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 4.83 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

Mass \(m=4.83 \times 10^{i}\) hg Density \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\) First, we have to find the diameter of the sphere The volume of water (V) in a spherical container is given by $$ \begin{array}{l} V=\frac{m}{\rho} \\ \text { or }=\frac{4.83 \times 10^{3} \mathrm{hg}}{1000 \mathrm{~kg} / \mathrm{m}^{3}} \\ \text { or }=4.83 \times 10^{2} \mathrm{~m}^{3} \end{array} $$ Or $$ \frac{4}{3} \pi r^{2}=4.83 \times 10^{2} m^{3} $$ Or \(\quad r=\sqrt[3]{\frac{\left(4.83 \times 10^{2} m^{2}\right) \times 3}{4 \pi}} \quad\) [use \(\left.\pi=\frac{22}{7}\right]\) $$ \begin{array}{l} \text { Or } \quad=4.87 \mathrm{~m} \\ \text { Diameter }=2 \mathrm{r}=9.74 \mathrm{~m} \end{array} $$ (A) Required gauge pressure in house "A" is given by Or $$ \begin{aligned} P-P_{0} &=\rho g h &(\text { Where } h=2 r+15 \mathrm{~m}) \\ &=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(9.74 \mathrm{~m}+15 \mathrm{~m}) \end{aligned} $$ Or \(=2.42 \times 10^{4} \mathrm{~Pa}\) (B) Required gauge pressure in house " \(\mathrm{B}^{\prime \prime}\) is given by $$ P-P_{0}=\rho g(2 r+15 m-7.30 m) $$ Or $$ =\left(1000 \mathrm{~kg} / \mathrm{m}^{2}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(9.74 \mathrm{~m}+15 \mathrm{~m}-7.3 \mathrm{~m}) $$ Or \(=1.71 \times 10^{5} \mathrm{~Pa}\)