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FUNDAMENTALS of FLUID MECHANICS Chapter 11 Analysis of Compressible Flow

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FUNDAMENTALS of FLUID MECHANICS Chapter 11 Analysis of Compressible Flow FUNDAMENTALSFUNDAMENTALS OFOF FLUIDFLUID MECHANICSMECHANICS ChapterChapter 1111 AnalysisAnalysis ofof CompressibleCompressible FlowFlow JyhJyh--CherngCherng ShiehShieh Department of Bio-Industrial Mechatronics Engineering National Taiwan University 1 MAINMAIN TOPICSTOPICS IdealIdeal GasGas RelationshipsRelationships MachMach NumberNumber andand SpeedSpeed ofof SoundSound CategoriesCategories ofof CompressibleCompressible FlowFlow IsentropicIsentropic FlowFlow ofof anan IdealIdeal GasGas NonisentropicNonisentropic FlowFlow ofof anan IdealIdeal GasGas TwoTwo--DimensionalDimensional CompressibleCompressible FlowFlow 2 IdealIdeal GasGas RelationshipsRelationships 3 IntroductionIntroduction Fluid compressibility is a very important consideration in numerous engineering applications of fluid mechanics. For example, ►►TheThe measurementmeasurement ofof highhigh--speedspeed flowflow velocitiesvelocities requiresrequires compressiblecompressible flowflow theory.theory. ►►TheThe flowsflows inin gasgas turbineturbine engineengine componentscomponents areare generallygenerally compressible.compressible. ►►ManyMany aircraftaircraft flyfly fastfast enoughenough toto involveinvolve aa compressiblecompressible flowflow field.field. In this study of compressibility effects, we mainly consider the steady, one-dimensional, constant (including zero) viscosity, compressible flow of an ideal gas. 4 IdealIdeal GasGas relationshipsrelationships 1/21/2 Before to develop compressible flow equation, we need to become more familiar with the fluid. The equation of state for an ideal gas p = ρRT (1) R is gas constant The gas constant, R, represents a constant for each distinct ideal gas or mixture of ideal gases λ R = (2) λis the universal gas constant. M is the molecular weight of the ideal gas or gas mixture. Mgas gas 5 IdealIdeal GasGas relationshipsrelationships 2/22/2 For an ideal gas, internal energy is a function of temperature only. Thus, the ideal gas specific heat at constant volume ⎛ ∂u( ⎞ du( cv = ⎜ ⎟ = (3) is function of temperature only ⎝ ∂T ⎠v dT Where the subscript v on the partial derivative refers to differentiation at constant specific volume, v=1/ρ. ( ( ( (5) du = cvdT ⇒ u2 − u1 = cv (T2 − T1 ) (5) ( ( T 2 u2 − u1 = cvdT (4) ∫T1 6 EnthalpyEnthalpy 1/21/2 The fluid property enthalpy ( p h = u( + (6) ρ For ideal gas u( = u((Ｔ) p From the equation of state = RT ρ ( ( ( h = u( + RT h = h(Ｔ) The ideal gas specific heat at constant pressure ( ( ⎛ ∂h ⎞ dh ( ( (8) c = ⎜ ⎟ = (7) h2 − h1 = cPdT P ⎜ ∂T ⎟ dT ⎝ ⎠p ( ( h2 − h1 = cP (T2 − T1) (9) 7 EnthalpyEnthalpy 2/22/2 ( h = u( + RT (10) ( ( ( dh du( ⎛ ∂h ⎞ dh ⎛ ∂u( ⎞ du( ( ( ⎜ ⎟ c = ⎜ ⎟ = ⇒ dh = du + RdT ⇒ = + R (11) cP = ⎜ ⎟ = v ∂T dT ⎝ ∂T ⎠v dT dT dT ⎝ ⎠p (12) cp − cv = R The specific heat ratio, k, is defined as c k = P (13) cv k 1 (12) + (13) cP = R (14) cV = R (15) k −1 k −1 8 ExampleExample 11.111.1 InternalInternal Energy,Energy, Enthalpy,Enthalpy, andand DensityDensity forfor anan IdealIdeal GasGas z Air flows steadily between two sections in a long straight portion of 4-in.