1.1. Laminar and Turbulent Flow (Contnd.)

Welcome back to this lecture. As we were talking, we are going to start with another problem of laminar flow in pipes. The question says that the crude oil of viscosity 0.9 poise and specific gravity 0.8 is flowing through a horizontal circular pipe of diameter 80 millimeters and length of 15 meter. Calculate the difference of pressure at the 2 ends of the pipe, if 50 kilograms of oil is collected in a tank in 15 seconds. So, we are knowing this is mu, this is specific gravity S is 0.8. So, we know the diameter is 80 millimeters and the length is 15 meters.

In this chunk, a scenario is presented involving the flow of crude oil through a pipe. The problem statement provides important data like viscosity (0.9 poise), specific gravity (0.8), diameter (80 mm), and length (15 m) of the pipe. We're tasked with calculating the pressure difference across the pipe based on the amount of oil collected over time (50 kg in 15 seconds). This sets the groundwork for applying fluid dynamics principles to find the solution.

Imagine a large straw through which you are trying to drink thick milkshake. The resistance you feel is akin to the viscosity of the milkshake. This problem illustrates how the thickness and properties of the substance flowing through a pipe can impact the ease of flow, similar to how you experience resistance when using different straws or liquids.

So, as always what we are going to do, we are going to write the things that are actually given. Given thing is mu is 0.9 poise or in SI unit is 0.09 Pascal seconds. We know it is specific gravity S is 0.8. Therefore, the density is given to be 0.8 and if we assume 1000 kilograms per meter cube density of water, so, the density of the fluid is 800 kilograms per meter cube, since S is rho / rho water.

Here, we convert the viscosity from poise to SI units (Pascal-seconds), which is crucial for calculations in SI units. Additionally, using the specific gravity of the crude oil, we calculate its density by comparing it to the density of water. Thus, the crude oil's density turns out to be 800 kg/m³. This understanding of fluid properties is fundamental in analyzing fluid flow.

Think of comparing different liquids in a glass, like oil and water. The oil floats above the water because it is less dense. In this bucket example, similar principles apply, where understanding the density helps us predict how these fluids behave under flow conditions.

So, first we are going to do, we are going to calculate the volume of oil collected in a tank in 15 seconds is equal to mass of oil collected in 15 seconds divided by density of oil and that is going to be 50 divided by 800 that is 0.0625 meter cube. This is the volume of the oil that is collected in a tank in 15 second. Therefore, discharge Q is equal to volume by time equal to 0.0625 divided by 15, Q is 4.17 into 10 to the power minus 3 meter cube per second.

In this chunk, we calculate the volume of oil collected by using the mass and density. This small liquid amount leads us to find the flow rate or discharge (Q) by dividing this volume by the time taken (15 seconds), resulting in a discharge of approximately 0.00417 m³/s. This step is essential as it informs the average velocity in the pipe.

Imagine you are pouring water into a cup. If you know how heavy the water is and how large the cup is, you can estimate how quickly the cup fills. This is similar to our oil calculation, where assessing the volume over time helps us understand the flow rate in the pipe.

See, calculate the area of the pipe, area of the pipe is pi D square / 4 and that is, pi by 4 into diameter was 80 into 10 to the power minus 3 and this gives us 5.026 into 10 to the power minus 3 meters square. So, the average velocity is Q by area and this will give us 4.17 into 10 to the power minus 3 divided by 5.026 into 10 to the power minus 3 is equal to 0.83 meters per second.

This segment deals with calculating the cross-sectional area of the pipe, which is essential for determining how fast the fluid moves. Using the formula for the area of a circle (πD²/4), we derive an area of approximately 0.005026 m². Subsequently, we find average velocity by dividing discharge (Q) by this area, yielding an average velocity of about 0.83 m/s.

Consider a garden hose. The wider the hose, the more water it can deliver at a time, which affects how fast you can water your plants. Here, the area of the pipe works similarly: smaller dimensions tend to have lower flow rates.

Therefore, we will find the Reynolds number that is, rho V average into D / mu and this rho is 800, V average is 0.83, diameter is 80 into 10 to the power minus 3, mu is 0.09. This is actually very important to check and this comes to around 590. So, as Reynolds number is less than 23, the flow is laminar.

Here, we calculate the Reynolds number using the formula (ρVD/μ). This number is pivotal in fluid mechanics as it indicates the flow regime: laminar or turbulent. In this case, a Reynolds number of approximately 590 confirms that the flow is indeed laminar, important for applying the correct equations later.

Think of a river. If the water flows smoothly in layers, it’s like our laminar flow; but when it swells and churns with waves, it’s turbulent. The Reynolds number helps us predict which 'river' we have, smoothing the way for calculations.

So, we have established that flow is laminar, therefore, Q is going to be the formula that we have derived minus 8 minus pi divided by 8 mu dp dx into R to the power 4. So, Q value we already know, that is, 4.17 into 10 to the power minus 3 is equal to minus pi 8 into 0.09 dp dx into R, we know, 40 into 10 to the power minus 3 to the power 4.

We apply the derived equation for laminar flow to relate discharge (Q) to the pressure gradient (dp/dx). Substituting known values helps isolate dp/dx, from which we can model the constant pressure difference driving the flow through this specific pipe section.

Imagine squeezing a tube of toothpaste; the harder you push, the more comes out. This is what we calculate here—the pressure needed to maintain a steady flow rate in the pipe.

This means, this is the dp dx and it is constant, so, the pressure difference per unit length is minus 373.37. If we assume P2 pressure at one end and P1 at 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 the final assessment, we calculate the actual pressure difference across the length of the pipe using relations derived through the steps before. Finding that P2 - P1 equals approximately -5599 N/m² indicates the pressure drop experienced due to resistance while flowing through the pipe.

Consider a waterslide; to keep the water flowing, you need a certain push at the top. The higher the slide (pressure difference), the faster and more smoothly everything flows down to the bottom. This example illustrates the significance of pressure difference in flow dynamics.

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