Buckingham s pi-theorem

Transcription

1 TMA495 Mathematcal modellng 2004 Buckngham s p-theorem Harald Hanche-Olsen hanche@math.ntnu.no Theory Ths note s about physcal quanttes R,...,R n. We lke to measure them n a consstent system of unts, such as the SI system, n whch the basc unts are the meter, klogram, second, ampere, and kelvn (m, kg, s, A, K). As t wll turn out, the exstence of consstent systems of measurement has nontrval consequences. We shall assume the fundamental unts of our system of unts are F,...,F m, so that we can wrte R = v(r )[R ] = ρ [R ] () where ρ = v(r ) s a number, and [R ] the unts of R. We can wrte [R ] n terms of the fundamental unts as a product of powers: [R ] = = F a ( =,..., n),. It s also mportant for the fundamental unts to be ndependent n the sense that = F x = x = = x m = 0. (2) We shall not be satsfed wth ust one system of unts: The whole crux of the matter hnges on the fact that our choce of fundamental unts s qute arbtrary. So we mght prefer a dfferent system of unts, n whch the unts F are replaced by ˆF = x F. Here x can be an arbtrary postve number for =,..., m. We can also wrte our quanttes n the new system thus: R = ˆv(R )[R ]ˆ = ˆρ [R ]ˆ. We compute R = v(r )F a F a m m = v(r )x a x a m m }{{} ˆv(R ) ˆF a ˆF a m m Perhaps we should also nclude the mole (mol) and candela (cd). Verson

2 Buckngham s p-theorem 2 from whch we deduce the relaton ˆρ = ρ = x a. (3) For example, f F = m and F s = s, and R s a velocty, then [R ] = ms = F F 2 and so a =, a 2 =. Wth ˆF = km and ˆF 2 = h, we fnd x = /000 and x 2 = /3600, and so ˆρ = ρ 3.6. Hence the example ρ = 0, ˆρ = 36 corresponds to the relaton 0m/s = 36km/h. We defne the dmenson matrx A of R,...,R n by a... a n A = a m... a mn It s now tme to ntroduce the heroes of our drama: The dmensonless combnatons of the varables R. A combnaton of these varables s merely a product of powers: R λ Rλ n n. We compute the unts of ths combnaton: [R λ Rλ n n ] = = F a λ + +a n λ n. (4) We call the combnaton dmensonless f ths unt s ; thus we arrve at the mportant result that ths s equvalent to Aλ = 0, where we wrte λ = (λ,...,λ n ) T. There s, therefore, a correspondence between the null space N(A) and the set of dmensonless combnatons of the varables. It may not come as a bg surprse that dmensonless combnatons have a value ndependent of the system of unts. We smply use (3) and compute: n ˆρ λ = = = n (ρ ( n = = ρ λ = ) x a ) λ = ( n n = ρ λ ) n n x a λ = ρ λ = = = = = x a λ snce Aλ = 0 mples n = xa λ =. Moreover, f you pck a bass for N(A) and take the correspondng dmensonless combnatons, π,...,π n r (here r s the rank of A), then any dmensonless combnaton can be wrtten as a product π c πc n r n r, where the exponents are unquely gven (they are the coeffcents of a member of N(A) n the chosen bass). We shall call ths a maxmal set of ndependent dmensonless combnatons. We can now state Verson

