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CHAPTER 1

Dimensional Analysis and Scaling

1. Dimensional analysis

Exercise 1.1 The variables are t, r, , e, P . We already know one dimensionlessquantity 1 = r

5/et2. Try to find another in terms of P , which is a pressure,or force per unit area, that is, mass per length per time-squared. By inspection,2 = t

2P/r2 is another dimensionless quantity. Thus we have

f(r5/et2, t2P/r2) = 0

Now we cannot isolate the r and t variables in one dimensionless expression; theymust occur in both. If we solve for the first dimensionless quantity we get

r5/et2 = F (t2P/r2)

Then

r =

(

et2

)1/5

F (t2P/r2)

Because the second dimensionless variable contains t and r in some unknown man-ner, we cannot conclude that r varies like t2/5. In this formulation, if one can arguethat the ambient pressure is small and can thus be neglected, then we can set P = 0and obtain the result

r =

(

et2

)1/5

F (0)

which does imply that r varies like t2/5.

Exercise 1.2 Here x = 0.5gt2; because g is an acceleration, we observe that =x/gt2 is dimensionless. Clearly this is the only dimensionless combination. So thephysical law is = 0.5. The law could not depend on mass because there are nodimensionless combinations of t, x, g that involve m.

2. The Buckingham Pi theorem

Exercise 2.1 Assume that f(v,, g) = 0. If is dimensionless

[] = [v12g3 ]

= (LT1)1L2(LT2)3

1

2 1. DIMENSIONAL ANALYSIS AND SCALING

Thus we have the homogeneous system

1 + 2 + 3 = 0, 1 23 = 0The rank of the coefficient matrix is one, so there is one dimensionless variable.Notice that (2, 1, 1) is a solution to the system, and thus

= g/v2

By the Pi theorem, F () = 0 or g/v2 = Const.

Exercise 2.2 Follow the idea in Example 2.3 on page 11. Pick length, time andmass as fundamental and write

x = 1x, t = 2t, m = 3m

Then write v = 112 v, and so on for the other variables. Show that

v 29r2 g 1(1 l/) = 112 (v

2

9r2g1(1 l/))

So, by definition, the law is unit free.

Exercise 2.6 We have A = f(c, ) = c2g() since area is a length-squared. Nowapply the rule in the hint to the other two smaller, similar triangles to find theirareas as A1 = f(a, ) = a

2g() and A2 = f(b, ) = b2g(). Thus A = A1 + A2

implies c2 = a2 + b2.

Exercise 2.7 The period cannot depend only on the length and mass. Thereis no way that length and mass can be combined to yield a time dimension. Ifwe assume, however, that f(T, l,m, g) = 0, then there is only one dimensionless

variable, = T

g/l. (Again, mass will not appear). So, by the pi theorem,

T

g/l =constant.

Exercise 2.9 Pick length L, time T , and mass M as fundamental dimensions.Then the dimension matrix has rank three and there are 5 3 = 2 dimensionlessvariables; they are given by 1 = and 2 = R

l/

P . Thus f(1, 2) = 0

implies

= R1

P/lG()

for some function G.

3. Scaling

Exercise 3.1 In (a) we have u = A sint and so u = A cost. Then M = Aand max |u| = A. Then we have tc = 1/. In (b) we have u = Aet and u =Aet. Then tc = max |u|/max |u| = 1/. In part (c) we have u = Atet andu = (1t)Aet. The maximum of u occurs at t = 1/ and is M = A/e. To findthe maximum of u we calculate the second derivative to get u = A(t 2)et.So the maximum derivative occurs at t = 2/ or at an endpoint. It is easily checkedthat the maximum derivative occurs at t = 0 and has value max |u| = A on thegiven interval. Therefore tc = (A/e)/A = 1/e.

3. SCALING 3

Exercise 3.2 Here u = 1 + exp(t/) and u = exp(t/)/. Then tc =max |u|/max |u| = 2/1 = 2. The time scale is very small, indicating rapidchange in a small interval. But a graph shows that that this rapid decrease occursonly in a small interval near t = 0; in most of the interval the changes occur slowly.Thus two time scales are suggested, one near the origin and one out in the intervalwhere t is order one.

