4 Fundamental Definitions: System And Property Basic Unit .

ES206 Fluid MechanicsUNIT A: Fundamental ConceptsROAD MAP . . . A-1: Engineering Fundamentals A-2: Fluid Properties and CharacteristicsES206 Fluid MechanicsUnit A-1: List of Subjects What is Fluid Mechanics? Basic Dimensions in Engineering Basic Unit System Fundamental Definitions: System and Property

Page 1 of 6Unit A-1What is Fluid Mechanics?Liquids – water, oil, . . .Gases – vapor, air, . . .Continuum assumptionHydrodynamics, Gas Dynamics, AerodynamicsWhat is Fluid? What is Fluid Mechanics? A fluid is defined as a substance that deforms continuously when acted by shearforces Fluid mechanics is the engineering science that deals with the action of forcesand associated motion of fluidsContinuum Mechanics In classic fluid mechanics, “fluids” are treated as a hypothetical continuoussubstance, called continuum Continuum assumption employs an idea that fluids are made up continuouslywith many (many) small particles. A continuum is a body that can be subdivided into many (many) infinitesimal elements with properties being those ofthe bulk material.Advanced Subjects in Fluid Mechanics Hydrodynamics: flow of fluids (liquids) in no change of density(incompressible) flows Gas Dynamics: flow of fluids (gases) that undergo significant change of density(compressible) flows Aerodynamics: flow of air

Page 2 of 6Unit A-1Basic Dimensions in EngineeringBasic Dimensions An engineering property can be described in terms of three basic dimensionsMass (M) / Length (L) / Time (T) MLT SystemForce (F) / Length (L) / Time (T) FLT System An engineering property (units) are all combinations of these three basicdimensions (plus temperature: ) So, is it possible to combine certain properties to create a “dimensionless”property? YES, and it is very important in engineering!For example . . . Reynolds number (Re):( Density )( Velocity )( Length ) ( ML )( LT ) ( L )Re ? 1 1MLT( Viscosity )() 3 1

Page 3 of 6Unit A-1Basic Unit System (1)SI (Systemé International) Unit SystemSI Unit SystemMass Kilogram (kg) / Length Meter (m) / Time Second (s)Temperature Kelvin (K) / Celsius ( C)Force Newton (N): 1 N (1 kg)(1 m/s2)Temperature unit conversion: K C 273Common Problem (KILOGRAM “FORCE”)Is 1 kilogram force 1 kilogram mass ?“An object of 1 kilogram MASS has WEIGHT (force) of 1 kilogram onEARTH”Kilogram “Force” (kgf) and kilogram “Mass” (kg) are different properties, andcannot be mixed-upForce unit conversion: 9.8 Newton 1 kgf

Page 4 of 6Unit A-1Basic Unit System (2)Traditional (U.S. Customary) British Gravitational (BG)or English Engineering (EE) Unit SystemSlug or Pound Mass (lbm)?US Customary Unit SystemMass Slug / Length Feet (ft) / Time Second (s)Temperature Rankine ( R) / Fahrenheit ( F)Force Pound (lb): 1 lb (1 slug)(1 ft/s2)Temperature unit conversion: R F 460Common Problem (POUND “MASS”)Is 1 pound force 1 pound mass ?“An object of 1 pound MASS has WEIGHT (force) of 1 pound on EARTH”Pound “Force” (lb) and Pound “Mass” (lbm) are different properties, and cannotbe mixed-upMass unit conversion: 1 slug 32.2 lbm

Page 5 of 6Unit A-1Class Example ProblemRelated Subjects . . . “Basic Unit System”Mass and weight are two different properties.Mass fundamental physical property (how much “matters” are packed into anobject: note that this is, in fact, “non-measurable” quantity . . .)Weight size of the gravitational pull (the “force” exerted on an object): this canbe measured, certainly.Newton’s 2nd law states: Weight Mass Gravitational Acceleration ( W mg )Hence, m W 300 N 2 30.6 kgg 9.8 m/sNote: this mass is a unique property of an object (will never change); therefore,under the gravitational acceleration of 4 ft/s, the weight of the same object becomes0.3048 mW mg ( 30.6 kg ) ( 4 ft/s 2 ) 37.3 N1 ft

