2.2 - Pressure Difference Calculation
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Interactive Audio Lesson
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Introduction to Laminar Flow Calculation
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Today we will discuss how to calculate the pressure difference in laminar flow through pipes. Can anyone tell me what 'laminar flow' means?
I think laminar flow is when the fluid moves in smooth paths, right?
Exactly! In laminar flow, the fluid flows in parallel layers. Now, what parameters must we know to perform these calculations?
We need viscosity, specific gravity, and pipe dimensions.
Very good! Also, understanding the discharge is crucial to calculating the pressure difference. Let's proceed to see how we derive these relationships.
Calculating Density from Specific Gravity
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To calculate pressure difference accurately, we need the fluid's density. If we have the specific gravity of crude oil, how do we calculate its density?
We multiply specific gravity by the density of water.
Correct! Since the specific gravity of oil is 0.8, the density would be 0.8 times 1000 kilograms per cubic meter, giving us 800 kilograms per cubic meter. Let’s find how to use this in our pressure difference formula.
Calculating Discharge Rate
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Now, let’s calculate the discharge Q of oil collected in a tank. If 50 kilograms are collected in 15 seconds, how would we find Q?
We divide the mass by density to find the volume, then divide by time.
Precisely! The volume is 0.0625 cubic meters, so Q equals this volume divided by 15 seconds, giving us approximately 4.17 times ten to the power of minus 3 cubic meters per second. Great work!
Utilizing Flow Equations
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To find the pressure difference using the Darcy-Weisbach equation for laminar flow, we apply our known values. How do we calculate the pressure gradient dp/dx once we find Q?
We can rearrange the equation and plug in Q, viscosity, and the radius of the pipe.
Excellent! By doing this, we can find dp/dx, and subsequently, the total pressure difference across the pipe. Can anyone share the final result from our example?
The final pressure difference we calculated was -5599 Newtons per square meter!
Conclusion and Recap
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Who can summarize what we learned today about pressure difference calculations in laminar flow?
We learned how to convert specific gravity to density, calculate discharge from mass flow, and utilize the flow equations to find pressure difference!
Fantastic summary! Understanding these calculations is fundamental in hydraulic engineering. Remember these key steps, and you’ll excel in practical applications.
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
The section explains the methodology for calculating the pressure difference in a laminar flow condition within a horizontal circular pipe. It incorporates given parameters, including viscosity, specific gravity, diameter, and flow rate, providing worked examples and the significance of various equations in hydraulic engineering.
Detailed
In this section, the various steps for calculating the pressure difference across the ends of a horizontally placed circular pipe under laminar flow conditions are presented. The problem involves fluids (in this case, crude oil) specified with viscosity and specific gravity, emphasizing the application of relevant hydraulic principles. The section walks through the calculation process, including determining the density of the fluid based on specific gravity, the flow rate based on mass and time, and the use of relevant equations such as the one relating discharge, viscosity, and pressure gradient. A worked example illustrates the practical application of these calculations, culminating in the determination of pressure difference, reinforcing the characteristics of laminar flow, and highlighting the importance of recognizing conditions such as Reynolds number.
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Given Data
Chapter 1 of 8
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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.
Volume Calculation
Chapter 2 of 8
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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.
Discharge Calculation
Chapter 3 of 8
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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.
Pipe Area Calculation
Chapter 4 of 8
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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.
Average Velocity Calculation
Chapter 5 of 8
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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.
Reynolds Number and Flow Type
Chapter 6 of 8
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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.
Pressure Gradient Calculation
Chapter 7 of 8
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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.
Final Pressure Difference Calculation
Chapter 8 of 8
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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.
Key Concepts
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Pressure Difference: It is a significant factor in fluid mechanics that determines how fluid moves through a pipe.
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Density: The mass per unit volume, crucial for converting specific gravity and needed for flow calculations.
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Viscosity: The measure of a fluid's resistance to flow, it is a key variable in determining flow behavior.
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Reynolds Number: A dimensionless quantity used to predict flow patterns in different fluid flow situations.
Examples & Applications
Calculating pressure difference in a pipe with known viscosity and density of the fluid.
Using mass flow rate and specific gravity to determine fluid density before calculation.
Memory Aids
Interactive tools to help you remember key concepts
Rhymes
Pressure is the force that pushes, through pipes, it crushes, in laminar flow it's smooth, like butter it rushes.
Stories
Imagine a flowing river where smooth stones guide the water's path. Just like laminar flow, the water glides without turbulence, demonstrating how layers flow efficiently past one another.
Memory Tools
For Pressure Difference, remember 'MVD' – Mass, Volume, and Density to calculate what’s needed.
Acronyms
LAMP
Laminar flow
Area
Mass
Pressure difference.
Flash Cards
Glossary
- Laminar Flow
A flow regime characterized by smooth and orderly fluid motion.
- Viscosity
A measure of a fluid's resistance to flow, typically denoted in Pascal seconds.
- Specific Gravity
A dimensionless quantity that expresses the ratio of the density of a substance to the density of a reference substance, usually water.
- Discharge (Q)
The volume of fluid that passes through a surface per unit time.
- Pressure Gradient (dp/dx)
The rate of change of pressure with respect to distance in a fluid flow.
Reference links
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