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流體力學 (Fluid Mechanics) (Spring 2009) 教科書 (Text Book)

流體力學 (Fluid Mechanics) (Spring 2009) 教科書 (Text Book) Munson, B. R., Young, D. F., and Okiishi, T. H., Fundamentals of Fluid Mechanics, 5th edition, John Wiley & Sons (Asia) Pte Ltd, 2006. Pte Ltd = Private Limited. Chapter 1 Basic Properties of Fluids.

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流體力學 (Fluid Mechanics) (Spring 2009) 教科書 (Text Book)

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  1. 流體力學(Fluid Mechanics) (Spring 2009) 教科書(Text Book) Munson, B. R., Young, D. F., and Okiishi, T. H., Fundamentals of Fluid Mechanics, 5th edition, John Wiley & Sons (Asia) Pte Ltd, 2006. Pte Ltd = Private Limited

  2. Chapter 1 Basic Properties of Fluids

  3. Chapter 1 Basic Properties of Fluids Fluid mechanics(流体力學)是什麼? －屬於applied mechanics(應用力學)之領域 －研究氣体及液体在靜止及運動之行為(Behavior). －所研究的流體種類包括，由血液至原油，其所流經 的管徑大小，由幾個microns(10-6m)至4 ft直徑之油 管(800 miles長)。 －研究問題範圍 為何飛機的表面需要平滑? 而高爾夫球的表面須要粗糙(或dimpled) 以增加其飛行效率?

  4. 有些很有趣的問題可以利用很簡單流體力學之觀念去回答，例如：有些很有趣的問題可以利用很簡單流體力學之觀念去回答，例如： －火箭為何能在外太空中沒有空氣裡可以飛行？ －何以超音速飛機，經過你之後才聽到聲音？ －何以在河床平坦之處的流水仍然流動很大？ －水龍頭流出之水，何以有時候平滑，有時候為不平 滑 之流動？ －何以能夠利用飛機模型去研究飛機建造之設計？ －何以流線型的汽車或卡車設計，能夠增加汽油使用的 里程數？ 流體力學是一門非常重要而實際的課題。

  5. §1.1 Some characteristics of fluids －What is a fluid? －What is the difference between a solid and a fluid?

  6. What is the difference between a solid and a fluid? • Ans: Descriptive answer: • Solid－"hard" ; not easily deformed • Fluid－"soft" ; easily deformed • From molecular structure of materials: • Solid－densely spaced molecules with large • intermolecular cohesive forces • to maintain its shape, and to not be easily deformed. • Fluid－the molecular are spaced further apart, • the intermolecular forces are smaller • than for solid, and the molecules have • more freedom of movement.

  7. 流體(Fluid)包括液体(Liquid)及氣體(Gases)，其區別如下：流體(Fluid)包括液体(Liquid)及氣體(Gases)，其區別如下： Liquids－easily deformed, not easily compressed, －free surface. Gases －greater molecular spacing and freedom of motion with negligible cohesive intermolecular forces. －no free surface

  8. What is a fluid? • Definition of fluid and how fluids deform : • A fluid is defined as a substance that deforms • continuously flow when acted on by a shearing • stress of any magnitude. • A shearing stress (F/A) is created whenever a • tangential force acts on a surface. • When common solids such as steel or other • metals are acted on by a shearing stress, they • will initially deform (usually a very small • deformation),but they will not continuously • deform(flow).

  9. Rheology: Some materials, such as slurries, tar, putty, toothpaste, and so on, are not easily classified since they will behave as a solid if the applied shearing stress is small. But if the stress exceeds some critical value, the substance will flow. The study of such material flow is called rheology.

  10. Fluid continuum Although the molecular structure of fluids is important in distinguishing one fluid from another, it is not possible to study the behavior of individual molecules when trying to describe the behavior of fluid at rest or in motion. Rather, we characterize the behavior by considering the average, or macroscopic, value of the quantity of interest, where the average is evaluated over a small volume containing a large number of molecules. The spacing for gases at normal P&T is on the order of 10-6 mm The spacing for liquid at normal P&T is on the order of 10-7 mm The number of molecules/mm3 is on the order of 1018 for gases. 1021 for liquids.

