Question

What is the elevation of the surface of a water reservoir given that the velocity in a 1m pipe used to drain the reservoir is v=10 m/s and the pressure in the pipe is p=200 kPa? The elevation of the drain is Z=250 m above sea level.

a)260 m
b)240 m
c)230 m
d)220 m

Answer 1

The elevation of the surface of the water reservoir is approximately 20.41 m above sea level.

Step-by-step explanation:

To determine the elevation of the surface of the water reservoir, we can use Bernoulli's equation, which relates pressure, velocity, and elevation.

According to Bernoulli's equation, the total pressure in the pipe is equal to the sum of the pressure due to elevation and the pressure due to velocity.

P_total = P_elevation + P_velocity

The pressure due to elevation is given by P_elevation = ρgh, where ρ is the density of water (1000 kg/m^3), g is the acceleration due to gravity (9.8 m/s^2), and h is the elevation difference in meters.

Since the pipe is draining the reservoir, the velocity in the pipe is 10 m/s, and the pressure in the pipe is 200 kPa, we can solve for the elevation h.

P_total = P_elevation + P_velocity

200 kPa = ρgh + ½ρv^2

Converting the pressure and solving for h:

h = (200 kPa - ½ρv^2) / (ρg)

Substituting the values ρ = 1000 kg/m^3, v = 10 m/s, and g = 9.8 m/s^2:

h = (200,000 N/m^2 - ½ x 1000 kg/m^3 x (10 m/s)^2) / (1000 kg/m^3 x 9.8 m/s^2)

h = 20.41 m

Therefore, the elevation of the surface of the water reservoir is approximately 20.41 m above sea level.