R. von Mises - Mathematical theory of compressible fluid flow, страница 4

Only theassumption that the fluid is inviscid must be retained since it was used inthe derivation of (1.1). T h e effect of viscosity on the energy equation willbe discussed in A r t . 3. I t is to be emphasized that the energy equation is amathematical consequence of (1.1) and ( l . I I ) and, therefore, does not imposeany restriction upon the system of functions defined by (1.1) and ( l . I I ) ;in particular, the energy equation, as such, cannot serve in place of thespecifying equation ( l .

I I I ) . If, however, the energy balance is subject tosome condition, which is equivalent to a specifying equation, one can usethe energy equation to express this condition in the form of a specifyingequation.Taking the scalar product with q on the two sides of the vector equation(1.1) we obtain the scalar equation(1)pq.^5 = pq-g -q-grad p.Each term will now be suitably transformed.12I. I N T R O D U C T I O N(a) From the fact that 2q*dq/dt is the derivative of q « q = q followspq.^ = p ldtdt(2)WP4^9( b ) From the geometrical definition of a scalar product, the value ofq»g is qg cos 0, where 0 is the angle between the directions of q and g (seeFig. 3 ) .

If h denotes the function of x, y, ζ giving the height of each point Ρabove some arbitrary horizontal plane of reference, then the height of Ρ ' ,reached after displacement from Ρ by an amount ds in the direction of q,is h + dh = h + (dh/ds) ds. Hence cos θ =—dh/ds from Fig. 3. Since h isa function of position only, dh/dt = 0, and the Euler rule of differentiation,(1.4) gives dh/dt = q dh/ds.

Thusdhto\(3)pq-g=- WTsdh=-f>9jd- P=t, ν(JtW .(c) From the analytic definition of a scalar product, we have,dp.dp.dpT h e first term on the right may be written asdpd ,νdq5 χ τ - = - (pq ) — Ρ -τ—,xxand similarly for the other two terms, so that(4)q-grad ρ = div (pq) -ρ div q = div (pq) + - ~ ,p atwhere the last term is obtained by substituting for div q from the equationof continuity ( 1 . Ι Γ ) ·2.2E N E R G Y EQUATION FOR PERFECTGAS13When ( 2 ) , (3), and (4) are substituted in ( 1 ) , the result isor, setting(5)= div( q) =W^dxP+«dy+«dz,rearranging terms and dividing by p,/Vι Λι&\2 \) ϊd+β+w=V dp~?dt=Vd (\\dt\- )pThis relation is valid for any functions p, p, and q of the variables x, y, z tsatisfying Eqs. (1.1) and ( l .

I I ) .y2 . The energy equation for an element of an inviscid perfect gasPhysical interpretation of the term w will show that (6) may rightlybe considered as a preliminary form of the energy equation for a fluidelement. Consider again an infinitesimal portion of the moving fluid in­cluded in a rectangular cell with sides of length dx, dy, dz (see Fig. 1). T h eforce due to pressure on the left-hand face has the magnitude ρ dy dz andthe direction of the positive x-axis, so that the work done per unit time bythis force is q p dy dz. On the opposite face the pressure has the directionof the negative rr-axis, so that the work done against pressure in unit timeis given by [q p + d{q p)]dy dz, where d(q p)= [d(q p)/dx] dx. Thus[d(q p)/dx] dx dy dz is an expression for the work done against pressure onthis pair of faces.

Adding the analogous expressions for the other two pairsof faces, the total work done per unit time against pressure is given bydiv (pq) dx dy dz. Thus w = div (pq) represents the work per unit volume,and w/p the work per unit mass, both referred to unit time, against pressureon the surface of an element of fluid.xxxxxxSince q/2 is the kinetic energy and gh the potential energy (except foran additive constant) per unit mass, the left-hand member of Eq. (6) maybe interpreted as the change per unit time of the sum of the kinetic andpotential energies, plus the work done per unit time against the pressure onthe surface of the fluid element, all per unit mass. T h e right-hand memberof (6) occurs in Eq. (1.8), which expresses the First L a w of Thermody­namics, from which pd(l/p)/dt= Q — c dT/dt.

Thus Eq. (6) may bewrittenf(7)vI.14INTRODUCTIONwhere it has been assumed that c is a constant, which is true for a perfectgas; the case of a variable c will be discussed in Sec. 3. For a perfect gas,the product c T is called the (specific) internal energy. T h e content of E q .(7) may then be expressed by this statement: For an inviscid perfect gas,the heat input per unit time, Q'> is equal to the time rale of change of the totalenergy {kinetic, potential, and internal) plus the work done per unit time atthe surface of the element against pressure) all quantities are expressed perunit mass.vvvUnder the assumption that the flow is strictly adiabatic, Q' = 0, theleft-hand member of (7) must vanish. I n this case (7) serves to expressthe condition Q = 0 in terms of mechanical variables [using (1.9')].

T h ereader may verify that equating the left-hand side of (7) to zero leads backto (1.11), which was given as the specifying equation for strictly adiabaticflow.fAnother expression for Q' has been obtained in (1.12), and (7) then be­comes£(£»*+* )r++7 - ' a -Either of the equations (7) or (8) is the classical energy equation for a fluidelement of a perfect gas. In Eq. (8) three thermodynamic quantities occur:T, S, c . If these are replaced by their values from the equation of state(1.6) and the definitions (1.7) and (1.9), respectively, (8) must reduceidentically to ( 6 ) .