-diameter pipe as is indicated in Figure E11.1. The uniformly distributed temperature and pressure at each section are T1=540°R, p1=100 psia, and T2=453°R, p2=18.4 psia. Calculate the (a) change in internal energy between sections (1) and (2), (b) change in enthalpy between sections (1) and (2), and (c) change in density between sections (1) and (2). 9 ExampleExample 11.111.1 SolutionSolution1/21/2 The change in internal energy between sections (1) and (2) ( ( u2 − u1 = cv (T2 − T1) 1 c = R (15) V k −1 From Table 1.7 k = 1.4 R = 1716(ft ⋅lb)/(slug ⋅°R) = 53.3(ft ⋅lb)/(lbm ⋅°R) ( ( u2 − u1 = cv ()T2 − T1 = 133(ft ⋅lb)/(lbm⋅°R) ×(453°R − 540°R) = −11,600ft ⋅lb/lbm ( ( cP h2 − h1 = cP (T2 − T1) (9) k = (13) cv 10 ExampleExample 11.111.1 SolutionSolution2/22/2 cP = kcv = ... = 186(ft ⋅lb)/(lbm ⋅°R) ( ( h2 − h1 = cP (T2 − T1) = ... = −16,200ft ⋅ lb/ lbm p = ρRT (1) p p 1 ⎛ p p ⎞ 2 1 ⎜ 2 1 ⎟ ⇒ ρ2 − ρ1 = − = ⎜ − ⎟ RT2 RT1 R ⎝ T2 T1 ⎠ = ... = −0.389lbm / ft3 11 EntropyEntropy 1/31/3 For any pure substance including ideal gases, the “first T ds equation” is ( ⎛ 1 ⎞ Tds = du + pd⎜ ⎟ (16) T is absolute temperature ⎝ ρ ⎠ ( ( ⎛ 1 ⎞ ⎛ 1 ⎞ dh = du + pd⎜ ⎟ + ⎜ ⎟dp (17) ⎝ ρ ⎠ ⎝ ρ ⎠ ( ⎛ 1 ⎞ (18) “second T ds equation” (16)+(17) Tds = dh − ⎜ ⎟dp ⎝ ρ ⎠ dT R ⎛ 1 ⎞ (1)+(3)+(16) ds = cv + d⎜ ⎟ (19) T yρ ⎝ ρ ⎠ dT dp (1)+(7)+(18) ds = cP − R (20) T P 12 EntropyEntropy 2/32/3 If cp and cv are assumed to be constant for a given gas T2 ρ1 (19)Ö s2 − s1 = cv ln + R ln (21) T1 ρ2 T2 p2 (20)Ö s2 − s1 = cP ln − R ln (22) T1 p1 13 EntropyEntropy 3/33/3 For the adiabatic and frictionless flow ds = 0, s1 = s2 T2 ρ1 T2 p2 Constant entropy flow ⇒ cv ln + R ln = cP ln − R ln = 0 (23) or isentropic flow T1 ρ2 T1 p1 k / k−1 k ⎛ T2 ⎞ ⎛ p2 ⎞ ⎛ ρ2 ⎞ (24) ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟ ⎝ T1 ⎠ ⎝ p1 ⎠ ⎝ ρ1 ⎠ p = cons tan t (25) ρk 14 ExampleExample 11.211.2 EntropyEntropy forfor anan IdealIdeal GasGas z For the air flow of Example 11.1, calculate the change in entropy, s2-s1, between sections (1) and (2). 15 ExampleExample 11.211.2 SolutionSolution T2 ρ1 T2 p2 s2 − s1 = cv ln + R ln (21) s2 − s1 = cP ln − R ln (22) T1 ρ2 T1 p1 ρ p T 100psia 453°R 1 = 1 2 = = 4.56 ρ2 T1 p2 540°R 18.4psia T ⎛ p T ⎞ 2 ⎜ 1 2 ⎟ s2 − s1 = cv ln + R ln⎜ ⎟ = ... = 57.5(ft ⋅lb)/(lbm ⋅°R) T1 ⎝ T1 p2 ⎠ T2 p2 s2 − s1 = cP ln − R ln = ... = 57.5(ft ⋅lb)/(lbm ⋅°R) T1 p1 16 MachMach NumberNumber andand SonicSonic SpeedSpeed 17 MachMach NumberNumber The Mach number, Ma, was a dimensionless measure of compressibility in a fluid flow. The Mach number is defined as the ratio of the value of the local flow velocity, V, to the local speed of sound, c. V Ma = c What we perceive as sound generally consists of weak pressure pulse that move through air. When our ear drums respond to a succession of moving pulse, we hear sounds. 18 SpeedSpeed ofof SoundSound 1/61/6 To better understand the notion of speed of sound, we analyze the one-dimensional fluid mechanics of an infinitesimally thin, weak pressure pulse moving at the speed of sound through a fluid at rest. Select an infinitesimally thin control volume that move with the pressure pulse. The speed of the weak Ahead of the pressure pulse is pressure pulse considered constant and in one direction only; Behind the thus, out control volume pressure pulse is inertial. 