Problem 56 - Fundamentals of Physics

Question

Suppose that two tanks, 1 and \(2,\) each with a large opening at the top, contain different liquids. A small hole is made in the side of each tank at the same depth \(h\) below the liquid surface, but the hole in tank 1 has half the cross-sectional area of the hole in tank 2 . (a) What is the ratio \(\rho_{i} / \rho_{2}\) of the densities of the liquids if the mass flow rate is the same for the two holes? (b) What is the ratio \(R_{n} / R_{V 2}\) of the volume flow rates from the two tanks? (c) At one instant, the liquid in tank 1 is \(12.0 \mathrm{~cm}\) above the hole. If the tanks are to have equal volume flow rates, what height above the hole must the liquid in tank 2 be just then?

Step-by-Step Solution

Answer

(a) Density ratio is 2. (b) Volume flow rate ratio is 0.5. (c) Height in tank 2 is 3 cm.

Step 1: Understand the Problem

We have two tanks with liquids of different densities but equal mass flow rates through the holes at depth \( h \). Tank 1 has a hole with area \( A_1 \) and tank 2 has a hole with area \( A_2 = 2A_1 \). We need to find the ratio of the densities, the ratio of volume flow rates, and the height in tank 2 for equal volume flow rates when tank 1 has liquid at a given height.

Step 2: Apply Torricelli's Law

The speed \( v \) of the fluid exiting a hole, according to Torricelli's law, is given by \( v = \sqrt{2gh} \), where \( g \) is the gravitational acceleration. The speed depends only on the height \( h \), not on the density or area of the hole.

Step 3: Determine the Mass Flow Rate

The mass flow rate for each tank is given by \( \dot{m} = \rho A v \), where \( \rho \) is the fluid density, \( A \) is the cross-sectional area of the hole, and \( v \) is the velocity of the fluid. For equal mass flow rates, \( \rho_1 A_1 v_1 = \rho_2 A_2 v_2 \), with \( v_1 = v_2 \).

Step 4: Solve for Density Ratio

Substitute \( v_1 = v_2 \) into the mass flow rate equation to get \( \rho_1 A_1 = \rho_2 A_2 \). Given \( A_2 = 2A_1 \), we find \( \rho_1 / \rho_2 = A_2 / A_1 = 2 \).

Step 5: Calculate Volume Flow Rate Ratio

The volume flow rate \( \dot{V} = A v \). Therefore, the ratio of volume flow rates \( R_{V1} / R_{V2} = (A_1 v_1) / (A_2 v_2) = A_1 / A_2 = 0.5 \).

Step 6: Find the Required Height in Tank 2

For equal volume flow rates, equate the volume flow rates: \( A_1 v_1 = A_2 v_2 \). Since \( A_2 = 2A_1 \) and \( v_1 = \sqrt{2gh_1} \) with \( h_1 = 12 \) cm, solve for \( v_2 \): \( v_2 = v_1 / 2 \). Thus, \( \sqrt{2g h_2} = \sqrt{2g \times 12} / 2 \), which gives \( h_2 = 3 \) cm.

Key Concepts

Torricelli's Law

Torricelli's Law is a fundamental principle in fluid dynamics that describes the speed of a fluid flowing out of an orifice under the influence of gravity. According to this principle, the speed, \( v \), is given by the equation:\[ v = \sqrt{2gh} \]where \( g \) is the acceleration due to gravity and \( h \) is the height of the fluid above the opening.
This means that the flow speed is dependent solely on the height of the fluid, making Torricelli's Law particularly useful in predicting how fluids will behave in containers with openings.
It is important to note that the area of the hole or the fluid's density does not affect the exit velocity according to Torricelli's Law, making it easier to calculate fluid speed where heights can be measured.

mass flow rate

The mass flow rate is a key concept in fluid dynamics that measures how much mass of fluid is passing through a section per unit time. It is denoted by \( \dot{m} \) and is calculated using the formula:\[ \dot{m} = \rho A v \]where:

• \( \rho \) is the density of the fluid,
• \( A \) is the cross-sectional area of the opening, and
• \( v \) is the velocity of the fluid flowing out.

If the mass flow rates are equal across different systems, changes in cross-sectional area or density will directly affect the required velocity for fluid to maintain equilibrium.
Solving problems involving mass flow rate often involve setting up these equations to balance different conditions, as demonstrated in the exercise where the flow from different tanks is analyzed.

volume flow rate

Volume flow rate is a measure of how much volume of fluid passes through a cross-section per unit time, given the symbol \( \dot{V} \). It is calculated by:\[ \dot{V} = A v \]With:

• \( A \) being the cross-sectional area of the hole,
• \( v \) as the velocity of fluid outflow.

This concept helps determine how quickly a tank or container loses its fluid through an opening or how fast another one fills. It is key in engineering applications to ensure proper flow rates are maintained.
In our original problem, the volume flow rate determines how the height difference between two tanks affects the system, specifically influencing what height is needed in one tank to achieve an equal volume flow rate with another.

density ratio

The density ratio between two fluids is a comparison measure indicating how much mass per unit volume one fluid has relative to another. In solving problems where fluid dynamics concepts like mass flow rate or volume flow rate are involved, understanding density ratio is crucial for balancing equations.
In our problem, we calculate the density ratio by leveraging the fact that mass flow rate is consistent across different systems. Substituting equal mass flow rates and using known area proportions, we find:\[ \frac{\rho_1}{\rho_2} = \frac{A_2}{A_1} \]In this specific case, upon substituting the given values, it was determined that \( \frac{\rho_1}{\rho_2} = 2 \).
This outcome means that for the same mass flow rate, the fluid in tank 1 is denser than the fluid in tank 2, provided the given conditions.