KUNDU Fluid Mechanics 2 Episode 2 potx

10. Static fipilibiiutn o/a (,impwwiblr! .Wediurn where r = dT/dz is thc tcmperature gradient; ra = -g/C,, is called the udiuburic lernperu~r~rc gradient and is the largest ratc at which the temperature can decrease with height without causing instability. For air at normal temperatures and pressures, the temperaturc or a neutral atmospherc dccreases with height at thc rate of g/C, 21 10 “C/km. Meteorologists call vertical temperature gradients the “lapse ratc,” so that in their terminology thc adiabatic lapse rate is IO“C/km. Figurc 1.9a shows a typical distribution of temperature in the atmosphere. Thc lower part has been drawn with a slope nearly equal to the adiabatic temperature Fa- dient becausc the mixing processes ncar thc ground tend to form a ncutral atmosphere, with its entropy “well mixed’ (that is, unirorm) with height. Observations show that the neutral atmosphere is “capped” by a layer in which the tempcraturc increases with height, signifying avery stablc situation. Meteorologists call this an inversion, because the ternpcrature gradient changcs sign here. Much of the atmospheric turbulence and mixing processes cannot pcnctralc this very stable laycr. Above this inversion layer thc temperature decreases again, but less rapidly than ncar the ground, which corrcsponds to stability. It is clcm that an isothermal atmosphere (a vertjcal linc in Figure 1.9a) is quitc stable. 19 Potential Temperature and Density The foregoing discussion of static stability of a comprcssible atmosphere can be expressed in terms d the concept ofpotential remperutum, which is generally denotcd by 19. Suppose the prcssure and temperaturc of a fluid particle at sl certain height arc p and T. Now if we takc the particle udiuhuticully to a standard pressure ps (say, thc sea level pressurc, nearly equal to 100 kPa), then the ternpcrature 0 attained by the particle is callcd its pnfenriul temperature. Using Eq. (1.26), it follows that thc actual temperature T and the potential tcmperslture 0 arc rclalcd by z I z , slope = - lh/lO”C stable 51 very stable neutral 3 10°C 20°C Temperature T (ai - 1 Potential temperature 0 (b) (1.31) / * Figure 1.9 Vcrlical variation ol‘ he (a) actual and (b) polcnlial lemperature in the a~mosphere. ’Thin straight lincs represent tcmpcratures for a nculral atmosphcrc. Taking the logarithm and differentiating, we obtain 1dT 1de y-lldp - - + T dz 0dz y pdz' Substituting dpldz = -pg and p = pRT, we obtain (1.32) Now if the temperature decreases at a rate r = ra, then the potential temperature e (and therefore the cntropy) is uniform with hcight. It follows that the stability of the atmosphere is determined according to de - > 0 (stable), dz (1.33) d6, - =O (neutral), dz de - < 0 (unstable). dz This is shown in Figure 1.9b. It is the gradient ofpofentiul temperature that determines the stability of a column of gas, not the gradient of the actual temperature. However, the di€fe=nce between the two is negligible for laboratory-scale phenomena. For example, over a height of lOcm the compressibility effects result in a decrease of temperatureintheairby only lOcm x (lOcC/km) = 10-30C. lnstead of using the potential temperature, one can use the concept of potentid density p~, defined as the density attained by a fluid particle if taken isentropically to a standard pressure pa. Using Eq. (1.26), the actual and potential densities are related by (1.34) Multiplying Eqs. (1.31) and (1.34), and using p = pRT, wc obtain epo = p,/R = const. Taking the logarithm and differentiating, wc obtain (1.35) The mcdium is stable, neutral, orunstable depcnding upon whctherdp#/dz is ncgative, zero, or positive, rcspectively. Compressibility effects are also important in the deep ocean. In the ocean thc density depcnds not only on the temperature