At a water-treatment plant, water at 20°C flows from tank A to tank B at a rate of 600 L/s through a 50 cm diameter asphalt lined iron pipe that is 150 m long. Determine the difference in water surface elevation between the tanks (open to the atmosphere) if there are three bends (R/D = 10.0) and a fully open lift type check valve in the pipeline.

The answer to the question is

The difference in water surface elevation between the tanks is 0.477 m

Explanation:

Bernoulli's Equation is given by

Volume flow rate = 600 L/s

P₁ + 1/2·ρ·v₁²+ρ·g·h₁ = P₂ + 1/2·ρ·v₂² + ρ·g·h₂

If we take the tanks as open to the atmosphere then we have

P₁ = P₂ Hence

1/2·ρ·v₁²+ρ·g·h₁ = 1/2·ρ·v₂² + ρ·g·h₂

From where we calculate the velocity of the water thus

Q = V × A  where A are of the pipe conveying the water, hence A = π × r² and  r = D/2 = (50 cm)/2 = 25 cm or 0.25 m and A = π × 0.25² = 0.196 m²

Then v = Q/A = (600 L/s)/(0.196 m²) = (0.6 m³/s)/(0.196 m²) =  3.06 m/s

If v₂ = 0 just before the valve is opened we have

1/2·ρ·v₁²+ρ·g·h₁  = ρ·g·h₂ or 1/2·ρ·v₁² =  ρ·g·h₂ -ρ·g·h₁  

That is the v₁² = 2 × g × (h₂ - h₁)

Therefore the difference in the fluid level is

(3.06 m/s)² = 2 × 9.81 m/s² × (h₂ - h₁)

and (h₂ - h₁) = 0.477 m

The water level difference between the two tanks is 0.477 m

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