2.5. Examples of Conservation of Mass

The flow out of the reservoir is 2 liters per second. The reservoir surface is 5 meters into 5 meters, the question is how fast is the reservoir surface dropping.

In this example, we start with a reservoir where water is flowing out at a rate of 2 liters per second. To analyze how fast the water level in the reservoir is dropping, we can apply the principle of conservation of mass, which states that mass cannot be created or destroyed. Using the equation of continuity derived from Reynolds transport theorem, we express the flow out of the reservoir as the mass flow rate (Q) which equals the velocity (V) of the water at the surface multiplied by the cross-sectional area (A) of the reservoir's surface.

We can set up the equation as: Q_out = -A * (dh/dt) where Q_out is the flow rate (in cubic meters per second), A is the area of the surface of the reservoir, and dh/dt is the rate of change of height. Using this relationship, we can isolate dh/dt to find out how fast the water surface is dropping.

Think of a bathtub with a drain. If you leave the tap running and the drain open, the water level will remain constant when the inflow equals the outflow. However, if you turn off the tap and continue draining water, the level will drop. In this case, if we know how much water is draining out per second, we can easily calculate how fast the water level is falling based on the size of the bathtub.

We can simply write $\int \rho V \cdot \hat{n} dA + \int \rho V \cdot \hat{n} dA = 0$ at two cross-sections.

The continuity equation for steady flow states that the mass flow rate entering a control volume must equal the mass flow rate leaving it. This can be expressed mathematically using integrals over two cross-sections of a flow. If we denote the first cross-section with a subscript 1 and the second with a subscript 2, the equation becomes: $\int_{cs_1} \rho V \cdot \hat{n} dA_1 + \int_{cs_2} \rho V \cdot \hat{n} dA_2 = 0$. This means that the mass flow rates across each section are equal in magnitude but opposite in direction. When the density $\rho$ is constant, we can simplify this further to show that $V_1 A_1 = V_2 A_2$. This is a direct consequence of the conservation of mass in fluid flow.

Imagine a crowded concert hallway that gets narrower as it leads to the exit. If a certain number of people (mass flow rate) can pass through the hallway while it's wide, when the hallway narrows, the same number of people must exit through, but they will be moving faster to accommodate the flow. The conservation of mass ensures that even as the available space decreases, the number of people moving remains constant, resulting in a higher speed.

The equation of continuity states: $dh/dt = -Q/A_{reservoir}$, where Q is the flow out.

To find out how fast the water level drops in the reservoir, we rearrange the flow rate equation derived earlier. We substitute the given outflow rate into the equation $dh/dt = -Q/A_{reservoir}$. Here, Q is the known flow rate (2 liters per second), which we must convert into cubic meters per second for consistency in units. The area of the reservoir can be calculated as the product of its dimensions (5m x 5m). By dividing the flow rate by the area, we get the rate of height reduction in meters per second, indicating how fast the water level is falling.

Think of pouring a bucket of water into a larger pool. If the drain (like the outflow of the reservoir) is more powerful than how quickly you pour in, the water level in the pool will drop. The exact drop rate can be calculated if you know how quickly you're pouring and the size of the pool, demonstrating conservation of mass as it relates to the volume of water.

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