Topic 1: Nondimensionalization, Scaling, and Units

Topic 1: Nondimensionalization, Scaling, and Units

Course Notes, Math 558 Spring 2012

Barenblatt.book Holmes.new.book

This section is a selection of material related to include Chapters 0?3 of [1], Chapters 1?3 of [2].

1 Dimensional Analysis

1.1 Example 1. Throw a ball

Consider the following example: We throw a ball directly upwards from the surface of the Earth. What is its maximum height?

We need two physical laws to describe this system:

1. Newton's Second Law. The acceleration of a body is directly proportional to its force and inversely proportional to its mass; moreover, the acceleration is parallel to the form. This is typically said as "Force equals mass times acceleration" or F = ma.

2. Newton's law of universal gravitation. Every body in the universe attracts every other body with a gravitational force, which force is proportional to each of the masses of the objects and inversely proportional to the square of the distance between the objects. In equations, if x is the vector from mass 1 to mass 2, then the force on mass 1 due to mass 2 is

F12

=

Gm1m2 x3

x

,

and the force on mass 2 due to mass 1 is

F21

=

-F12

=

-

Gm1m2x x3

.

Since this problem is one-dimensional, we can think of the force, acceleration, and position as scalars (distance from the ground). So if we define mB as the mass of the ball, mE as the mass of the earth, then we have the differential equation for x(t), the height above the ground, as

mB

d2x dt2

=

-

GmB mE d2

,

(1)

where d = R + x is the distance between the center of the ball and the center of the earth (R is the radius of

eq:d

the earth). Simplifying (1) we obtain

d2x

gR2

dt2 = - (R + x)2 ,

(2)

where we write g = GME/R2. Clearly this problem, being second-order, requires two integration constants, but these can be specified by x(0) and x (0), the initial position and velocity. Let us denote the maximal height by xmax.

eq:d eq:x

1

This equation can actually be solved exactly, but it is tedious and requires special functions (inverse sinh, for example) and in general is a big mess. Let us delay this solution for now.

Idea 1. The first idea we might have is to say that x(t) R for all t, and thus

x(t) + R R.

eq:x

If we make this approximation, then (2) becomes

(3) eq:approx

d2x

dt2 = -g,

(4)

which we can easily solve as

x(t) = - g t2 + x (0)t + x(0). 2

Writing x(0) = 0 and x (0) = v0, it is easy to see that the time of maximal height is t = v0/g, and therefore the maximal height is xmax = v02/2g.

Of course, what we have done here is what is known as an uncontrolled approximation! We have no idea,

eq:approx

a priori, how the error we have introduced in the approximation in (3) percolates into errors in the solution

eq:x

of (2). Studying the effects of this approximation is exactly what we will study in Topic 2 of this course in a

few weeks.

eq:constcoeff

Idea 2. Let us imagine that we didn't know how to solve (4), and look at the units of the problem.

There are three physical dimensions that appear in this problem, length, time, and mass; we denote these as L, T , M . Clearly, the units of velocity are L/T , acceleration is L/T 2, force is M L/T 2, etc. (If we didn't

know these, we could deduce the first two from the definitions and the third from Newton's Second Law!)

Now, let us make the Ansatz that the maximal height depends only on g, mB, and v0, so that

eq:constcoe

xmax = f (g, mB, v0).

If this is true, then these quantities must have the same units, i.e.

[xmax] = [f (g, mB, v0)],

and let us make the further Ansatz that this function can be written as a monomial, so that [xmax] = [mav0bgc].

(We will justify this second Ansatz later.) This then becomes

L = Ma

Lb T

L T2

c

= M aLb+cT -b-2c.

Equating powers, we obtain the system

a = 0, b + c = 1, -b - 2c = 0, or, a = 0, b = 2, c = -1.

Therefore we have

xmax

=

v02 . g

This is consistent with the exact answer derived above, although is a weaker statement (here we only know that there is a constant out front and not that this constant is = 1/2). However, notice that we had to insert much less information into the problem to obtain this solution and did not have to know how to solve an ODE.

2

1.2 Example 2. Drag on a sphere

Imagine a sphere moving through a fluid, we want to compute the force due to drag on the sphere. We postulate that it should depend on the dimensional quantities R, v, , ?: R is the radius of the sphere, v is the velocity of the sphere, is the density of the fluid, and ? is the (dynamic) viscosity of the fluid. The units of these quantities are

L

M

M

[R] = L,

[v] = , T

[] = L3 ,

[?] = . LT

By the previous logic, we should write

[DF ] = [Ravbc?d],

or

ML T2

=

La

Lb T

M L3

c

M LT

d

Expanding these out and equating powers, we obtain the system

a + b - 3c + d = 1, c + d = 1, b + d = 2.

There are only three equations, but four unknowns! So there will not be a unique solution and there is (at least) one free variable.

For now, let us make the choice of d as the free variable, and then we obtain

a = 2 - d, b = 2 - d, c = 1 - d,

giving us

DF = R2-dv2-d1-d?d = R2v2

?d ,

Rv

where is a scalar. Let us denote

? =

Rv

and we have

DF = R2v2d.

