The rate of flow of water through a hole is fundamentally determined by the velocity at which the water exits the hole and the cross-sectional area of that hole. The velocity of the water flowing through a hole, under the influence of gravity, is described by a specific physical law.
Understanding Water Flow Through a Hole
The velocity of water flowing out of an orifice (hole) in a container is directly related to the height of the water above the hole. This phenomenon is a practical application of fluid dynamics principles, often associated with Torricelli's Law.
The instantaneous velocity of flow through the hole, denoted as $v(t)$, is given by the formula:
$v(t) = k\sqrt{2gh(t)}$
Where:
- $v(t)$ is the velocity of the flow through the hole at a specific time $t$ (e.g., in meters per second, m/s).
- $k$ is a discharge coefficient, a dimensionless value typically less than 1 (0 < k ≤ 1). This coefficient accounts for energy losses due to friction, the contraction of the fluid jet (vena contracta) as it exits the hole, and the shape of the hole. For an ideal, frictionless flow, k would be 1.
- $g$ is the acceleration due to gravity (approximately 9.81 m/s² or 32.2 ft/s²).
- $h(t)$ is the instantaneous height of the water level above the center of the hole at time $t$ (e.g., in meters or feet).
While this formula provides the velocity of the water, the "rate of flow" most commonly refers to the volumetric flow rate—the volume of water passing through the hole per unit of time.
The Fundamental Formula for Velocity
The relationship $v(t) = \sqrt{2gh(t)}$ (when k=1) is derived from Torricelli's Law, which states that the speed of efflux from a hole in a tank under gravity is the same as the speed that a body would acquire in falling freely from the surface of the liquid to the hole. The coefficient 'k' adjusts this ideal velocity for real-world conditions.
Calculating the Volumetric Flow Rate
To determine the actual volumetric flow rate (often denoted as $Q$), you need to multiply the velocity of the flow by the cross-sectional area of the hole.
$Q(t) = A \times v(t)$
Substituting the velocity formula:
$Q(t) = A \times k\sqrt{2gh(t)}$
Where:
- $Q(t)$ is the volumetric flow rate at time $t$ (e.g., in cubic meters per second, m³/s, or liters per second, L/s).
- $A$ is the cross-sectional area of the hole (e.g., in square meters, m²). For a circular hole with radius $r$, $A = \pi r^2$.
This formula provides the complete rate of flow (volumetric flow rate) of water through a hole, taking into account the water height, gravity, the hole's dimensions, and real-world losses.
Factors Influencing the Flow Rate
Several key factors determine the rate of water flow through a hole:
- Height of Water ($h(t)$): The higher the water level above the hole, the greater the pressure and, consequently, the higher the flow velocity and volumetric flow rate.
- Area of the Hole ($A$): A larger hole allows more water to pass through, directly increasing the volumetric flow rate.
- Acceleration Due to Gravity ($g$): This is a constant on Earth, influencing the speed at which the water is driven out.
- Discharge Coefficient ($k$): This factor, specific to the hole's design and fluid properties, accounts for inefficiencies. For example, a sharp-edged hole will have a lower 'k' than a well-rounded or re-entrant hole.
- Fluid Properties: While not explicitly in the formula, properties like viscosity can indirectly affect 'k'.
Practical Applications and Considerations
Understanding the rate of flow through a hole is crucial in various fields:
- Hydraulics and Civil Engineering: Designing drainage systems, culverts, spillways in dams, and irrigation channels.
- Water Management: Estimating reservoir depletion rates or flow through sluice gates.
- Manufacturing: Controlling fluid dispensing, such as in chemical processing or food production.
- Everyday Situations: Understanding how quickly a pool or tub drains.
For more in-depth information on the underlying principles, explore resources on Torricelli's Law and basic fluid dynamics.
Examples and Solutions
Let's consider a hypothetical scenario:
Problem: A tank has a circular hole with a diameter of 5 cm at its base. The water level in the tank is maintained at a constant height of 1.2 meters above the hole. Assuming a discharge coefficient ($k$) of 0.98 for this hole, what is the volumetric flow rate of water?
Solution:
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Calculate the radius of the hole:
- Diameter = 5 cm = 0.05 m
- Radius ($r$) = Diameter / 2 = 0.05 m / 2 = 0.025 m
-
Calculate the area of the hole ($A$):
- $A = \pi r^2 = \pi \times (0.025 \text{ m})^2 \approx 0.00196 \text{ m}^2$
-
Identify knowns:
- $h = 1.2 \text{ m}$
- $g = 9.81 \text{ m/s}^2$
- $k = 0.98$
-
Calculate the velocity of flow ($v$):
- $v = k\sqrt{2gh} = 0.98 \times \sqrt{2 \times 9.81 \text{ m/s}^2 \times 1.2 \text{ m}}$
- $v = 0.98 \times \sqrt{23.544 \text{ m}^2/\text{s}^2}$
- $v \approx 0.98 \times 4.852 \text{ m/s} \approx 4.755 \text{ m/s}$
-
Calculate the volumetric flow rate ($Q$):
- $Q = A \times v = 0.00196 \text{ m}^2 \times 4.755 \text{ m/s}$
- $Q \approx 0.00932 \text{ m}^3/\text{s}$
Therefore, the volumetric flow rate of water through the hole would be approximately 0.00932 cubic meters per second.