  11. Thus, we assume that all the fluid characteristics we are interested in (pressure, velocity, etc) vary continuously throughout the fluid－that is, we treat the fluid as a continuum. One area of fluid mechanics, for which the continuum concept breaks down, is in the study of rarefied gases such as would be encountered at very high altitudes. Spacing between air molecules can become large The continuum concept is no longeracceptable

  12. §1.2 Dimensions, Dimensional Homogeneity, and Units It is necessary to develop a system for describing fluid characteristics both qualitatively and quantitatively. The qualitative aspect serves to identify the nature, or type of the characteristics -- such as length, time, stress, and velocity. The quantitative aspect provides a numerical measure of the characteristics. The quantitative description required both a standard (or a unit) and a number. A standard (or unit) for a length might be a meter or foot for time might be an hour or second for mass might be a slug or kilogram

  13. Basic dimensions (or primary quantity): • The qualitative description is conveniently given in terms of certain primary quantities, such as • Length (L), Time (T), Mass (M), and Temperature(θ) •  MLT θ system • or • Length (L), Time (T), Force (F), and Temperature(θ) •  FLT θ system • or • Length (L), Time (T), Mass (M),Force (F), and Temperature(θ) •  FMLT θ system

  14. Secondary dimensions(or secondary quantity): Secondary quantities are based on primary quantities (or basic dimensions) , such as area [=] L2, velocity [=] LT-1, density [=] ML-3 and so on v ≐ LT-1 or v[=]LT-1 "The dimensions of a velocity equal length divided by time"

  15. Secondary quantities v (velocity) = L/T ≐ LT-1  (stress) = F/A [=] FL-2 or  (stress) = F/A[=]ML-1T-2 Table 1.1(p.4)a list of dimensions

  16. Table 1.1 (p. 4)Dimensions Associated with Common Physical Quantities (continued on next slide)

  17. Table 1.1 (p. 4)continued

  18. Dimensional Homogeneity: All theoretically derived equations are dimensionally homogeneous. Dimensional homogeneity unit [=] unit Same unit [=] unit + unit Same Same

  19. For example: v = v0 + at -----(1.1) dimensional homogeneity LT-1 [=] LT-1 + LT-2 T LT-1 [=] LT-1 + LT-1 d = ------(1.2)dimensional homogeneity where g = 32.2 ft/s2 d = 16.1 t2 ------------------(1.3) 16.1 is a constant with dimensions of LT-2 Eq(1.1) & Eq(1.1) are general homogeneous equation Eq(1.3) is restricted homogeneous equation Dimensional analysis (in Chapter 7)

  20. Example 1.1 Investigate the dimensional homogeneity of the following formula: • Q = 0.61A • Solution: • On next page

  21. Q = 0.61A • Left Side [=] L3T-1 • Right Side [=] 0.61(L2) • Right Side [=] Left Side • = pure constant • Thus Q = 0.61A is a general dimensional homogeneity • (valid for any constant system of units) Q = volume rate of flow(volume/time) [=] L3T-1 A = area [=] L2 g = acceleration of gravity [=] LT-2 h = height [=] L

  22. Alternately, • Q = 0.61A • = 0.61A where g = 32.2 ft/s2 • Q = 4.90A • Left Side [=] L3T-1 • Right Side [=] 4.90L2L1/2 • [=] 4.90L5/2 • for left side = right side • L3T-1[=] 4.90L5/2 • 4.90 [=] L3T-1L-5/2[=] L1/2T-1 • it is not a constant, instead it has a dimension of L1/2T-1 • (or ft1/2/s) • Thus Q = 4.9A is a restricted homogeneous equation • (valid for a specific system of units)

  23. §1.2.1 System of Units • For example, if we measure the width of a black board and say • 10 units wide --------no meaning • 10 meters wide ------meaningful • “meter”--- standard length • A unit must be established for each of the remaining basic quantities( length, force, mass, time, and temperature). • There are several systems of units in use and we shall consider three systems that are commonly used in engineering. • British Gravitational (BG) system • International System (SI) • English Engineering (EE) System

  24. *British Gravitational (BG) System • -- BG units (FLT ) • Basic Length (L) --- foot (ft) • Dimensions Time (T) --- second (s) • Force (F) --- pound (lbf or lb) • Mass (M) --- slug (slug) • Mass is defined from Newton's Second law • force = mass * acceleration • or 1 lbf = 1 slug * 1 ft/s2 • or 1 slug = 1 lbf / 1 ft/s2 • Temperature() ---- • Fahrenheit(0F) or Rankine(0R) • 0R = 0F + 459.67

  25. *International System (SI) ---- SI units (MLT ) • In 1960 the Eleventh General Conference on Weights and Measures, the international organization responsible for maintaining precise uniform standards of measurements, formally adopted the international system of Units as the international standard • Length (L) ---- meter(m) • Time (T) ----- second (s) • Mass (M) ---- kilogram (kg) • Temperature() ----- kelvin (k) ºk wrong • k = 0C+273.15

  26. Force(F) ------ newton (N) • F = m a • 1 N = (1kg)(1m/s2) • weight W = mg • W(N) = m(kg)g(m/s2) • where g = 9.807 m/s2 = 9.81m/s2 • work W = FS • W(Joule, J) = F(N)S(m) • power P = FV • P(watt) = F(N)V(m/s)

  27. *English Engineering (EE) System • --- EE Units (FMLT ) • Length(L) --- foot(ft) • Time(T) --- second(s) • Mass(M) --- pound mass(lbm) • Temperature(θ) --- degree Rankine(0R) • Force(F) --- pound(lbf) • Newton's Second law • weight W=mg/gc where gc=32.174 lbm ft / s2 / lbf

