A pump developing a pressure of 800 kN/m² is used to pump water through a 150 mm pipe 300 m long to a reservoir 60 m higher. With the valves fully open, the flowrate obtained is 0.05 m³/s. As a result of corrosion and scaling the effective absolute roughness of the pipe surface increases by a

Question

A pump developing a pressure of 800 kN/m² is used to pump water through a 150 mm pipe 300 m long to a reservoir 60 m higher. With the valves fully open, the flowrate obtained is 0.05 m³/s. As a result of corrosion and scaling the effective absolute roughness of the pipe surface increases by a factor of 10. By what percentage is the flowrate reduced? Viscosity of water = 1 mN s/m².

Accepted Answer

800 kN/m² is equivalent to a head of 80,000/(1000× 9.81) = 81.55 m of water. If the pump is required to raise the water through a height of 60 m, then neglecting kinetic energy losses, the head loss due to friction in the pipe = (81.55 - 60) = 21.55 m.

The flowrate under these conditions is 0.05 m³/s.

The cross-sectional area of the pipe = (π/4)0.15² = 0.0177 m².
Velocity of the water = (0.05/0.0177) = 2.82 m/s.
Head loss due to friction is:

[latex]h_{f}[/latex] = 8(R/ρu²)(l/d)(u²/2g) (equation 3.20)

Thus: 21.55 = 8(R/ρu²)(300/0.15)(2.82²/(2× 9.81))
and: R/ρu² = 0.0033

Re = ρud/μ = (1000 × 2.82 × 0.15)/[latex]10^{-3}[/latex] = 4.23 × [latex]10^{5}[/latex]
From Fig. 3.7, e/d is 0.003.
If, as a result of scaling and fouling, the roughness increases by a factor of 10, the new value of e/d = 0.03. Fig. 3.7 can no longer be used since the new velocity, and hence the Reynolds number, is unknown. Use is made of equation 3.23 and Fig. 3.8 to find the new velocity.
The maximum head loss due to friction is still equal to 21.55 m as the pump head is unchanged.
Thus: 21.55 m = (21.55 × 1000 × 9.81) = 211,410 N/m²

(R/ρu² ) Re² = -Δ[latex]P_{f}[/latex]d³ ρ/4lμ² (equation 3.23)

= (211,410 × [latex]0.15^3[/latex] × 1000/4 × 300 × [latex]10^{-6}[/latex] )= 6.0 × [latex]10^{8}[/latex]
From Fig. 3.8, Re = 2.95 × [latex]10^{5}[/latex] when e/d= 0.03.
Hence the new velocity = (2.95 × [latex]10^5[/latex] × [latex]10^{-3}[/latex])/(1000 × 0.15) = 1.97 m/s
Reduction in flow = (100(2.82 - 1.97)/2.82) = 30.1 per cent.

Question 8.22: A pump developing a pressure of 800 kN/m² is used to pump wa...

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