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**Introduction to Fluid Mechanics**

Bellagio Fountain

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**Lecture 11 - Introduction to Fluid Mechanics**

Approximate Running Time - 21 minutes Distance Learning / Online Instructional Presentation Presented by Department of Mechanical Engineering Baylor University Procedures: Select “Slide Show” with the menu: Slide Show|View Show (F5 key), and hit “Enter” You will hear “CHIMES” at the completion of the audio portion of each slide; hit the “Enter” key, or the “Page Down” key, or “Left Click” You may exit the slide show at any time with the “Esc” key; and you may select and replay any slide, by navigating with the “Page Up/Down” keys, and then hitting “Shift+F5”.

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**Lecture 11 - Introduction to Fluid Mechanics**

Lecture 8 Topics Outline Measuring Devices for Measuring Drag Basics of Fluid Mechanics Flight Characteristics of Baseballs & Golf Balls This lecture and the next lecture will serve as an introduction to fluid mechanics. These two lecturers will prepare you for the fluid mechanics experiment that will be performed using the wind tunnel during the laboratory component of this course that will take place on the Baylor University campus. First, I will introduce measuring devices that are used to measure drag on an object in the wind tunnel. These devices include the manometer and the Pitot static tube. Then I will discuss the basics of fluid mechanics including the definition of and the properties of fluids, the relationship of pressure to velocity, laminar versus turbulent flow, and drag and lift. Finally, we will use fluid mechanics topics to discuss how a curveball curves and why a golf ball is dimpled. This lecture and the next lecture will serve as an introduction to fluid mechanics. These 2 lectures will prepare you for the fluid mechanics experiment that will be performed using the wind tunnel during the laboratory component of this course that will take place on the Baylor University campus. First, I will introduce measuring devices that are used to measure drag on an object in the wind tunnel. Manometer Pitot-static tube Basics of fluid mechanics Definition of and the properties of fluids The relationship of pressure to velocity Laminar vs turbulent flow Drag, Lift Finally, will use fluid mechanics topics to discuss how a curve ball curves and why a golf ball is dimpled. Dr. Carolyn Skurla Speaking

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**Lab: Drag Force Experiment**

Lecture 11 - Introduction to Fluid Mechanics Lab: Drag Force Experiment Performing a fluid mechanics experiment Collect experimental data Perform integration of experimental data Equipment: Wind tunnel Cylinder Pressure transducer Pitot-static tube When you come to the Baylor University campus to perform the on-campus laboratory experiment component of this course, one of the experiments will be performed using a wind tunnel to measure the drag force on a cylinder. You will collect experimental data, and then later in the course, you will use this data to integrate and calculate the total force over the surface of the cylinder. The equipment that will be used includes a wind tunnel, a cylinder that is the object to be measured, a pressure transducer, also called a manometer, and a pitot static tube. So, the first part of the fluid mechanics lecture topic will be spent providing you with the information necessary to perform this experiment. When you come to the Baylor University campus to perform laboratory experiments, One of the experiments will be performed using a wind tunnel to measure the drag force on a cylinder. You will Collect experimental data Later in the course, you will use this data to integrate the total force over the surface of the cylinder The equipment to be used Wind tunnel Cylinder (object to be measured) Pressure transducer (aka manometer) Pitot-static tube So, the first part of this topic will be spent providing you with the information necessary to perform this experiment

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**So, What is Fluid Mechanics?**

