3.3. Equation for the Flow

Given 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.

In this chunk, we identify the important parameters involved in the flow problem. The viscosity of the fluid (mu) is noted as 0.9 poise, which is converted to SI units (Pascal seconds) for calculations. The specific gravity (S) tells us about the fluid's density relative to water, leading us to calculate the actual density of crude oil. We also define the dimensions of the pipe, namely its diameter and length, which are crucial for further calculations related to flow rate and pressure drop.

Think of viscosity as the thickness of a syrup. Just like thicker syrups pour slower, fluids with higher viscosity flow slowly through pipes. The specific gravity helps us compare the densities of different liquids, similar to how we can tell if one fruit is heavier than another based on their sizes when submerged in water.

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.

Here, we compute the volume of crude oil collected in a tank over a given time. Using the mass of the oil (50 kg) and its density (800 kg/m³), we can find the volume of the oil. Following this, we calculate the discharge (Q), which represents the rate of flow, or how much volume of oil is flowing through the pipe every second. This gives us important insight into the flow dynamics.

Imagine filling a bottle with water from a sink. The amount of water you collect in a certain time represents the volume just like how we calculated volume of the oil. The speed at which your bottle fills up reflects the discharge of water from the tap, showing a clear relationship between quantity and time, like our calculations here.

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. 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.

We calculate the cross-sectional area of the pipe using its diameter, which is essential for determining the average speed (velocity) of the fluid. The average velocity is derived from the previously calculated flow rate (discharge) and cross-sectional area. Furthermore, we compute the Reynolds number, which helps determine the nature of flow—if it's laminar or turbulent. A Reynolds number below 2000 indicates laminar flow, which is lower than the calculated value of 590, confirming that our flow is indeed laminar.

Imagine water flowing through a garden hose. The diameter of the hose affects how fast the water can flow out. If the hose is too narrow (smaller diameter), less water (lower discharge) flows out more quickly (higher velocity), while a wider hose (larger diameter) allows more water to flow out slowly. The Reynolds number is like a traffic rule for flow – it tells us whether the traffic (flow of water) moves smoothly or gets jammed whether the flow is laminar or turbulent.

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, what we get is dp dx is equal to minus 373.32 Newton per meters square per meter.

In this section, we utilize the flow rate and the derived equation for laminar flow to find the pressure drop (dp/dx) per unit length of the pipe. By substituting the known values into our formula, we calculate that the pressure drop is approximately -373.32 N/m² per meter, which indicates a consistent drop in pressure across the length of the pipe under laminar conditions.

Think of a water slide: if the slide (our pipe) is straight and smooth, people (fluid particles) slide down easily with minimal resistance, showcasing laminar flow. However, gravity pulls them down (pressure drop), which is analogous to the pressure decreasing as the fluid moves along the pipe.

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 both ends is - 5599 Newton per meter square.

Here, we calculate the total pressure difference between the two ends of the pipe by multiplying the pressure gradient (dp/dx) with the length of the pipe (15 meters). By rearranging our terms, we analyze that the overall pressure difference amounts to -5599 N/m², indicating a significant drop in pressure as the fluid travels along the lengthy pipe.

Picture two ends of a long hose – the water at one end comes out under a certain pressure, but as it travels all the way to the other end, some pressure is lost due to the friction along the inner walls of the hose. This loss means that the water pressure reduces significantly by the time it reaches the end.

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