8.3.3. Integration of Pressure Functions

Now if I go for the next ones that how to get the pressure field when fluid is at rest. That means I am just looking the 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 it when the fluid is at rest the shear stress become zero. So there is no shear stress component on this fluid plane. Only this normal stress which is as equivalent to the pressure will act over this control surface.

In a resting fluid, the pressure can be described as a function of three spatial dimensions: P = P(x, y, z). The focus here is on a simple model, a control volume, that helps visualize how pressure functions in a static fluid. Since the fluid is at rest, it experiences no shear stress, meaning that the only force acting on any surface within the fluid is normal stress, which effectively equates to pressure. This simplifies the calculations because we need to take into account only the pressure acting on the control surface, without worrying about shear forces.

Imagine a calm pond where the water is completely still. If you were to drop a pebble, you wouldn't see any ripples from the sides where the pebble lands until it creates disturbance. This is similar to how in a fluid at rest, there are no additional stress forces acting on surfaces, only the pressure from water directly beneath the surface.

At that point, the gravity force is acting it which is the body force component. The gravity force what will be there? It will be, unit weight multiply the volume of the control volume, which is very simple things. The control volumes, the volume is this much and the unit is ρg and ρ stands for here the density and g stands for accelerations due to gravity.

In the control volume at the centroid where pressure is measured, the effects of gravity must be considered. The gravity creates a body force in the fluid which is calculated by multiplying the unit weight (density multiplied by gravitational acceleration, ρg) with the volume of the control volume. This calculation helps determine how much force is acting on the fluid due to its weight, contributing to the pressure exerted by the fluid.

Think of this as taking a balloon filled with water. The water inside exerts pressure on the walls of the balloon due to its weight. When you hold the balloon at a certain height, gravity is pulling the water down, creating more pressure at the bottom of the balloon than at the top. This pressure difference is essential in understanding how fluids behave under static conditions.

Now if you look it over this control volume, all these pressure is going to act it as the Pascal law says that pressure acts normal to the surface. So that what we can consider pressures act there normal to the surface.

According to Pascal's law, pressure in a fluid acts perpendicular to the surface at every point. This means that when examining the control volume, all pressures can be considered as acting normally (or perpendicularly) to its surface. This is crucial because it enables us to understand how pressure operates within fluids at rest and allows us to derive equations predicting forces on the control volume accurately.

Imagine standing in a swimming pool. As you dive into the water, you can feel the water pressing against your body from all sides. This effect demonstrates Pascal's principle in action: no matter where you touch the water, the pressure you feel acts straight out from where you are submerged, regardless of the angle.

Then we need to define now as a pressure as a x, y, z is a pressure field what we are looking it which is a function of x, y, z I can use Taylor series to approximate distance of this surface which is a far away from.

To analyze the pressure field in three-dimensional space, we can employ a Taylor series expansion. This mathematical tool allows us to express the pressure at a certain point in terms of its neighboring points. In essence, it provides a way to approximate the pressure function across distances by expanding it in power series, which reflects how pressure may vary due to changing location in the x, y, and z directions.

Consider making a cake where you start with a basic recipe but apply different layers or flavors to it as you bake. Each layer or flavor represents a different point in our pressure approximation, just as the Taylor series builds on variations of pressure from nearby locations to create a full picture of the pressure field.

So now as we derive pressure distribution equations which in vector forms and let we simplify that equations which earlier we consider the acceleration due to gravity is a vector which will have, If you need to compute it what will be the absolute pressure, then it is a very easy. You just use whether the gauge pressure or this the vacuum pressure that difference what we can get it that what you just add with atmospheric pressure to get the absolute pressure.

When determining pressures, two important concepts are gauge pressure and vacuum pressure. Gauge pressure measures pressure against atmospheric pressure; if you exceed atmospheric pressure, you get a positive gauge pressure, while if you fall below it, you get a vacuum pressure. To find the absolute pressure, simply add the gauge pressure to the atmospheric pressure. This process simplifies determining the true pressure at any point in a fluid system since it takes into account the baseline atmospheric pressure that affects all measurements.

Using a car tire as an example, when you check the tire's pressure, you might read 30 psi on the gauge. This number is the gauge pressure, not considering the atmospheric pressure of, say, 14.7 psi. Hence, the tire's absolute pressure would be 30 psi + 14.7 psi, giving a complete understanding of how pressurized the tire truly is for safety and functionality.

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