The image of the water tower and the houses is in the attachment.
Answer: (a) P = 245kPa;
(b) P = 173.5 kPa
Explanation: Gauge pressure is the pressure relative to the atmospheric pressure and it is only dependent of the height of the liquid in the container.
The pressure is calculated as: P = hρg
where
ρ is the density of the liquid, in this case, water, which is ρ = 1000kg/m³;
When it is full the reservoir contains 5.25×10⁵ kg. So, knowing the density, you know the volume:
ρ = [tex]\frac{m}{V}[/tex]
V = ρ/m
V = [tex]\frac{5.25.10^{5}}{10^{3}}[/tex]
V = 525 m³
To know the height of the spherical reservoir, its diameter is needed and to determine it, find the radius:
V = [tex]\frac{4}{3}.\pi.r^{3}[/tex]
[tex]r = \sqrt[3]{ \frac{3}{4\pi } .V}[/tex]
r = [tex]\sqrt[3]{\frac{525.3}{4\pi } }[/tex]
r = 5.005 m
diameter = 2*r = 10.01m
(a) Height for House A:
h = 15 + 10.01
h = 25.01
P = hρg
P = 25.01.10³.9.8
P = 245.10³ Pa or 245kPa
(b) h = 25 - 7.3
h = 17.71
P = hρg
P = 17.71.1000.9.8
P = 173.5.10³ Pa or 173.5 kPa
The gauge pressure at house A, and B, is the measurement of pressure
relative to the pressure of the atmosphere.
Reasons:
Question parameters;
Mass of water in the tower, m = 5.25 × 10⁵ kg
Shape of the reservoir = Spherical
Height of the bottom of the reservoir above the main ground level = 15.0 m
Height of the house A above the ground = 0
Height of the house B above the ground = 7.30 m
(a) Required:
The gauge pressure at house A
Solution:
The density of the water, ρ ≈ 997 kg/m³
[tex]\mathrm{Volume \ of \ water \ in \ the \ tank,} \ V = \dfrac{5.2 \times 10^5 \, kg}{997 \, kg/m^3} \approx 526.58 \, m^3[/tex]
Volume of water in the filled tank = Volume of the tank ≈ 526.28 m³
[tex]\mathrm{Volume \ or \ a \ sphere, } V = \dfrac{4}{3} \cdot \pi \cdot r^3[/tex]
Therefore;
[tex]\mathrm{Volume \ or \ the \ spherical \ tank} = \dfrac{4}{3} \cdot \pi \cdot r^3 \approx 526.28[/tex]
Which gives;
[tex]r = \sqrt[3]{\dfrac{526.58}{\pi} \times \dfrac{3}{4} } \approx 5[/tex]
The diameter of the tank = The height of the tank, h = 2 × r
∴ h = 2 × 5 m = 10 m
The gauge pressure is the pressure above the prevailing atmospheric
pressure;
Therefore;
The gauge pressure at house A, [tex]P_{gA}[/tex] = 997 × 9.81 × (10 + 15) = 244514.25
The gauge pressure at house A, [tex]P_{gA}[/tex] = 244514.25 N/m² ≈ 244.5 kPa
(b) Required:
Gauge pressure at house B
At house B, we have;
Gauge pressure at house B, [tex]P_{gB}[/tex] = 997 × 9.81 × (10 + 15 - 7.3) = 173116.089
The gauge pressure at house B = 173116.089 N/m² ≈ 173.116 kPa
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