3 3 Buckngham s p-theorem Theorem. (Buckngham s p-theorem) Any physcally meanngful relaton Φ(R,...,R n ) = 0, wth R 0, s equvalent to a relaton of the form Ψ(π,...,π n r ) = 0 nvolvng a maxmal set of ndependent dmensonless combnatons. The mportant fact to notce s that the new relaton nvolves r fewer varables than the orgnal relaton; ths smplfes the theoretcal analyss and expermental desgn alke. We are not qute ready to prove ths, however. Amongst other thngs, we must gve a precse meanng to the phrase physcally meanngful. Frst of all, Φ must also have unts, and a value: [Φ] = F b. (5) = The value s gven by ust nsertng the values of R n the formula for Φ and computng: v ( Φ(R,...,R n ) ) = Φ ( v(r ),..., v(r n ) ) Furthermore, when we change to a dfferent set of unts, the value of Φ must change accordng to a law smlar to (3). Thus we get and therefore Φ ( ˆv(R ),..., ˆv(R n ) ) = ˆv ( Φ(R,...,R n ) ) = x b xb m m v ( Φ(R,...,R n ) ) = x b xb m m Φ ( v(r ),..., v(r n ) ) Φ(x a x a m m ρ,..., x a n x a mn m ρ n ) = x b xb m m Φ(ρ,...,ρ n ) (6) for all real ρ,...,ρ n and postve x,..., x m. It s ths relaton we shall thnk of when we say physcally meanngful n Buckngham s p-theorem. We shall, however, have to nsst on one more feature: Snce Φ s supposed to combne the quanttes R, the unts of Φ must be the unts of some combnaton of the varables R. We now begn the proof. Frst note that, by the fnal statement of the above paragraph, we may as well replace Φ by R c Rc n n Φ(R,..., c ) where the coeffcents c,..., c n are chosen so that the new functon s dmensonless that s, b = = b m = 0 n (5). Verson

4 Buckngham s p-theorem 4 The dmenson matrx A, havng the rank r, has r lnearly ndependent columns. We may as well assume these are the frst r columns, correspondng to the varables R,...,R r. 2 Then R,...,R r are dmensonally ndependent n the sense that ther only dmensonless combnaton s the trval one: R λ Rλ r r s dmensonless only f λ = = λ r = 0 (ths follows mmedately from (4)). I clam a natural correspondence: (R,...,R n ) (R,...,R r,π,...,π n r ) Clearly, the only possble dffculty here s expressng R k (where k > r) n terms of the quanttes on the rght-hand sde. But lnear algebra tells us that column k of A s a lnear combnaton of the frst r columns, and so [R k ] = [R c Rc r r ] for a sutable choce of c,..., c r. But then R k R c R c r r s dmensonless, so t can be wrtten π d πd n r n r. Therefore we can wrte R k = R c Rc r r π d πd n r n r. Now, usng the above correspondence, wrte Φ(R,...,R n ) = ψ(r,...,r r,π,...,π n r ) (7) for a sutable functon ψ. In a moment, I shall prove that ψ(r,...,r r,π,...,π n r ) does n fact not depend on R,...,R r. Thus we may wrte ψ(r,...,r r,π,...,π n r ) = Ψ(π,...,π n r ) and the proof of Buckngham s p-theorem wll be complete. To prove the ndependence of R,...,R r, replace each R n (7) by ts value ρ and substtute ths n both sdes of (6), and remember that b = 0: ψ(x a x a m m ρ,..., x a r x a mr m ρ r,π,...,π n r ) = ψ(ρ,...,ρ r,π,...,π n r ). Now, I clam that, gven postve numbers ρ,...,ρ r, we can pck postve numbers x,..., x m so that the numbers x a x a m m ρ (for =,..., r) on the left-hand sde of the above equaton can be any gven postve numbers. To be specfc, we can make them all equal to. That s, we can solve the equatons = x a = /ρ, =,..., r 2 If not, we ust renumber the varables to make t thus. Verson

