignore the diamter of the delivery pipes.

2. Relevant equations

3. The attempt at a solution

My first concern is that last sentence. Is it correct to assume that the pressure at the water tower is simply equal to it’s weight? I proceeded under that assumption…

$P_{1} + \frac{1}{2}\rho v^{2}_{1} + \rho g y_{1} = P_{2} + \frac{1}{2}\rho v^{2}_{2} + \rho g y_{2}$

I’m using $g = 10ms^{-2}$ to keep the numbers neat.

$v_{1} = v_{2} = 0$

$P_{B} > P_{WT}$ Pressure at faucet B is greater than the pressure at the water tower, since it is lower.

$P_{WT} + \rho g y_{WT} = P_{B} + \rho g y_{B}$

$P_{WT} + \rho g y_{WT} – \rho g y_{B}= P_{B}$

(I’ve plugged in the numbers, but i’m more interested in whether or not my method is correct)

$P_{B} = 5327000 Pa$

For faucet A;

$P_{A} > P_{B} > P_{WT}$ Due to being lowest of all.

$P_{WT} + \rho g y_{WT} = P_{A} + \rho g y_{A}$

$P_{WT} + \rho g y_{WT} – \rho g y_{A}= P_{A}$

A’s height is 0m.

$P_{A} = P_{WT} + \rho g y_{WT} – 0$

$P_{A} = 5400000 Pa$

My answers do agree with the inequalities I was expecting, but that first assumption is troubling me.

Is this okay?

Thanks for taking the time to read!

Attached Images
 bernoulli.png (3.4 KB)

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