For a nozzle with area A and head h, the ideal discharge Q_th = A sqrt(2 g h). If Cd = 0.95 and h doubles, how does Q change?

• A. It remains unchanged.
• B. It doubles.
• C. It increases by a factor of sqrt(2) ≈ 1.414.
• D. It decreases by half.

Answer: (C)

The main idea is that discharge through a nozzle (with a given area) scales with the square root of the driving head. The actual discharge is Q = Cd × A × sqrt(2 g h), so Cd just reduces the ideal value but stays constant if the nozzle and conditions don’t change.

If the head doubles, replace h with 2h in the expression: Q_new = Cd × A × sqrt(2 g (2h)) = Cd × A × sqrt(4 g h) = 2 × Cd × A × sqrt(g h). The original ideal discharge is Q_old = Cd × A × sqrt(2 g h) = sqrt(2) × Cd × A × sqrt(g h). Taking the ratio shows Q_new / Q_old = (2) / sqrt(2) = sqrt(2). So the actual discharge increases by a factor of sqrt(2), i.e., about 1.414.