2. Calculation Problem: 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.

In this chunk, we summarize the problem statement where a liquid (crude oil) flows through a pipe. Here, viscosity and specific gravity of the oil are given, as well as the dimensions of the pipe and the amount of oil collected over a specific time. This sets up the problem for calculating the pressure difference, which is a critical aspect of fluid mechanics.

Imagine you're pouring syrup from a bottle into a glass. The properties of the syrup, such as thickness and the speed at which it flows, can affect how long it takes to fill the glass, just as the viscosity and flow rate of the oil affect pressure in the pipe.

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, 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 details all the important parameters for the calculation: the viscosity of the oil converted to SI units, specific gravity leading to the calculation of density, as well as the dimensions of the pipe. Knowing these values is crucial as they are utilized in the equations that follow to determine the pressure drop.

Think of a straw in a drink. The thickness of the drink and the size of the straw determine how easily you can sip. In our case, the viscosity and size of the pipe influence how fluid flows through it.

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. This is the volume of the oil that is collected in a tank in 15 seconds. 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 based on the mass and density. The discharge (flow rate) is also calculated by dividing the volume by the time, revealing how much fluid is moving through the pipe per second. This is a critical step as it directly impacts the pressure calculations.

Imagine filling a water balloon. The faster you fill it (higher discharge), the more pressure builds inside. Similarly, in our problem, knowing how quickly the oil is flowing helps us understand the pressure in 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. 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.

Here we find the cross-sectional area of the pipe, which is essential for further calculations. Knowing the area and the flow rate enables us to calculate the average velocity of the fluid. This average velocity is necessary for final equations used to determine the pressure drop due to friction in the pipe.

Think of squeezing a garden hose. If the opening is larger (greater area), you can spray a lot of water quickly. In our case, the size of the pipe affects how quickly the oil can flow.

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 23 the flow is laminar.

The Reynolds number indicates the flow regime—whether it is laminar or turbulent. By calculating this number based on speed, size, and viscosity, we confirmed that the flow in this pipe is laminar, meaning the oil moves in smooth and orderly layers.

Picture a calm stream vs a rushing river. In a calm stream, water flows smoothly (laminar), while in a rushing river, the flow is chaotic. Our calculations tell us which kind of flow we’re dealing with.

Flow is laminar, therefore, Q is going to be the formula that we have derived: Q = - (pi / 8 mu) (dp/dx) R^4. We already know Q, 4.17 into 10 to the power minus 3 is equal to - (pi / 8) * 0.09 * (dp/dx) * (40 * 10^(-3))^4. dp/dx is equal to - 373.32 Newton per meters square per meter.

In this chunk, we derive the pressure gradient (dp/dx) using the flow rate equation for laminar flow. By substituting Q, viscosity, and radius into the equation, we find the pressure drop per unit length of the pipe, which tells us how pressure decreases along the length of the pipe.

Imagine water flowing down a slide. The steeper the slide (greater pressure gradient), the faster it goes. Similarly, a strong pressure gradient means the oil moves quickly.

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 ends is - 5599 Newton per meter square.

This final calculation translates the pressure gradient into an actual pressure difference between the two ends of the pipe. Knowing the pressure at each end helps to understand how the pipe influences fluid flow.

Think of a balloon: the difference in air pressure inside and outside determines if it’s inflated or deflated. In our problem, the oil's pressure differences dictate how it flows through the pipe.

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