7.4.3. Simplified Problems Involving Fluid at Rest

Now let us come to the very basic concept what we are talking about the fluid at rest. So the basically we are talking about now, the fluid at rest, okay? If it is a fluid is at rest, it is a very simplified problem now. Like as I said it any fluid flow problems we look at either the pressure field, the velocity field, or the density and temperature field. When the fluid is rest now, very simply way the velocity vectors becomes zero and if I consider incomprehensible the density is a constant and if the temperature is not very much I need not need a thermodynamics first laws to define the problems. Then only the left is that the pressure field.

When we discuss fluid at rest, we mean that the fluid isn't moving. In this case, all the fluid's velocities are zero, which simplifies our analysis. Since the density of the fluid is constant in incompressible fluids and the temperature changes are negligible, we primarily focus on how pressure varies throughout the fluid. This understanding helps us analyze problems involving fluids without complexity from velocity or changes in density.

Think of a glass of water sitting on a table. The water is at rest; it doesn't flow, and the surface is calm. In this scenario, the only forces at play are the pressure from the weight of the water above and the atmospheric pressure around it. This setup allows us to easily measure and calculate the water pressure at different depths.

Since there is no velocity, there is no velocity gradient, definitely as Newton's second law says that there is no velocity gradients that means no shear stress. So when a fluid is at rest, there is no shear stress acting on that. So you can take a surface or take a control volume. Over that control surface you can define it the shear stress components become zero. That is very simplified now that any control volume you consider it over that control volume surface as the fluid is at rest conditions, there is no shear stress acting on the control surface. So that is the very easy problems what we have now.

Without motion, fluids don't have velocity gradients, which eliminates shear stress from the equations. In simple terms, shear stress is the force that causes layers of fluid to slide past one another, but with no movement, this stress does not exist. Thus, when we analyze a control volume—an imaginary box that contains fluid—we find that the only significant forces we need to consider arise from the pressure within the fluid and any external forces like gravity.

Imagine a thick syrup resting in a jar. Because it's not moving, the syrup doesn't exert shear stress in the traditional sense; rather, it's just exerting pressure downwards due to the weight of the syrup. If you try to tilt the jar, you'll notice the syrup remains where it is, demonstrating the lack of shear stress.

And what the two forces we have? The gravity force and the force due to the pressure distribution. So the whatever the pressure distribution forces that what is equate with the gravity force. Very simple things now, and since is a fluid is at the rest, so you can say that there is no mass flux is coming into the any control volumes or the momentum flux or no external work done it. So this is the what the simplified case.

When considering fluids at rest, we examine two primary forces: the force of gravity acting downwards and the pressure forces acting on the surface of the fluid. In a state of equilibrium (like when water is still in a dam), these forces counterbalance each other. Since there's no movement, mass isn't flowing in or out, and there’s no additional work being done on the system, leading to this simplified analysis.

Think about a swimming pool. The water is at rest, and the pressure at any point in the pool is balanced by the weight of the water above it. As you go deeper, the pressure increases due to the added weight of the water above, but the pool's stillness means all the forces are relatively simple to understand and calculate.

Like for example, if you take this dam, which is 100-meter-high and we have a reservoir, let you consider this is what 90 meter height from the bottom. That is what the water levels is 90 meter from the bottom. And you can understand it because of these in the reservoirs the fluid is at the rest conditions and that rest condition exerting the pressures on these surface.

In engineering, understanding how fluid pressure works is crucial for designing structures like dams. For example, a dam holds a large volume of water, and the pressure exerted by this water changes with depth. By calculating the pressure due to the 90 meters of water, engineers can determine the stress on the dam’s walls and ensure it does not fail.

Consider filling a balloon with water. The deeper you go into the balloon (the more water you add), the stronger the pressure feels at the bottom from the column of water pushing down. This principle is the same for dams—the deeper the water, the more pressure there is against the dam wall, and engineers must factor this into their designs to ensure safety.

Second thing is that fluid act as a rigid body motions. Let us take an example here that I have a tank which is a half filled liquid is there. Let us consider it may have water and this tank is accelerated with a constant acceleration a. As you start accelerating this tank with a constant acceleration a, you can understand it, this free surface is going to change it.

When we accelerate a tank filled with liquid, the liquid surface will adjust to maintain equilibrium. Even though the tank moves, the behavior of the fluid can be analyzed as if it were a rigid body—this means we can still apply hydrostatic principles. Thus, we can find out the new height of the liquid surface when the tank is in motion without complexity from shear forces since there are no gradients in velocity.

Imagine you're in a car that suddenly accelerates. If you have a cup of water in hand, the water will tilt back as you accelerate. Though the car moves, the surface of the water remains flat, showcasing that the fluid behaves as if it's part of a rigid system, making hydrostatic calculations still valid.

Now whenever I am talking about that I am looking for a pressure field as the fluid is at rest. So basically I am looking at the pressure is a function of the positions. Time is not there, as the fluid is at rest condition.

Pressure in a fluid at rest is dependent on the positions within the fluid instead of time. This means we can analyze how pressure changes with depth or across different areas without worrying about how time affects this system. The focus is solely on the spatial distribution of pressure and how it correlates to the fluid's weight above.

Think of a beaker filled with different colored layers of fluids. Each layer creates pressure at the bottom due to its weight, and we can easily measure how this pressure changes from the top layer to the bottom without considering any time factors.

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