Question
Asked Jul 22, 2019

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 shows a spherical reservoir that contains 7.16 x 105 kg of water when full. The reservoir is vented to the atmosphere at the top. For a full reservoir, find the gauge pressure that the water has at the faucet in (a) house A and (b) house B. Ignore the diameter of the delivery pipes.

Vent
BE
15.0 m
7.30m
A
Faucet
BB
Faucet
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Vent BE 15.0 m 7.30m A Faucet BB Faucet

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Step 1

Mass of water stored in the reservoir is 7.16×kg, and density of water is 1000 kg/m3

The volume of the container is.

 

here
p is the density
V is the volume
m is the mass
V ="
substitute,7.16x 105 kg for m
7.16x10 kg
1000kg/m3
1000kg/m' for p
3
=716m3
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here p is the density V is the volume m is the mass V =" substitute,7.16x 105 kg for m 7.16x10 kg 1000kg/m3 1000kg/m' for p 3 =716m3

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Step 2

The container is spherical. Hence the volume will be 4/3πr3, compare this volume equation with the above obtained answer to find the radius of the spherical container and then the diameter.

4
= 716m
3
3
3
716m3 x
47
= 5.54 m
2r 11.08m
Il
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4 = 716m 3 3 3 716m3 x 47 = 5.54 m 2r 11.08m Il

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Step 3

(a)

Write the expression for required gauge p...

here, g is the acceleration due to gravity,
h is the height at which reservoir
P- P gh
palced from house A
substitute, 2r 15m for h in equation (I)
substitute, 11.08m for 2r,
9.80 m/s for g.
1000kg/m3 for p in equation (II)
=pg(2r+15m)
(II)
= 1000kg/m3x 9.80 m/s2(11.08m+15m)
2.556 x10 Pa
help_outline

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here, g is the acceleration due to gravity, h is the height at which reservoir P- P gh palced from house A substitute, 2r 15m for h in equation (I) substitute, 11.08m for 2r, 9.80 m/s for g. 1000kg/m3 for p in equation (II) =pg(2r+15m) (II) = 1000kg/m3x 9.80 m/s2(11.08m+15m) 2.556 x10 Pa

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