Lecture 11 - Introduction to Fluid Mechanics So, What is Fluid Mechanics? The study of fluids in motion Solid -> Can resist a shear stress by a static deformation Fluid -> Cannot resist a shear stress Any shear stress applied to a fluid will result in motion of that fluid There are two classes of fluids: Liquids Gases So, what exactly is fluid mechanics? It's defined as the study of fluids in motion. First I would like to discuss the difference between a solid and fluid. A solid is an object that can resist a shear stress by a static deformation. What I mean by that is that you can, take for example, your hand and place it on a tabletop and then you can try to push against that tabletop, and if there is any deformation it is going to be probably on the microscopic level. Then once you remove the pressure that you're applying with your hand, the object goes back to its original shape. That is how a solid acts. A fluid, on the other hand, cannot resist a shear stress. Any shear stress that is applied to a fluid will result in motion of that fluid. So, let's think about the child's building block that is a cube. If you were to take that cube and tilt it at a 45° angle what happens? Not a lot. It will retain its shape, and then when you take it back to laying it down flat it will retain its shape. On the other hand, if you were to take a spoonful of molasses and put it on a plate and then tilt the plate at a 45° angle what is going to happen? That molasses is going to flow very slowly because it's pretty thick and viscous. Then when you lay the plate down flat again is the molasses going to go back to its original shape? No, it's going to stop moving where it is. The same thing if you think about a beaker full of water and you tilted at 45°. You are going to see that water move around inside that beaker. It's not going to maintain the shape that it had when the beaker was laying on the table. So, there are two classes of fluids as defined in fluid mechanics, and those are liquids and gases. If you tilt a solid by 45 degrees, what happens? If you tilt a beaker of fluid by 45 degrees, what happens? What about molasses on a plate? (White, 1994)

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**Thermodynamic Properties of a Fluid**

Lecture 11 - Introduction to Fluid Mechanics Thermodynamic Properties of a Fluid Pressure, p Compression stress at a point in a fluid Differences, or gradients, of pressure often drive a fluid flow Temperature, T Measure of internal energy level of a fluid There are a number of thermodynamic properties that a fluid has, and we will discuss a few of them here. The first is pressure which is defined with the variable P, and that is the compression stress at a point in a fluid. If there are differences or gradients of pressure, that will drive the fluid into motion. Temperature, denoted with the variable capital T, is a measure of the internal energy of the atoms in the fluid

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**Thermodynamic Properties of a Fluid**

Lecture 11 - Introduction to Fluid Mechanics Thermodynamic Properties of a Fluid Density, Mass per unit volume Highly variable in gases (i.e., =f(p)) Nearly constant in liquids Almost incompressible Assumed to be imcompressible to make analysis easier Specific Weight, Weight per unit volume Density is denoted with the Greek letter ρ. This is the mass per unit volume. In SI units for example, the units would be kilograms per meter cubed. The density of a gas is highly variable because a gas can be compressed. So, the more you compress it the higher the pressure and the higher the density. Therefore, density in a gas is proportional to pressure. However, liquids are almost incompressible so the density remains virtually constant, and we can assume that liquids are incompressible in order to make our analysis easier. Specific weight is simply density times the acceleration due to gravity, and you will wind up with weight per unit volume. Specific weight is denoted with the Greek letter gamma. Now, one thing I would like you to be careful of are the units here. So, the unit of kilograms per meter cubed for density would be multiplied times the units for the acceleration of gravity, which would be meters per second squared, and so what you would wind up with for the units for specific weight would be kilograms divided by meters times seconds quantity squared. Density is proportional to pressure For example, SI units would be kg/m3 Specific weight is the mass times gravitational constant or WEIGHT per unit volume Units of kg/m3 times m/s2 yields units for specific weight of kg/m2s2

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**Pressure Transducer: Manometer**

Lecture 11 - Introduction to Fluid Mechanics Pressure Transducer: Manometer How do we measure pressure, p ? Change in elevation of a liquid is equivalent to a change in pressure Therefore, a static column of liquid can be used to measure pressure difference between 2 points So how do we measure pressure? We use a pressure transducer. One form of a pressure transducer is a manometer, and what a manometer does is measure a pressure differential between 2 points. What a simple manometer looks like is a U-shaped tube that is filled with a static column of liquid. In this example the column is water. The difference in the height of the column of fluid is used to calculate the difference in the pressure at the two points. You can see in this formula that you have the pressured point two minus the pressure point 1, or the change in pressure between point 2 and point 1, is equal to the specific weight times the difference in height of that column of water. You would wind up with kilograms per meter times seconds quantity squared times meters. So, your final units for pressure would be kilograms divided by meters time second squared A manometer is a form of pressure transducer (or pressure measuring device) that measures a pressure differential Simplest manometer is a u-shaped tube filled with a static column of liquid. The difference in height of the column of fluid can be used to calculated the difference in pressure at 2 points (White, 1994)

