19.4. Reynolds Transport Theorem

Now, if you come back to the surface forces, like for example, for tetrahedral structures like this where you have dx, dy, and dz components are there, and your axis is like this, x direction, y direction, and z direction. So, you can define the surface forces as a stress tensor. Stress means what, force per unit area. So, you can define as a stress tensor. So, this stress will have nine components.

The concept of stress in mechanics is crucial for understanding forces acting on objects. A stress tensor is a mathematical construct that represents the internal forces (stresses) within a material. It quantifies how these forces are distributed across different surfaces within the object. The stress tensor has nine components in three-dimensional space, which correspond to the forces acting in different directions: two for normal forces (perpendicular to a surface) and the others for shear forces (parallel to a surface).

Think of a sponge under pressure, like when you squeeze it with your hand. The pressure applied creates stresses inside the sponge, pushing against different internal surfaces. Just like the stress tensor captures how the internal forces are distributed inside the sponge, it helps us understand how forces act within materials under various conditions.

So, you have the stress tensor coordinate systems defining nine stress components. If you look at that, if you take this diagonal component which is the normal component to this surface like σ₁, σ₂, these are all normal components. That means these are compositions of the pressure force and the viscous force component. But the diagonal component what we have is σ₃, σ₄, and all, which is acting tangentially.

The nine components of the stress tensor can be categorized into two main types: normal stresses, which act perpendicular to the surfaces, and shear stresses, which act parallel to the surfaces. Normal stresses are related to how much pressure the material is under, while shear stresses arise from the viscosity of the material, representing how it resists flow. These components help define how the material will behave when forces are applied.

Imagine a set of books stacked on a shelf. The weight of the books pressing down creates a normal stress on the shelf. If someone pulls a book sideways, that creates shear stress on the books and the shelf. Understanding both types of stress helps us design stronger shelves that can hold more weight without bending or breaking.

Now, if I have the stress component there and I have the normal vectors, if I resolve the force components, I will have the scalar product between the stress tensor and the n vectors, that is how we do it. And for the total surface area we do surface integrals to compute it.

To determine the total forces acting on a surface, we use the concept of surface integrals, where we calculate the forces on each infinitesimally small piece of the surface. By taking the scalar product of the stress tensor and the normal vector to the surface, we can find out how much force acts in each direction and then sum it up across the entire surface area.

Consider a flat, thin pizza. To find out how much weight the pizza dough can bear before it collapses, you'd evaluate how much pressure is applied by various toppings across every slice of pizza. By aggregating the forces from each topping on the dough, you can determine the total pressure experienced by different areas.

The total force acting on the control volume will have the body force component and surface force component. The body force component will have volume integrals of rho g dV, g is the vector quantity as we consider the g, acceleration due to gravity.

Forces acting on a control volume can be divided into body forces and surface forces. Body forces, like gravity, act throughout the material (e.g., the weight of fluid in a tank), while surface forces act at the boundaries (e.g., pressure forces on the walls of a tank). By using volume integrals for body forces and surface integrals for surface forces, we can compute the total forces on a volume.

Imagine a water balloon. The gravity pulling down on the water (body force) affects every drop inside, while the pressure exerted on the surface of the balloon (surface force) affects only the material of the balloon itself. Understanding both types of forces helps engineers design stronger and safer containers for liquids.

Now, if I divide the force components, that means, the total force will be one component from this body force, it is gravity force component. The surface force component we can resolve it into the force due to the pressure, force due to viscosity.

When applying the linear momentum equations, we can simplify the total force by splitting it into its components: the gravitational body force and the various surface forces due to pressure and viscosity. This separation allows for a clearer analysis of each force's contribution to the overall behavior of the system, making it easier to apply principles of fluid mechanics.

Think of a car driving down a slope. The weight of the car (gravity) pulls it down the slope (body force), while the friction between the tires and the road (surface force) resists that motion. By understanding both forces, we can predict how fast the car will go and how much stopping power it will need on the way down.

If you can understand it, if I take a control volume, everywhere I will have atmospheric pressure, then the absolute pressure from some locations. So, if I consider the atmospheric pressure is acting throughout this control surface and do surface integral, if the pressure is having the direction, and if I do a surface integrals over this, it will be cancelled out and becomes 0.

In many engineering applications, the effects of atmospheric pressure can be neglected when analyzing forces within a control volume. Since atmospheric pressure acts equally in all directions and integrates out to zero in enclosed spaces, we can focus on gauge pressures instead, which simplifies calculations while still capturing the essential physics of the problem.

Imagine a sealed balloon at sea level. The pressure inside the balloon is influenced by atmospheric pressure, but when calculating how much the balloon expands or contracts, the atmospheric pressure can be ignored since it affects all parts equally. Instead, we can focus solely on the additional pressure generated by the air inside the balloon.

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