Question

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 7.41 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.

Answer 1

Complete Question

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Answer:

The gauge pressure that water has at the House A
`P_A = 257020.68 Pa`

The gauge pressure that water has at the House B
`P_B = 188454 \ Pa`

Step-by-step explanation:

From the question we are told that

The mass of water when full is
`m_f = 7.41* 10^(5) kg`

Generally the volume of water in this tank is mathematically represented as

`V = (m )/(\rho)`

Where
`\rho` is the density of water with a value of with a value of
`\rho = 1000 kg /m^3`

substituting values

`V = (7.41 *10^5)/(10^3)`

`V = 741 m^3`

This volume is the volume of a sphere since the tank is spherical so

`V = (4 \pi ^3)/(3)`

making r the subject of the formula

`r =\sqrt[3]{ (741 *3 )/(4\pi) }`

`r = 5.6134 m`

Now we can use this parameter to obtain the diameter

So

`d = 2 * r`

substituting values

`d = 2 * 5.6134`

`d = 11.23m`

The pressure the water has at faucet in House A is mathematically evaluated as

`P_A = \rho g h_A`

This height is obtained as follows

`h_A = d+ 15`

The value 15 is gotten from the diagram

so

`h_A = 15 + 11.23`

`h_A = 26.22 m`

Now substituting values

`P_A = 26.23 * 9.8 * 1000`

`P_A = 257020.68 Pa`

The pressure the water has at faucet in House B is mathematically evaluated as

`P_B = \rho g h_B`

This height is obtained as follows

`h_B = d+ 15`

The value 15 is gotten from the diagram

so

`h_B = d + 15 -h`

substituting values

`h_B =11.23 + 15 -7`

`h_A = 19.23 m`

Now substituting values

`P_B = 19.23 * 9.8 * 1000`

`P_B = 188454 \ Pa`