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**Pressure Transducer: Manometer**

Lecture 11 - Introduction to Fluid Mechanics Pressure Transducer: Manometer Manometer units are in·H2O How do I convert in·H2O to more standard units for pressure? SI Units English The units for a manometer that is filled with water are inches of water. Now this is a little on the strange side and not a standard unit for either SI or English unit systems. So, I am providing you here in this slide with conversions to the standard units of measure for pressure. So, we need to convert it to something that it's a bit more familiar. The conversion factors are both provided here and you'll also notice that pressure units in SI are called Pascal's. I have taken Pascals and broken it down as force over area or Newtons over meter squared, and then that can be simplified. A Newton is a kilogram meter per second squared. So, a Pascal can be further simplified to a kg per meter times seconds squared and this takes you down to the basic SI units so that your future calculations are easy to verify for the correct units in your final answer. The units for a manometer filled with water are inches of water. But inches of water is not a standard unit of measure for pressure So, we need to convert it to something a bit more familiar Conversion factors for both SI and English units are provided on this slide Also, the SI units are simplified to basic SI units to make future calculations easy to verify for units in the final answer.

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**Pressure – Velocity Relationship**

Lecture 11 - Introduction to Fluid Mechanics Pressure – Velocity Relationship A 1 2 v = Flow velocity x y ds Now let's explore the relationship between pressure and velocity of a fluid. First, let's define a volume through which a fluid is flowing. This is the blue volume shown that has a surface with area A and a thickness of ds. The coordinate system that will be used for this problem is described with the x-axis parallel to the side of the volume labeled DS. The direction of flow is perpendicular to the surfaces with area A and in the positive x direction. The surface of area A that is upstream is labeled as surface one and the surface of area A that is downstream is labeled as surface two. Stress and pressure have the same units and are defined as force over area. Therefore, in order to calculate the force on surface one, we can multiply area A times the pressure on the fluid at surface one. By the same definition, the force on surface two is the pressure at surface two times the area A. Force is also defined as mass times acceleration according to Newton’s second law. Acceleration is defined as the change in velocity divided by the change in time. The mass of the fluid can be defined as the volume of the fluid or area A times ds multiplied by the density of the fluid rho. Now, ds is the change in position and can also be defined as the velocity times the change in time. So, our formula for mass of the fluid volume would be rho times area times velocity times change in time. The net force on the fluid is the force at surface one minus the force at surface two. So, now let’s explore the relationship between pressure and velocity of a fluid First, let’s define a volume through which a fluid is flowing This is the blue volume shown that has a surface with area A and a thickness of ds The coordinate system that will be used for this problem is described with the x-axis parallel to the side of the volume labeled ds The direction of flow is perpendicular to the surfaces with area A and in the positive x-direction The surface of area A that is upstream is labeled as surface 1 The surface of area A that is downstream is labeled as surface 2 Stress, which is force divided by area has the same units as pressure. Therefore, the force on surface 1 can be defined as the pressure on the fluid at surface 1 times area A. By the same definition, the force on surface 2 is the pressure at surface 2 times A. Force is also defined as mass times acceleration Acceleration is also defined as the change in velocity divided by change in time And the mass of the fluid is the volume of the fluid (Ads) times the density of the fluid, rho Ds, which is the change of position is also defined as the velocity times the change in time. And the net force on the fluid in the volume is F1 minus F2

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**Pressure – Velocity Relationship**

