Given A=2.0×10^-3 m^2, h=4 m, g=9.81 m/s^2, and Cd=0.68, what is Q? (Cd = Q / (A sqrt(2 g h)))

• A. 0.020 m^3/s
• B. 0.010 m^3/s
• C. 0.012 m^3/s
• D. 0.008 m^3/s

Answer: (C)

Discharge through an orifice is described by Q = Cd A sqrt(2 g h). Here Cd is the discharge coefficient (dimensionless), A is the opening area, g the gravitational acceleration, and h the head. Start by calculating the head term: 2 g h = 2 × 9.81 × 4 = 78.48, and its square root is about 8.859 m/s. Multiply by the area to get the theoretical flow without Cd: A × sqrt(2 g h) = 0.002 × 8.859 ≈ 0.017718 m^3/s. Now apply Cd: Q = 0.68 × 0.017718 ≈ 0.0120 m^3/s. Rounding to three significant figures gives 0.012 m^3/s, which matches the provided value. This shows how the flow depends on the opening area, the head, and the efficiency factor Cd.