Velocity of approach for a tank–orifice system A large tank (area a1) drains through a sharp-edged orifice of area a2 located on its side/bottom. If the head on the orifice is h (measured to the centre of the orifice), the mean velocity in the tank due to drawdown—called the velocity of approach—is:

Introduction / Context:
When a tank empties through an orifice, the free surface drops. The finite, though often small, downward velocity of the free surface is termed the “velocity of approach” and slightly increases the actual discharge beyond the simple Torricelli estimate if accounted for.

Given Data / Assumptions:

• a1 = plan area of tank free surface (assumed constant).
• a2 = area of the orifice.
• h = head on the orifice, measured to its centre.
• Neglect coefficient effects for the relation between Va and h.

Concept / Approach:
Continuity requires the volumetric flow leaving through the orifice to equal the rate of decrease of volume in the tank. If v is the jet velocity at the orifice, and Va is the downward velocity of the free surface, then Q = a2 * v = a1 * Va. For ideal conditions, v ≈ √(2 g h). Therefore, Va = (a2 / a1) * √(2 g h).

Step-by-Step Solution:

Verification / Alternative check:
If a1 ≫ a2, Va is very small, justifying its neglect in many basic problems while recognizing its existence for accuracy.

Why Other Options Are Wrong:

• √(2 g h) is the orifice jet velocity, not the free-surface approach velocity.
• Inverted or squared area ratios give wrong dimensions or magnitudes.

Common Pitfalls:
Confusing jet velocity with surface velocity; ignoring that Va scales with a2/a1.

Final Answer:
Va = (a2 / a1) * √(2 g h)