Lecture 11 - Introduction to Fluid Mechanics Pressure – Velocity Relationship 1 2 A v = Flow velocity ds We can now rewrite the formula force equals mass times acceleration by substituting in for the net force. So, we would substitute force with P1 times A for force one, and P2 times A for force two. Then we can substitute dvdt for A. Then we can rearrange and simplify the left side of the equation and substitute the formula from the previous slide for mass. dt cancels on the right side of the equation and we can then divide through both sides of the equation by A to further simplify our formula. Then we can integrate vdv from the velocity of surface one to the velocity at surface two. For those of you who have not taken calculus yet, please don't freak out at this notation. I am going to define it for you in the final version of this equation and you won't have to look at this integration sign again. The integral of vdv is v square divided by two. So, by substituting in the velocities at surface one and surface two we now have our formula which relates pressure to velocity, and this is the formula that we will use to measure fluid velocity using a pitot static tube in our wind tunnel laboratory experiment. We can rewrite F=ma By substituting p1A for F1 and p2A for F2 By substituting dv/dt for a Then we can rearrange and simplify the left side of the equation and substitute the formula from the previous slide for mass. Dt cancels on the right side of the equation and We can divide through both sides of the equation by A to further simplify We can then integrate vdv from the velocity at surface 1 to the velocity at surface 2 For those of you who have not taken calculus, please don’t freak out at this notation. I’m going to define it for you in the final version of this equation. The integral of vdv is v2/2. By substituting in the velocities at surface 1 and surface 2, we have our formula relating pressure to velocity. This is the formula that we will use to measure fluid velocity using a pitot-static tube.

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**Lecture 11 - Introduction to Fluid Mechanics**

Pitot-Static Tube Static Point Static Pressure, (pS ) Static Velocity, (vS) Stagnation Point Stagnation Pressure, (p0 ) Alright we finished with the manometer. Now let's move on to the pitot static tube. This is a device that is used in airplanes and wind tunnels to measure velocity. In a wind tunnel, it is aligned with the flow of the air through the wind tunnel, and it's also mounted on the exterior of aircraft in order to measure wind speed for the aircraft. The figure shown is a schematic drawing of a pitot static tube. The first point of interest is the stagnation point at the front of the tube that is facing directly into the fluid flow. The pressure that builds up inside the pitot tube is the stagnation pressure or Po. The reason that this point is called the stagnation point is that the velocity of the fluid flow at this point is zero. Therefore, Vo is equal to zero. The second point of interest is the static point. The outer tube has a number of holes, somewhere between four and eight, drilled around the outer tube of the static tube. The fluid, as it flows past the static points, causes a pressure inside the static tube. This pressure is labeled Ps, and the velocity of the fluid flow past the static point is defined as the static velocity or Vs. The difference between these pressures is measured using the differential pressure transducer or manometer. A pitot-static tube is a device that is used in airplanes and wind tunnels to measure velocity It is aligned with flow in the wind tunnel or on the exterior of the aircraft. The figure is a schematic drawing of a pitot-static tube. The first point of interest is the Stagnation point at the front of the tube that is facing directly into the fluid flow. The pressure that builds up inside the pitot tube is the stagnation pressure, p0 The reason that this point is called the stagnation point is that at this point, the velocity of fluid flow is zero. The second points of interest are the static point. The outer tube has a number of holes drilled around the outer tube into the static tube. The fluid as it flows past the static points causes a pressure inside the static tube, labelled ps And the velocity of the fluid flow past the static point is defined as the static velocity, vs The difference between these pressures is measured using a differential pressure transducer, or manometer. Stagnation Velocity, (v0) Differential Pressure Transducer (Manometer)

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**Lecture 11 - Introduction to Fluid Mechanics**

Pitot-Static Tubes ps = Static pressure (in the moving stream) Nominal air pressure in atmosphere p0 = Stagnation pressure Air pressure in the pitot tube vs = Static velocity Speed of air passing the pitot tube Equivalent to speed of plane through the air v0 = Stagnation velocity = 0 So, let's review with regard to the formula that was developed on slide 10. The static pressure, Ps, is the pressure in the moving stream. The stagnation pressure, Po, is the air pressure in the pitot tube. The static velocity is the speed of the fluid as it passes the static point, and the stagnation velocity is equal to zero. By substituting these values into the formula from slide 10 and then rearranging the formula we come up with a formula for calculating the velocity of the fluid flow as a function of the pressure and density of the fluid. So, let’s review in light of the formula that was developed on slide 10. The static pressure, ps, is the pressure in the moving stream The stagnation pressure, p0, is the air pressure in the pitot tube The static velocity is the speed of the fluid as it passes the static point The stagnation velocity is 0. By substituting these values into the formula from slide 10 and then rearranging the formula, we come up with the formula for calculating the velocity of the fluid flow as a function of the pressure and density of the fluid.

