Dimensions represent classes of units we use to describe a physical quantity. Most fluid problems involve four primary dimensions


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1 BEE 5330 Fluids FE Review, Feb 24, A fluid is a substance that can not support a shear stress. Liquids differ from gasses in that liquids that do not completely fill a container will form a free surface in a gravitational field (and mix minimally with any atmosphere) while a gas will form an atmosphere (and eventually mix with an existing atmosphere). 1 Dimensions Dimensions represent classes of units we use to describe a physical quantity. Most fluid problems involve four primary dimensions Mass [M] Length [L] Time [T] Temperature [Θ] For example velocity has the dimensions of LT System of Units Units are the bane of the United States! Remember the NASA Jet Propulsion Lab (JPL) satellite disaster?! In September 1999 we lost a $125,000,000 Mars Orbiter because a subcontractor to NASA was working in English units while NASA had converted to metric units in A system of units is a particular method of attaching a number to a dimension. A major source of calculation error is units errors check your units! Use your engineering common sense, you should always have a rough estimate of the answer you expect, at
2 BEE 5330 Fluids FE Review, Feb 24, least to an order of magnitude. If the answer is outside this range there is a good chance you have made a units error British Gravitational (BG) Length [L] foot Mass [M] slug; F = ma 1 lbs = 1 slug 1 ft/s 2 Time [T] second Temperature [Θ] R (degrees Rankine absolute temperature scale)= F International System (SI) Length [L] meter Mass [M] kilogram; F = ma 1 Newton = 1 kg 1 m/s 2 Time [T] second Temperature [Θ] K (Kelvin absolute temperature scale)= C Thermodynamic Properties Temperature Measure of internal energy level. Pressure Measure of compressive (normal) stress at a point. P = F A Density ρ = Mass Volume
3 BEE 5330 Fluids FE Review, Feb 24, Specific Weight γ water =62.4 lbs/ft 3 γ = ρg = weight volume 2.2 Specific Gravity The specific gravity is the density of a substance normalized by the density of water at a certain temperature, often 4 C, the temperature of maximum density at normal pressures. Hence we write S.G. = ρ ρ 4 C S.G. of sands and gravels is about = ρ in S.I. units 1000 kg/m3 2.3 Viscosity Therefore µ = τ du dz τ = µ du dz = [MLT 2 L 2 ] [LT 1 ] [L] = [M] [LT] Kinematic Viscosity If we normalize the viscosity by the density we have the kinematic viscosity. ν = µ ρ = [L2 ] [T] At 20 C water has an absolute or dynamic viscosity of N s m 2 (or Pa s) and a kinematic viscosity of m 2 s 1.
4 BEE 5330 Fluids FE Review, Feb 24, Example  A block sliding down an inclined plane If the block has mass 1 kg: 1. Determine the viscosity, µ, of the lubricant fluid in the gap. 2. What speed will the block travel if the angle, θ, is adjusted to 10 and the gap, δ, is decreased to 0.5 mm 1) µ = kg m s (= N s m 2 = Pa s); 2) V = m s
5 BEE 5330 Fluids FE Review, Feb 24, Surface Tension The water molecule is polar. The O attracts the H +. Within the fluid this attraction is in balance, i.e., the net force due to all of the polar pairs is zero. However, at the surface half of this force is missing and the surface is pulled toward the fluid interior with a certain energy. surface energy = J m 2 = N m m 2 = N m = force length = tension hence we refer to this energy as the surface tension. 2.6 Example Rise/drop in a capillary tube 3 Hydrostatics In many fluid problems the velocity is zero or the velocity is constant τ = 0. Pressure as a scalar quantity as it is a quantity with no dependence on direction (as opposed to velocity, which is a vector). P γ k = ρ a
6 BEE 5330 Fluids FE Review, Feb 24, Now, if the fluid is at rest (or at least moving at a constant velocity): a = 0 P = γ k Hence we can write: P x = 0, P y = 0 P z = γ We see that P = P(z) only, hence the pressure at a given elevation (z position) is constant. In the vertical we have dp dz = γ where we have replaced with d now. Hence as z P Incompressible Fluids It is usually reasonable to assume that all liquids essentially have a constant γ, certainly true for all fluids at constant temperature and pressure. Under this condition we have P 1 = γh + P 2 or h = P 1 P 2 γ where we refer to h as the pressure head as it is a pressure measured in units of length. 3.1 Measurement of Pressure Absolute pressure is the pressure relative to a perfect vacuum, hence it is always a positive (or zero) value. Gage pressure is the pressure relative to the local atmospheric value P > P a P gage = P P a > 0 P < P a P gage = P P a < 0 P vacuum = P suction = P a P > 0
7 BEE 5330 Fluids FE Review, Feb 24, Manometry A manometer is a vertical or inclined tube used to measure pressure. There are three fundamental types of manometers: Piezometer tube, Utube, and inclined tube. Manometer analysis is straight forward hydrostatics  the key is to keep track of the signs of the pressure terms! Consider the following Utube manometer: P 2 P 1 = γ(z 2 z 1 ) Now we track the pressure from A to 1 and from 1 to 2. A simple rule, based on our understanding that as we move down in a static fluid the pressure increases, is: P down = P up + γ z This obviates the need to keep track of signs of directions in the vertical. Hence for our Utube problem we have P A + γ 1 z A z 1 γ 2 z 2 z 1 = P atm and we solve for P A
8 BEE 5330 Fluids FE Review, Feb 24, Example  Manometer 3.3 Hydrostatic Force on a Plane Surface Note: The approach presented here is different than the approach presented in the text but I feel it is more powerful and makes more sense! Consider F R = P C A Therefore the magnitude of the force depends on P C the pressure acting at the centroid of the surface S. A the area of the surface S.