19 SpeedSpeed ofof SoundSound 2/62/6 For an observer moving with this control volume, it appears as if fluid is entering the control volume through surface area A with speed c at pressure p and densityρand leaving the control volume with speed…. Entering CV Leaving CV 20 SpeedSpeed ofof SoundSound 3/63/6 Apply the continuity equation to the flow through this control volume ρAc = (ρ + δρ)A(c − δV) (26) small ρc = ρc − ρδV − (δρ)(δV) (27) ρδV = cδρ (28) 21 SpeedSpeed ofof SoundSound 4/64/6 Apply the linear momentum equation to the flow through this control volume − cρcA + (c − δV)(ρ + δρ)(c − δV)A = ρA − (p + δp)A (29) − cρcA + (c − δV)ρAc = −δpA (26)+(29) δp (30) ρδV = c δp δp (28)+(30) c2 = c = (31) δρ δρ This expression for the speed of sound results from application of the conservation of mass and conservation of linear momentum principles to the flow through the control volume. 22 SpeedSpeed ofof SoundSound 5/65/6 Apply the conservation of energy principle to the flow through this control volume 2 δp ⎛ V ⎞ 2 (32) (δV) << cδV + δ⎜ ⎟ + gδz = δ(loss) (32) ρ ⎝ 2 ⎠ δp (c − δV)2 c2 δp gδz≒0 δ(loss)= 0 + − = 0 ρδV = (33) ρ 2 2 c 2 δp δp (28)+(33) c = c = (31) δρ δρ This expression for the speed of sound results from application of the conservation of mass and conservation of energy principles to the flow through the control volume. 23 SpeedSpeed ofof SoundSound 6/66/6 Further assume that the frictionless flow through the control volume is adiabatic, δp then the flow is isentropic. ⎛ ∂p ⎞ c = c = ⎜ ⎟ (34) δρ δp → ∂p → 0 ⎝ ∂ρ ⎠s For isentropic flow of an ideal gas Used to designate that partial differentiation p = (constan t)(ρk ) (25) occurs at constant entropy. ⎛ ∂p ⎞ p p ⇒ ⎜ ⎟ = (constan t)kρk−1 = kρk−1 = k = RTk (35) ⎜ ⎟ k ⎝ ∂ρ ⎠s ρ ρ E c = RTk (36) c = RTk = v (38) ρ dp ⎛ ∂p ⎞ Bulk modulus of elasticity for any fluid Ev = = ρ⎜ ⎟ (37) dρ /ρ ⎝ ∂ρ ⎠s 24 ExampleExample 11.311.3 SpeedSpeed ofof SoundSound // SolutionSolution z Verify the speed of sound for air at 0℃ listed in Table B.4. In Table B.4, we find the speed of sound of air at 0℃ given as 331.4 m/s. c = RTk k = 1.4 R = 286.9J /(kg ⋅ K) c = RTk = [286.9J /(kg ⋅ K)](273.15K)(1.401) = 331.4(J / kg)1/ 2 = 331.4m /s 25 CategoriesCategories ofof CompressibleCompressible FlowFlow 26 CompressibilityCompressibility andand MachMach NumberNumber 1/21/2 The effects of compressibility become more significant as the Mach number increase. The incompressible flows can only occur at low Mach number. The compressibility has a large influence on other important flow variables. 27 CompressibilityCompressibility andand MachMach NumberNumber 2/22/2 Experience has demonstrated that compressibility can have a large influence on other important flow variables. For example. The variation of the drag coefficient of a sphere with Reynolds number and Mach number. Ö D Compressibility effects can be of considerable importance. 28 ToTo illustrateillustrate somesome curiouscurious featuresfeatures ofof compressiblecompressible flowflow EmissionEmission ofof weakweak pressurepressure plusespluses fromfrom aa pointpoint sourcesource 29 EmissionEmission ofof PressurePressure PulsePulse 1/51/5 Image the emission of weak pressure pulse from a point source.
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