4.. Mass Conservation Equation

Now let us apply the mass conservation equation where inflow minus outflow and this case will be, Outflow = Inflow.

The mass conservation equation is fundamental in fluid mechanics, asserting that mass cannot be created or destroyed in a closed system. In this context, we are stating that the mass flowing into a system (inflow) must equal the mass flowing out (outflow). This means that if we keep track of the mass flow rates, we can ensure that they balance out over time.

Consider a swimming pool with a hose filling it up while an overflow drain is open. The water flowing in from the hose is the inflow, while the water draining out is the outflow. If the flow rates are equal, the water level in the pool remains constant.

Incompressible flow: $\rho_1 A_1 V_1 = \rho_2 A_2 V_2$

In incompressible flow, the mass density ($\rho$) remains constant, which leads to a simplified form of the mass conservation equation where the mass flow rate entering equals the mass flow rate exiting. This equation states that if the area (A) and velocity (V) change at different points, the product of these quantities must remain constant, indicating that the flow rate does not change across different sections of the flow.

Imagine a garden hose. If you put your finger over the end, the area at the end decreases, but the speed of the water increases to keep the flow rate constant. This shows the relationship between area and velocity in a flow system.

This is what the discharge is given to us; the volumetric discharge is given to us. Divide by the area, we will get the velocity.

The volumetric flow rate (discharge) can be used to calculate the velocity of a fluid in a certain area. To find the velocity, we take the volumetric flow rate and divide it by the cross-sectional area through which the fluid flows. This gives us a clear indication of how fast the fluid is moving in that specific area.

Think of a drinking straw. When you suck a drink through it, the volume of liquid is being drawn up at a certain rate (volumetric discharge). If the straw is narrow, the liquid moves faster compared to a wider straw, where the same drink would flow slower. This demonstrates how area impacts velocity.

Now we have to find out what is the force acting in the x and y directions, applying the Reynolds transport theorems and simplifying the tops.

In order to find the forces acting on a control volume, we can apply the Reynolds Transport Theorem, which relates the rate of change of momentum within a control volume to the flux of momentum across its boundaries. By analyzing both inflow and outflow momentum, we can deduce the net force acting on the system, which is critical in determining how the flow interacts with the surroundings.

Consider a car moving in a straight line. The car experiences forces from the engine pushing it forward as well as air resistance acting against it, along with other factors like friction with the road. By analyzing these forces, you can determine how much acceleration the car will have.

Now take another example which is a moving control volume... Let us have a water jet, if you look at this is the water jet, is impinging this normal to a flat plate.

In this scenario, a water jet strikes a flat plate moving in the same direction. To find the force required to keep the plate moving at a constant velocity, we must analyze the pressure and momentum change due to the jet striking the plate. This problem illustrates how a moving control volume behaves under these conditions and emphasizes the need to apply conservation equations appropriately.

Think about a water hose spraying onto a moving car. The water impacts the car's surface as it moves, creating a force that affects how much pressure is felt by the car’s body. Just like analyzing this water jet on the car, we can analyze forces acting on the moving plate.

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