3 Determining equations

Let us suppose that we want to work out how the time period t for a pendulum depends on its length l and the accelleration due to gravity g. Our principle, that each term in an equation has to have the same dimensions, can be used to good effect. The time period t has dimensions [T], so if we write this as some function of l and g

t = f(l,g)

then f has to have these dimensions as well. If l and g are the only other variables in the problem, then we need to find some combination of the two which has dimensions [T]. Since the dimensions of l are [L], and the dimensions of g (an accelleration) are [LT^-2], then

l has dimensions [T 2] g

This suggests that the time period is given by

-- V~ l t = - g

But wait a minute! The time period could also be given by

V~ -- l t = 2 g

which is still dimensionally correct, since 2 is a pure number (no dimensions). In fact we could multiply the rhs by any dimensionless number, so from considering dimensions we can conclude that

V ~ - t = k l g

with k some dimensionless constant whose value has to be determined in some other way, generally experimentally. In this case experiment shows that k = 2

In this case, we could work out the relationship easilly from the dimensions of the variables. However the relationship is not always this straightforward, and so we need a general technique for doing this.

3.1 General approach

Now : the details. This approach works if we want to determine how a particular parameter in the problem depends on 3 other parameters. For example, suppose we want to work out how the flow Q of an ideal fluid through a hole of diameter D depends on the pressure difference p. It seems plausible that Q might also depend on the density of the fluid , so we look for a relationship of the form

Q = f(D, Dp,r)

Specifically, our function f(...) should look like this

a b c Q = kD Dp r

(1)

Here a, b, c are exponents whose form we want to determine. We can start by writing down the dimensions for each of these quantities :

[Q]	[L³T^-1]
[D]	[L]
[p]	[ML^-1T^-2]
[]	[ML^-3]

Our postulated expression becomes

[L3T- 1] = k[L]a[M L -1T-2]b[M L- 3]c

Now the dimensions of mass [M] have to balance on both sides of this expression, so the exponents have to satisfy

0 = b+ c ==> b = -c

Also, the dimensions of [L] and [T] must balance. For this to be the case

3	= a - b - 3c for [L]
- 1	= -2b for [T]

By writing this out like this, we have produced 3 simultaneous equations for the 3 unknowns a, b and c, from which we can see

1 1 1 1 b =- , c = --, and 3 = a- - + 3× - 2 2 2 2

so a = 2. Putting these values for the exponents back into the original expression (1),

V ~ ---- Q = kD2Dp1/2r -1/2 or Q = kD2 Dp- r

(2)

Again, k is a dimensionless constant whose value has to be determined via some other method typically experimentally. Knowing that the relationship has to be of the form (2), we could design an experiment where, for instance, Q is measured for different values of

p, and a plot of Q versus

would give a straight line with slope kD²/

3.2 More variables

What happens if we realise that there are more variables in the problem we are considering? The previous example assumed the flow to be ideal, i.e. the viscosity is zero. For a real fluid, Q may well depend on the viscosity. In this case equation (1) would have to be

Q = kDaDpbrcnd

i.e. (inserting the appropriate dimensions)

[L3T -1] = k[L]a[M L-1T- 2]b[M L -3]c[L2T-1]d

Equating dimensions gives

[M] :	0	= b + c
[L] :	3	= a - b - 3c + 2d
[T] :	- 1	= -2b - d

Now we have 3 equations and 4 unknowns, so there is no unique solution. The best we can hope for will be to express everything in terms of one of the exponents, say b :

c	= -b
d	= 1 - 2b
3	= a - b + 3b + (2 - 4b) a = 1 + 2b

In this case the equation will become

Q = kD1+2bDpbr-bn1-2b

This could be rearanged into the form

( 2 )b -Q- = k D-Dp2-- Dn rn

We note that both

2 -Q- and D--D2p- Dn rn

are dimensionless combinations of the variables. In fact, since

Q ---2-- pD /4

is the mean velocity, Q/D

is a Reynolds number for the problem.

3.3 Summary

In fluid mechanics, generally we deal with 3 base dimensions, M, L, T. If a problem involves 4 variables A, B, C, D, we can relate these as an equation of the form $A = kBbCcDd$ and determine the coefficients b, c, d
If there are more than 4 variables - say E, F, then we cannot uniquely determine their coefficients. The best we can hope for is to find groups of variables that are themselves dimensionless. There is a general method for doing this called the Buckingham theorem, (section 4).
There are exceptions to both of these statements. For example, if there are 4 dimensional quantities involved - maybe a heat transfer problem involving the dimension of temperature [] - we can determine a relation between up to 5 different variables.

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