Various alternative forms of the energy equation can begiven if thermodynamic variables other than c , T, and S are used. Forexample, if the specific heat at constant pressure, c , is introduced (for aperfect gas) by c = c + gR* Eq. (1.8) with Τ = p/pgR becomesvvppO' =vvL(e\gR dt \pj4- «^p1(i\ = _L (cdt \p)pgR \dt*R - 25? ^ \ρ dt J'This expression may replace the right-hand member of Eq. ( 8 ) .3. Nonperfect (inviscid) gasEquations (7) and (8) have been derived under the hypotheses thatthe equation of state is (1.6) and that S and c are given by (1.7) and (1.9).Slightly generalized forms of Eqs. (7) and (8) hold also in the case of a nonperfect gas.

I t is assumed that some equation of state holds, in the formv(9)Τ = T(p,).PT h e nature of the gas is determined by the form of the function T(p, p)and that of two other functions: the entropy S(p, p) and the internal energy* I t then follows from (1.9) that y = c /c , the ratio of the specific heats.pv2.3N O N P E R F E C T(INVISCID)15GASU(p, p ) . These two functions, however, cannot be chosen independently ofT(p, p) and of each other. I t is required that for a nonperfect gas, too,E q . (1.12) is valid—which means that the heat input Q' per unit time andmass equals Τ dS/dt—and that the First L a w of Thermodynamics holdsfor any change in ρ and ρ in the formcorresponding to (1.8).

SincedU=dt "dUdp,dUdpdp dtdSdp dtadtndSdp=dSdp^~ dp dtdp dt 'the relation (10) can be correct for arbitrary dp/dt and dp/dt only if(11)τψ-ψdpτψ-ψ-Έ..anddpdpdpdpp/pzEliminating U, we obtaindp \dp)dp \2This leads to the condition(12)dQS, DdSdT _θ ( ρ , ρ) ~ dp dp=dSdTdp dp=l_p2>or(12')//χ = 1.I t can easily be verified that (12) is satisfied by the expressions for Τ andS given in Eqs. (1.6) and (1.7) for a perfect gas, while the correspondingvalue of U is found to be, up to an additive constant(13)U = -1Vlp'and this equals c T by (1.9').For a given equation of state ( 9 ) , there exists an infinity of functionsS(p, p) satisfying Eq.

(12). When one of these has been chosen, the internalenergy U(p, p) is determined, except for an additive constant, by (11).T h e right-hand side of (6) is then to be replaced from (10) b y Q' — dU/dtv16or T(dS/dt)I.— dU/dt, and the resulting equation differs from (7) or ( )only in having U in place of(.4)INTRODUCTION$.(ί +c T:v+ υ+) ϊ.<,.τ«+T h e condition of adiabatic flow is again dS/dt = 0, and this becomesa (p, p)-relation is the value of S is supposed to be the same for all particles.For example, if the equation of state is taken asT =AlV+ ^ ,pwhere Ai and Bi are constants, then an admissible choice for S isS = Ap + -withABX-ABi= 1,Ρand the corresponding expression for U isU = ^V + AB, ? + ψ±2+constant,as can easily be verified by differentiation.4.

Energy equation for an elastic fluidIf the specifying equation is a (p, p)-relation, the energy equation maybe given another form. Since only a single fluid element is under considera­tion, it is not necessary to know that a (p, p)-relation holds for the entirefluid; it is sufficient to assume that such a relation holds for the particle inquestion, i.e., that for any two states of the particle with the same valuefor ρ the value of ρ is also the same.Since ρ is a function of p, the right-hand member of E q . (6) can be ex­pressed entirely in terms of p.

Let e be a function of ρ (or of p) whose de­rivative with respect to ρ is p/p :2(15)e =fj dp.VThend /l\de _ ρ dp _dt ~ ~p dt ~2~Vso that Eq. (6) can be written in the formdt \ p / '2.517BERNOULLI EQUATIONThe quantity e may be called the expansion energy or strain energy] thenEq. (16) states: The time rate of change of the sum of the kinetic,potential,and strain energies is equal to the work done by the pressure forces on the surfaceper unit time and mass. I t is to be noted that the strain energy exists onlyin the case of a relation between ρ and ρ for the particle in question.

Forthe important examples of over-all (p, p)-relationships given in Sec:1.4the corresponding strain energies are:(a) Incompressible flow, or ρ = p :0e = constant;(b) Isothermal flow of a perfect gas, or ΓροJ— = — log ρ +Po ρ(c) Polytropic flow, or — =PK(17)e =fH1—:Poconstant;—:Po*Po*p-dp-+ constantPo κ — 1poJκiPo=-η— 1ρ+constant;(d) Linearized condition, ρ = p:0Ρe =Po ι Βι ,,— — + - — + constant.Ρ2p2T h e lower limit in (15) has been omitted, which means that there re­mains an undetermined additive constant in e. This is of no importancesince only the derivative of e appears in (16).5.