We might want to know the dimensions of , so we compute

(5) eq:defofPi (6) eq:ad

M/LT [] = L(L/T )(M/L3) = 1.

eq:ad

We say that is dimensionless. Now, since is dimensionless, (6) is dimensionally correct no matter what the choice of , or d. Therefore we could write

DF = R2v2(1d1 ),

or DF = R2v2(2d2 ),

or in fact any linear combination of such terms, namely

n

DF = R2v2

kdk .

k=1

In general, given any function of , this expression is dimensionally correct, so we can write

DF = R2v2f ()

3

for some unknown function f . Since is nondimensional, f () is nondimensional for any function f , and therefore we cannot proscribe the solution further.

Now, notice that we made a choice of d as a free parameter in the derivation above. Would anything have changed had we made another choice there? For example, let's now say that c is free. Solving, we would obtain

a = 1 + c, b = 1 + c, d = 1 - c,

giving

DF = R1+cv1+cc?1-c = Rv

Rv ?

c

.

Thus defining

Rv

= , ?

(7)

we have

DF = Rv?c,

and by the same argument of arbitrary powers, we have

eq:defofPit

DF = Rv?g().

Are these two different expressions really different? Note, first of all, that = 1/. Moreover, if we set them equal, we obtain

R2v2f () = Rv?g(1/), Rv f () = g(1/), ? f () = g(1/).

Thus there is a functional relationship between f and g; if we know one, we know the other. So these expressions are equivalent even though they do not seem so at first glance.

Now, one might ask how one determines theHoulnmkenso.nwenw.fbuonocktion f (or g), and this is something that should be done by experiment. See Figure 1.3 of [2].

The nondimensional quantities and show up so often in fluid dynamics that they are given names: is known as the Reynolds number, and is the Pe?clet number.

1.3 Using dimensional analysis for scale models

Let us imagine that we have specific values for the quantities R, v, , ? in mind, but we want to know the

drag a sphere would experience without building the object itself and making a measurement. Can we

figure out how to build a scale model of the sphere, and then embed it in another physical experiment to

get the measurement.

We know that

DF = R2v2f (),

? = .

Rv

As long as we know f (), then we are done. So let us build a model, with model parameters Rm, vm, m, ?m

so that m = . Then f (m) = f (), and we can measure the former to get the value for the latter. The

restriction that m = means that

? =

?m .

Rv Rmvmm

Assuming that we're using the same fluid, so that m = , ?m = ?, we then have

Rv

Rv

=

Rmvm,

or,

vm

=

. Rm

4

As an example, let us say that we wanted to measure the drag force on a sphere of radius 1000m at a given velocity v. This is much to large to build, but we could, for example, build a sphere with radius Rm = 2m and then choose velocity vm = 500v.

1.4 Buckingham Pi Theorem

The questions we have from the example above are clear. Will such a procedure always work? Will we have choices? When we are given a choice during the procedure, will this affect matters significantly? The answer to this is given in a theorem that we prove below.

We will state and prove the theorem in the case that the only dimensions available to us are mass, length, and time, for concreteness. There could be other dimensional quantities (e.g. charge) but it will be easy to see at the end how to modify the statements when there are other dimensions.

Let us consider a physical quantity q which depends on the n physical quantities p1, p2, . . . , pn. We have the relationship

q = f (p1, . . . , pn),

and let us assume a monomial dependence of the units as

[q] = [pa11 pa22 . . . pann ]

(8)

Let us assume that the dimensions of each quantity are known, and denote them as

eq:mono

[pi] = Lli M mi T ti , [q] = Ll0 M m0 T t0 .

eq:mono

Plugging these into (8) and equating powers, we obtain the three equations

n

aili = l0,

i=1

n

aimi = m0,

i=1

n

aiti = t0.

i=1

We can write these equations efficiently in matrix form as follows. Define the matrix and vectors

(9) eq:lmt

l1 l2 ? ? ? ln

a1 a2

l0

A = t1 m1

t2 m2

??? ???

tn , mn

a=

...

,

b = t0 , m0

an

eq:lmt

then (9) becomes

Aa = b.

(10)

eq:A

So, the question remains: does (10) have a solution? If so, is it unique?

eq:A

Definition 1. We say that p1, . . . , pn are dimensionally complete if (10) has a solution for every q, and dimensionally incomplete if it does not. Equivalently, we say that p1, . . . , pN are dimensionally complete if the matrix A has rank 3.

eq:A

We are now in position to state the theorem:

Theorem 1. Assume that q = f (p1, . . . , pn) is a dimensionally homogeneous relation and p1, . . . , pn are dimensionally complete. Then there exists a function F such that q = QF (1, . . . , k), where i are dimensionless products of the pi, [Q] = [q], and k is the dimension of the kernel of A.

eq:A

Proof. We know that all solutions to (10) can be written in the form

k

a = a + (i)a(i),

i=1

5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download