2.3. Conservation of Mass

So, now conservation of mass, how can we apply, what we have learned in Reynolds transport theorem to the conservation of mass. So, B is total amount of mass in the system, when we do the conservation of mass B which we saw in Reynolds transport theorem is the total amount of mass in the system and b will be mass per unit mass which is equal to 1. So, we have determined B and b for the conservation of mass. So, B is M, capital B is M and small b is 1.

The principle of conservation of mass states that mass cannot be created or destroyed in a closed system. In the context of fluid mechanics, when we use Reynolds transport theorem, we define two quantities: B (the total mass in the system) and b (the mass per unit mass). Here, B is represented by capital M (the total mass), and b is equal to 1, indicating that we are dealing with mass itself. This sets the foundation for how we analyze mass flow in systems.

Think of a closed container filled with water. No matter how much you shift or stir the water inside, the total amount of water (mass) in that container remains constant. This is similar to how mass behaves in a fluid system according to conservation of mass.

So, we can simply write in the control volume sys ∂/∂t ∫ ρV⋅n̂ dA = − ∫ ρ dV. This is the equation that we get by substituting B system as mass B as mass system and b as 1 as we have seen. So, this is actually nothing but a continuity equation.

In fluid dynamics, a control volume is an imaginary box where we analyze the flow of fluid. The equation ∂/∂t ∫ ρV⋅n̂ dA = − ∫ ρ dV states that the rate of change of mass within the control volume is equal to the mass flow out of the control volume. This equation is essential as it is known as the continuity equation, which ensures that mass flow is conserved, meaning mass entering a system must equal mass leaving the system plus any mass that accumulates.

Imagine a water tank with a faucet at the bottom. If water flows out through the faucet at a certain rate, and no water is entering the tank, the water level will drop. The continuity equation helps us understand that the rate of water leaving (mass flow out) corresponds directly to the change in water level (mass accumulation) in the tank.

So, we can simply write the velocity, average velocity as here this quantity integral V dot n cap dA divided by the whole area of cross section. So, V bar is the spatially averaged velocity normal to the cross section, this is important to note.

When we consider flow through a cross-section, the continuity equation can be expressed in terms of velocities. The average velocity (V bar) can be derived by integrating the flow velocity across the cross-section area and dividing it by the area. This averaged value is crucial as it represents the effective velocity at which mass moves through a given section of the fluid flow.

Think of a garden hose. If you narrow the hose's nozzle, the water flows faster out of the end. However, if you measure the average speed of water across the entire cross-section of the hose, it accounts for varying flow rates as water moves. This is similar to how we calculate the average velocity in fluid flow.

continuity equation for constant density and uniform velocity, so, if there are 2 cross sections as we have seen in this, you know, this problem. We can simply write ∫ ρV ⋅ n̂ dA + ∫ ρV ⋅ n̂ dA = 0.

In the flow between two cross-sections of varying shapes or sizes, the conservation of mass states that the mass flow rate must remain constant. This leads us to write the equation ∫ ρV ⋅ n̂ dA + ∫ ρV ⋅ n̂ dA = 0, which indicates that mass leaving one section must equal mass entering another. This is derived from the continuity principle where the density is constant.

Consider a highway with two lanes merging into one. The number of cars (mass flow) entering the single lane must equal the number of cars leaving, assuming no cars are stopping or leaving the highway. This analogy illustrates how mass flow rates should balance out in fluid systems.

So, the flow out of reservoir, for example, 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 analyze how the water level in a reservoir changes as water flows out. Given a flow rate of 2 liters per second and the dimensions of the reservoir, we can apply the continuity equation to determine how quickly the surface level drops. By understanding how mass flow translates into changes in height, we can calculate the rate of descent of the water level.

Think of filling a bathtub with water. If you drain the water at a consistent rate, you can predict how quickly the water level will decline based on the drain's flow rate and the bathtub's dimensions. This same principle applies to our example of the reservoir.

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