John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 23
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Tounderstand Buckingham’s theorem, we must first overcome one conceptual hurdle, which, if it is clear to the student, will make everything thatfollows extremely simple. Let us explain that hurdle first.Suppose that y depends on r , x, z and so on:y = y(r , x, z, . . . )We can take any one variable—say, x—and arbitrarily multiply it (or itraised to a power) by any other variables in the equation, without alteringthe truth of the functional equation, like this:y 2y=x r , x, xzxxMany people find such a rearrangement disturbing when they first see it.That is because these are not algebraic equations — they are functionalequations. We have said only that if y depends upon r , x, and z that itwill likewise depend upon x 2 r , x, and xz.
Suppose, for example, thatwe gave the functional equation the following algebraic form:y = y(r , x, z) = r (sin x)e−z151152Analysis of heat conduction and some steady one-dimensional problems§4.3This need only be rearranged to put it in terms of the desired modifiedvariables and x itself (y/x, x 2 r , x, and xz):x2rxzy= 3 (sin x) exp −xxxWe can do any such multiplying or dividing of powers of any variablewe wish without invalidating any functional equation that we choose towrite. This simple fact is at the heart of the important example thatfollows.Example 4.2Consider the heat exchanger problem described in Fig.
3.15. The “unknown,” or dependent variable, in the problem is either of the exittemperatures. Without any knowledge of heat exchanger analysis, wecan write the functional equation on the basis of our physical understanding of the problem:⎡⎤⎢⎥ Cmax , Cmin , Thin − Tcin , U , A ⎥Tcout − Tcin = fn ⎢⎣ ⎦KW/KW/K(4.14)W/m2 K m2Kwhere the dimensions of each term are noted under the quotation.We want to know how many dimensionless groups the variables ineqn. (4.14) should reduce to. To determine this number, we use theidea explained above—that is, that we can arbitrarily pick one variable from the equation and divide or multiply it into other variables.Then—one at a time—we select a variable that has one of the dimensions.
We divide or multiply it by the other variables in the equationthat have that dimension in such a way as to eliminate the dimensionfrom them.We do this first with the variable (Thin − Tcin ), which has the dimension of K.⎡Tcout − Tcin⎢= fn ⎣Cmax (Thin − Tcin ), Cmin (Thin − Tcin ), Th − Tc in inWW⎤dimensionless⎥(Thin − Tcin ), U(Thin − Tcin ), A ⎥ ⎦KW/m2m2Dimensional analysis§4.3153The interesting thing about the equation in this form is that the onlyremaining term in it with the units of K is (Thin − Tcin ). No suchterm can exist in the equation because it is impossible to achievedimensional homogeneity without another term in K to balance it.Therefore, we must remove it.⎡⎤⎢⎥Tcout − Tcin= fn ⎢Cmax (Thin − Tcin ), Cmin (Thin − Tcin ), U(Thin − Tcin ), A ⎥⎣ ⎦Th − Tc in in2m2WWW/mdimensionlessNow the equation has only two dimensions in it—W and m2 .
Next, wemultiply U (Thin −Tcin ) by A to get rid of m2 in the second-to-last term.Accordingly, the term A (m2 ) can no longer stay in the equation, andwe have⎡⎤Tcout − Tcin⎢⎥= fn ⎣Cmax (Thin − Tcin ), Cmin (Thin − Tcin ), U A(Thin − Tcin ), ⎦ Thin − TcinWWWdimensionlessNext, we divide the first and third terms on the right by the second.This leaves only Cmin (Thin −Tcin ), with the dimensions of W. That termmust then be removed, and we are left with the completely dimensionless result:Tcout − TcinCmax U A= fn,Thin − TcinCmin Cmin(4.15)Equation (4.15) has exactly the same functional form as eqn.
(3.21),which we obtained by direct analysis.Notice that we removed one variable from eqn. (4.14) for each dimension in which the variables are expressed. If there are n variables—including the dependent variable—expressed in m dimensions, we thenexpect to be able to express the equation in (n − m) dimensionlessgroups, or pi-groups, as Buckingham called them.This fact is expressed by the Buckingham pi-theorem, which we stateformally in the following way:154Analysis of heat conduction and some steady one-dimensional problems§4.3A physical relationship among n variables, which can be expressed in a minimum of m dimensions, can be rearranged intoa relationship among (n − m) independent dimensionless groupsof the original variables.Two important qualifications have been italicized. They will be explainedin detail in subsequent examples.Buckingham called the dimensionless groups pi-groups and identifiedthem as Π1 , Π2 , ..., Πn−m .
Normally we call Π1 the dependent variableand retain Π2→(n−m) as independent variables. Thus, the dimensionalfunctional equation reduces to a dimensionless functional equation ofthe formΠ1 = fn (Π2 , Π3 , . . . , Πn−m )(4.16)Applications of the pi-theoremExample 4.3Is eqn. (2.24) consistent with the pi-theorem?Solution.