9 BEE 5330 Fluids FE Review, Feb 24, Note the above is a bit different than most books present this material. If we make two more assumptions 4. The column of fluid above the surface S is exposed to atmospheric pressure. 5. The density of the entire fluid, from the deepest part of the surface S all the way to the free surface of the fluid, is constant. Then we arrive at the form our book (and most books) present, namely F R = γ sin θ y da A where θ is the angle between vertical and our surface S (e.g., θ = 0 for a vertical surface). But from mechanics we recognize 1 y da as the first moment of the area A with respect A A to the x axis which we will denote y c since this is the position along the y axis of the centroid of the area. Hence we have y c = 1 y da and hence A A F R = γ sin θ y c A but we can write h c = sin θ y c hence F R = γh c A where h c is the depth of the centroid (e.g., now in a direction parallel to gravity). But this assumes assumptions 4 and 5 are true! Examples of problems that violate these assumptions will appear in the problem sets and Lab #2 so proceed with caution if you like the books approach Where is the force located on the surface? where y R = I xcγ cosθ P C A (1) I xc = y 2 da (2)
10 BEE 5330 Fluids FE Review, Feb 24, Pressure Prism Consider the following example: We can solve this directly by applying the analysis presented above (left as an exercise for the student), however, for many situations (or just for many people who prefer to think in a different manner!) a decomposition of the pressures into a series of pressure prisms is often easier. Consider the decomposition such that F yr = F 1 y 1 + F 2 y 2 ( = (γbh 2 h 1 ) h 1 + h ) ( h 1 h 1 + h ) 2 + h Therefore y R = = h 1 + h 2 2 = ( h 1 + h 2 2 h 1 + 2h 2 3 h 1 + h 2 ) 2 = 7.43 ( ) ( γb h2 2 2 ) ( h 1 + 2h 2 3 h 1 2h 2 3 ) 3.5 Buoyancy  Archimedes Principal F B = Weight of fluid displaced by a body or a floating body displaces its own weight of the fluid on which it floats.
11 BEE 5330 Fluids FE Review, Feb 24, Example  Buoyancy 4 Conservation of Mass If we have onedimensional inlets and outlets (ρi A i V i ) in or (ṁ i ) in = (ρ i A i V i ) out = (ṁ i ) out Incompressible onedimensional flow then we can write (Vi A i ) out = (V i A i ) in or (Q)out = (Q) in Example  Pipe Entrance Flow ( ) u = U max 1 r2 R 2 If the inlet flow is uniform and denoted U 0, what is U max
12 BEE 5330 Fluids FE Review, Feb 24, U max = 2U 0 5 Conservation of Linear Momentum for a constant Control Volume 1D system FC.V. = (ṁ i v i ) out (ṁ i v i ) in Example
13 BEE 5330 Fluids FE Review, Feb 24, Conservation of Energy and Bernoulli Equations Incompressible 1D Flow With No Shaft Work in head form: ( ) ( ) P γ + v2 P 2g + z = γ + v2 2g + z h f where we can think of h f as the friction losses and we see that h f > 0 Example Gas Pipeline out in Consider the following pipe flow: If Q = 75 m 3 /s, the pipe radius is r = 6 cm, the inlet pressure is maintained at 24 atm by a pump, the outlet vents to the atmosphere, the pipe rises 150 m from inlet to outlet and the pipe length is 10 km, what is h f? What is the velocity head? h f =198 m Therefore the friction loss is greater than the z and the pump must drive against both! The velocity head is only 0.17 m! Note that the length did not come into our solution. h f includes the total losses along
14 BEE 5330 Fluids FE Review, Feb 24, the pipe due to friction effects and hence includes the effect of length implicitly. 6.1 Bernoulli Equation ( ) ( ) P γ + v2 P 2g + z = out γ + v2 2g + z in Clearly anywhere along a streamline, as long as no work is done between analysis points and the assumption of frictionless flow is good, we can write P γ + v2 2g + z = h 0 where the constant h 0 is referred to as the Bernoulli constant and varies across streamlines. Bernoulli Equation Assumptions Flow along single streamline different streamlines, different h 0. Steady flow (can be generalized to unsteady flow). Incompressible flow. Inviscid or frictionless flow, very restrictive! No w s between analysis points on streamline. No q between points on streamline. 6.2 Pressure form of Bernoulli Equation If we multiply our head form of the Bernoulli equation by the specific weight we arrive at the pressure form of the Bernoulli Equation: P + ρ v2 2 + γz = P t
