Problem 68 - Physics

Question

Two circular holes, one larger than the other, are cut in the side of a large water tank whose top is open to the atmosphere. The center of one of these holes is located twice as far beneath the surface of the water as the other. The volume flow rate of the water coming out of the holes is the same. (a) Decide which hole is located nearest the surface of the water. (b) Calculate the ratio of the radius of the larger hole to the radius of the smaller hole.

Step-by-Step Solution

Answer

The smaller hole is nearest the surface. The radius ratio is approximately 0.84.

Step 1: Identifying Variables

Let's denote the smaller depth as \( h_1 \) and the larger depth as \( h_2 \). Given \( h_2 = 2h_1 \). Let the area of the smaller hole be \( A_1 \) and the area of the larger hole be \( A_2 \). The volume flow rate, \( Q \), for each hole is the same.

Step 2: Applying Torricelli's Law

Torricelli's Law states \( v = \sqrt{2gh} \), where \( v \) is the velocity, \( g \) is the acceleration due to gravity, and \( h \) is the depth of the fluid. For the smaller hole, the velocity \( v_1 = \sqrt{2gh_1} \), and for the larger hole, \( v_2 = \sqrt{2gh_2} \).

Step 3: Equating Flow Rates

The flow rate \( Q \) is given by \( Q = Av \), where \( A \) is the area and \( v \) is the velocity. Since \( Q_1 = Q_2 \), we have \( A_1v_1 = A_2v_2 \). Using the velocities from Torricelli's Law, we have \( A_1\sqrt{2gh_1} = A_2\sqrt{2gh_2} \). Simplifying, \( A_1\sqrt{h_1} = A_2\sqrt{2h_1} \).

Step 4: Simplifying the Equation

Simplify the equation \( A_1\sqrt{h_1} = A_2\sqrt{2h_1} \) by dividing both sides by \( \sqrt{h_1} \): \( A_1 = A_2\sqrt{2} \).

Step 5: Relating Areas to Radii

Since the area \( A \) of a hole is \( \pi r^2 \), where \( r \) is the radius, we can express \( A_1 = \pi r_1^2 \) and \( A_2 = \pi r_2^2 \). From our previous equation, we have \( \pi r_1^2 = \pi r_2^2 \sqrt{2} \). Simplifying, \( r_1^2 = r_2^2 \sqrt{2} \).

Step 6: Calculating the Ratio of Radii

Take the square root of both sides: \( r_1 = r_2(2^{1/4}) \). To find the ratio \( \frac{r_2}{r_1} \), rearrange the equation to \( \frac{r_2}{r_1} = \frac{1}{2^{1/4}} \). Calculating this gives approximately \( 0.84 \).

Step 7: Verifying Depth Relation

Given that \( h_2 = 2h_1 \), the smaller hole is at \( h_1 \) which is closer to the surface. Hence, the larger radius corresponds to the hole at \( h_2 = 2h_1 \).

Key Concepts

Torricelli's Law

Torricelli's Law is a fundamental concept in fluid mechanics that describes the speed of a fluid flowing out of an orifice under the force of gravity. It is mathematically represented as \( v = \sqrt{2gh} \), where \( v \) is the velocity of the fluid, \( g \) represents the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \) on Earth), and \( h \) is the height or depth of the fluid above the opening.
This law helps us understand how fluids behave under the influence of gravity. In essence, the deeper the point in the fluid, the faster the fluid will be traveling when it exits from a hole.
In practical applications like the exercise above, Torricelli's Law was used to determine velocities at different depths, simplifying the analysis of flow rates from two openings at different positions.

Volume Flow Rate

The volume flow rate is a measure of the quantity of fluid that passes through a surface per unit of time. It is often described using the symbol \( Q \) and calculated by multiplying the velocity of the fluid \( v \) by the area \( A \) of the opening through which it flows. Therefore, the formula is \( Q = Av \).
In scenarios where two holes in a tank contribute to the same flow rate, understanding volume flow rate helps equate and compare the behavior of fluid through different sized openings, despite varying conditions like depth or hole radius.

• For our specific exercise, ensuring equal flow rates from two different holes despite different depths and diameters required using Torricelli's Law alongside the flow rate equation.
• The focus was on maintaining the equilibrium of \( Q_1 = Q_2 \), balancing both the velocity obtained from the fluid depth and the area represented by its circular opening.

Circular Holes

Circular holes are often used in fluid mechanics problems because they provide a straightforward geometry for calculations and theoretical applications.
In our exercise, the holes were described as having two distinct radii, affecting how fluids exited from the tank. Let’s expand on this:

• A circular hole’s area is calculated using \( \pi r^2 \), where \( r \) represents the radius.
• The difference in radii impacts the amount of fluid that can flow through. A larger hole allows more fluid to pass per unit time, hence affecting the flow rate, depending on the fluid's velocity.
• By relating the areas to the problem's depth condition, the exercise guided students to find the radii's ratio by using basic principles of area from geometry.

This geometric relationship is pivotal in equating the flow rate despite various dimensions, as different radii create different performance in fluid dynamics.

Depth Relation

The depth relation fundamentally alters how fluids behave when exiting from an opening. In this exercise, understanding the role of depth is crucial for determining flow dynamics.
Since depth directly influences velocity through Torricelli's Law \((v = \sqrt{2gh})\), it plays a vital role in how different holes with the same flow rate must relate in terms of size and position.

• In our exercise, the holes were positioned at different depths, precisely \( h_1 \) and \( h_2 = 2h_1 \). This difference means the hole twice as deep facilitates a faster fluid exit velocity due to gravity.
• To maintain equal flow rates from both openings, the hole closer to the surface needed a different radius, large enough to compensate for the slower fluid speed due to its shallower depth.
• This creates an intuitive link between position and size, where the placement of each hole influences its corresponding radius based on velocity requirements.

By aligning the problem's requirement for equal flow rates with depth differences, students can more deeply understand this linkage in fluid mechanics situations.