Question

Determine the minimum gauge pressure needed in the water pipe leading into a building if water is to come out of a faucet on the fourteenth floor, 44 m above that pipe.

Answer 1

The minimum amount of pressure needed is the difference in hydrostatic pressure at h=0m and at h=44m. 
This difference can be calculated using following formula:
$\Delta P=\rho g \Delta h$
Where $\rho$ is the density of the fluid, g is the gravitational acceleration, and $\Delta h$ is the height difference. 
The density of water is:
$\rho=1000\frac{kg}{m^3}$
In our case $\Delta h$ is 44m. When we plug in the numbers we get:
$\Delta P= 1000\cdot 9.81 \cdot 44=431640 Pa$

Answer 2

The minimum gauge pressure needed in the water pipe leading into the building with the given parameters is;

P_g,min = 431.2 KPa

We are given the height the faucet above the pipe; h = 44 m

       Now, the formula for minimum gauge pressure in this case is;

P_g,min = ρgh

Where;

ρ is density of fluid

g is acceleration due to gravity = 9.8 m/s²

h height of fluid column

        Now, in this question, the fluid we are dealing with is water and from tables of density of fluids, density of water is;

ρ_water = 1000 kg/m³

Thus, plugging in the relevant values into the Pressure equation gives;

P_g,min = 1000 × 9.8 × 44

P_g,min = 431200 Pa

Converting to KPa gives;

P_g,min = 431.2 KPa

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