8.. Pressure Field when Fluid is at Rest

Now if I go for the next ones that how to get the pressure field when fluid is at the rest. That means I am just looking at what could be the functions of the P, P = P (x, y, z). If it that it is now again I am considering a very simple case, let us consider a control volume like this okay. This is what my control volume. As I said when the fluid is at rest the shear stress becomes zero. So there is no shear stress component on this fluid plane.

When a fluid is at rest, its pressure distribution can be defined as a function of its coordinates, represented by P(x, y, z). In a theoretical scenario, we consider a control volume — a fixed space within the fluid — where we analyze how pressure operates. With fluid at rest, the shear stress (a stress that occurs due to the movement of fluid layers sliding past each other) is zero. Therefore, only normal stress, which is equivalent to pressure, acts on the surfaces of this control volume.

Imagine a book lying flat on a table. The weight of the book presses straight down on the table without any sliding or movement. This is similar to how pressure acts in a resting fluid, where only 'downward' pressure is experienced without any forces trying to cause a flow or movement.

Only this normal stress which is as equivalent to the pressure will act over this control surface. Over the control surface only the pressure is going to act it because the shear stress is zero okay. The pressure at the center of the fluid control volume is assumed to be P (x, y, z). At that point, the gravity force is acting which is the body force component.

Since the fluid is not moving, the only kind of force being considered in our control volume analysis is the pressure acting at the surfaces, along with the gravitational force acting within the fluid's volume. The pressure at the center of our control volume is a function of its spatial coordinates, P(x, y, z). Additionally, we also consider the weight of the fluid above this point, which contributes to the total pressure due to the gravitational effect on the fluid.

Think about the fluid in a large swimming pool. The water exerts pressure down on deeper layers due to both the weight of the water above and the fact that it’s not moving at all—just like how gravity pulls down on everything. The deeper you go, the more pressure you feel, just like you would feel more weight if you had several people standing on your shoulders.

The body force components if you look at it that it will be unit weight multiplied by the volume of the control volume, which is simple things. The volume is this much and the unit is where the density and g stands for accelerations due to gravity.

Body forces in a fluid arise due to the weight of the fluid itself acting throughout its volume. This can be calculated by multiplying the specific weight (density times gravitational acceleration) by the volume of the control volume. This relationship helps determine how pressure varies with depth, as the deeper layers support the weight of the layers above them.

If you have a tall glass of water, the pressure at the bottom of the glass is greater than at the top because of all the water above it pressing down. The body force due to gravity is what causes this increased pressure at greater depths.

Now if you look over this control volume, all these pressures are going to act as Pascal's law says that pressure acts normal to the surface. So we can consider pressures act there normal to the surface. Then we need to define now as a pressure as x, y, z is a pressure field which we are looking at which is a function of x, y, z.

According to Pascal's law, fluid pressure is exerted equally in all directions when contained. Each point in the fluid reacts to pressures in its local environment, resulting in a pressure field that is consistent in any horizontal layer if the fluid is at rest. This field can be described mathematically as a function of spatial coordinates x, y, z.

Consider a sealed balloon. If you were to squeeze one side, you would feel the pressure distribute evenly throughout the balloon. This is similar to how pressure in a resting fluid adjusts evenly across a horizontal layer.

I can use Taylor series to approximate distance of this surface which is far away from. The pressure at point on the surface can be expressed as ...

Using a Taylor series expansion helps in predicting or approximating how pressure varies in the fluid as you move from one point to another. As the pressure values change across the fluid, this mathematical tool allows us to model and understand those variations more precisely, particularly when distances are relatively small.

Imagine climbing a hill. As you ascend, the incline and the landscape may change gently or sharply. If we were to map these changes, we could use a series of smaller steps to ensure we accurately capture the terrain's slopes—similar to how the Taylor series approximation helps us understand small changes in pressure across distances.

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