Question 8.17: 60% sulphuric acid is to be pumped at the rate of 4000 cm³/s...
60% sulphuric acid is to be pumped at the rate of 4000 cm³/s through a lead pipe 25 mm diameter and raised to a height of 25 m. The pipe is 30 m long and includes two right-angled bends. Calculate the theoretical power required. The density of the acid is 1531 kg/m³ and its kinematic viscosity is 4.25 × 10^{-5} m²/s. The density of water may be taken as 1000 kg/m³.
Cross-sectional area of pipe = (π/4)(0.025)² = 0.00049 m²
Velocity, u = (4000 × 10^{-6}/0.00049) = 8.15 m/s.
Re = ρud/μ = ud/(μ/ρ) = (8.15 × 0.025)/(4.25 × 10^{-5}) = 4794
If e is taken as 0.05 mm from Table 3.1, e/d = 0.002 and from Fig. 3.7, R/ρu² = 0.0047.
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 given by:
h_{f} = 4(R/ρu² )(l/d)(u²/g) (equation 3.20)
= (4 × 0.0047)(30/0.025)(8.15²/9.81) = 152.8 m and Δz = 25.0 m
From Table 3.2, 0.8 velocity heads (u²/2g) are lost through each 90° bend so that the loss through two bends is 1.6 velocity heads or (1.6 × 8.15²)/(2 × 9.81) = 5.4 m.
Table 3.2. Friction losses in pipe fittings
| Number of pipe diameters | Number of velocity heads (u² /2g) | |
| 45° elbows (a)* | 15 | 0.3 |
| 90° elbows (standard radius) (b) | 30-40 | 0.6-0.8 |
| 90° square elbows (c) | 60 | 1.2 |
| Entry from leg of T-piece (d) | 60 | 1.2 |
| Entry into leg of T-piece (d) | 90 | 1.8 |
| Unions and couplings (e) | Very small | Very small |
| Globe valves fully open | 60-300 | 1.2-6 |
| Gate valves: fully open | 7 | 0.15 |
| \frac{3}{4} open | 40 | 1 |
| \frac{1}{2} open | 200 | 4 |
| \frac{1}{4} open | 800 | 16 |
* See Figure 3.17.
Total head loss = (152.8 + 25 + 5.4) = 183.2 m.
Mass flowrate = (4000 × 10^{-6} × 1.531× 1000) = 6.12 kg/s.
From equation 8.61 the theoretical power requirement = (6.12 × 183.2 × 9.81) = 11,000 W or 11.0 kW.
