Question

The flow rate in the pipe system below is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. Point 1 is 0.60 m higher than point 2. All the pipes are galvanized iron with roughness value of 0.015 mm.

Answer 1

Answer:

Explanation:

The rate of flow in the pipe system in Figure P4.5.2 is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. All the pipes are galvanized iron with roughness value of 0.15 mm. Determine the pressure at point 2. Take the loss coefficient for the sudden contraction as 0.05 and v = 1.141 × 10−6 m2/s.

The answer to the above question is

The pressure at point 2 = 75.959 kPa

Explanation:

Bernoulli's equation with losses gives

hL = z₁ - z₃ +(P₁-P₃)/(ρ×g) + (v₁²-v₃²)/(2×g)

Between points 1 and 2, z₁ = z₃ + 0.6 m therefore

hL = 0.6 m +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

hL = (f₁×L₁×v₁²)/(D₁×2×g) + (f₂×L₂×v₂²)/(D₂×2×g) + (f₃×L₃×v₃²)/(D₃×2×g) + k×V₃₂/(2×g) = 0.6 +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

But v = Q/A

or  since A = π×D²/4 we have

A₁ = 1.77×10-2 m² , A₂ = 5.73×10-2 m², A₃ = 3.8×10-2 m²  

Therefore from v = Q/A we have v₁ = 2.83 m/s v₂ = 0.87 m/s and v₃  = 1.315 m/s  from there we find the friction coefficient from Moody Diagram as follows

ε = $\frac{Roughness _. value}{ Diameter}$ Which gives

the friction coefficients as f₁ = 0.02, f₂ = 0.017 and f₃ =0.0175

Substituting he above values into the $h_{l}$ equation we get $h_{l}$ = 19.761 m

Combined head loss = 19.761 m

Hence 19.743 m  = 0.6 m +(260 kPa-P₃)/(ρ×9.81) + (6.276)/(2×9.81)

or 260 kPa-18.82 m × 9.81 m/s²×ρ=  P₃

Where ρ = density of water, we have

260000 Pa - 18.82 m×9.81 m/s²×997 kg/m³ = 75958.598 kg/m·s² = 75.959 kPa