Question

Water towers store water above the level of consumers for times of heavy use, eliminating the need for high-speed pumps. How high above a user must the water level be to create a gauge pressure of 3.00 x 10^5 Pa?

Answer 1

Answer:

$h=305.8103\ m$

Explanation:

Given:

• gauge pressure desired, $P_g=3\times 10^5\ Pa$

As we know that the layers of the liquid column exert pressure due to their weight which is given as:

$P=\rho.g.h$

where:

$P=$ pressure due to liquid column

$\rho=$ density of the liquid $=1000\ kg.m^{-3}$ for water

$g=$ acceleration due to gravity

$h=$ height of the liquid column

Similarly:

$P_g=\rho.g.h$

$3\times 10^{6}=1000\times 9.81\times h$

$h=305.8103\ m$

Answer 2

Belive the answer to your question is 30.6

got it right on the test