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What is fluid mechanics?

Introduction to Fluid Mechanics. What is fluid mechanics?. Three states of matter: solid; liquid and gas ( or plasma). Liquid and gas are both fluids. Distinction between solids and fluids :

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What is fluid mechanics?

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  1. Introduction to Fluid Mechanics What is fluid mechanics? Three states of matter: solid; liquid and gas ( or plasma) Liquid and gas are both fluids. • Distinction between solids and fluids: • According to our experience: A solid is “hard” and not easily deformed. A fluid is “soft” and deforms easily. • Fluid is a substance that alters its shape in response to any force however small, that tends to flow or to conform to the outline of its container, and that includes gases and liquids and mixtures of solids and liquids capable of flow. • A fluid is defined as a substance that deforms continuously when acted on by a shearing stress of any magnitude.

  2. shearing forces : forces which act tangentially to a surface. the shear stress :force per unit area,

  3. Newton's law of viscosity Using the experimental result that shear stress is proportional to rate of shear strain then, where is the velocity. The constant of proportionality() is known as the dynamic viscosity, , of the fluid.

  4.  : shearing stress(전단응력)  : viscosity(점성계수) du/dy : velocity gradient(속도구배) We can then say: A Fluid is a substance which deforms continuously, or flows, when subjected to shearing forces. and conversely this definition implies the very important point that: If a fluid is at rest there are no shearing forces acting.

  5. Dimensions and Units • In fluid mechanics we must describe various fluid characteristics in terms of certain • basic quantities such as length, time and mass • A dimension is the measure by which a physical variable is expressed qualitatively, i.e. length is a dimension associated with distance, width, height, displacement. • Basic dimensions: Length, L • (or primary quantities) Time, T • Mass, M • Temperature, Q • We can derive any secondary quantity from the primary quantities i.e. Force = (mass) x (acceleration) : F = M L T-2 • A unit is a particular way of attaching a number to the qualitative dimension: Systems of units(SI unit) can vary from country to country, but dimensions do not • All theoretically derived equations are dimensionally homogeneous: dimensions of the left • side of the equation must be the same as those on the right side. • All equations must use consistent units: each term must have the same units. Answers will • be incorrect if the units in the equation are not consistent. Always chose the system of units prior to solving the problem.

  6. Dimensions and Units • Conversion factors are available in the textbook.

  7. Properties of Fluid Mass, force and Weight • Mass : m (M) • Force : F (M L T-2) • Newton’s 2nd. Law : F=ma • In general, F=ma/gc • gc = gravitational conversion factor = ma/F • If F= 1N, m=1kg, and a=1m/s2, then gc = 1(kg)(m/s2)/N=1N/N =1 • If F= 1kgf, m=1kg, and a=9.8m/s2, • then gc = 1(kg)(9.8m/s2)/kgf =9.8kg m/s2/kgf=9.8N/kgf • Weight : w = mg (In general, w=mg/gc)

  8. Specific volume(비체적) Defined as the volume of substance per unit mass. Vs= V / m = 1 / ρ Specific weight(비중량) Defined as the amount of weight per unit volume of a substance r = w / V = ρ g Units: Newton's per cubic metre, (or ) Dimensions: . Typical values: Water =9814 , Mercury = 132943 , Air =12.07 Paraffin Oil =7851 Specific Gravity (비중)

  9. Dynamic(Absolute) Viscosity(점성도) Viscosity, m, is the property of a fluid, due to cohesion and interaction between molecules, which offers resistance to sheer deformation. Different fluids deform at different rates under the same shear stress. Fluid with a high viscosity such as syrup, deforms more slowly than fluid with a low viscosity such as water. All fluids are viscous, "Newtonian Fluids" obey the linear relationship given by Newton's law of viscosity. , which we saw earlier. Units of Viscosity, ; kg/m/s, g/cm/s (poise, p), or N-s/m2, 1 poise=100 cp(centi-poise), 10 P = 1 Dimensions : ML-1T-1 is the velocity gradient(속도구배) or rate of shear strain. Time * Typical values of viscosity : Water =1.14 , Air =1.78

  10. Viscosity in Gasses and Liquids • Gases : • If temperature of a gas increases the momentum exchange between layers will increase thus increasing viscosity. • Viscosity will also change with pressure - but under normal conditions this change is negligible in gasses. • Liquids : • Increasing the temperature of a liquid reduces the cohesive forces. Reducing cohesive forces reduces shear stress and hence viscosity decreases. • High pressure can also change the viscosity of a liquid. As pressure increases the relative movement of molecules requires more energy hence viscosity increases. * Dynamic(absolute) Viscosity : 점성계수, * Kinematic Viscosity : 동점성계수, Units: , or Stoke(St), where St = 1 , 1St=100 centi-Stoke Dimensions :

