Question

The main water line enters a house on the first floor. The line has a gauge pressure of 2.10 x 105 Pa. (a) A faucet on the second floor, 7.30 m above the first floor, is turned off. What is the gauge pressure at this faucet? (b) How high could a faucet be before no water would flow from it, even if the faucet were open?

Answer 1

To answer this question is necessary to apply the concepts related to Bernoulli's equation. The Bernoulli-related concept describes the behavior of a liquid moving along a streamline. Pressure can be defined as the proportional ratio between height, density and gravity:

$P = h\rho g$

Where,

h = Height

$\rho$ = Density

g = Gravity

Our values are

$\rho = 1000kg/m^3 \rightarow$ density of water at normal conditions

h = 7.3m

$g = 9.8m/s^2$

PART A) Replacing these values to find the total pressure difference we have to

$P_1 = h_1 \rho g$

$P_1 = (7.3)(1000)(9.8)$

$P_1 = 71540Pa$

In this way the pressure change would be subject to

$\Delta = P_2-P_1$

$\Delta = 2.1*10^5Pa- 0.7154*10^5Pa$

$\Delta = 138460Pa$

$\Delta = 0.135Mpa$

PART B) Considering the pressure gauge of the group as the ideal so that at a height H the water cannot flow even if it is open we have to,

$P_2 = H\rho g$

$2.1*10^5 = H (1000)(9.8)$

$H = 21.42m$

Therefore the high which could a faucet be before no water would flow from it is 21.42m