Design of distribution main: compute the approximate internal diameter required to convey 7.2 MLD (million litres per day) at an average velocity of 1.2 m/s. Choose the nearest standard size.

Introduction / Context:
In water-supply engineering, a quick way to size a distribution main is to convert the demand to a discharge in m^3/s, select a practical flow velocity (often 0.8–1.5 m/s), and then compute the pipe area and equivalent internal diameter. This checks hydraulic feasibility before detailed head-loss analysis.

Given Data / Assumptions:

• Total flow = 7.2 MLD (million litres per day).
• Velocity V = 1.2 m/s.
• Steady flow; circular pipe flowing full.
• 1 m^3 = 1000 litres; 1 day = 86,400 s.

Concept / Approach:
First convert 7.2 MLD to m^3/s, then compute required area A = Q / V. Finally, determine internal diameter D from A using D = sqrt(4A / pi). Round to the nearest standard size.

Step-by-Step Solution:

Verification / Alternative check:
Back-check velocity using D = 0.30 m: A = pi * D^2 / 4 ≈ 0.0707 m^2; V = Q / A ≈ 0.08333 / 0.0707 ≈ 1.18 m/s, consistent with the assumed 1.2 m/s.

Why Other Options Are Wrong:

• 24–28 cm (smaller) would increase velocity above the target and raise head loss.
• 32 cm (larger) would lower velocity and increase capital cost unnecessarily at this stage.

Common Pitfalls:
Forgetting to convert MLD to m^3/s; using pi incorrectly; not checking that the selected diameter gives a reasonable velocity band.

Final Answer:
30 cm