5 5 Buckngham s p-theorem wth respect to x. In fact f we wrte x = exp(ξ ) the above equaton s equvalent to m a ξ = lnρ, =,..., r. = Ths equaton s solvable because the left m r submatrx of A has rank r, and therefore ts rows span R r. Ths proves the clam above, and therefore the theorem. Practce Ppe flow. We consder the problem of determnng the pressure drop of a flud flowng through a ppe. If the ppe s long compared to ts dameter, we shall assume that the pressure drop s proportonal to the length of the ppe, all other factors beng equal. Thus we really look for the (average) pressure gradent P, and presume the length of the ppe to be rrelevant. Varables that are relevant clearly nclude other propertes of the ppe: Its dameter D, and ts roughness e. To a frst approxmaton, we ust let e be the average sze of the unevennesses of the nsde surface of the ppe; thus t s a length. Also relevant are flud propertes. We shall use the knematc vscosty ν = µ/ρ together wth the densty ρ. In a Newtonan flud n shear moton, the shear tenson (a force per unt area) s proportonal to a velocty gradent, and the dynamc vscosty µ s the requred constant of proportonalty: Thus the unts of µ are Nm 2 /s = kgm s, and therefore the unts of ν are m 2 s. Fnally, the average flud velocty v s most defntely needed. The dmenson matrx can be wrtten as follows. P v D e ν ρ m kg s We can fnd the null space of ths, and hence use t to fnd the dmensonless combnatons. However, t s n fact easer to fnd dmensonless combnatons by nspecton. It s easy to see that the matrx has rank 3, so wth 6 varables, we must fnd 6 3 = 3 ndependent dmensonless combnatons. There are, of course, an nfnte number of possbltes, snce the choce corresponds to choosng a bass for the nullspace of A. In ths case we may be guded by common practce, Verson

6 Buckngham s p-theorem 6 however, and pck dmensonless quanttes as follows: Reynold s number Re = vd ν Relatve roughness (no name) ε = e D P D Snce we expect the P to be a functon of the other varables, we should have a relatonshp between the above quanttes whch has a unque soluton for the only varable contanng P: whch we wrte as P D ρv 2 = f (Re,ε), P = 2ρv2 D f (Re,ε). The extra factor 2 s there because then f s known as Fannng s frcton factor. Presumably, Fannng used the radus of the ppe rather than the dameter as the bass for hs analyss. (A copy and descrpton of the Moody dagram should be ncluded here.) One fnal remark. We dd not really need to wrte down the dmenson matrx. It s qute clear that the three dmensonless quanttes we found are ndependent, snce each of them contans at least one varable whch s not present n the two others. Snce there were only three fundamental unts nvolved, the dmenson matrx could not possbly have rank greater than 3, and therefore there could not exst more than three ndependent dmensonless combnatons. Stll, the dmenson matrx provdes a convenent way to summarze the dmensons and to reduce everythng to a problem n lnear algebra. Water waves. We consder surface waves n water. These waves can be convenently characterzed by a wave number k = 2π/λ (where λ s the wavelength) and an angular frequency ω. We seek a dsperson relaton expressng ω as a functon of ω. Presumably, the depth d plays a role, as well as the wave heght h, the acceleraton of gravty g, and the flud propertes: the densty ρ and (for very small waves) the surface tenson τ. We shall assume that the vscosty s neglgble. The dmensons of all these varables can be summarzed as follows. ρv 2 Varable ω k h d ρ τ g Unts s m m m kgm 3 Nm = kgs 2 ms 2 Verson

7 7 Buckngham s p-theorem Wth no less than seven varables, and three fundamental unts, we expect to fnd four ndependent dmensonless combnatons. One reasonable choce s (Bo s the Bond number): hk, dk, ω 2 g k, ρg Bo = τk 2. A relatonshp between all these, solved for the one combnaton that nvolves ω, then leads to a relatonshp of the form ( ) ω 2 = g k Ψ hk, dk,bo. We see that, for example, when waves are long, Bo s large, so we may gnore the nfluence of surface tenson. If the water s deep compared to the wave length then dk, whle f the wave heght s small compared to wave length, hk 0. When all of these approxmatons hold, then, we expect Ψ to be roughly constant, so ω 2 s proportonal to g k. In fact we fnd ω 2 = g k n the lmt, but ths requres more detaled analyss. For very short waves (rpples) n deep water, t seems reasonable to assume that only surface tenson s responsble for the wave moton, so that g does not enter the problem. You could do a new dmensonal analyss under ths assumpton, but t s easer to see drectly that Ψ must be a lnear functon of ts last argument for g to cancel out. If we stll assume dk and hk, we end up wth a relatonshp of the form ω 2 = τk3 ρ except the rght-hand sde should be multpled by a dmensonless constant. But agan, ths constant turns out to be. Verson