8.3. Pressure Distribution Equations

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 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 fluid at rest, we analyze how pressure behaves. The function P depends on three variables: x, y, and z, which represent the fluid's positional coordinates in a 3D space. When a fluid is static, shear stress (which often manifests in flowing conditions) becomes zero, meaning there's no sliding or deformation within the fluid. The only stress acting on any surface within this control volume is the normal stress, which we often refer to as pressure. Thus, pressure acts uniformly on the surface of the control volume.

Imagine a glass of water sitting on a table. The water is at rest, and the only force acting on the bottom of the glass is the weight of the water above it, pushing down due to gravity. The pressure at the bottom of the glass is just the weight of the water divided by the glass’s area, and there are no other forces trying to move the water sideways.

There are three Cartesian directions of x, y, and z directions and I am considering a point which is just a centroid of the parallelepiped is this thing. The pressure at the centre of the fluid control volume is assumed to be P (x, y, z). At that point, the gravity force is acting it which is the body force component.

We define a control volume as a three-dimensional space (the parallelepiped) where we measure certain properties of the fluid, including pressure. The centroid of this shape is where we specifically analyze pressure, denoted as P(x,y,z). In addition to normal stress, we also consider gravity, which acts as a body force. This means it influences the fluid at all points within our control volume by causing the fluid to exert downward pressure proportional to its weight.

Think of a cube of jelly resting on a plate. The jelly's weight causes it to press down on the plate, creating pressure at the plate's surface beneath the jelly. Here, the jelly's weight is analogous to the gravity acting on our control volume, and the pressure at the plate represents the normal stress acting on the surface.

So we can have two force components. One is the surface force component and other is body force. The body force components if you look it that it will be unit weight multiply the volume of the control volume, which is simple things. The control volumes, the volume is this much and the unit is is the density and g stands for accelerations due to gravity.

In our control volume, we identify two forces acting on the fluid: surface forces (like pressure) and body forces (like gravity). The body force acting on the control volume due to gravity is calculated by multiplying the fluid's unit weight (density times acceleration due to gravity) by the volume of the control volume. This relationship aligns with Pascal's Law, which states that pressure applied to a confined fluid will be transmitted undiminished throughout the fluid.

Consider a balloon filled with water. If you squeeze the balloon at one point, the water inside doesn't just compress at that point; it pushes out evenly in all directions. This illustrates Pascal's Law: the pressure is distributed uniformly throughout the fluid in the balloon.

Now if you look it over this control volume, all these pressure is going to act it as Pascal law says that pressure acts normal to the surface.

According to Pascal’s Law, pressure at any point within a fluid at rest states that the pressure acts perpendicularly to the surfaces it contacts. This applies to our control volume at all faces where pressure is acting, ensuring that the net effect of these pressure forces affects the fluid's behavior. The gradient of pressure can be described mathematically as the change in pressure over the distance, providing insight into how the pressures are distributed inside and outside the volume.

Imagine you're swimming underwater. As you go deeper, you notice increased pressure against your body, which evenly presses on all sides. This is an application of Pascal’s Law, where pressure increases with depth in the water, acting perpendicularly against all surfaces of your body continuously.

We can derive the 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.

When we derive the equations for pressure distribution, we conceptualize gravitational acceleration as a vector quantity that can point in the direction opposite to fluid motion. By aligning gravity along one of our coordinate axes (commonly the z-axis), we can separate the hydrostatic pressure equations. This allows us to analyze how pressure behaves with depth in the fluid, illustrating that pressure increases linearly with the elevation within the fluid's vertical direction due to gravitational pull.

Think of a tall glass of soda. As you drink the soda down, the liquid at the top exerts pressure due to gravity on the soda below. The deeper you go in that soda, the more pressure you feel, equivalent to opening a soda can at high altitudes where less pressure builds up compared to when you're at sea level.

Now let me come to the mercury barometer case with a very simplified same capability concept what we used to measure the atmospheric pressure.

The mercury barometer utilizes the capillary effect to measure atmospheric pressure. It consists of a tube filled with mercury, inverted into a dish of mercury. The height of the mercury column indicates the pressure of the atmosphere pushing down on the mercury in the dish, balancing the weight of the mercury column. This relationship evidences how pressure can be quantified using the height of a liquid column — linking the physical properties of liquids to pressure measurement.

Imagine a straw that you put into a glass of water. When you put your finger over the straw's top and lift it out, the water stays inside the straw because of atmospheric pressure holding it up. This is similar to how a mercury barometer works; the weight of the mercury must be balanced by the atmospheric pressure exerted on the mercury in the dish.

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