Bernoulli's equation, static fluid, gauge pressure problem.

[BOAS]

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.

Homework Statement

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 5.25x10^{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.

Homework Equations

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!

 

[SteamKing]

It's not clear how you are calculating Pa and Pb. What are you using for the density of water, ρ?

In other words, you can't just take the mass of water in the tower and multiply by g.

The pressures you have calculated would indicate that turning on a faucet in either house would run the risk of causing massive flooding, since it is unlikely you would be able to close the valves by hand.

 

[BOAS]

It's not clear how you are calculating Pa and Pb. What are you using for the density of water, ρ?

In other words, you can't just take the mass of water in the tower and multiply by g.

The pressures you have calculated would indicate that turning on a faucet in either house would run the risk of causing massive flooding, since it is unlikely you would be able to close the valves by hand.

I'm using 1000kgm^-3 for the density of water.

I do agree with you that my numbers seem rather high. I am confused about how to quantify the pressure from the water tank. I re-read the question and see that it is specified that the tank is spherical.

Knowing the mass of the water, and it's density, I can calculate a volume. From this, I can calculate the 'height' of water the water above where it exits the tank. This works out to be 10m.

Since it is the pressure gauge and not the absolute pressure I'm dealing with, the pressure at the bottom of the water tank should be equal to ρgh = 100000 Pa.

Which is ≈ 5 times less.

I shall rework the question using this info. Thanks for the help, often it's just a little nudge I need to see where to go. (here's to hoping I haven't spoken too soon :) )

 

[SteamKing]

That's OK. The problem as written in the OP did not contain sufficient information to determine the static pressure in the water lines, i.e. having the height of the bottom of the tank is not enough info to do a calculation. The use of a spherical tank should provide the missing info.