A horizontal pipeline (diameter = 60 cm) carries oil at the rate of 10⁵ m³/day (specific weight = 9000 N/m³. The frictional head loss of fluid during flow is observed as 8.5 m per 1000 m of pipe run. It is planning to place pumping stations at every 20 km along the pipe, what will be the pressure drop between two pumping stations?

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BHEL Engineer Trainee Mechanical 24 Aug 2023 Official Paper

Answer 1

Concept:

The pressure drop in a pipeline due to frictional head loss can be calculated using:

\(Δ P = γ \cdot h_f\)

Where, ΔP = pressure drop, γ = specific weight of the fluid, h_f = head loss over the given length.

Calculation:

Given:

Specific weight of oil, γ = 9000 N/m³

Head loss = 8.5 m per 1000 m of pipe

Distance between pumping stations = 20 km = 20,000 m

Total head loss over 20 km:

\(h_f = \frac{8.5}{1000} \times 20000 = 170~\text{m}\)

Now, calculate pressure drop:

\(Δ P = γ \cdot h_f = 9000 \times 170 = 1,530,000~\text{N/m}^2 = 1.53~\text{MN/m}^2\)