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