15 BEE 5330 Fluids FE Review, Feb 24, where we call the first term the static pressure, the second the dynamic pressure, the third the hydrostatic pressure, and the righthandside the total pressure. Hence the Bernoulli Equation says that in inviscid flows the total pressure along a streamline is constant. If we remain at a constant elevation the above equation reduces to P + ρ v2 2 = P s where we refer to P s as the stagnation pressure. Thus by definition the stagnation pressure is the pressure along horizontal streamlines when the velocity is zero. 6.3 PitotStatic Tube The static and stagnation pressures can be measured simultaneously using a Pitotstatic tube. Consider the following geometry: The equation for the Pitot tube is known as the Pitot formula V 1 = 2 P 2 P 1 ρ or in terms of heads V = 2g(H h)
16 BEE 5330 Fluids FE Review, Feb 24, Example Flow accelerating out of a reservoir 2gh V 2 = 1 ( ) 2 A2 A and if A 1 A 2 1 ( A2 A 1 ) 2 1 V 2 2gh, again! This was first noted by Torricelli and is known as Torricelli s equation. 6.5 Energy and the Hydraulic Grade Line As we have seen we can write the head form of the Energy equation as P γ + v2 + z = H = Energy Grade Line (EGL) 2g In the case of Bernoulli flows the energy grade line is simply a constant since by assumption energy is conserved (there is no mechanism to gain/lose energy). For other flows it will drop due to frictional losses or work done on the surroundings (e.g., a turbine) or increase due to work input (e.g., a pump). Note that this is the head that would be measured by a Pitot tube. We can also write P γ + z = Hydraluic Grade Line (HGL) and we see that the HGL is due to static pressure the height a column of fluid would rise due to pressure at a given elevation or in other words the head measured by a static
17 BEE 5330 Fluids FE Review, Feb 24, pressure tap or the piezometric head. Example Venturi Flow Meter Consider 2g h Q = A 2 V 2 = C v A 2 ( A2 1 where C v is known as the coefficient of velocity and lies in the range 0.95 and 1.0 typically. It accounts for the minor energy losses relative to the ideal Bernoulli flow. A 1 ) Frictional Effects If we have abrupt losses, say at a contraction, a simple way of accounting for this is through a discharge coefficient. We can write a modified form of Torricelli s formula for incompressible flow Q = A = C d A 2gh where A is the area of the orifice and C d is the discharge coefficient and is 1 for frictionless (inviscid) flow and can range down to about 0.6 for flows strongly effected by friction. Note we can handle nonuniform (violation of 1D assumption) flow effects with a C d as well. Note C d = C c C v where C c is defined below
18 BEE 5330 Fluids FE Review, Feb 24, Vena Contracta Effect For a flow to get around a sharp corner there would need to be an infinite pressure gradient, which of course does not happen. Hence if the boundary changes directions too rapidly at an exit, the flow separates from the exit and forms what is known as a vena contracta Clearly A j /A 1. For a round sharpcornered exit the coefficient is C c = A j /A = 0.61 and typical values of the coefficient fall in the range 0.5 C c Dimensional Analysis & Similitude Dimensional analysis is a method for reducing the number of variables describing the physics if k dimensional variables are important we seek to reduce (condense) them to n dimensionless variables. 7.1 Buckingham Pi Theorem If k dimensional variables are described by r physical dimensions then k r independent dimensionless variables completely describe the physics. Before resorting to dimensional analysis try by inspection. Almost always will find a Reynolds number and if there is a free surface or a ship riding on a free surface will likely find a Froude number.
19 BEE 5330 Fluids FE Review, Feb 24, Similitude & Experiments 7.3 Types of Similarity Total Similarity = Geometric Similarity + Kinematic Similarity + Dynamic Similarity Geometric Similarity length scale ratios the same. Kinematic Similarity velocity scale ratios the same. Dynamic Similarity force scale ratios the same. If total similarity can not be achieved we accept partial similarity but we let some parameters go (usually the Reynolds number) and make sure they are in a range (i.e., Re is high enough to be turbulent at model scale if it is turbulent at prototype scale) that is consistent with the physics at prototype scale.
