Problem 42 - College Physics

Question

At a point where an irrigation canal having a rectangular cross section is 18.5 \(\mathrm{m}\) wide and 3.75 \(\mathrm{m}\) deep, the water flows at 2.50 \(\mathrm{cm} / \mathrm{s} .\) At a point downstream, but on the same level, the canal is 16.5 \(\mathrm{m}\) wide, but the water flows at 11.0 \(\mathrm{cm} / \mathrm{s} .\) How deep is the canal at this point?

Step-by-Step Solution

Answer

The downstream canal is approximately 0.955 meters deep.

Step 1: Understanding the Continuity Equation

The continuity equation for fluid dynamics states that the mass flow rate must remain constant for an incompressible fluid. This leads to the equation: \(A_1v_1 = A_2v_2\), where \(A\) is the cross-sectional area and \(v\) is the velocity of water flow. Subscripts 1 and 2 denote upstream and downstream points, respectively.

Step 2: Calculating the Upstream Cross-Sectional Area

At the upstream point, the canal has a width of 18.5 m and a depth of 3.75 m. So, the cross-sectional area \(A_1\) is calculated by \(A_1 = 18.5 \, \text{m} \times 3.75 \, \text{m} = 69.375 \, \text{m}^2.\)

Step 3: Calculating Downstream Flow Velocity in Meters

Convert the water flow velocity from the downstream side into meters per second. The flow velocity \(v_2\) is given as 11.0 cm/s, which is \(11.0 \, \text{cm/s} = 0.11 \, \text{m/s}.\)

Step 4: Using Continuity Equation to Find Downstream Depth

Set up the equation using the continuity principle: \[18.5 \, \text{m} \times 3.75 \, \text{m} \times 0.025 \, \text{m/s} = 16.5 \, \text{m} \times d_2 \, \times 0.11 \, \text{m/s}.\]Calculate \(d_2\), the depth at the downstream point by solving for \(d_2\): \[d_2 = \frac{18.5 \, \text{m} \times 3.75 \, \text{m} \times 0.025 \, \text{m/s}}{16.5 \, \text{m} \times 0.11 \, \text{m/s}}.\]

Step 5: Solving for Downstream Depth

Compute the value by evaluating the expression:\[d_2 = \frac{69.375 \, \text{m}^2 \times 0.025 \, \text{m/s}}{16.5 \, \text{m} \times 0.11 \, \text{m/s}} = \frac{1.734375 \, \text{m}^2/s}{1.815 \, \text{m/s}} \approx 0.955 \, \text{m}.\]Thus, the depth of the canal at the downstream point is approximately 0.955 meters.

Key Concepts

Continuity Equation

In fluid dynamics, the continuity equation is a fundamental principle that describes the conservation of mass in a fluid flow. The basic idea is that the mass of fluid entering a system must equal the mass exiting if the fluid is incompressible and there's no accumulation in between. This is mathematically expressed as \(A_1v_1 = A_2v_2\), where \(A\) represents the cross-sectional area and \(v\) denotes the fluid velocity. The subscripts 1 and 2 refer to two different points along the flow. This relationship ensures that for a constant flow, any decrease in the cross-sectional area results in an increase in fluid velocity, and vice versa. Applying this concept helps predict how changes in a system impact flow characteristics.

Cross-Sectional Area

The cross-sectional area of a canal or pipe is critical in determining the flow rate of a fluid. It is essentially the surface area that is perpendicular to the flow of the fluid. In the context of the exercise, the upstream cross-sectional area \(A_1\) is calculated as the product of the canal's width and depth, giving \(18.5 \, \text{m} \times 3.75 \, \text{m} = 69.375 \, \text{m}^2\). Similarly, the cross-sectional area downstream can be affected by changes in either width or depth. By understanding and calculating the cross-sectional area, one can easily determine the changes needed to maintain a consistent flow as per the continuity equation.

Incompressible Fluid

An incompressible fluid is one whose density remains constant regardless of pressure changes within the fluid flow. Water is often treated as an incompressible fluid in many engineering and physics problems unless extreme conditions apply. In the problem at hand, treating water as an incompressible fluid allows us to use the continuity equation effectively. This assumption implies that as water flows through different sections of the canal, its volume remains unchanged even if the shape of the canal changes, facilitating simpler calculations and predictions of flow behavior.

Mass Flow Rate

Mass flow rate is a quantitative measure of the mass of fluid flowing through a cross-section per unit of time. It is often expressed in units such as kilograms per second (kg/s). However, when dealing with incompressible fluids, mass flow rate is not explicitly calculated if only flow velocity and cross-sectional area calculations are needed. Instead, volume flow rate, which involves multiplying cross-sectional area by velocity, becomes the primary focus since the density is constant. For practical purposes, understanding that the mass flow rate is proportional to the product of these two parameters provides insight into how changes in velocity or cross-sectional dimensions impact the flow dynamics.