and prcssure, but also on the salinity, defined as kilograms of salt per kilogram of water. (The salinity of sea water is ~3%) Here, one defines the potential density as the density attained if a particle is Laken to a reference pressure isentropically mid at constant salinity. The potential density thus defined must decrcase with height in stable conditions. Oceanographers automatically account for the compressibility of sea water by converting their density measurements at any depth to the sea lcvcl pressure, which serves as the reference pressure. From (1.32), the temperature al a dry neutrally stable atmosphere decreases upward at a ratc dT,/dz = -g/C,, due to the decrease of pressure with height and the comprcssibility ol the medium. Static stability of the atmosphcrc is dctcrmincd by whcther the actual temperature gradient dT/dz is slowcr or faster than dTa/dz. To determine the static stability of the ocean, it is more convcnicnt to formulale the criterion in tcrms ol density. The plan is to compare the density gradient of the actual static state with that of a neutrally stable reference state (denoted here by the subscript “a”). The pxssure or the reference state decreases vertically as - -/Jag- dP, dz (1.36) Tn the occan he speed of sound cis &fined by c2 = ap/ap, where the partial derivative is taken at consmt values of entropy and salinity. Tn the reference state these variables arc uniform, so that dpa = c2dpa. Therefon, the density in the neutrally stable state varies due to thc compressibility effect at a rate (1.37) where the subscript “a” on p has been dropped because pa is ncarly equal to the actual density p. The static stability of thc ocean is determined by the sign of the potenrial densirj gradient (1 38) &pol d~ dpa d~ ~g - + dz dz dz dz c2 The medium is statically stable if the potcntial density gradient is ncgative, and so on. For a perfect gas, it can be shown that Eqs. (1 30) and (1.38) are equivalent. - Scale Height of the Atmosphere Expressions for pressure distribution and “thickness” of the atmosphere can be obtained by assuming that they are isothermal. This is a good assumption in the lower 70 km of the atmospherc, where the absolutc tcrnperature remains within 15% of 250 K. The hydrostatic distribution is PR _- - -pg = dz RT dP Integration givcs where po is the pressurc at z = 0. The pressure therefore falls to e-‘ of its surface value in a height RT/g. Thc quantity RTIg, called the scale height, is a good measure of the thickness of the atmosphere. For an average atmospheric tcmperature of T = 250 K, the scale height is RTIg = 7.3km. = e-RzlRT fhIVkCS 1. Estimate the height to which water at 20°C will rise in a capillary glass tube 3mm in diameter exposed to the atmosphcre. For water in contact with glass the wetting angle is nearly 90’. At 20 “C and water-air combination, d = 0.073 N/m. (Answer: h = 0.99cm.) 2. Consider the viscous flow in a channel of width 2h. The channel is aligned in the n direction, and the velocity at a distance y from the centerline is given by the parabolic distribution u(y) = uo [ 1 - $1. In terms of the viscosity p, calculatc the shear stress at a distance of y = h/2. 3. Figure 1-10 shows ammameter, which is a U-shaped tube containing mercury of density p,,,. Manometers arc used as pressurc measuring dcvices. If the fluid in the tank A has a pressure p and density p, then show that the gauge pressure in the tank is Note that the last term on the right-hand side is negligible if p << pm. (Hint: Equate thc pressures at X and Y .) 4. A cylinder contains 2kg of air at 50°C and a pressure of 3 bars. The air is compressed until its pressure rises to 8 bars. mat is the initial volume? Find the final volume for both isothermal compression and isentropic compression. 