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PART IA: DIMENSIONAL ANALYSIS
HANDOUT 3
16.2 The “by inspection” method
After some experience has been gained in solving dimensional analysis problems, the
dimensionless groups can usually be found most easily “by inspection”. This approach is so
called, because the only working shown (e.g. in an examination answer) is the words “by
inspection”. Sometimes it is genuinely possible to write down the answer by inspection;
sometimes a bit of “back of an envelope” working is required.
Let us demonstrate the “by inspection” method through an example.
Example: A sphere of diameter d moves horizontally through a liquid of density ρ and
dynamic viscosity μ. Find a suitable functional expression for the force D
required to keep it in motion with uniform velocity V. ( D is the drag on the
sphere in uniform motion.)
Using the information given, our starting point is:
Quantity D depends on { d , ρ, μ, V }
Dimensions 2 T
ML
L
3 L
M
LT
M
T
L
In the “by inspection” method we start by considering Buckingham’s rule:
Buckingham’s rule: N = 5 , M = 3 " K! 2
So, we can form at least 2 dimensionless groups. We know one of them, the dependent
dimensionless group (DDG), must involve the dependent variable [ D ] and one (an
independent dimensionless group, IDG) must not. Let’s start by trying to find a suitable
DDG.
Looking at what we have to work with, we can see that d can eliminate any L’s we have, V
is the simplest thing to use to eliminate any T’s (as it only involves L and T, whereas μ
involves M, L and T), and ρ is the simplest thing to use to eliminate any M’s. It is also
sensible to leave the elimination of L till last, because we know we can deal with any
residual L’s using d.
To form a suitable DDG we start by writing down (on our envelope) the dependent variable
and its dimensions:
Quantity Dimensions
D
2 T
ML
We first aim to eliminate T or M (it doesn’t matter which). Let’s pick T. We can eliminate
T by dividing by 2 V :
Quantity Dimensions
2 V
D
L
M
L
T
T
ML
2
2
Next we eliminate M (remember we’re saving L till last) by dividing by ρ:
Quantity Dimensions
2 2
V
D
V
D
2
3 L L
M
M
L
Finally we eliminate L by dividing by
2 d :
Quantity Dimensions
2 2 2 2
V d
D
V d
D
L
L
2 2 ! = "
Now we can try to form an IDG. Remember that we can only use the independent variables
[ d , ρ, μ and V ] to do so. Looking at these we can see that we can create a quantity from
which M has been eliminated by dividing ρ by μ (this is a sensible place to start because we
know we can eliminate T and L using V and d ):
Quantity Dimensions
μ
3 2 L
T
M
LT
L
M
Next we can eliminate T by multiplying by V :
Quantity Dimensions
μ
μ
! V
" V =
L
T
L
L
T
2
Finally we eliminate L by multiplying by d :
Quantity Dimensions
μ
!
μ
! Vd d
V
" = L 1
L
This IDG should be familiar to you from section 15.3. It is Reynolds Number, Re.
As Buckingham’s rule only predicts the minimum number of dimensionless groups we can
form, in principle it might be possible to form further IDGs, but a bit of thought shows that
it is not possible to form any others that are independent of Re. Thus, we arrive at the result:
Quantity 2 2 V d
D
!
depends on { μ
! Vd }
Dimensions - -
17. DIMENSIONLESS GRAPHS
Usually the results from experiments cannot be presented as algebraic expressions, instead
they must be shown in graphical form. Using dimensionless graphs enables data to be
presented very concisely.
We saw in section 15.2 that, using a dimensionless graph, the relationship between v , u , a
and t for an accelerating car can be represented by a single line. Let us highlight just how
concise this is by considering some of the alternatives.
If we were asked to present the relationship [ v = u + at ] in graphical form, so that, given
values of u and a , the value of v can be read off at various times t , we could produce a series
of graphs of the form:
t (s)
0 5
v (m/s)
20
60 u^ =^10 ms
! 1
a =^6 ms
! 2
40
To cover the range of possible values of u and a adequately, we would require a lot of such
graphs!
If we wanted to reduce the number of graphs, we could plot a family of lines for different
values of a , all for the same value of u :
t (s)
0 5
v (m/s)
20
60
u =^10 ms
! 1
a =^6 ms 40!^2
a =^0 ms
! 2
a =^2 ms
! 2
a = 10 ms
! 2
a =^8 ms
! 2
a =^4 ms
! 2
Such graphs are still easily read, but we would require several of them to cover the range of
possible values of u , i.e. a small booklet.
