Question 8.20: Calculate the power required to pump oil of density 850 kg/m...
Calculate the power required to pump oil of density 850 kg/m³ and viscosity 3 mN s/m² at 4000 cm³/s through a 50 mm pipeline 100 m long, the outlet of which is 15 m higher than the inlet. The efficiency of the pump is 50%. What effect does the nature of the surface of the pipe have on the resistance?
Cross-sectional area of pipe = (π/4)0.05² = 0.00196 m²
Velocity of oil in the pipe = (4000 × 10^{-6})/0.00196 = 2.04 m/s.
Re = ρud/μ = (0.85 × 1000 × 2.04 × 0.05)/(3 × 10^{-3}) = 2.89 × 10^{4}
If the pipe roughness e is taken to be 0.05 mm, e/d = 0.001, and from Fig. 3.7,
R/ρu² = 0.0031.
Head loss due to friction is:
h_{f} = 4(R/ρu²)(l/d)(u²/g) (equation 3.20)
= (4 × 0.0031)(100/0.05)(2.04²/9.81) = 10.5 m
The total head = (10.5 + 15) = 25.5 m
The mass flowrate = (4000 × 10^{-6} × 850) = 3.4 kg/s
Power required = (25.5 × 3.4 × 9.81/0.5) = 1700 W or 1.7 kW
The roughness of the pipe affects the ratio e/d. The rougher the pipe surface, the higher will be e/d and there will be an increase in R/ρu². This will increase the head loss due to friction and will ultimately increase the power required.
