2.2. Pressure Difference Calculation

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. Diameter we know, it is 80 millimeters or 80 into 10 to the power minus 3 meter, R we know 40 into 10 to the power minus 3 meter and length L is given as 15 meter.

This chunk introduces the provided data needed for the calculation. The viscosity (mu) of the fluid is given in poise and converted to SI units (Pascal seconds). The specific gravity (S) provides information about the fluid's density relative to water, where the density of the oil is calculated to be 800 kg/m³ by multiplying the specific gravity by the density of water. The dimensions of the pipe are also stated, with diameter and radius noted in meters.

Imagine you are filling a balloon with water. The balloon's size can be compared to the diameter of the pipe, and the thickness of the water inside can be related to the fluid's viscosity, affecting how quickly it flows into the balloon.

First, 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.

To calculate the volume of oil collected in the tank, we use the formula: Volume = Mass / Density. Here, 50 kg of oil is divided by its density (800 kg/m³), yielding a volume of 0.0625 cubic meters over a duration of 15 seconds.

Think about how much water you can collect in a container when you pour it. Just as you measure how much water (in liters) you have, here we calculate the amount of oil collected over a period.

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.

Discharge (Q) measures the flow rate and is calculated by dividing the volume of fluid collected by the time taken. Here, the volume (0.0625 m³) is divided by the time (15 seconds), resulting in a discharge Q of approximately 4.17 x 10^-3 m³/s.

Imagine a faucet dripping water—the rate at which the water fills a bucket represents discharge. This calculation shows how quickly the oil is flowing through the pipe.

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.

The area of the circular pipe is calculated using the formula for the area of a circle (A = πD² / 4). By substituting the diameter into the formula, we find the cross-sectional area of the pipe to be approximately 5.026 x 10^-3 m².

Consider the cross-section of a straw. The area of the straw determines how much liquid can flow through it at once—similarly, the pipe's area affects the flow of oil.

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.

The average velocity of oil in the pipe is calculated by dividing the discharge (Q) by the area of the pipe. This gives us an average velocity of approximately 0.83 m/s, which indicates how fast the oil is moving through the pipe.

If you blow air through a wide flexible tube versus a narrow one, the speed of the air changes based on the width—the same principle applies to fluids flowing through pipes.

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 comes to around 590. So, as Reynolds number is less than 2300 the flow is laminar.

The Reynolds number (Re) helps determine the flow regime. Calculated as Re = (ρVD)/μ, it indicates whether the flow is laminar or turbulent. A value below 2300 signifies laminar flow, and in this case, the calculated Reynolds number is approximately 590, confirming laminar flow.

Imagine a river: when the flow is slow and smooth, it's like laminar flow. But if the current picks up speed and starts to swirl, it becomes turbulent flow—just like air can flow steadily or in chaotic gusts.

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.

Using the derived formula for laminar flow, we equate the discharge (Q) to the formula involving pressure gradient (dP/dx) and the other known quantities. Rearranging gives us the pressure gradient value to be -373.32 N/m²/m, indicating the pressure drop per meter.

Think about how much harder you have to push water through a narrow pipe compared to a wider one. This pressure difference is crucial to understand how fluids behave in various scenarios.

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 both ends is -5599 Newton per meter square.

The pressure difference (P2 - P1) across the two ends of the pipe is calculated by multiplying the pressure gradient by the length of the pipe. Thus, the total pressure difference is determined to be -5599 N/m²; the negative value indicates a drop in pressure from one end to the other.

Imagine a hill where one side is higher than the other causing a waterfall effect. The pressure at the top is higher and drops dramatically as the water cascades down, similar to the pressure drop across the pipe.

--