20 BEE 5330 Fluids FE Review, Feb 24, Example Similitude 7.5 Summary of Dimensionless Parameters Here are some of the most common dimensionless numbers that show up in fluid mechanics: Re (Reynolds Number) forces. ρv L µ = V L ν The ratio of intertial forces to frictional Fr (Froude Number) V gl The ratio of intertial forces to gravitational forces We (Weber Number) ρv 2 L σ Eu (Euler Number) The ratio of intertial forces to surface tension forces. P The ratio of pressure forces to intertial forces. ρv St (Strouhal Number) fl The ratio of event frequency (often vortex shedding) V and the advective frequency (inverse of the advective, or inertial, time scale). V Ma (Mach Number) The square root of the ratio of intertial forces to c compressibility effect forces can be thought of as a Froude number for compressible flows.
21 BEE 5330 Fluids FE Review, Feb 24, Viscous Flow in Pipes Re < 2100 Laminar Re > 4000 Turbulent 2100 < Re < 4000 Transition where for circular pipes we define Re as Re D Re D = ρv D µ These values are approximate and depend on the smoothness of the pipe wall and the quietness of the initial flow (e.g., if the pump supplying the pipe is vibrating the pipe turbulence is apt to set in at lower Re). In general we take an energy equation perspective and use the Darcy Weisbach equation. For laminar flows we have P = γ z + f L D Eu = 64 Re D L D and for turbulent flows we use the Moody Chart or an explicit equation V 2 2g 8.1 Moody Chart Laminar flow f independent of ǫ/d Fully turbulent flow f independent of Re inertially dominated Turbulent flow at moderate Re f a function of both Re and ǫ/d There are curve fits to the Moody chart, the most famous of which is the Colebrook formula:
22 BEE 5330 Fluids FE Review, Feb 24, Haaland has worked out an explicit relationship that is accurate to within 2%: ǫ = 1.8 log D f 3.75 Re D
23 BEE 5330 Fluids FE Review, Feb 24, Three Types of Pipe Flow Problems Type Given Find I D, L, V, ρ, µ, g h L P II D, L, h L, ρ, µ, g V (or Q) III V (or Q), L, h L, ρ, µ, g D 8.3 Explicit Solutions for 3types of Pipe Flow  Swamee & Jain Swamee and Jain extended the concepts of Haaland to find explicit forms of the solutions for the three types of pipe flow, like Haaland, accurate to within 2% of the Moody diagram determined iterative solution. They are: { [ h L = 1.07 Q2 L 1 log gd ǫ D ( ) ]} νd (3000 < Re < ; 10 6 < ǫ Q D < 0.01) ( ) [ gd ( ) ] h L 1 ǫ 3.17ν Q = log L 3.7 D + L (Re > 2000) gd 3 h L [ ( ) LQ D = 0.66 ǫ ( ) ] L + νq 9.4 (5000 < Re < ; 10 6 < ǫ gh L gh L D < 0.01) Example  pipe flow 9 Drag  (Lift is similar) In general we rely on experiments to determine the total drag on an object (the sum of the pressure and friction drags). We express the drag (D) nondimensionally as a drag coefficient, C D C D = D 1 ρv 2 2 A where A is an appropriately chosen area with respect to the stresses generating the drag force.
24 BEE 5330 Fluids FE Review, Feb 24, Characteristic Area There are three standard area types used in drag coefficients: 1. Frontal Area the area seen by the flow stream. For drag coefficients dominated by pressure drag (e.g., blunt objects such as cylinders and spheres) this is often the appropriate choice of area. 2. Planform Area area as seen from above (e.g., perpendicular to flow stream). For drag coefficients dominated by friction drag (e.g., wide flat bodies such as wings and hydrofoils) this is often the appropriate choice of area. 3. Wetted Area area of wetted surface. Often used for boats and other surface water vessels where friction drag is dominant. 10 Drag Coefficient Dependencies Geometry, Re & Roughness In general C D = φ(re, Fr, ǫ/l, Ma, Geometry). There are nomographs and tables for regions where the C D is about constant (higher Re). Example Drag Consider a parachute with diameter 4 m carrying a person and gear such that the total mass is 80 kg. If the parachuter jumps from high enough that terminal
25 BEE 5330 Fluids FE Review, Feb 24, velocity (e.g., steady state  no acceleration) is achieved, what is this velocity? U=8.6 m/s 11 OpenChannel Flow In uniform flow y 1 = y 2, V 1 = V 2 = V, therefore the energy equation becomes: z 1 z 2 = S 0 L = h f Manning s Equation V α n R2/3 h S1/2 0 and Q α n AR2/3 h S1/2 0
26 BEE 5330 Fluids FE Review, Feb 24, where R h = A P = hydraulic radius where (P) is the length of the wetted perimeter, A is the wetted area, and n is Manning s n which varies by a factor of 15 and is tabulated. Example  Open Channel Flow
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