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**Pitot-Static Tube Sample Problem**

Lecture 11 - Introduction to Fluid Mechanics Pitot-Static Tube Sample Problem Okay sample problem time. Let's work a sample pitot static tube problem similar to the type that you will see in lab and in your next homework. A sample problem has been provided in your lecture notes workbook. Please turn to that page now. The problem statement reads as follows; in the figures shown air is flowing and water is in the manometer. For R prime equals 1.2 inches, calculate the velocity of air. The density of air is given as 1.2 kg per meter cubed. So, how do we solve an engineering problem such as this one? Well here's my method for working engineering problems in a very structured manner. The first thing is to draw a picture because a picture is worth a thousand words. However in this case, one has already been provided for you so a picture is not necessary. The second step is to write down everything that is given. The density of air and the height difference in the manometer, R prime, were given in the problem statement. Now, write down any governing equations that you may need. This is the equation from slide 12 and I will not repeat here. Please refer back to slide 12. Next, identify what it is that we need to know. What is our unknown? In this case, it is static velocity. Once you've done all this, then it's easy to solve the problem. So, your final step is to solve for your unknown. First, we need to convert the differential pressure to SI units that are more common, pascals. Use the conversion factors that were supplied earlier in this lecture then write your formula after substituting in the values for the variables. Here's another piece of advice from me; I think that keeping units straight is 90% of getting engineering calculations right, so simplify pascals down to the most basic units so that you can perform a reality check to make sure that you are getting the units in your final answer that makes sense. So, even if you can't remember the formula you can quite often get to the correct answer by using units analysis in this way. Then the final step is to punch the numbers in your calculator and calculate the final answer. The answer to this problem of the velocity of the air at the static point is 22.3 m per second. So, now let’s work a sample pitot-static tube problem similar to the type you will see in lab and in your next homework. In the figure, air is flowing and water is in the manometer. For R’=1.2 inches, calculate the velocity of air. The density of air is given as 1.2 kg/m3 So, how do we solve an engineering problem such as this one. Here’s my method for working engineering problems First, draw a picture. Since in this case one has been provided for you, this is not necessary. Next write down everything that is given The density of air and the height difference in the manometer, R’, were given in the problem statement. Now, write down any governing equations that you may need. This is the equation from Slide 12, and I will not repeat it here. Please refer back to slide 12. Next, identify what it is that we need to know. What is our unknown. In this case, it is static velocity. Finally, solve for the unknown First, convert the differential pressure to SI units that are more common, Pascals Then write your formula after substituting in the values for the variables. Here’s another piece of advice from me. I think that keeping units straight is 90% of getting engineering calculations right. So, simplify Pa down to the most basic units so that you can perform a reality check to make sure that you are getting units in your final answer that make sense. Even if you can’t remember the formula, you can quite often get to the correct answer by using units analysis. Finally, calculate the final answer. The velocity of the air at the static point is 22.3 m/s. 2

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**Lecture 11 - Introduction to Fluid Mechanics**

Velocity When there is friction between the fluid and the solid surface No slip of the fluid at the boundary Velocity = 0 A boundary layer forms near the solid surface Shear stress is greatest adjacent to the boundary layer at the surface What happens when fluid is flowing past a solid surface and there is friction between the fluid and the surface? At the interface between the solid and the fluid there will be no split. By that I mean that the velocity of the fluid at the surface is zero. As you get further and further from the solid surface, you will see a velocity profile such as that shown in the figure until the fluid velocity reaches its maximum. So, what is happening is that a boundary layer forms near the solid surface and the shear stress is greatest at the solid surface-fluid interface. 1 Now, what happens when fluid is flowing past a solid surface and there is friction between the fluid and the surface? At the interface between the solid and the fluid, there will be no slip. By that, I mean that the velocity of the fluid at the surface is zero. As you get further and further from the solid surface, you will see a velocity profile such as that shown in the figure until the fluid velocity reaches its maximum So, what is happening is that a boundary layer forms near a solid surface. The shear stress is greatest at the solid surface/fluid interface. (White, 1994)