Exercise 3.3 This exercise is a project on which students may work independentlyor on a team. (a) The constant a must be budworms-squared since it is added tosuch a term in the denominator. The entire predation term must be budwormsper time, and so b must have dimensions budworms per time. (b) The parameter adefines the place where the predation term makes a significant rise. Thus it indicatesthe threshold where the number of budworms is plentiful so that predation kicksin; there are enough budworms to make the birds interested. (c) The dimensionlessequation is

dN

d= sN(1 N/q) N

2

1 +N2

(d) To find equilibrium solutions we set

sN(1 N/q) N2

1 +N2= 0

At this point we can use a calculator or a computer algebra program like Maple orMathematica to solve the equation for N . Observe that we obtain a fourth degreepolynomial equation when we simplify this algebraic equation:

sN(1 N/q)(1 +N2) N2 = 0

When s = 12 and q = 0.25 the equilibrium populations are N = 0, 0.261. Whens = 0.4 and q = 35 the equilibrium populations are N = 0, 0.489, 2.218, 32.29. Theplot in figure 1.1 shows the Maple generated solution curves in the case n(0) =2, 25, 40.

Exercise 3.4 There are two possible time scales, one based on frequency 1, andone based on period,

l/g. If we scale by the inverse of frequency, we let = t.Then the governing equation becomes

d2

d2+

(

3g

2l2 1)

sin = 0

If we scale by period, then pick the dimensionless time as T = t/

l/g. In this casethe governing equation becomes

d2

dt2+

(

3

2 l

2

g

)

sin = 0

When is small we should scale by

l/g because that places the small term in theequation, at the rotational force term, where it belongs; when is large we shouldscale by 1 because then the small term appears in the gravitational force term,where it belongs.

4 1. DIMENSIONAL ANALYSIS AND SCALING

Figure 1. Budworm populations for problem 3.3.

Exercise 3.5 To solve P = P (1 P ), P (0) = we separate variables andintegrate using partial fractions:

1

P (1 P )dP =

d + c

or(

1

P+

1

1 P

)

= + c

This gives

ln |P | ln |p 1| = + c

3. SCALING 5

orP

P 1 = Ce

The initial condition forces C = /(1). This last expression can be manipulatedalgebraically to obtain equation (23) in the text.

Exercise 3.6 Introduce the following dimensionless variables:

m = m/M, x = x/R, t = t/T, v = v/V

where T and V are to be determined. In dimensionless variables the equations nowtake the form

m = TM

x = V TRv

v =T

MV

1

m TgV (1 x)2

To ensure that the terms in the velocity and acceleration equations are the sameorder, with the gravitational term small, pick

V T

R=T

MV

which gives

V =

R/M

as the velocity scale.

Exercise 3.7 The differential equation is

c = qV

(ci c) kc2, c(0) = c0

Here k is a volume per mass per time. Choosing dimensionless quantities via

C = c/ci, = t/(V/q)

the model equation becomes

dC

d= (1 C) bC2, C(0) =

where = c0/ci and b = kV ci/q. Solve this initial value problem using separationof variables or a computer algebra package.

Exercise 3.8 The quantity q is degrees per time, k is time1, and is degrees (onecan only exponentiate a pure number). Introducing dimensionless variables

T = T/Tf , = t/(Tf/q)

we obtain the dimensionless model

dT

d= eE/T (1 T ), T (0) =

where = T0/Tf , E = /Tf , and = kTf/q.

6 1. DIMENSIONAL ANALYSIS AND SCALING

Exercise 3.9 Let h be the height measured above the ground. Then Newtonssecond law gives

mh = mg a(h)2, h(0) = 0, h(0) = VChoose new dimensionless time and distance variables according to

= t/(V/g), y = h/(V 2/g)

Then the dimensionless model is

y = 1 (y)2, y(0) = 0, y(0) = 1where prime is a derivative and = aV 2/mg.

Exercise 3.10 Scale time by

l/g and angular displacement by 0. Introducingthe dimensionless variables

= t/

l/g, = /0

gives the scaled model

d2

d2+ 10 sin(0) = 0, (0) = 1,

(0) = 0

CHAPTER 2

Perturbation Methods

1. Regular Perturbation

Exercise 1.1 Since the mass times the acceleration equals the force, we havemy = ky a(y)2. The initial conditions are y(0) = A, y(0) = 0. Here, a isassumed to be small. The scale for y is clearly the amplitude A. For the timescale choose

m/k, which is the time scale when no damping is present. Letting

y = y/A, = t/

m/k be new dimensionless variables, the model becomes

y + (y)2 + y = 0

where aA/m. The initial data is y(0) = 1, y(0) = 0. Observe that the smallparameter is on the resistive force term, which is correct.

Exercise 1.2 The problem is

u u = tu, u(0) = 1, u(0) = 1A two-term perturbation expansion is given by

y(t) = et +1

8(et et(1 + 2x+ 2x2))

A six-term Taylor expansion is

y(t) = 1 t+ 12t2 +

1 6

x3 +1 2

24x4 1 4

120x5

The plots in figure 1 show the superior performance of the two-term perturbationapproximation.

Exercise 1.3 (a) We have

t2 tanh t

t2= tanh t < 1

for large t. Thus t2 tanh t = 0(t2) as t .

(b) We have

limt