17.4.1. Control Volume Application

Now, let me come in to the, if the flow is incompressible. Again, we have a lot of simplifications. Again, I can repeat it. The flow systems when you have mac number less than 0.3, okay, whether it is gas, whether it is a liquid or any flow system, if you think that the within the flow system the flow becomes less than the mac number less than 0.3, then there will be density variation, but that variation of density is much much negligible comparing to other components. So, we can assume the flow is incompressible nature. Again, I am going to summarise that. When you have any flow systems, mac number is less than 0.3, so we can use flow as incomprehensible flow, density does not vary significantly. So, density becomes constant, as density becomes constant, as you know it, it is very simplified problem what we are going to solve.

In fluid dynamics, we often deal with different types of flows. When the Mach number (a measure of the speed of flow relative to the speed of sound) is less than 0.3, we can typically treat the fluid as incompressible. This means that the fluid density remains essentially constant throughout the flow, which significantly simplifies our calculations. In other words, even if there are small density changes, they are so minor that they can be ignored without impacting the overall problem significantly.

Think of a garden hose: when water flows through it at low velocities, the water's density remains constant and does not compress much. However, if you speed up the water too much (like a rocket taking off), then you might have to consider compressibility effects, just like how air behaves under different speeds.

So, density varies negligible, as the density variation is not significant and beta equal to 1, so only this equation is left for us. Simple thing. This is very simple equation. DB sys = density * velocity * area + density * (density vector * normal vector) dA. Now, there, the scalar product of V and n and d A, okay, and density can come out. So, instead of the mass flux we are now talking about volumetric flux. That means, if you multiply the velocity into area, then what you get is unit volume per unit height, volumetric flux, okay? So, please do not confuse, this is a different equation.

When dealing with incompressible flows, we transition from discussing mass flow to volumetric flow. In volumetric flow, we consider how much volume of fluid passes through a particular area in a given time. The equation incorporates velocity and the area through which the fluid passes, allowing us to estimate how much fluid is flowing without needing to include density, which remains constant.

Imagine filling a pool with a garden hose. The amount of water flowing (volumetric flow rate) can be calculated simply by measuring the diameter of the hose and how fast the water flows out. You don’t need to worry about the density of water in this case because it’s consistent throughout the flow.

If the inlet and outlet are one-dimensional, where, Q = A V. Only that we are not showing the density multiplication. If you multiply with a Q, the V into A is Q is discharge. So, Q = V alpha , is the discharge. So, most of the conservation of mass you write, it, since density is a constant, you make it come out from that equation. So, it looks like volumetric level we are comparing but all are mass conservation equations.

In control volume analysis, we often look at one-dimensional flow where the mass flow rate (Q) can be simply expressed as the product of the cross-sectional area (A) and the average flow velocity (V). Here, we can simplify matters by treating density as a constant, allowing us to focus on volumetric relationships rather than mass relationships, thus simplifying our analysis.

Think of pouring a specific amount of sand through a funnel. The area of the funnel and the speed at which you pour the sand dictate how quickly the sand flows out. If the density of sand were to change dramatically, it would complicate this flow, but if we treat it as constant (which is usually valid), we can focus on how fast it pours out relative to the area of the funnel.

So, what I am looking at is again summarised here. To solve this mass conservation equation, I should have knowledge on the velocity field. I should know how the velocity varies or I should know whether the velocity is a constant or the velocity varies. If I know the velocity variations on this control surface, then I can solve the problem.

Understanding how velocity varies within the flow is crucial for applying mass conservation principles. In pipe flow, for instance, the velocity is highest at the center and decreases towards the walls. This velocity distribution impacts calculations and assumptions we can make about the flow behavior.

Picture a highway where cars move fastest in the center and slow down as they approach the edges. If you only pay attention to the average speed of all cars instead of their position, your understanding of traffic flow becomes less accurate—similarly, understanding velocity distribution in a pipe is essential for accurate flow analysis.

So, basically, when you apply the mass conservation equation, your knowledge of velocity variation is important to you, how you are simplifying the velocity field on the control surface. That way we solve the problem. If you can understand properly, applying this Reynolds transport theorem and the simplification and put the numerical value, it does not take much time or much problem to solve any problem.

Assuming uniform velocity distribution simplifies calculations but is not always realistic. It is crucial to assess if the flow can be approximated as uniform before applying mass conservation equations. This assessment helps ensure that our final conclusions and calculations are valid and realistic.

Consider a bathtub where water is added evenly. If you assume all parts of the water are level (uniform), you can easily calculate how much time it takes to fill it. But if the water is splashing towards one side, then your assumption will lead to incorrect results. You must account for variations in flow to get accurate results.

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