3.7. Final Answers from the Problem

This means, this is the dp/dx and it is constant, so, the pressure difference per unit length is -373.37. If we assume P2 pressure at one end and P1 at the other and length is 15 this should be equal to dp/dx equal to -373.32 implies P2 – P1 the pressure difference between the both end is -5599 Newton per meter square.

In laminar flow, the pressure difference across a section of the pipe is calculated using the derivative of the pressure (dp/dx). In this segment, we learn that dp/dx is a constant value which allows us to express the pressure difference, P2 - P1, over a given length of the pipe. Specifically, the value of dp/dx calculated is -373.32 N/m²/m, which when multiplied by the length of 15 meters gives the total pressure difference of -5599 N/m². This negative sign indicates a drop in pressure from P1 to P2.

Think of the pressure difference in a pipe like water flowing down a slope. The higher end of the slope has greater pressure, while the lower end has less. As water travels down, it loses pressure—much like how the oil in our pipe loses pressure from one end to the other.

So, this is how we solved a detailed solution. Let me take down this ink. So, now we have done the laminar flow through the circular pipe. Now we are going to see the laminar flow between the parallel plates.

This final segment summarizes the entire calculation process for laminar flow in a circular pipe. It reaffirms understanding that the solution has been reached through step-by-step calculations and demonstrations. It also indicates a transition to the next concept of laminar flow between parallel plates, highlighting the continuity of the lecture and the material covered.

Imagine solving a puzzle: each piece must fit perfectly before unveiling the picture. In this case, each step in the calculation represents a piece of that puzzle. Now that the pipe flow is 'complete,' we’re moving on to another interesting puzzle—how fluids behave between parallel plates!

So, final answers is going to be one is the first one is pressure gradient, that was, required dP/dx came out to be minus 3924 Newton per meter square per meter. Second one was shear stress at the plates, that is, tau not equal to 196.2 Newton per meter square. And the third one was discharge per unit width Q is equal to 0.1333 meter cube per second.

In this summary, the final answers to the problems at hand are presented clearly. The pressure gradient (dP/dx) is calculated to be -3924 N/m²/m, shear stress (τ0) at the plates is 196.2 N/m², and the discharge per unit width (Q) is 0.1333 m³/s. These figures are critical for understanding the behavior of fluid flow under laminar conditions and serve as benchmarks for similar calculations in engineering.

Consider the flow of cars on a highway: the pressure gradient can be compared to the rate at which cars slow down in a traffic jam. If traffic were to flow smoothly, the shear stress could relate to the effort it takes for a car to change lanes. Discharge, akin to how many cars travel past a specific point per minute, highlights efficiency in flow.

--