5.1. Reservoir case with V = 0

So, now, proceeding forward first simple case, very simple case, where V = 0. So, there is point 1 and there is point 2, z direction is shown as above, you know, this is pressure datum and this is elevation datum. Reservoir means V = 0. So, now, as I said this point 1 has been put on surface and we can put point 2 anywhere we have decided to put at this random point, point 2.

In this chunk, we are introducing a scenario where the velocity (V) of fluid is zero, which is typically seen in a reservoir. Here, we have defined two points: Point 1 at the surface of the reservoir and Point 2 at an arbitrary location below it. This establishes the start for applying the Bernoulli equation, which will allow us to understand how pressure and elevation relate in a fluid at rest.

Imagine a still lake where the water surface is calm. The water at the surface (Point 1) is still, and you’re looking at the water at different depths down to Point 2. Just like how pressure increases as you dive deeper into a lake, the Bernoulli equation helps us relate those pressure changes in fluid static conditions.

So, we apply the Bernoulli equation between these 2 points. So, in at in the reservoir the velocity is 0. So, V square / 2g is going to be eliminated here. So, the between these 2 points we can write p1 / gamma + z1 = p2 / gamma + z2, whatever it is going to be...

When applying Bernoulli's equation in the context of a reservoir where V = 0, we eliminate the kinetic energy term (V^2/2g) because there is no fluid motion at the surface. The equation simplifies to a relationship between the pressures and elevations at the two points—showing how pressure differences correspond to differences in elevation.

Think of a drinking straw submerged in water. If you sip without any effort, the water doesn’t move much, similar to how the fluid at rest does not have velocity. As you analyze how much deeper you need to sip to draw water (Point 2), it helps understand how pressure and depth work together in a simple and relatable way.

we also know that p1 is a gauge pressure and this is a pressure datum atmospheric pressure, so, p1 will be 0. In our second case, so, we can write z1 – z2 is = P2 / gamma.

In this chunk, we dive into how the pressure at Point 1 is defined as gauge pressure (which is relative to atmospheric pressure), resulting in p1 being zero. Consequently, the difference in elevation between Point 1 and Point 2 can be directly related to the pressure at Point 2 using the hydrostatic pressure formula, P2 = γ(z1 - z2).

Imagine you are using a barometer, which measures atmospheric pressure. If you remove it from the air to take it underwater, the pressure reading changes based on the height of the water column. Similarly, by understanding how pressure works relative to elevation, you can grasp how pressure changes affect fluid heights.

So, proceeding forward, so we take another simple case, where pressure is 0 or constant. What is an example of fluid experiencing a change in elevation but remaining at a constant pressure? It is an example is a free jet...

This chunk introduces a scenario where pressure remains constant even while the fluid changes elevation, using a free jet as an example. Here, when fluid exits a nozzle or an opening, it maintains an atmospheric pressure, further simplifying the Bernoulli relation since pressure does not change across points in the jet.

Think of a garden sprinkler. When the water comes out of the nozzle, it sprays out uniformly. The pressure created by the water at the source remains constant, allowing the spray to form. Recognizing this can help understand how fluid dynamics work in practical applications.

So, we can write z1 + V1 square / 2g = z2 + V2 square / 2g and therefore, the velocity at point 2 will be V = 2g (z1 - z2) + V2.

Here, we further explore how to find the velocity at Point 2 using the elevation differences calculated. This relationship showcases how changes in elevation can determine the flow speed when pressure is constant in a free jet scenario, highlighting practical applications of Bernoulli's equation in fluid dynamics.

Imagine water gushing out from a garden hose, directed downward to create a jet. The height difference between the water source (hose end) and the ground (where the water lands) will result in how quickly the water reaches the ground. This simple logic can help understand how to apply calculations in engineering and design.

--