Page 6 of 6Fundamental Definitions:System and PropertyUnit A-1Extensive and Intensive PropertiesSystem and Property System: an object of focus, enclosed by system boundaries Property: characteristics that can be measuredExtensive and Intensive Properties Extensive properties (B): depends on the extent of the system (area, weight,length, mass, . . .)Mass (m)Linear Momentum (m V )Energy (E) Intensive properties (b): independent of the extent of the system (density,temperature, . . .)BB mb or b mBB Mass(m) b 1mmVB Linear Momentum (m V ) b VmEB Energy (E) b em

ES206 Fluid MechanicsUNIT A: Fundamental ConceptsROAD MAP . . . A-1: Engineering Fundamentals A-2: Fluid Properties and CharacteristicsES206 Fluid MechanicsUnit A-2: List of Subjects Fluid Properties Ideal Gas Model Compressibility of Fluid Vapor Pressure Surface Tension Viscosity Newtonian and Non-Newtonian Fluids Dynamic Viscosity of Common Fluids

Page 1 of 12Unit A-2Fluid Properties gSpecific Gravity (SG)Fluid Density Density (or “Mass” Density) mass per unit volume (kg/m3 or slug/ft3) Specific Weight (or “Weight” Density) weight per unit volume (N/m3 or lb/ft3) Specific weight and density are related: gDensity of water: 1,000 kg/m3 or 1.94 slug/ft3Specific weight of water: 9,800 N/m3 or 62.4 lb/ft3Specific Gravity Specific Gravity (SG) is the ratio of the density (or specific weight) of a givenfluid to the waterSG fluid or fluid water water Specific gravity indicates “how heavy” the liquid is, compared against water:SGHg 13.6

Page 2 of 12Unit A-2Ideal Gas Modelp RTEquation of State Ideal Gas Equation of State: p RT R is the specific gas constant (air: 287 J/kg K or 1,716 ft lb/slug R)Note) specific gas constant depends on the type of gas (gas specific):Helium: 2,077 J/kg K or 12,420 ft lb/slug RHydrogen: 4,124 J/kg K or 24,660 ft lb/slug RNitrogen: 296.8 J/kg K or 1,775 ft lb/slug ROxygen: 259.8 J/kg K or 1,554 ft lb/slug R Pressure (p) must be the absolute pressure (not “gage” pressure)Note) conversion of pressures:pabsolute pgage patm Temperature (T) must be in absolute scale (not “relative scale”)Note) conversion of temperatures:K C 273and R F 460

Page 3 of 12Unit A-2Class Example ProblemRelated Subjects . . . “Ideal Gas Model”Let us assume Nitrogen as an ideal gas. From Table 1.8 (textbook),RNitrogen 296.8 J/(kg K)Note) local standard atmosphere depends on altitude: Prescott (5,000 ft): 12.228 psi Daytona Beach (sea-level): 14.696 psiApplying the equation of state (ideal gas law):p RT (1.5 kg/m3 ) 296.8 J/(kg K) ( 25 oC 273 K ) 132,670 Pa (132.67 kPa)Note: this is the “absolute” pressure. The purpose of this problem is to determinethe “gage” pressure.The gage pressure is the pressure above (or below, if vacuum) the atmosphericpressure, where the given atmospheric pressure here is: patm 97 kPaTherefore:pgage pabs patm 132.67 kPa 97 kPa 35.67 kPaNote that the atmospheric pressure is equal to “ZERO GAGE” pressure:patm 0 (gage)

Page 4 of 12Unit A-2Compressibility of FluidEv dpdp dV V d p constantEv pp k constantEv kpc kRTCompressibility of Liquids Bulk Modulus of Elasticity (Ev) indicates “compressibility” of liquids:Water: Ev 2.2 GN/m2 (almost “incompressible”)Compressibility of Gases Mach Number (M) indicates “compressibility” of gases: M Speed of Sound (c) of an ideal gas: c VckRT (k 1.4 for air)Expansion and Compression of Gases (Thermodynamic Processes) Isobalic process (the process is under the constant pressure):p constant Isothermal process (the process is under the constant temperature):p constant Isentropic process of gases (the process is frictionless and no heat exchangewith surroundings): “reversible” / “no heat exchange”p k p 1.4 constant