5. Assume that the temperature of the atmosphere varies with hcight z as Show that the prcssure varies with height as P = PO [ where g is gravity and R is the gas constant. Figure 1.10 A mercury msnomcbr. 6. Suppose the atmospheric temperature varies according to T = 15 - 0.001~ where T is in degrees Cclsius and height z is in mcters. Is lhis atmosphere stable? 7. Provc that if e(T, u) = e(T) only and if h(T, p) = h(T) only, then the (thermal) equation of state is Eq. (1.2 1 ) or pv = kT. 8. For a reversible adiabatic process in a perfect gas with constant specific heats, dcrive Eqs. (1.25) and (1.26) startingfrom Eq. (1.18). 9. Considcr a heat insulatcd enclosure kat is separated into two compartments of volumes VI and VZ. containing perfect gases with pressurcs and temperatures of pl, pz, and Ti, Tz, respectively. The compartments are separated by an imperme- able membrane that conducts heat (but not mass). Calculate the final steady-state tcmperature assuming each of the gases has constant specific heats. 10. Consider the initial state of an enclosure with two compartments as dcscribed in Exercise 9. At t = 0, the membrane is broken and the gases are mixed. Calculate the final tcmperature. 1 1. A heavy piston of weight W is dropped onto a thcrmally insulated cylinder of cross-sectional area A containing a perfect gas of constant specific heats, and initially having thc cxternal pressure p1, temperature q, and volume VI. After some oscillations, the piston reaches an equilibrium position L meters below the equilibrium position of a weighlless piston. Find L. Is thcre an entropy increase? Llilr?ratuurc Wed Taylor, G. I. (1974). Thc intcrwtion between experimcnl md theory in fluid mwhtmics. Annuul Review o/ Von Ktirmm, T. (1954). Aerodynamics, New York: McGraw-Hill. l;luia Mechanics 6 1-16. Supplcmenlal Reading Batchelor, G. K. (1967). “An fnfdf~tion IO Fluid Dynumics,” landon: Cambridge University Prclnln, (A detailed discussion ol classical thermodynamics: kinctic theory of gases, surfacc tcnsion efiects, and transport phcnomena is given.) Hu~rcipoulos, G. N. and J. H. Kccnan (1981). Principles of Geneml Thr?nnoCtyrrumics. Mclhoumc, FL: Kricgcr Publishing Co. (This is a god text on thermodynamics.) Prandtl, L. and 0. G. Tictjcns (1934). k~undainenrals uJHydn,- and Asmmechanics, Ncw York Dover Publications. (A clear and simplc discussion of potential and adiibalic temperature gradients is given.) Chapkr 2 Cartesian Tensors 1. 2. 3. 4. 5. a 7. 8. 9. IO. ScalarsandVwhm 24 11. Holalion of Axes: hmd UefiRilion oJnV~:tw 25 12. :Wult@diaihri of :Wairices 28 Conhuhn (mtiMull+licdion. 3 1 fime on (1 Surf;. 32 Exnmple2.1 34 14. .Second- O&r %ww. 29 13. Kn,rrder Jkh and Allermdng Ii?mr 35 15. IM Ihriuct 36 16. O~nmittir V: (:mclienl, Uicetpnce, wid Ciul 37 (,'NhW hhhct. 36 .SytnmelniC and An&ynm.lrie nmom 38 l?i~ywulwx mi iT&ruxxtom of (1 ,Yymmc!ttik Tm.sor 40 Kxamplc 2.2 40 Gauss' 'l'heorm 42 Examplc 2.3. 43 %hx ' Thmm 45 hrripk 2.4 46 Comma :Valuliorr 46 Ilol&ce m Indicia1 !Voia.iiori. 41 I-kerckw 47 Idikrahn (lid 49 supplemenla1 Headkg 40 1. Scalam and Vichm In fluid mechanics we need to deal with quantities of various complexities. Some of these are defined by only one component and are called scalars, some others are defined by three components and are called vectors, and certain other variables called tensors need as many as nine components for a complete description. We shall assume that the reader is familiar with a certain amount of algebra and calculus of vectors. The concept and manipulation of tensors is the subject or this chapter. A scakur is any quantity that is completely specified by a magnitude only, along with its unit. It is independent of the coordinate system. Examples cd scalars are temperature and density of the fluid. A vector is any quantity that has a magnitude and a direction, and can be completely