Question

A water reservoir is drained through a 1m pipe with a velocity (v = 10 m/s) and a pressure (p = 200 kPa). If the drain is located (Z = 250 m) above sea level, what is the elevation of the surface of the water reservoir?

a) Z + 1/2 * v²/g

b) Z + p/rhog + 1/2 * v²/g

c) Z - p/rhog + 1/2 * v²/g

d) Z - 1/2 * v²/g

Answer 1

The elevation of the surface of the water reservoir is given by Z - 1/2 * v²/g, where Z is the elevation of the drain, v is the velocity of the water, and g is the acceleration due to gravity.

Step-by-step explanation:

According to Bernoulli's equation, the pressure at the bottom of the reservoir (P₁) is equal to the pressure at the drain (P₂) plus the kinetic energy of the water (1/2 * ρ * v²) and the potential energy due to elevation (ρ * g * h). Since both P₁ and P₂ are equal to atmospheric pressure, they cancel out of the equation. Therefore, the elevation of the surface of the water reservoir is given by Z - 1/2 * v²/g, where Z is the elevation of the drain, v is the velocity of the water, and g is the acceleration due to gravity.