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**Laminar vs. Turbulent Flow**

Lecture 11 - Introduction to Fluid Mechanics Laminar vs. Turbulent Flow Laminar -> smooth and steady. Turbulent -> fluctuating and agitated. Why do we care about boundary layers? Well this has a lot to do with the study of laminar versus turbulent flow. If you look at the photo in the slide you will see a fluid flowing past a cylinder and the fluid has had dye injected into it at discrete points. Laminar flow is defined as fluid flow that is smooth and steady. So, you can see that there are areas near the top and bottom of the photograph where the flow is very smooth and steady and you can still see the distinct lines of dye, but if you look directly behind the cylinder you can observe areas where the dye is swirling and agitated. This area is experiencing turbulent flow. Why do we care about boundary layers? Well, this has a lot to do with the study of laminar vs. turbulent flow. Laminar flow is defined as fluid flow that is smooth and steady. This is a photo of fluid flowing past a cylinder with dye that has been injected into the fluid at discrete points. You can see that there are areas near the top and bottom of the picture where the flow is very smooth and steady. This is laminar flow Then directly behind the cylinder, you can observe areas wehre the dye is swirling and agitated. This type of flow is called turbulent flow.

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**Lecture 11 - Introduction to Fluid Mechanics**

Reynolds Number Dimensionless parameter Correlates viscous behavior of all newtonian fluids = density = viscosity V = characteristic velocity of flow L = length scale of flow Most important parameter in fluid mechanics Governs transition from laminar to turbulent flow According to Frank White in his third edition of fluid mechanics, the dimensionless Reynolds number is the primary parameter correlating viscous behavior of all Newtonian fluids. This is a fancy way of saying that we can use this parameter to scale fluid and size of flow chambers to allow the same experiments to be run in water tunnels and in wind tunnels. We have both types of tunnels in our laboratory at Baylor University and you will see that they are very different in size, but we can run the same experiment in each type of tunnel and get the same results provided the experiment has been scaled properly using the Reynolds number. The formula for Reynolds number is provided in the upper right-hand corner. The variables in this equation are density, viscosity, the characteristic velocity of the flow, and the length scale of the flow. With regard to density, consider that water is more dense than air and water is also more viscous than air so it has a higher viscosity. With regard to velocity of flow, which you think, air or water, would flow faster providing each was subjected to the same pressure differential. I would think that air would more easily flow faster than water. Finally, the length scale of the flow. This length scale would allow us to use the Reynolds number for an experiment and run it with a smaller scale for water than air. The Reynolds number is the most important parameter that a mechanical engineer uses when dealing with fluid mechanics, and the Reynolds numbers is also measure that allows us to estimate when laminar flow will transition to turbulent flow. According to Frank White in his 3rd edition of “Fluid Mechanics”, the dimensionless Reynolds number is the “primary parameter correlating viscous behavior of all newtonian fluids.” p23 This is a fancy way of stating that we can use this parameter to scale fluid and size of flow chambers to allow the same experiments to be run in water tunnels and in wind tunnels. We have both types of tunnels in our lab, and you will see that they are very different in size, but we can run the same experiment in each type of tunnel and get the same results provided the experiment has been scaled properly. This parameter takes into account the density (consider that water is more dense than air) The viscosity (water is more viscous than air) The velocity of flow (which would flow faster providing each was subjected to the same pressure differential?) I would think that air would more easily flow faster than water. Finally, the length scale of the flow. This length scale would allow us to set the Reynolds number for an experiment and run it with a smaller scale for water than air. This is the most important parameter that a mechanical engineer uses when dealing with fluid mechanics The Reynolds number is a measure that allows us to estimate when laminar flow will transition to turbulent flow.

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**This Concludes Lecture 8**

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