2 - Calculation Problem: Laminar flow in pipes
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Understanding Laminar Flow
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Today, we're exploring laminar flow in pipes. Who can explain what laminar flow means?
Isn't it when the fluid flows in parallel layers without mixing?
Exactly! It's characterized by smooth and orderly flow. Now, can anyone tell me how we determine if a flow is laminar?
The Reynolds number can tell us, right? If it's below about 2000, the flow is considered laminar.
Great job! The Reynolds number helps us quantify the flow type. Let's remember: "Reynolds Rules Laminar" as a mnemonic to recall this concept. What were the key fluid parameters we use in our calculations?
Viscosity and density are important for determining the flow characteristics.
Correct! Viscosity affects the flow's resistance. Let's move on to a practical calculation based on these principles.
Calculating Volume and Discharge
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Let's take a look at the crude oil flow through our pipe. We collected 50 kg of oil in 15 seconds. How do we convert this into volume?
We need to divide the mass by the density.
Exactly! What density do we use here?
800 kg/m³, based on the specific gravity of 0.8.
Good! After finding the volume, how do we calculate the discharge Q?
Discharge is volume divided by time.
Right! This gives us Q = 4.17 x 10⁻³ m³/s. Let’s summarize this step for future reference: ‘Volume to Discharge - Divide & Conquer!’ Now, let's continue to calculate the average velocity.
Average Velocity and Area.
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With the discharge calculated, what comes next in our flow analysis?
We need to calculate the average velocity using the area of the pipe.
Exactly! The area formula is πD²/4. Can anyone tell me the diameter and radius in meters?
The diameter is 0.08 m, so the radius is 0.04 m.
Correct! Therefore, the area would be about 5.026 x 10⁻³ m². Now, how do we find the average velocity?
We divide Q by the area!
Yes! This gives us approximately 0.83 m/s. You’re all doing great! Remember to always validate your calculations with fundamental equations. Let’s recap: ‘Area is πD², then velocity is Q/A!’
Pressure Drop Calculation
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Now that we have the average velocity, let's focus on calculating the pressure difference. Recall the formula involving dp/dx?
Yes! We use Q = -8π/(8μ)(dp/dx)*R^4.
Excellent! What does this equation represent?
It describes the relationship between flow rate and pressure drop for laminar flow.
Exactly right! Plugging the values in, we find dp/dx to be about -373.32 N/m²/meters. What does this tell us?
It indicates how the pressure drops along the length of the pipe.
Correct! The total pressure difference then becomes -5599 N/m². Great understanding here. Let's apply the lessons we've learned by reviewing: ‘Pressure Drop is key for flow impacts!’
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
In this section, we analyze a problem involving the laminar flow of crude oil through a circular pipe, exploring the calculation of pressure difference based on given fluid properties and flow characteristics. Several formulas and key concepts, such as Reynolds number and flow discharge, are applied to solve the problem.
Detailed
Detailed Summary
In this section, we investigate a calculation problem focused on laminar flow within pipes, specifically the flow of crude oil characterized by its viscosity and specific gravity. The objective is to determine the pressure difference at both ends of an 80mm diameter pipe over a length of 15 meters, given that 50 kilograms of oil is collected in 15 seconds.
Key Steps and Concepts Covered:
- Given Parameters: Viscosity (0.9 poise), specific gravity (0.8), diameter (80mm or 0.08m), length (15m), and mass flow.
- Density Calculation: With a specific gravity of 0.8, the fluid's density can be calculated as 800 kg/m³, based on the density of water.
- Volume and Discharge: The volume of oil collected and the discharge are calculated, leading to a discharge rate of approximately 4.17 x 10⁻³ m³/s.
- Flow Velocity: The average velocity is determined using the area of the pipe calculated with the formula (πD² / 4).
- Reynolds Number: Checking the flow regime by calculating the Reynolds number to confirm the laminar flow conditions, established as < 2000 (in this case, ~590).
- Pressure Drop Calculation: Applying the equation for flow in laminar conditions, we derive the pressure gradient and subsequently the total pressure difference between the two ends of the pipe, arriving at a result of -5599 N/m².
The importance of understanding laminar flow, pressure differences, and the relationships between various physical properties is emphasized through practical calculations relevant in hydraulic engineering.
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Problem Statement
Chapter 1 of 7
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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 Data
Chapter 2 of 7
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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, 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 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.
Examples & Analogies
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.
Volume Calculation
Chapter 3 of 7
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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. 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.
Detailed Explanation
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.
Examples & Analogies
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.
Area and Velocity Calculation
Chapter 4 of 7
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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. 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.
Detailed Explanation
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.
Examples & Analogies
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.
Reynolds Number and Flow Type
Chapter 5 of 7
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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.
Pressure Gradient Calculation
Chapter 6 of 7
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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.
Final Pressure Difference Calculation
Chapter 7 of 7
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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.
Key Concepts
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Viscosity affects flow behavior: Higher viscosity means greater resistance to flow.
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Reynolds number indicates flow regime: It's used to determine whether the flow is laminar or turbulent based on its value.
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The pressure drop across a pipe depends on the flow rate and fluid properties: Understanding this relationship is key for designing piping systems.
Examples & Applications
To calculate the oil flow rate, we derived discharge Q using volume collected over time, confirming the application of fluid dynamics principles learned.
In the determination of average velocity through the pipe, we used geometric considerations of the pipe's dimensions, demonstrating practical applications of theoretical concepts.
Memory Aids
Interactive tools to help you remember key concepts
Rhymes
Reynolds number is the key, less than 2000, flow is free.
Stories
Imagine a river flowing smoothly (laminar) vs. a disturbed stream (turbulent) to visualize flow types.
Memory Tools
Remember 'Quality Viscous Oil' to recall the key parameters of Q, V, and μ in flow analysis.
Acronyms
L.V.Q. - Laminar, Viscosity, and Q (Discharge) to connect the concepts.
Flash Cards
Glossary
- Laminar Flow
A type of fluid flow where the fluid moves in parallel layers with minimal disruption between them.
- Reynolds Number
A dimensionless number that helps predict flow patterns in different fluid flow situations.
- Discharge (Q)
The volume of fluid which passes through a surface per unit time.
- Viscosity (μ)
A measure of a fluid's resistance to deformation or flow.
- Specific Gravity
A dimensionless quantity that represents the ratio of the density of a substance to the density of a reference substance, typically water.
Reference links
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