For a sharp-edged orifice with area A = 1.0×10^-4 m^2, Cd = 0.62, upstream head h = 3.0 m, compute discharge Q using Q = Cd A sqrt(2 g h).

• A. 4.50×10^-4 m^3/s
• B. 4.75×10^-4 m^3/s
• C. 5.25×10^-4 m^3/s
• D. 6.00×10^-4 m^3/s

Answer: (B)

Discharge through a sharp-edged orifice uses Q = Cd A sqrt(2 g h), which links the flow rate to the velocity produced by the head h and the discharge coefficient Cd, with the area A setting how much fluid can pass.

Compute the velocity term: sqrt(2 g h) = sqrt(2 × 9.81 × 3.0) ≈ sqrt(58.86) ≈ 7.67 m/s.

Now multiply by area and the discharge coefficient: Q ≈ 0.62 × (1.0 × 10^-4 m^2) × 7.67 m/s ≈ 4.76 × 10^-4 m^3/s.

So the discharge is about 4.75 × 10^-4 m^3/s.