SYMBOLS )
Step 7. Use Physical Reasoning or Additional Knowledge to Simplify the Dimensionless Relationship
Sometimes there are only two extensive thermodynamic variables involved and these must be proportional in the final scaling law. An overall conservation law can be applied that restricts one or more parametric dependencies, or a phenomena may be known to be linear, quadratic, etc., in one of the parameters and this dependence must be reflected in the final scaling law. This seventh step may not always be possible, but when it is, significant and powerful results may be achieved from dimensional analysis.
EXAMPLE 1.2
Use dimensional analysis to find the parametric dependence of the scale heightHin a static isothermal atmosphere at temperatureTocomposed of a perfect gas with average molecular weight Mwwhen the gravitational acceleration isg.
Solution
Follow the six steps just described.
1. The parameter list must includeH,To,Mw, andg. Here there is no velocity parameter, and there is no need for a second specification of a thermodynamic variable since a static pressure gradient prevails. However, the universal gas constantRumust be included to help relate the thermal variableToto the mechanical ones.
1. INTRODUCTION
26
2. The dimensional matrix is:
Note that thekmolee1specification ofMwandRuis lost in the matrix above since akmoleis a pure number.
3. The rank of this matrix is four, sor¼4.
4. The number of dimensionless groups is:ner¼5e4¼1.
5. UseHas the solution parameter, and the others as the repeating parameters. Proceed with exponent algebra to find the dimensionless group:
M0L0T0q0¼ ½P1 ¼h
HToaMbwgcRdu
i¼ ðLÞðqÞaðMÞb LT2Þc
ML2T2q1d
¼MbþdL1þcþ2dT2c2dqad:
Equating exponents yields four linear algebraic equations:
bþd¼0, 1þcþ2d¼0, 2c2d¼0, andad¼0,
which are solved by:a¼ 1,b¼1,c¼1, andd¼ 1. Thus, the lone dimensionless group is:
P1¼HgMw/RuTo.
6. Because there is only a single dimensionless group, its most general behavior is to equal a constant, soHgMw=RuTo ¼ 4ð.Þ ¼ const:, orH¼const. (RuTo/gMw). Based on the finding at the end of the previous section andR¼Ru/Mw from (1.22), this parametric dependence is correct and the constant is unity in this case.
EXAMPLE 1.3
Use dimensional analysis andFigure 1.10to prove the Pythagorean theorem based on a right triangle’s areaa, the radian measurebof its most acute angle, and the lengthCof its longest side (Barenblatt, 1979).
C A
A
B β
β
FIGURE 1.10 A right triangle with areaa, smallest acute angleb, and hypotenuseC. The dashed line is perpendicular to sideC.
1.11. DIMENSIONAL ANALYSIS 27
Solution
Follow the six steps given earlier and then consider similarity between the main triangle and two sub-triangles.
1. The parameter list (a,b,C) is given in the problem statement son¼3.
2. The dimensional matrix is:
3. With no M or T units, the rank of this matrix is one, sor¼1.
4. The number of dimensionless groups is:ner¼3e1¼2.
5. Let the triangle’s areaabe the solution parameter. By inspection,P1¼a/C2, andP2¼b.
6. Therefore, the dimensionless relationship is:a=C2 ¼ 4ðbÞora ¼ C24ðbÞ.
7. When the dashed line is perpendicular to sideC, then the large triangle is divided into two smaller ones that are similar to the larger one. These sub-triangles have A andB as their longest sides and both have the same acute angle as the large triangle. Therefore, the sub-triangle areas can be written as A24ðbÞ andB24ðbÞ. Summing the sub-triangle areas produces: A24ðbÞ þB24ðbÞ ¼ C24ðbÞ orA2þB2 ¼ C2 when 4ðbÞs0.
EXAMPLE 1.4
Use dimensional analysis to determine the energyEreleased in an intense point blast if the blast- wave propagation distanceDinto an undisturbed atmosphere of densityris known as a function of timetfollowing the energy release (Taylor, 1950; seeFigure 1.11).