  11. Newtonian and non-Newtonian Fluids Newtonian fluids: Fluids which obey the Newton's law of viscosity are called as Newtonian fluids. Newton's law of viscosity is given by Non-Newtonian fluids: Fluids which do not obey the Newton's law of viscosity are called as non-Newtonian fluids. Generally non-Newtonian fluids are complex mixtures: slurries, pastes, gels, polymer solutions etc. ,where A, B and n are constants. For Newtonian fluids A = 0, B =  and n = 1. There is also one more - which is not real, it does not exist - known as theideal fluid. This is a fluid which is assumed to have no viscosity. This is a useful concept when theoretical solutions are being considered - it does help achieve some practically useful solutions.

  12. Toothpaste Latex Paint Corn Starch Newtonian Fluids are Linear Relationships between stress and strain: Most common fluids are Newtonian. Non-Newtonian Fluids are Non-Linear between stress and strain

  13. Bulk modulus of elasticity (체적탄성계수) : Compressibility(압축성) Bulk modulus (E) = (change in pressure) / (volumetric strain) Volumetric strain =(change in volume) / (original volume) Typical values of Bulk Modulus: (표 1-7, p 37) for water : E = 2.05 x 104 kgf/m2 for oil : E = 1.62 x 104 kgf/m2 • Large values of the bulk modulus indicate incompressibility • The concept of the bulk modulus is mainly applied to liquids, since for gases the compressibility is so great that E is not constant.

  14. Compressibility of Fluids: Compression of Gases Ideal Gas Law: P is pressure, r is the density, R is the gas constant, and T is Temperature Isothermal Process (constant temperature): Isentropic Process (frictionless, no heat exchange): k is the ratio of specific heats, cp (constant pressure) to cv (constant volume), K=cp/cv, and R = cp – cv.

  15. Compressibility of Fluids: Speed of Sound Pressure disturbances in the fluid propagate as sound, and their velocity is known as the speed of sound or the acoustic velocity(음속), c. Isentropic Process (frictionless, no heat exchange): Ideal Gas and Isentropic Process: • Speed of Sound in Air at 60 °F 1117 ft/s or 300 m/s • Speed of Sound in Water at 60 °F  4860 ft/s or 1450 m/s • If a fluid is truly incompressible, the speed of sound is infinite, however, all fluids compress slightly.

  16. Mach Number, M M=V/c M<1 : 아음속(subsonic velocity) M=1 : 음속 (sonic velocity) M>1 : 초음속(supersonic velocity) 비압축성 유체(incompressible fluid) : M<<0.3 Example: A jet aircraft flies at a speed of 250 m/s at an altitude of 10,700 m, where the temperature is -54 °C. Determine the ratio of the speed of the aircraft, V, to the speed of sound, c at the specified altitude. Assume k = 1.40 Ideal Gas and Isentropic Process:

  17. The above ratio is known as the Mach Number, Ma • For Ma < 1 Subsonic Flow • For Ma > 1 Supersonic Flow

  18. Surface Tension (표면장력) The surface tension (s sigma) of a liquid : the work per unit area of that surface Units : Joule/m2 = N/m, or dynes per centimeter (1 dyn/cm = 0.001 N/m). There is a natural tendency for liquids to minimize their surface area. Surface tension will cause an increase of internal pressure p in order to balance the surface force.

  19. Applied to Circumference Applied to Area Surface Tension: Liquid Drop The pressure inside a drop of fluid can be calculated using a free-body diagram: Real Fluid Drops Mathematical Model R is the radius of the droplet, s is the surface tension, Dp is the pressure difference between the inside and outside pressure. The force developed around the edge due to surface tension along the line: This force is balanced by the pressure difference Dp:

  20. Surface Tension: Liquid Drop Now, equating the Surface Tension Force to the Pressure Force, we can estimate Dp = pi – pe: This indicates that the internal pressure in the droplet is greater that the external pressure since the right hand side is entirely positive. Is the pressure inside a bubble of water greater or less than that of a droplet of water? Prove to yourself the following result:

  21. Force balances in spherical droplet : Δp ∏r2 = 2 ∏ rσ i.e. Δp = 2 σ /r or pi-po = 2 σ /r In general, pi-po = σ(1/r1 +1/r2)

  22. Adhesion Cohesion Adhesion Cohesion Surface Tension: Capillary Action Capillary action in small tubes which involve a liquid-gas-solid interface is caused by surface tension. The fluid is either drawn up the tube or pushed down. “Wetted” “Non-Wetted” Cohesion > Adhesion Adhesion > Cohesion h is the height, R is the radius of the tube, q is the angle of contact. The weight of the fluid is balanced with the vertical force caused by surface tension.

