6. Linear Momentum Equations

And here as you know it we have considered it is a horizontal surface. So z equal to z. That is what we cancelled out. And we are just substituting the mass conservation equation and the Bernoulli’s equations to compute what will be the Q value. So once I know the Q value to estimate whatever will be the force component will going to apply linear momentum equations.

This chunk introduces the context for applying linear momentum equations in fluid mechanics. It emphasizes that in this analysis, the surface is horizontal, and heights (z values) are considered equal, allowing for simplifications. The next step involves substituting values related to mass conservation and Bernoulli’s equations to determine the flow rate (Q). This flow rate is essential for later calculations of forces using momentum equations.

Imagine water flowing smoothly out of a garden hose held horizontally. Since the hose is at a constant elevation, we don’t need to worry about gravitational effects altering the water pressure above or below the hose. Instead, we focus on how the speed and amount of water (flow rate) affect how far that water shoots out.

Based on Reynolds transport theorems, we can write for linear momentum equations. We can write that the sum of the force is acting is equal to same path, change of... momentum outflux and this is what momentum influx components.

This chunk focuses on the Reynolds transport theorem, which provides a way to relate changes in momentum within a control volume to the forces acting on that volume. It states that the sum of the forces acting on a fluid is equivalent to the change in momentum flux (both incoming and outgoing) across the control volume. This principle is foundational in fluid dynamics when analyzing flow through a system.

Think of a crowded subway station. The people entering (influx) and leaving (outflux) create a net flow of people. If more people leave than enter, the crowd thins out. In fluid dynamics, similar logic applies: the forces acting on the fluid relate directly to how much momentum flows in and out of a defined space.

If I apply in the x direction, we will get the sum of the force F x will be,... F will be 2.94 Newton.

In this segment, the focus shifts to applying the momentum equations specifically in the horizontal (x) and vertical (y) directions. By calculating the momentum flux and applying the appropriate equations, the resultant forces in these directions are derived. For example, the calculation leads to an estimate of 2.94 Newtons in the x-direction. Understanding the breakdown along these axes is crucial for determining the overall effect of the fluid flow.

Picture pushing a swing. If you push from the side (x-direction), it swings left and right. If you push from below (y-direction), it goes up and down. Similarly, fluid forces can be analyzed in different directions — understanding the contribution from each direction helps predict the swing's movement.

we can compute resultant force which will be... the resultant force is equal to 3.29 kilo Newton.

This chunk discusses how to combine the computed forces from the x and y directions to get a resultant force. Here, the addition of 2.94 Newton (from the x-direction) and the force in the y-direction gives a final force of 3.29 kilo Newton, showing how both components work together to give an overall picture of the force exerted by the fluid.

Consider pulling an object using ropes from two different angles; one from the left (x) and one from above (y). The total force acting on the object can be found using vector addition, just as the fluid forces combine to create a resultant force in our calculations.

In this case there is a horizontal jet of the flow with angle theta strike on a vertical plate... we can classify the flow, the flow classifications like the steady, incompressible and one dimensional flow even if it looks a two dimensional.

In this example, a horizontal jet of fluid impacts a vertical plate. The flow is analyzed under assumptions of being steady and incompressible, which simplifies calculations. The momentum equations are applied to determine the angle (theta) of the jet's impact that keeps the system in equilibrium, emphasizing the practical application of the linear momentum equations discussed earlier.

Think of a fireman's hose aiming water at a wall. The angle at which the water hits the wall is crucial for understanding how the water behaves. By adjusting the angle, the fireman can ensure that the water flows effectively without splashing back, analogous to using momentum equations to quantify the impact.

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