75% sulphuric acid, of density 1650 kg/m³ and viscosity 8.6 mN s/m² , is to be pumped for 0.8 km along a 50 mm internal diameter pipe at the rate of 3.0 kg/s, and then raised vertically 15 m by the pump. If the pump is electrically driven and has an efficiency of 50%, what power will be required?

Question

75% sulphuric acid, of density 1650 kg/m³ and viscosity 8.6 mN s/m² , is to be pumped for 0.8 km along a 50 mm internal diameter pipe at the rate of 3.0 kg/s, and then raised vertically 15 m by the pump. If the pump is electrically driven and has an efficiency of 50%, what power will be required? What type of pump would you use and of what material would you construct the pump and pipe?

Accepted Answer

Cross-sectional area of pipe = (π/4)(0.05)² = 0.00196 m².
Velocity, u = 3.0/(1650 × 0.00196) = 0.93 m/s.
Re = ρud/μ = (1650 × 0.93 × 0.05)/8.6 × [latex]10^{-3}[/latex] = 8900
If e is taken as 0.046 mm from Table 3.1, e/d = 0.00092.
From Fig. 3.7, R/ρu² = 0.0040.
Table 3.1. Values of absolute roughness e

	(ft)	(mm)
Drawn tubing	0.000005	0.0015
Commercial steel and wrought-iron	0.00015	0.046
Asphalted cast-iron	0.0004	0.12
Galvanised iron	0.0005	0.15
Cast-iron	0.00085	0.26
Wood stave	0.0006-0.003	0.18-0.9
Concrete	0.001-0.01	0.3-3.0
Riveted steel	0.003-0.03	0.9-9.0

Head loss due to friction is:

[latex]h_{f}=-\Delta P_{f}/ ho g=4(R/ ho u^{2})(l/d)(u^{2}/g)[/latex] (equation 3.20)

= (4 × 0.004)(800/0.05)(0.93²/9.81) = 22.6 m

Total head = (22.6 + 15) = 37.6 m.

Power = (mass flowrate × head × g) (equation 8.61)
= (3.0 × 37.6 × 9.81) = 1105 W
If the pump is 50% efficient, power required = (1105/0.5) = 2210 W or [latex]\underline{\underline{2.2 ext{ kW}}}[/latex].
For this duty a PTFE lined pump and lead piping would be suitable.

Question 8.16: 75% sulphuric acid, of density 1650 kg/m³ and viscosity 8.6 ...

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