To find out, we first write the dimensional functionalequation for Example 2.6:T − Ti = fn r , ri , ro , h , k , (T∞ − Ti ) KmmmW/m2 K W/m·KKThere are seven variables (n = 7) in three dimensions, K, m, and W(m = 3). Therefore, we look for 7 − 3 = 4 pi-groups. There are fourpi-groups in eqn. (2.24):Π1 =T − Ti,T∞ − TiΠ2 =r,riΠ3 =ro,riΠ4 =hro≡ Bi.kConsider two features of this result. First, the minimum number ofdimensions was three. If we had written watts as J/s, we would havehad four dimensions instead.
But Joules never appear in that particularproblem independently of seconds. They always appear as a ratio andshould not be separated. (If we had worked in English units, this wouldhave seemed more confusing, since there is no name for Btu/sec unless§4.3Dimensional analysiswe first convert it to horsepower.) The failure to identify dimensionsthat are consistently grouped together is one of the major errors that thebeginner makes in using the pi-theorem.The second feature is the independence of the groups.
This meansthat we may pick any four dimensionless arrangements of variables, solong as no group or groups can be made into any other group by mathematical manipulation. For example, suppose that someone suggestedthat there was a fifth pi-group in Example 4.3:2hrΠ5 =kIt is easy to see that Π5 can be written as222 2Π2hrorriΠ5 == Bikri roΠ3Therefore Π5 is not independent of the existing groups, nor will we everfind a fifth grouping that is.Another matter that is frequently made much of is that of identifyingthe pi-groups once the variables are identified for a given problem. (Themethod of indices [4.1] is a cumbersome arithmetic strategy for doingthis but it is perfectly correct.) We shall find the groups by using eitherof two methods:1. The groups can always be obtained formally by repeating the simpleelimination-of-dimensions procedure that was used to derive thepi-theorem in Example 4.2.2. One may simply arrange the variables into the required number ofindependent dimensionless groups by inspection.In any method, one must make judgments in the process of combiningvariables and these decisions can lead to different arrangements of thepi-groups.
Therefore, if the problem can be solved by inspection, thereis no advantage to be gained by the use of a more formal procedure.The methods of dimensional analysis can be used to help find thesolution of many physical problems. We offer the following example,not entirely with tongue in cheek:Example 4.4Einstein might well have noted that the energy equivalent, e, of a rest155156Analysis of heat conduction and some steady one-dimensional problems§4.3mass, mo , depended on the velocity of light, co , before he developedthe special relativity theory. He would then have had the followingdimensional functional equation:kg· m2= fn (co m/s, mo kg)e N·m or es2The minimum number of dimensions is only two: kg and m/s, so welook for 3 − 2 = 1 pi-group. To find it formally, we eliminated thedimension of mass from e by dividing it by mo (kg). Thus,e m2=fnco m/s,mo s2mo kgthis must be removedbecause it is the onlyterm with mass in itThen we eliminate the dimension of velocity (m/s) by dividing e/moby co2 :e= fn (co m/s)mo co2This time co must be removed from the function on the right, since itis the only term with the dimensions m/s.
This gives the result (whichcould have been written by inspection once it was known that therecould only be one pi-group):Π1 =e= fn (no other groups) = constantmo co2ore = constant · mo co2Of course, it required Einstein’s relativity theory to tell us that theconstant is unity.Example 4.5What is the velocity of efflux of liquid from the tank shown in Fig. 4.4?Solution. In this case we can guess that the velocity, V , might depend on gravity, g, and the head H.
We might be tempted to includeDimensional analysis§4.3157Figure 4.4 Efflux of liquidfrom a tank.the density as well until we realize that g is already a force per unitmass. To understand this, we can use English units and divide g by theconversion factor,4 gc . Thus (g ft/s2 )/(gc lbm ·ft/lbf s2 ) = g lbf /lbm .ThenV = fn H , g m/smm/s2so there are three variables in two dimensions, and we look for 3−2 =1 pi-groups. It would have to beV= fn (no other pi-groups) = constantΠ1 = 3gHor4V = constant · gHThe analytical study of fluid √mechanics tells us that this form iscorrect and that the constant is 2.
The group V 2/gh, by the way, iscalled a Froude number, Fr (pronounced “Frood”). It compares inertialforces to gravitational forces. Fr is about 1000 for a pitched baseball,and it is between 1 and 10 for the water flowing over the spillway ofa dam.4One can always divide any variable by a conversion factor without changing it.158Analysis of heat conduction and some steady one-dimensional problems§4.3Example 4.6Obtain the dimensionless functional equation for the temperaturedistribution during steady conduction in a slab with a heat source, q̇.Solution. In such a case, there might be one or two specified temperatures in the problem: T1 or T2 . Thus the dimensional functionalequation is⎡⎤⎢⎥(T2 − T1 ), x, L, q̇ , k , h ⎥T − T1 = fn ⎢⎣ ⎦KKmW/m3 W/m·K W/m2 Kwhere we presume that a convective b.c.
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