Page 5 of 12Unit A-2Class Example ProblemRelated Subjects . . . “Compressibility of Fluid”The speed of sound for an ideal gas can be calculated by: c kRT .This equation is, however, cannot be applied for liquids (near incompressible).Extremely high bulk modulus of elasticity of liquids leads to considerably highspeed of sound.EThe speed of sound of a liquid can be determined by: c v . Bulk modulus of elasticity can be found from Table 1.6 (textbook).For (a) gasoline:1.3 109 N/m 2c 1,382.67 m/s 3,092.94 mph3680kg/mFor (b) mercury:2.85 1010 N/m 2c 1,447.61 m/s 3,238.21 mph431.36 10kg/mFor (c) seawater:2.34 109 N/m 2c 1,507.26 m/s 3,371.64 mph1.03 103 kg/m 3In case of air (at sea-level): c 760 mph

Page 6 of 12Unit A-2Vapor PressureVapor Pressure and Cavitation The pressure at which a change of phase (liquid gas) occurs is called vaporpressure Water boils at 212 F at 14.7 psi at sea-level Alternatively, vapor pressure is raised to 14.7 psi by increasing the temperatureto 212 F Water boiling occurs at a room temperature, if the pressure of the water isreduced to its vapor pressure. This is called cavitation Therefore, the property “vapor pressure” of a liquid is not a constant value:rather, it depends on the temperature.For example: the vapor pressure of water is 1.77 103 Pa (absolute) at 15.6 C.However, you can also alternatively say that the vapor pressure of water isexactly 1 atm (sea-level) at 100 C (this is called the “boiling”).

Page 7 of 12Unit A-2Surface TensionSurface Tension Each portion of the liquid surface exerts tension on adjacent portions of the solidsurface that are in contact. This is called surface tension Surface tension ( ) for a water-air interface is: 0.0734 N/mCapillary MotionLiquid in a glass tube: for wetting liquid (for example, water), the liquid will riseup in the glass tube (figure a). For non-wetting liquid (for example, Mercury), theliquid will be depressed (figure c)Applying the equation of equilibrium (figure b), 2F 0( ) Rh 2 R ( cos ) y 2 cos h ( 0 for water, 130 for Mercury) R

Page 8 of 12Unit A-2Class Example ProblemRelated Subjects . . . “Surface Tension”Due to the surface tension, a capillary action (raising or depressing of a column ofliquid within a tube) will occur. The height of capillary action can be determinedby:2 cos h RProperties of mercury can be found from Table 1.6 (textbook).Mercury is a “non-wetting” liquid ( 130 o ): therefore,2 ( 4.66 10 1 N/m ) cos130oh 3.003 10 3 m (3.003 mm)33 3(133 10 N/m )(1.5 10 m )Negative sign means the column is depressed.

Page 9 of 12Unit A-2Viscositydydudu dy : Dynamic orAbsoluteViscosityShear Stress in a Fluid Shear stress of a viscous fluid is proportional to rate of shear strain: dudydu/dy represent “how much velocity change” is expected in the directionperpendicular to the solid surface. The “velocity” can be interpreted as “howmuch deformation of a fluid” in the direction parallel to the solid surface (sheardirection) is happening (shear strain). The proportional constant ( ) is called dynamic (or absolute) viscosityViscosity Dynamic (or Absolute) viscosity ( ): units in N s/m2 or lb s/ft2 Kinematic viscosity ( ): (units in m2/s or ft2/s)

Page 10 of 12Newtonian andNon-Newtonian Fluids Unit A-2dudyNewtonian and Non-Newtonian Fluids Newtonian fluids: shear stress and rate of shear strain are directly proportional du(Newtonian fluid)dy Non-Newtonian fluids: shear stress and rate of shear strain are not directlyproportional: Bingham plastic, shear thinning / thickeningNon-Newtonian fluids shows non-linear behavior (relationship between shearstress and rate of shear strain cannot be described like Newtonian fluid). NonNewtonian fluids are not the main scope of this course: ES206 Fluid Mechanicsis a Newtonian Fluid Mechanics.Examples of Non-Newtonian fluids include:Shear thinningShear thickeningBingham plastic

Page 11 of 12Dynamic Viscosity ofCommon FluidsUnit A-2 De B / TCT 3 / 2 T SViscosity of Gases and Liquids Viscosity is a function of temperature Empirical equation of viscosity for gases: Southerland’s equation:CT 3 2 T S kg For air: C 1.458 10 6 and S 110.4 (K) m s K Empirical equation of viscosity for liquids: Andrade’s equation: De B T Empirical equations (often also called, “correlations”) are the equations obtainedby typically experimental data based “curve-fit,” thus the equation is not basedon theory (based on the “fact”).