To reduce the number of graphs to one, we could combine such plots to produce a single
graph of this form:
t (s)
0 5
v (m/s)
20
60
a =^6 ms 40!^2
a =^0 ms
! 2
a =^2 ms
! 2
a = 10 ms
! 2
a =^8 ms
! 2
a =^4 ms
! 2
u
However, it is now quite difficult to read information off the graph, and, if the values of u
and a we are given are not ones for which lines are plotted, we have to interpolate.
The advantages of the dimensionless graph, which is a single line, and which works for any
values of u and a without requiring interpolation, are clear:
at / u
0 2
v / u
1
3
2
Thus, the two results are, in fact, equivalent. The new DDG!
"
Vd
D
μ
is merely a combination
of the old DDG!
"
2 2 V d
D
and the IDG! "
μ
' Vd .
Note that provided that there is at least one IDG, there are infinite possibilities for such
combinations.
Note also that no one combination is more “correct” than any other, but one may be more
commonly used.
Elimination Method Worked Example
Our starting point is:
Quantity D depends on { d , ρ, μ, V }
Dimensions 2 T
ML
L
3 L
M
LT
M
T
L
First, let’s target M by dividing both D and ρ by μ:
Quantity μ
D
depends on { d , μ
, μ, V }
M
LT
T
ML
2
M
LT
L
M
3
Dimensions T
L
2 L 2 L
T
LT
M
T
L
We can now eliminate μ because it is the only quantity with dimension M:
Quantity μ
D
depends on { d , μ
, V }
Dimensions T
L
2 L 2 L
T
T
L
Next we target T by dividing D μ and multiplying! μ by V :
Quantity V
D
μ
depends on { d , μ
! V
, V }
L
T
T
L
2 ! T
L
L
T
2
Dimensions L L L
T
L
continued…
We can now eliminate V because it is the only quantity with dimension T:
Quantity V
D
μ
depends on { d , μ
! V
Dimensions L L L
Finally, we can target dimension L by using d appropriately:
Quantity Vd
D
μ
depends on { d , μ
! Vd }
Dimensions 1 L
L! = L L 1
L
Now we can eliminate d (the only quantity with dimension L), reaching our final result:
Quantity Vd
D
μ
depends on { μ
! Vd }
Both the lift L and drag D are dependent variables, because they both depend on the aircraft
design and flight conditions. It is not possible to vary either L or D independently.
As this example shows, care is sometimes needed in identifying the dependent and
independent variables. It is possible that you will end up with two (or more) situations to be
investigated.
21. GEOMETRICAL SIMILARITY
Complex engineering situations require many lengths to be defined in order to specify the
geometry completely.
l
W
Side view
d
b
End view
Example: Consider the elastic deflection δ of a beam of length l , breadth b and depth d
subjected to a force W at its centre. Denote the Young’s modulus (modulus of
elasticity) by the symbol E.
The dependent variable is δ. Thus we have:
Quantity δ depends on { W , E , l , d , b }
Dimensions L 2 T
ML
2 LT
M
L L L
Buckingham’s rule: N = 6 , M = 3 "^ K!^3
By inspection, a suitable DDG is! l or! d or! b — each is an equally good choice,
although! l would be the most common one.
Looking at the independent variables, two IDGs are again obvious by inspection: d l and
bl. (Other pairs, e.g. db and lb , are possible, but only two independent dimensionless
groups can be formed from l , b and d .)
We have now formed one DDG and two IDGs, i.e. three DGs in all — the minimum
number predicted by Buckingham’s rule. But, looking again at the independent variables,
we still have not used W or E. By inspection (or after scribbling on the back of an
envelope), it should be apparent that we can form another IDG, for instance 2 El
W
or 2 Eb
W
(and several other possibilities). Note that only one of these two can be an independent
dimensionless group if we have already chosen bl as one, because
2
2 2
b
l
El
W
Eb
W
Thus, an appropriate dimensional analysis solution to this problem is that “by inspection”:
Quantity l
depends on { 2 El
W
l
d , l
b }
Note that this is an example of a case where equality does not hold for Buckingham’s rule.
But try solving this problem using the F-L-T system rather than the M-L-T system. What do
you find?
Now,
Two objects are geometrically similar
when one object is an exact scale model of the other.
So, for a set of geometrically similar beams, the groups bl and d l will have fixed values,
i.e. they are not independent dimensionless groups as they cannot be changed. Thus, in
considering a set of geometrically similar beams, just one geometric measurement ( l , b or d )
will be sufficient to define the geometry, and the problem reduces to:
Quantity δ depends on { W , E , l }
Dimensions L 2 T
ML
2 LT
M
L
to which the solution is, “by inspection”:
Quantity l
depends on { 2 El
W
Note that in any situation involving geometrically similar objects,
e.g. in a scale-modelling problem, just one length scale needs to be specified