  28. Compare • In BG unit In EE unit • 1slug=32.174 lbm • Units in this textbook • BG Units ***Using a consistant system of units • SI Units throughout a given solution. • EE Units ---sparingly • *Conversion tables: • Table 1.3;1.4; 1.5~1.8 • (inside of back & front covers)

  29. System of Units BG (FLT ) SI (MLT ) EE (FMLT ) British Gravitational System International System English Engineering System Length(L) ft(foot) m(meter) ft(foot) Time(T) S(second) S(second) S(second) Mass(M) slug(slug) kg(kilogram) lbm(pound mass) Temperature() Raukine(0R) 0R=0F+459.67 Kelvin(k) k=0c+273.15 Raukine(0R) or Fahrenhert(0F) Lbf (Pound force) Lbf (Pound force) Force N (Newton)

  30. 1lbf=1slug1ft/s2 1slug = 1 ft/s2 / 1 lbf 1Newton=1kg1m/s2 gc=1 gc=1 BG (FLT ) SI (MLT ) EE (FMLT )

  31. 1 pound-force = 0.453 kilogram-force 1 pound-force = 4.448 newton 1 newton = 0.224 pound-force 1 newton = 0.101 kilogram-force

  32. w • Example 1.2 • Given: • tank • Ff=?

  33. Solution: Newton's second law of motion to the water body table 1.3 table 1.3

  34. Table 1.3 (back cover)Conversion Factors from BG and EE Units to SI Units

  35. Table 1.4 (back cover)Prefixes for SI Units

  36. §1.3 Analysis of Fluid Behavior • Fundamental laws involved in the study of fluid mechanics • (same as you have encountered in physics and other mechanics courses) • ---Newton's laws of motion (the 1st, 2nd, and 3rd laws) • ---Conservation of mass • ---the first and second laws of thermodynamics

  37. Fluid statics ------- the fluid is at rest • Fluid mechanics • Fluid dynamics --- the fluid is moving • Fluid properties Fluid behavior • For example • Compressible fluid • Incompressible fluid • Viscous fluid • Inviscous fluid

  38. Fluid properties: • Heaviness • –density() • –specific weight( γ ) • –specific gravity(SG) • Fluidity • –viscosity • Compressibility • Vapor pressure • Surface tension

  39. §1.4 Measures of Fluid Mass and Weight • §1.4.1 Density() •  = m/v  [=] slug/ft3 ( in BG units ) •  [=] kg/m3 ( in SI units ) •  = f ( different fluids ) Table 1.5 & 1.6 • Δliquid = f (ΔPressure, Δtemp. ) = small • Table B.1 p. 761

  40. = 1.94 slug/ft3 • = 999.9 kg/m3 • ( or 1000 kg/m3 ) Table B.1 & B.2 p. 761 • This property is not commonly • The specific volume, υ = 1/used in fluid mechanics but • is used in thermodynamics. at 600F

  41. §1.4.2 Specific weight (γ ) • γ = weight/volume = mg/ = g • γ = g [=] lbf/ft3 in BG Units Table 1.5 • [=] N/m3 in SI Units 1.6 • §1.4.3 Specific Gravity ( SG ) • SG = at some specified temperature • SG = independent of the system of units used

  42. §1.5 Ideal Gas Law • Ideal (or perfect) gas law, or Equation of state • p = RT ------------------------ (1.8) • where P = absolute pressure • = Normal force on a plan surface/area • [=] lbf/ft2 abs ( psfa ) or lbf/in2 abs( psia ) in BG Units • [=] N/m2 ( Pascal, Pa ) in SI Units • absolute pressure --- relative to absolute 0 pressure • gauge pressure --- relative to the local atmospheric pressure • absolute pressure (psia) • = gauge pressure (psig) + local atmospheric pressure

  43. Standard Sea – level atmospheric pressure • = 14.696 psia or 101.33 kpa ( abs. ) • or 14.7 psia or 101 kpa ( abs. ) •  = density • T = absolute temperature • R = gas constant • in BG Units R [=] • in SI Units R[=] • table 1.7 & 1.8 • (front page)

  44. Table 1.7 (front cover)Approximate Physical Properties of Some Common Gases at Standard Atmospheric Pressure (BG Units)

  45. Table 1.8 (front cover)Approximate Physical Properties of Some Common Gases at Standard Atmospheric Pressure (SI Units)

  46. Example 1.3 • Given: Compressed air tank = 0.84 ft3 ; p = 50 psig, • T = 700F, Patm = 14.7 psia • Question:  = ? w = ? • Solution: •  = p/RT • R = 1716 ft-lb/slug0R • T = 700F = 70+459.6 = 529.60R

  47. Table 1.3 (back cover)Conversion Factors from BG and EE Units to SI Units

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