A petroleum fraction is pumped 2 km from a distillation plant to storage tanks through a mild steel pipeline, 150 mm in diameter, at the rate of 0.04 m³/s. What is the pressure drop along the pipe and the power supplied to the pumping unit if it has an efficiency of 50%? The pump impeller is eroded

Question

A petroleum fraction is pumped 2 km from a distillation plant to storage tanks through a mild steel pipeline, 150 mm in diameter, at the rate of 0.04 m³/s. What is the pressure drop along the pipe and the power supplied to the pumping unit if it has an efficiency of 50%? The pump impeller is eroded and the pressure at its delivery falls to one half. By how much is the flowrate reduced? Density of the liquid = 705 kg/m³ . Viscosity of the liquid = 0.5 mN s/m². Roughness of pipe surface = 0.004 mm.

Accepted Answer

Cross-sectional area of pipe = (π/4)0.15² = 0.0177 m².
Velocity in the pipe = (0.04/0.0177) = 2.26 m/s.
Reynolds number = (0.705 × 1000 × 2.26 × 0.15)/(0.5 ×[latex]10^{-3}[/latex]) = 4.78 × [latex]10^{5}[/latex]

e = 0.004 mm, e/d = (0.004/150) = 0.000027 and from Fig. 3.7, R/ρu² = 0.00165
The pressure drop is: [latex]-Δ P_{f}[/latex] = 4(R/ρu²)(l/d)(ρu²) (equation 3.18)
[latex]-Δ P_{f}[/latex] = (4 × 0.00165)(2000/0.15)(705 × 2.262 ) = 316,900 N/m² or 320 kN

If the pump efficiency is 50%,
power = (head ×mass flowrate × g)/0.5
= pressure drop (N/m²) × volumetric flowrate (m³s)/0.5
= (316,900 × 0.04)/0.5 = 25,350 W or 25.4 W
If, due to impeller erosion, the delivery pressure is halved, the new flowrate may be found from:
(R/ρu²) Re² = [latex]-Δ P_{f}[/latex]d³ ρ/4lμ² (equation 3.23)
The new pressure drop = (316,900/2) = 158,450 N/m² and:
(R/ρu²) Re² = (158,450 × 0.153 × 705)/(4 × 2000 × 0.5² × [latex]10^{-6}[/latex]) = 1.885 × [latex]10^{8}[/latex]
From Fig. 3.8, when (R/ρu²) Re² = 1.9 × [latex]10^{8}[/latex] and e/d = 0.000027, Re = 3.0 × [latex]10^{5}[/latex]
and: (3.0 × [latex]10^{5}[/latex]) = (705 × 0.15 × u)/(0.5 × [latex]10^{-3}[/latex]) and: u = 1.418 m/s
The volumetric flowrate is now: (1.418 × 0.0177) = 0.025 m³/s

Question 8.19: A petroleum fraction is pumped 2 km from a distillation plan...

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