6.1.1. Applying Reynolds Transport Theorems

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.

In fluid mechanics, we often want to find out how various forces act on a fluid as it moves. Here, we're beginning to apply linear momentum equations after understanding the flow rate (Q value) through mass conservation and Bernoulli's principle. The Q value represents the volume of fluid passing through a specific area in a given time, and knowing this helps us evaluate the forces exerted by the fluid on surrounding structures.

Imagine a garden hose: when you turn on the water, it flows out of the nozzle at a certain rate (Q value). If you wanted to understand how that water affects plants or structures nearby (like pushing against a wall), you would use linear momentum equations to calculate those forces based on the flow rate.

So 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 is equal to since it is a one inflow and one outflow, so we will have, momentum outflux and this is what momentum influx components.

Reynolds Transport Theorem describes how the momentum of a fluid changes within a control volume. It states that the sum of the forces acting on this volume equals the change in momentum. If we have one inlet (inflow) and one outlet (outflow) of fluid, we need to account for the momentum that enters and the momentum that exits to evaluate the net effect, leading to an understanding of how forces are distributed in the fluid.

Think of a busy highway: cars entering (inflow) and leaving (outflow) create traffic patterns. The total 'momentum' can be thought of as the number of cars multiplied by their speed. If more cars enter than leave, traffic jams might occur (net force acting), while the opposite reduces congestion.

And here we can assume it the momentum conditions become zero, as it is a steady flow conditions.

In steady flow conditions, the liquid's velocity remains constant at any point over time. This means that the flow properties (like velocity, pressure, etc.) do not change as time progresses. In such situations, for the analysis of forces, we can simplify our equations as we don't need to consider changing conditions.

Imagine water flowing steadily from a faucet: the flow rate is consistent, and it splashes out in a smooth stream without interruption. This predictability simplifies calculations related to how the water impacts surfaces.

So we will play along the x directions to get it. If I apply in the x direction, we will get it the sum of the force F x will be, momentum flux components will come it. That what will be the minus rho Q square by A 1.

When applying forces in the x-direction, we derive formulas based on the characteristics of the fluid (like density, denoted rho) and the flow parameters (Q and A1). Momentum flux is essentially the quantity of momentum passing through an area per unit of time. This allows us to compute the forces in that direction caused by the fluid flow.

Consider a river pushing against a dam. The force exerted by the flowing water can be likened to momentum flux. By knowing how fast the river flows and its width, we can calculate how hard it pushes against the dam, just like how we compute forces using fluid density and flow rates.

So we will also get it the F y value, then we can compute resultant force which will be... Thus, the value comes out to be 3.29 kilo Newton.

After obtaining both the x and y force components from the momentum calculations, we can determine the resultant force using vector addition. The final computation gives us a precise amount of force acting on a surface, which in this scenario is a water jet impact quantified at 3.29 kilo Newton.

Imagine two people pushing a heavy box from different angles. To find out how hard they're pushing together, we would consider the direction and force of each push (the x and y components) combining these helps us see the total effort applied to move the box.

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