7.3. Hydrostatics

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. That means when fluid is at rest conditions only we need to know it how the pressure variations is going over these fluid domains. That is what the simplified problems.

When we talk about a 'fluid at rest', it means that there are no movements in the fluid. This causes the velocity to be zero, meaning there’s no flow. For incompressible fluids, like most liquids, the density remains constant. In this state, the only factor we need to study is the pressure in the fluid, which can vary depending on where you measure it. The rest simplifies our calculations since we don't have to worry about changes in velocity or temperature, allowing us to focus solely on how pressure works within the fluid.

Think of a calm lake on a still day. The water is not moving (it's at rest), which means the pressure at different depths just depends on how deep you go. If you were to dive into the lake, the deeper you go, the more pressure you feel – and that’s because the weight of the water above you is pushing down.

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. And what the two forces we have? The gravity force and the force due to the pressure distribution.

In a fluid at rest, since there’s no flow, there’s no change in velocity – this means there are no shear stress forces acting on the fluid. Shear stress is the resistance to flow, and it occurs when layers of fluid are in motion relative to each other. Therefore, we can simplify our analysis by focusing on two forces: gravity, which pulls the fluid downwards, and the pressure of the fluid acting outward. This makes it much easier to understand how pressures behave in a stationary fluid.

Think of a swimming pool that’s not being disturbed by waves or people. The water isn’t moving; the only forces at play are the weight of the water (gravity) and the pressure it exerts on the walls and bottom of the pool. Imagine holding a ping pong ball under the water - the pressure will be greater at greater depths because of the weight of the water above it, but the ball remains still as long as there are no external disturbances.

So 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. 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.

In simple terms, we can balance the forces in a fluid at rest by equating the pressure forces to the gravitational forces. For example, consider a dam. The height of the water behind a dam (90 meters in this case) creates pressure that acts on the dam walls. We can calculate how much pressure is acting at different depths based on the weight of the water above. Since the fluid is at rest, we also know that no additional forces (like movement of the water or external work) are acting against this balance.

Imagine a tall glass of water sitting on a table. The water is at rest, and the pressure at the bottom of the glass increases as more water is added. You can feel this increase if you tap your finger against the bottom of the glass. Just like that glass of water, the pressure due to the depth of water applies the same way behind a dam, ensuring we can design it to withstand that pressure without collapsing.

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. The many of the times when you consider the pressure field, you try to look it from one point to other point. ... So fluid flow what we are considering here all the cases the pressure is a scalar quantity and that what they must stated by Pascal’s law.

Pascal's law states that when a fluid is at rest, the pressure exerted on an enclosed fluid is transmitted equally in all directions. This means that if you push on one part of a contained fluid, that pressure is felt throughout the fluid uniformly. Mathematically, you can represent changes in pressure at different locations using functions indicating positions in space. Essentially, it confirms that pressure does not depend on the direction of application; it acts as a scalar quantity.

Imagine filling a balloon with air. If you squeeze one side, you can feel the pressure on the opposite side increase even though you didn’t touch that area at all. This is because the air compresses and thus transmits that pressure throughout the entire balloon. Similarly, in a fluid at rest, any pressure applied at one point is felt everywhere.

--