described by its components dong threc specified coordinate directions. A vector is usually denoted by a boldlace symbol, for example, x for position and u for velocity. We can take a Cartesian coordinatc systemxl , x2, ~3, with unit vectors al, a2, and a3 in the three mutually perpendicular directions (Figurc 2.1). (In texts on vector analysis, the uni t vectors are usually denoted 24 2. Ibtatiwi 0JArn.v: finmu1 Ihfiiilion oJa Vitar 25 1 Figure 2.1 a’, a?, and a3. by i. j, and k. We cannot use this simple notation here because we shall use ijk to denote coinponenls of a vector,) Then the position vector is writtcn as x = alx, + 8% + a3x3, where (XI, x2, x3) are the components of x along the coordinate directions. (The superscripts on the unit vectors a do not denote the components of a vector; the a’s are vectors themselves.) Instead of writing all three components explicitly, we can indicate the threc Cartesian components of a vector by an index ha1 lakes all possible values of 1, 2, and 3. For example, the components of the posilion vcctor can bc dcnoted by xi. where i takes all of its possible values, namely, 1,2, and 3. To obcy the laws of algcbra that wc shall present. the components of a vector should be writlen as a column. For example, Position vcctor OP and its three Cartesian componcnts (XI. xz, x:j). The three unit vectors arc X= [::I. x3 In matrix algebra, one defines the trculspose as the matrix obtained by interchanging rows and columns. For cxample, the transposc of a column matrix x is the row matrix xT = [XI ~2 x~J. 2. .Itotution ofAmx fiwmul llcfiitiim (fa ktor A vector can be formally defined as any quantity whose components change similarly to the components of a position vector under thc rotation of thc coordinate system. Let x1 x2 x3 be the original axes, and xi xi xi be thc rotatcd system (Figure 2.2). The 3 3’ t 1 1’ Figure 2.2 Rotation of coordinate system 0 I 2 3 10 0 1‘ 2’ 3’. components of the position vector x in the original and rotated systems are denoted by xi and xf, respectively. The cosine of the angle between the old i and new j axes is represented by Cij. Here, theJirst indcx of the C matrix refers to thc old axes, and the second index of C refers to the new axes. It is apparent that Cij # Cji. A little geomelry shows that the components in the rotated system are related to the components in the original system by For simplicity, we shall verify the validity of Eq. (2.1) in two dimensions only. Referring to Figure 2.3, let aij be the angle between old i and new j axes, so that Cij = cosq. Then As a1 1 = 90” - ~21, we havc sin a1 1 = cos a21 = CZI . Equation (2.2) then becomes 2 In a similar manner i 1' 0 &_I A Figurc 2.3 Roldlion ora coordinate system in two dimensions. As ull = a22 = - 90 (Figure 2.3), this becomes 7- x; =x2cosa~2 +XI cosa12 = Cx;Ci,. (2.4) In two dimensions, Eq. (2.1) reduces to Eq. (2.3) for j = 1, and to Eq. (2.4) for j = 2. This completes our verification d Eq. (2.1 ). Note that thc indcx i appcars twicc in the samc term on the right-hand side of Eq. (2.1), and a summation is carricd out over all valucs of this rcpcated index. This type of summation over repeated indiccs appcars frequently in tensor notation. A convention is thcrcforc doptcd that, whenever an index occurs twice in Q term, Q smation over the repeated index is implied, although no summation sign is explicitly writreen. This is frequently called the Einstein summation convention. Equation (2.1) is then simply written as x'. = x; c. IJ ' (2.5 j where a summation over i is understood on the right-hand sidc. The free index on both sides of Eq. (2.5) is j, and i is the rcpeated or dummy index. Obviously any letter (other than j) can be used as the dummy index with- out changing thc mcsuzing of this equation. For example, Eq. (2.5) can be written