Question 8.16: 75% sulphuric acid, of density 1650 kg/m³ and viscosity 8.6 ...
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?
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 × 10^{-3} = 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:
h_{f}=-\Delta P_{f}/\rho g=4(R/\rho u^{2})(l/d)(u^{2}/g) (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 \underline{\underline{2.2 \text{ kW}}}.
For this duty a PTFE lined pump and lead piping would be suitable.
