Answer: hL/L = 37.05/2000 ≈ 0.0185 m/m or 18.5 mm/m
Step-by-step explanation:
The rate of head loss in the pipe can be determined using the Bernoulli's equation, which relates the pressure, velocity, and elevation of fluid flowing through a pipe. The Bernoulli's equation is given by:
P/ρ + V^2/2g + Z = constant
where P is the pressure, ρ is the density of the fluid, V is the velocity, g is the acceleration due to gravity, Z is the elevation, and the constant represents the total energy of the fluid.
Assuming the flow in the pipe is steady and the pipe is horizontal, the elevation term can be ignored, and the Bernoulli's equation can be simplified as:
P1/ρ + V1^2/2g = P2/ρ + V2^2/2g + hL
where P1 and V1 are the pressure and velocity at the tank, P2 and V2 are the pressure and velocity at the meter location, and hL is the head loss in the pipe.
Converting the given values to SI units:
Diameter of the pipe, d = 1,200 mm = 1.2 m
Radius of the pipe, r = d/2 = 0.6 m
Cross-sectional area of the pipe, A = πr^2 = π(0.6)^2 ≈ 1.13 m^2
Flow rate, Q = 0.126 m^3/s
Density of water, ρ = 1000 kg/m^3
Gravity, g = 9.81 m/s^2
Pressure at the tank, P1 = ρgh1, where h1 is the water surface elevation = 540 m
Pressure at the meter location, P2 = 586 kPa = 586000 Pa
Distance between the tank and meter location, L = 2 km = 2000 m
Using the continuity equation, Q = AV1, we can find the velocity of the water at the tank:
V1 = Q/A = 0.126/1.13 ≈ 0.1117 m/s
Substituting the values in the Bernoulli's equation and solving for hL:
hL = (P1 - P2)/ρg + (V2^2 - V1^2)/(2g)
= (ρgh1 - P2)/ρg + (Q^2/A^2 - V1^2)/(2g)
≈ (1000×9.81×540 - 586000)/(1000×9.81) + (0.126^2/1.13^2 - 0.1117^2)/(2×9.81)
≈ 37.05 m
Therefore, the rate of head loss in the pipe is 37.05 m over a distance of 2000 m, which gives the average rate of head loss per unit length as:
hL/L = 37.05/2000 ≈ 0.0185 m/m or 18.5 mm/m