Solution
Again follow the six steps given earlier.
1. The parameter list (E,D,r,t) is given in the problem statement son¼4.
2. The dimensional matrix is:
D E
FIGURE 1.11 In an atmosphere with undisturbed densityr, a point release of energyEproduces a hemi- spherical blast wave that travels a distanceDin timet.
1. INTRODUCTION
28
3. The rank of this matrix is three, sor¼3.
4. The number of dimensionless groups is:ner¼4e3¼1.
5. Let the point-blast energy be the solution parameter and construct the lone dimensionless group by inspection. First useEandrto eliminate M: [E/r]¼L5Te2. Next useDto eliminate L: [E/rD5]¼Te2. Then use tto eliminate T: [Et2/rD5]¼dimensionless, so P1¼Et2/rD5.
6. Here there is only a single dimensionless group, so it must be a constant (K). This produces:
Et2=rD5 ¼ 4ð.Þ ¼ Kwhich implies:E ¼ KrD5=t2, whereKis not determined by dimensional analysis.
7. The famous fluid mechanician G. I. Taylor was able to estimate the yield of the first atomic-bomb test conducted on the White Sands Proving Grounds in New Mexico in July 1945 using: 1) the dimensional analysis shown above, 2) a declassified movie made by J. E. Mack, and 3) timed photographs supplied by the Los Alamos National Laboratory and the Ministry of Supply.
He determined the fireball radius as a function of time and then estimatedEusing a nominal atmospheric value forr. His estimate ofE¼17 kilotons of TNT was very close to the actual yield (20 kilotons of TNT) in part because the undetermined constantKis close to unity in this case. At the time, the movie and the photographs were not classified but the yield of the bomb was entirely secret.
EXAMPLE 1.5
Use dimensional analysis to determine how the average light intensityS(Watts/m2) scattered from an isolated particle depends on the incident light intensityI(Watts/m2), the wavelength of the lightl (m), the volume of the particleV(m3), the index of refraction of the particle ns(dimen- sionless), and the distance d (m) from the particle to the observation point. Can the resulting dimensionless relationship be simplified to better determine parametric effects whenl[V1=3?
Solution
Again follow the six steps given earlier, knowing that the seventh step will likely be necessary to produce a useful final relationship.
1. The parameter list (S,I,l,V,ns,d) is given in the problem statement son¼6.
2. The dimensional matrix is:
3. The rank of this matrix is 2 because all the dimensions are either intensity or length, sor¼2.
4. The number of dimensionless groups is:ner¼62¼4.
5. Let scattered light intensitySbe the solution parameter. By inspection the four dimensionless groups are:
1.11. DIMENSIONAL ANALYSIS 29
P1¼S=I, P2¼d=l, P3¼V=l3, andP4¼ns: 6. Therefore, the dimensionless relationship is:S=I ¼ 41ðd=l,V=l3,nsÞ.
7. There are two physical features of this problem that allow refinement of this dimensional analysis result. First, light scattering from the particle must conserve energy and this implies:
4pd2S¼const. so Sf1=d2. Therefore, the result in step 6 must simplify to:S=I ¼ ðl=dÞ242 ðV=l3,nsÞ. Second, whenl is large compared to the size of the scatterer, the scattered field amplitude will be produced from the dipole moment induced in the scatterer by the incident field, and this scattered field amplitude will be proportional toV. Thus,S, which is proportional to field amplitude squared, will be proportional toV2. These deductions allow a further simplification of the dimensional analysis result to:
S I ¼
l d
2 V l3
2
43ðnsÞ ¼ V2 d2l443ðnsÞ:
This is Lord Rayleigh’s celebrated small-particle scattering law. He derived it in the 1870s while investigating light scattering from small scatterers to understand why the cloudless daytime sky was blue while the sun appeared orange or red at dawn and sunset. At the time, he imagined that the scatterers were smoke, dust, mist, aerosols, etc. However, the atmospheric abundance of these are insufficient to entirely explain the color change phenomena but the molecules that compose the atmosphere can accomplish enough scattering to explain the observations.