Question 8.18: 1.3 kg/s of 98% sulphuric acid is to be pumped through a 25 ...
1.3 kg/s of 98% sulphuric acid is to be pumped through a 25 mm diameter pipe, 30 m long, to a tank 12 m higher than its reservoir. Calculate the power required and indicate the type of pump and material of construction of the line that you would choose. Viscosity of acid = 0.025 N s/m² . Density = 1840 kg/m³.
Cross-sectional area of pipe = (π/4)(0.0025)² = 0.00049 m².
Volumetric flowrate = 1.3/(1.84 × 1000) = 0.00071 m³/s.
Velocity in the pipe, u = (0.00071/0.00049) = 1.45 m/s
Re = ρud/μ = ((1.84 × 1000) × 1.45 × 0.025)/0.025 = 2670
This value of the Reynolds number lies within the critical zone. If the flow were
laminar, the value of R/ρu² from Fig. 3.7 would be 0.003. If the flow were turbulent, the value of R/ρu² would be considerably higher, and this higher value should be used in subsequent calculation to provide a margin of safety. If the roughness is taken as 0.05 mm, e/d = (0.05/25) = 0.002 and, from Fig 3.7, R/ρu² = 0.0057.
The head loss due to friction,
h_{f} = 4(R/ρu²)(l/d)(u²/g) (equation 3.20)
= (4 × 0.0057(30/0.025)(1.45²/9.81) = 5.87 m
Δz = 12 m so that the total head = 17.87 m.
The theoretical power requirement, from equation 8.61, is:
\pmb{P}=\frac{1}{η}Ghg (8.61)
power = (17.87 × 1.3 × 9.81) = 227 W
If the pump is 50% efficient, actual power = (227/0.5) = 454 W
A PTFE lined centrifugal pump and lead or high silicon iron pipe would be suitable for this duty.
