Draining Time for a Rectangular Swimming Pool
Problem Statement
A rectangular swimming pool is 1 m deep at one end and increases uniformly in depth to 2.6 m at the other end. The pool measures 8 m in width and 32 m in length. It is emptied through an orifice of area 0.224 m2 located at the lowest point on the deep end. Taking the coefficient of discharge (Cd) as 0.6, determine:
- (a) The time for the water depth to fall by 1 m (from 2.6 m to 1.6 m).
- (b) The total time to empty the pool completely.
Given Data
| Width of Pool | 8 m |
| Length of Pool | 32 m |
| Area of Pool (Apool) | 8 × 32 = 256 m2 (for the region of constant cross‐section) |
| Orifice Area (a0) | 0.224 m2 |
| Coefficient of Discharge (Cd) | 0.6 |
| Initial Depth at Deep End | 2.6 m |
| Intermediate Depth for (a) | 1.6 m |
| Gravitational Acceleration (g) | 9.81 m/s2 |
| Note for (b) | For depths below 1.6 m, the pool area varies with depth. |
(a) Time for Depth to Fall by 1 m
For a constant cross-sectional area, the continuity equation gives:
With the orifice discharge given by Q = Cd · a0 √(2gh), separating variables and integrating from H1 = 2.6 m to H2 = 1.6 m, we obtain:
Substituting the values:
Evaluating the expression yields:
(b) Time to Empty the Pool Completely
For depths below 1.6 m, the pool area is not constant because the depth increases uniformly from the shallow end. Given that the shallow end is 1 m deep when the deep end is 1.6 m, the ratio of the pool length to depth in this region is:
Therefore, the cross-sectional area in terms of h becomes:
Applying the continuity equation:
Integrating from H1 = 1.6 m to H2 = 0 m gives:
Evaluating the integral (∫ h1/2 dh = 2/3 h3/2):
This yields:
The total time to empty the pool is:
Detailed Explanation
The problem is divided into two parts because the cross-sectional area of the pool changes as the water level falls. In part (a), the pool has a constant area (256 m2) since the water level is above the sloping region. The continuity equation combined with Torricelli’s law allows us to relate the change in water level to the discharge through the orifice.
In part (b), as the water level drops below 1.6 m, the geometry of the pool changes—the length of the water surface becomes a function of the water depth. By deriving a relationship (L = 20h), we obtain a variable cross-sectional area (A = 160h), which requires integration to determine the time for the water level to fall from 1.6 m to 0 m.
Physical Meaning
The analysis of the draining pool illustrates several fundamental fluid mechanics concepts:
- Continuity Equation: This represents the conservation of mass. The rate at which the water level drops in the pool is balanced by the discharge through the orifice.
- Torricelli’s Law: The discharge Q = Cd · a0 √(2gh) indicates that the exit velocity is governed by the potential energy (height) of the water above the orifice.
- Variable Cross-sectional Area: When the pool has a sloping bottom, the effective area through which the water drains changes with depth, leading to a non-linear relationship between time and water depth.
- Integration: The integration process accumulates the small changes in water level over time, providing the overall drainage time for each section.
Conclusion
(a) The time for the water depth to fall by 1 m (from 2.6 m to 1.6 m) is approximately 299 s.
(b) The total time to empty the pool completely is approximately 662 s.
