3.5. Average Velocity and Discharge Calculations

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, which is calculated as 50 divided by 800, yielding 0.0625 meter cube. Therefore, discharge Q is equal to volume by time, calculated as 0.0625 divided by 15, resulting in Q = 4.17 × 10^(-3) meter cube per second.

In this section, we start by calculating the discharge of oil flowing through the pipe. Given that 50 kg of oil is collected in 15 seconds, and knowing the density of the oil is 800 kg/m³, we compute the volume of oil as follows: Volume = Mass / Density = 50 kg / 800 kg/m³ = 0.0625 m³. To find the discharge (Q), we divide the volume by the time it takes to collect it: Q = Volume / Time = 0.0625 m³ / 15 seconds = 4.17 × 10^(-3) m³/s. This gives us the rate at which oil moves through the pipe.

Imagine you are filling a bathtub with water. If you know how much water you have added after a specific time (say you add 50 liters of water in 15 minutes), you can easily calculate the flow rate of water by using a similar method to determine the discharge. Just like you convert the volume of water to a flow rate by dividing the total volume by the time taken, we do the same with oil in the pipe.

The area of the pipe is calculated as area = πD² / 4, which gives us area = π / 4 × (80 × 10^(-3))² = 5.026 × 10^(-3) m². The average velocity is then calculated as V average = Q / Area = 4.17 × 10^(-3) / 5.026 × 10^(-3) = 0.83 m/s.

Following the calculation of discharge, we proceed to find the average velocity of the oil. First, we calculate the area of the pipe's cross-section using the formula: Area = πD² / 4. For our pipe with a diameter of 80 mm (0.08 m): Area = π × (0.08 m)² / 4 = 5.026 × 10^(-3) m². Next, we find the average velocity (V average) by using the discharge we previously calculated: V average = Q / Area = 4.17 × 10^(-3) m³/s / 5.026 × 10^(-3) m², which simplifies to 0.83 m/s. This means the average speed of oil flowing through the pipe is 0.83 meters per second.

Think of a garden hose spraying water. If you know how fast the water is discharged from the hose, you can think about how wide the hose is to determine how quickly the water travels through it. This is akin to calculating the average speed of the oil moving through the pipe: the amount of oil flowing out per second is the discharge, while the hose width helps us find how fast it is moving by calculating the area.

To determine flow type, we calculate Reynolds number as Re = ρV average D / μ, where ρ is 800 kg/m³, V average is 0.83 m/s, D is 80 × 10^(-3) m, μ is 0.09. The Reynolds number calculated is around 590, indicating a laminar flow since it is less than 2300.

Next, we assess the type of flow using the Reynolds number, which helps us determine if the flow is laminar or turbulent. The formula for the Reynolds number is given by: Re = ρV average D / μ. Here, we substitute ρ = 800 kg/m³, V average = 0.83 m/s, D = diameter of the pipe = 0.08 m, and μ = viscosity = 0.09 Pa.s. Substituting these values, we find Re to be approximately 590. Since this value is significantly lower than 2300, we conclude that the flow is laminar, which means it flows in smooth, parallel layers.

Consider water flowing smoothly in a straight river. If it moves steadily without chaos, you can think of it as a laminar flow, similar to the oil in our pipe example. On the other hand, if you saw sections where water splashes and swirls chaotically, that would represent turbulent flow, which happens when the Reynolds number exceeds certain thresholds.

For laminar flow, the flow rate Q is derived from the formula: Q = - (π / 8) (μ)(dP/dx)(R^4). Replacing Q with 4.17 × 10^(-3) m³/s yields: 4.17 × 10^(-3) = - (π / 8)(0.09) (dP/dx)(40 × 10^(-3))^4. Solving gives dP/dx = -373.32 N/m²/m. The pressure difference between both ends equates to -5599 N/m².

Having established that the flow is laminar, we now calculate the pressure drop across the length of the pipe. We use the formula: Q = - (π / 8) (μ)(dP/dx)(R^4). Given our earlier calculation of discharge and substituting known values, we solve for dP/dx, arriving at approximately -373.32 N/m²/m. This term signifies the pressure drop per unit length of the pipe. To find the total pressure difference from one end of the pipe to another, we multiply dP/dx by the length of the pipe (15 m), leading to a total pressure difference of -5599 N/m², indicating a decrease in pressure along the flow direction.

Imagine a water pipe where water slows down due to a blockage. As water moves through the pipe, the pressure decreases due to the blockage's resistance, much like how the oil experiences a pressure drop while flowing through the pipe.

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