equivalently as because they all mean x,; = Cljxl + C2jx2 -F C3jx). Likewise, any letter can also be used for thc frcc index, as long as the same free index is used on both sides of thc cquation. For example, denoting the free indcx by i and the summed index by k, Eq. (2.5) can be written as (2.6) This is bccausc: the set of three equations reprcsented by Eq. (2.5) corresponding to all values of j is the same set of equations represented by Eq. (2.6) for all valucs o€i. ;=I J XiCij =xkckj =xmcmj xi = xk ck; . It is ea3y to show that the components of x in thc old coordinate system are related to those in the rotated system by xj = cjjx;. (2.7) Note that the indicia1 positions on the right-hand sidc of this relation are dilferent rrom those in Eq. (2.3, because the first index of C is summed in Eq. (2.5), whereas the second index of C is summed in Eq. (2.7). We can now formally define acartesian vector as any quantity that transforms like a position vector under the rotation of the coordinate system. Therefore, by analogy with Eq. (2.5), u is a vcctor if its components transform as I I 3. Mu1lC;nliculion of:?Iu&ices In this chapter we shall gcnerally follow the convention that 3 x 3 matriccs are repre- sented by uppercase lettcrs, and column vectors arc represented by lowcxase letters. (An exception will be the usc of lowercase t far thc stress matrix.) Le1 A and R be two 3 x 3 matrices. The product of A and R is defined as the matrix P whosc clements are related to those of A and R by 3 k=l or, using the summation convention Symbolically, this is written as P=A-B. (2. IO) A single dot between A and B is included in Eq. (2.10) to signify that a single index is summed on thc right-hand side of Eq. (2.9). The important thing to note in Eq. (2.9) is that the elements are summed over the inner or adjacent index k. It is sometimcs useful to writc Eq. (2.9) as p -A. B . - IJ - Ik kJ - (A R)ij, where thc last term is to be read as the "ij-clement of thc product of matrices A and B." In explicit form, Eq. (2.9) is written as [...]... double sum in Fq (2. 21) result from i = 2, j = 3, and from i = 3, j = 2 This follows from the definition equation (2. 18) that the permutation symbolis zero if any two indices are equal Then JZq (2. 21) gives (u x v)] = & i j ] u i v j = &23 1U2t'3 +c 321 u3v2 = u 2 v 3 - u3v2, 10 Qterulor 31 V Ciwdiwit, Ilirewince, mid Curl : which agrees with @ (2. 20) Note that h e sccond form oTEq (2. 2 1 ) is obtaincd... in the rotated frame are ;I =CllC21t 12 ti2= C l l C B t 1 2 + + C2ICllt21 4 = T 3Z1 U ~21 ~12t21 = $4 +I& T l U -43 -T U - IIa 22 = la 2 , Thc normal stress is thenfore &u /2, and h e shear stress is a /2 This gives a magnitude u and a direction 6 0 or 24 0' depending on the sign of a 0 7 Kn,nc?cker Della und Altkrnahg 7kmor Thc Kronecker delta is de6ncd as if i = j 1 6 - (2. 16) which is written in the matrix... Aii All A 12 A 22 A23 All AI3 = A21 A 22 + A 32 A33 + A31 A33 13 = det(Aij) Z1 I 1I II I [Hint: the result of Exercise 4 and the transformation rule (2. 12) to show that Use 11' - A'i i - Aii = f 1 Then show that AijAji and AijAjkAki are also in~ariant~ In fact, ull contracted scalars of the form Aij Ajk Ami are invariants Finally, verify that - f2 = 1 [ 1 2 - A A 'I Z3 = AijA,jkAki - IiAijAji 2 1 IJ... of(.hd o y . .;I = CllC21t 12 + C2ICllt21 = TZU + TlU - TU, ti2 = CllCBt 12 + ~21 ~12t21 = $4. - IIa = la. Thc normal stress is thenfore &u /2, and he shear stress is a /2. This gives. 90 (Figure 2. 3), this becomes 7- x; =x2cosa ~2 +XI cosa 12 = Cx;Ci,. (2. 4) In two dimensions, Eq. (2. 1) reduces to Eq. (2. 3) for j = 1, and to Eq. (2. 4) for j = 2. This completes. equation (2. 18) that the permutation symbol is zero if any two indices are equal. Then JZq. (2. 21) gives (u x v)] = &ij]uivj = & ;23 1U2t'3 + c 321 u3v2 = u2v3 - u3v2, 10.