  23. Vapor Pressure: Evaporation and Boiling Evaporation occurs in a fluid when liquid molecules at the surface have sufficient momentum to overcome the intermolecular cohesive forces and escape to the atmosphere. Vapor Pressure is that pressure exerted on the fluid by the vapor in a closed saturated system where the number of molecules entering the liquid are the same as those escaping. Vapor pressure depends on temperature and type of fluid. Boilingoccurs when the absolute pressure in the fluid reaches the vapor pressure. Boiling occurs at approximately 100 °C, but it is not only a function of temperature, but also of pressure. For example, in Colorado Spring, water boils at temperatures less than 100 °C. Cavitation is a form of Boiling due to low pressure locally in a flow.

  24. Weight, 46,800 N = mg • Mass, m = 46,800 N/ 9.81 (m/s2)= 4770.6 kg • Density, r = Mass / volume = 4770.6 kg / 5.6 m3= 852 kg/m3 • Specific gravity, SG SG Fluid Properties Examples (1) 5.6m3 of oil weighs 46,800 N. Find its density, r, and specific gravity, SG. (2) From table of fluid properties the viscosity of water is given as 0.01008 poises. What is this value in Ns/m2 and Pa s units? (Pascal, Pa=N/m2) • m = 0.01008 poise • 1 poise = 0.1 Pa s = 0.1 Ns/m2 • = 0.01008 poise/(0.1 Pa s/1 poise)= 0.001008 Pa s • = 0.01008 poise/(0.1 Ns/m2/1 poise)= 0.001008 Ns/m2

  25. (Pa s)(m/s)/(m) ( or N/m2) (3) The density of an oil is 850 kg/m3. Find its specific gravity and Kinematic viscosity if the dynamic viscosity is 5 x 10-3 kg/ms. • roil = 850 kg/m3 , rwater = 1000 kg/m3 • SGoil = 850 / 1000 = 0.85 • Dynamic viscosity = m = 5x10-3 kg/m/s • Kinematic viscosity = n = m / r • = 5x10-3 kg/m/s / 850 kg/m3 • = 5.88x10-6 m2/s (4) In a fluid the velocity measured at a distance of 75mm from the boundary is 1.125m/s. The fluid has absolute viscosity 0.048 Pa s (Ns/m2) and specific gravity 0.913. What is the velocity gradient and shear stress at the boundary assuming a linear velocity distribution. m = 0.048 Pa s, SG = 0.913, and at y=75 mm=0.075 m, u= 1.125 m/s

  26. - 2(0) = 0.68 s-1 At the plate face y = 0m, At y = 0.34m, t = 0 (5) The velocity distribution of a viscous liquid (dynamic viscosity m = 0.9 Ns/m2) flowing over a fixed plate is given by u = 0.68y - y2 (u is velocity in m/s and y is the distance from the plate in m). What are the shear stresses at the plate surface and at y=0.34m? Calculate the shear stress at the plate face. * As the velocity gradient is zero at y=0.34 then the shear stress(t) must also be zero.

  27. du/dy (s-1) 0.0 0.2 0.4 0.6 0.8 t (N m-2) 0.0 1.0 1.9 3.1 4.0 Using Newton's law of viscosity, where m is the viscosity. So viscosity is the gradient of a graph of shear stress against velocity gradient of the above data, or (6) Explain why the viscosity of a liquid decreases while that of a gas increases with a temperature rise. The following is a table of measurement for a fluid at constant temperature. Determine the dynamic viscosity of the fluid.

  28. du/dy (s-1) 0.0 0.2 0.4 0.6 0.8 Plot the data as a graph: t (N m-2) 0.0 1.0 1.9 3.1 4.0 Gradient(m) - 5.0 4.75 5.17 5.0 Calculate the gradient for each section of the line Thus the mean gradient = viscosity = 4.98 N s / m2

  29. Homework #1 (2003년 3월 25일 오후 5시까지 제출) Textbook 제 1장 연습문제 (p. 54~58) 중에서, 1-1, 1-3, 1-6, 1-9, 1-24, 1-25 * 강의노트 및 HW solution 다운로드 site http://san.hufs.ac.kr/~leesk * 3월 20일(목) 출석 확인 예고 !

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