The dimensionless discharge Cd for an orifice is defined as Cd = Q / (A sqrt(2 g h)). If Q=0.012 m^3/s, A = 2.0×10^-3 m^2, h = 4 m, compute Cd.

• A. Cd ≈ 0.50
• B. Cd ≈ 0.75
• C. Cd ≈ 0.60
• D. Cd ≈ 0.679

Answer: (D)

The dimensionless discharge coefficient tells us how the actual discharge through an orifice compares to the ideal, inviscid flow. It’s defined as Cd = Q / [A sqrt(2 g h)], so Cd represents the fraction of the ideal flow rate that is actually achieved.

Compute the term sqrt(2 g h): with g ≈ 9.81 m/s² and h = 4 m, 2 g h = 2 × 9.81 × 4 = 78.48, so sqrt(78.48) ≈ 8.862 m/s. Multiply by the area A = 2.0 × 10^-3 m² to get the ideal discharge denominator: A sqrt(2 g h) ≈ 2.0×10^-3 × 8.862 ≈ 0.01772 m³/s.

Now Cd = Q / (A sqrt(2 g h)) = 0.012 / 0.01772 ≈ 0.677, which rounds to about 0.679. This value, less than 1, reflects real-world losses (viscosity, vena contracta, etc.) through the orifice.