Question

Chapter 11, Problem 027 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 3.45 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 15,0 m Faucet Faucet (a) Number (b) Number Click if you would like to Show Work for this question: Units Units Open Show Work

Answer 1

(27)
(a) First of all we have to find the radius of the sphere
mass of water in sphere = 3.45*10⁵ kg
we know that ,
mass = density*Volume
Density of water =1000 kg/m³
Volume 0f sphere= (4/3)$\pi$R³
where R is the radius of sphere , therfore
3.45*10⁵ = 1000*[(4/3)$\pi$R³ ]
R = 4.35 m
Hence the diameter of the sphere = 2R = 8.7 m
Therfore the required gauge pressure at house A
P - P_o = pgZ₁
P- P_o = 1000*9.81*(8.7+15)
= 2.325*10⁵ Pa
(b) Similarly for B
The required gauge pressure
P- P_o = pgZ₂ = 1000*9.81*(8.7+15 - 7.3)
= 1.61*10⁵ Pa

(26) Let say pressure in the artery is P₁ and pressure in the heart is P₂
therefore
P₁ =P₂ - pgH
where H is the height of the artery in the brain above the heart = 0.54 m
p is density of blood = 1060 kg/m³
P₁ =(1.8*10⁴) - [(1060)*9.81*0.54]
P₁ = 1.24*10⁴ Pa

Vent 1 2R 8.7 m 15.0 m Faucet กา 2 Faucel For (a)