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PART IA: DIMENSIONAL ANALYSIS
HANDOUT 2
10. DIMENSIONLESS RELATIONSHIPS
Obviously, if the algebraic relationship between variables is known in the form of an
equation, then it is easy to rearrange the equation into dimensionless form. However, the
mathematics of engineering problems is often too complex to be solved algebraically.
Alternatively, if we are investigating a novel situation, the underlying physics may not be
fully understood, so that governing equations have yet to be established. In either of these
circumstances, we can use dimensional analysis, especially dimensionless groups, to help in
the reduction of the problem to a more manageable form.
We will use the accelerating car problem to show how it is possible to produce the relevant
dimensionless groups by knowledge only of the dimensions of the variables involved.
Assume we know only that:
v depends on u , a and t
So, we can call v the dependent variable , while u , a and t are independent variables —
the values of u , a and t can be varied independently of each other (e.g. you can change u
without having to change a or t ).
We may write this dependency as:
Quantity v depends on { u , a , t }
Dimensions LT
Note that this is a much more general statement than the equation used before. It merely
says that u , a and t (and only those) are the variables upon which v depends.
To produce dimensionless groups we need to rearrange the variables to “eliminate” the
physical dimensions. If we first seek to eliminate L, our first rearrangement might be to
divide v by u :
Quantity u
v depends on { u , a , t }
Dimensions (LT
- 1 ) ÷ (LT - 1 ) = 1 LT - 1 LT - 2 T
It is very important to note that no information has been “lost”. (Knowledge of vu and u
implies the value for v .) However, the above rearrangement has produced one
dimensionless group by the “elimination” of LT
We can now eliminate the dimension of length from the acceleration variable a by dividing
by u :
Quantity u
v depends on { u , u
a , t }
(LT
Dimensions - LT
Again the rearrangement has lost no information at all. (Knowledge of a u and u implies
the value for a .)
Now there is only one term [ u ] that explicitly involves the dimensions of length. If we refer
back to the PDC (i.e., “In a complete statement of a physical law, it is possible to rearrange
the terms so that all groups or quantities are dimensionless.”), then the term u cannot be
used to eliminate any other physical dimensions (no others include L). It must therefore be
dropped from the relationship:
Quantity u
v depends on { u
a , t }
Dimensions - T
Now we can rearrange the terms to eliminate the dimensions of time from the term a u by
multiplying by t :
Quantity u
v depends on { u
at , t }
Dimensions - T
Examining the terms in the relationship shows that only one [ t ] explicitly involves the
dimensions of time. Therefore, using the PDC, we can drop the t term:
Quantity u
v depends on { u
at }
Dimensions - -
Thus, we have deduced that the fundamental relationship for the accelerating car problem
must be between the two dimensionless groups vu and atu. These are exactly the same
dimensionless groups as we obtained by rearranging the (known) algebraic equation.
What we have not obtained is the exact form of the relationship between vu and atu.
This can only be obtained through theoretical analysis or experimental observation. If the
relationship can be obtained through theoretical analysis, then the result produced by
dimensional analysis really only has value as a means of confirming that the theoretical
analysis is correct. Thus, the major use of dimensional analysis, in practice, is in
conjunction with experimentation.
12. BUCKINGHAM’S PI THEOREM
Let us assume we have N variables in circumstances where the statement
Quantity v 1 depends on { v 2 , v 3 , v 4 , …, vN }
expresses a complete relationship between the variables v 1 , …, vN and M is the number of
dimensions required to describe v 1 , …, vN.
If this is so, then the relationship can be reduced to one between K dimensionless groups
denoted by D 1 , …, DK , so that
Quantity D 1 depends on { D 2 , D 3 , …, DK }
where:
K < N
In the production of the dimensionless groups by the elimination method , as demonstrated
in section 10, if one variable is dropped each time a dimension is eliminated, we find that
the number of the dimensionless groups is:
K = N – M
If eliminating one variable ‘costs’ more than one dimension, then:
K > N – M
So that in general:
K ≥ N – M
This result is known as “Buckingham’s Pi theorem”.
In other words:
If we have N variables and M dimensions, we should expect
at least N – M dimensionless groups.
Note that, in practice, equality holds for all but the simplest problems.
13. THE ELIMINATION METHOD
In section 10 we saw an example of the elimination method. Let us now review and
formalise what the method entails.
- Identify all the relevant variables
v 1 , v 2 , v 3 , v 4 ,…
- Identify the dependent variable
v 1 depends on { v 2 , v 3 , v 4 ,…}
v 2 , v 3 , v 4 ,… are the independent variables
- Identify the dimensions of each variable.
In section 10, this brought us to this point:
Quantity v depends on { u , a , t }
Dimensions LT
- Select a dimension to eliminate.
- Use one of the independent variables containing this dimension to eliminate it from the dependent variable (if it occurs there) and any other independent variables in which it occurs.
In section 10, using u to eliminate L brought us to this point:
Quantity u
v depends on { u , u
a , t }
Dimensions - LT
- If 5 is done correctly, then the independent variable used to do the eliminating will now be the only one containing the “targeted” dimension, and can be eliminated using the principle of dimensional consistency.
In section 10, we therefore dropped u bringing us to:
Quantity u
v depends on { u
a , t }
Dimensions - T
- If further dimensions remain, go to 4; if not, stop.
14. HOW TO DO DIMENSIONAL ANALYSIS I
We have seen (in section 10) an example of dimensional analysis in action, and reviewed
the process (in section 11). Let us now summarise the process:
- Identify all the relevant variables
v 1 , v 2 , v 3 , v 4 ,…
- Identify the dependent variable
v 1 depends on { v 2 , v 3 , v 4 ,…}
v 2 , v 3 , v 4 ,… are the independent variables
- Transform
v 1 depends on { v 2 , v 3 , v 4 ,…}
into
D 1 depends on { D 2 , D 3 ,…}
D 1 (the dependent dimensionless group , DDG) MUST contain v 1 (the
dependent variable) and can contain any of v 2 , v 3 , v 4 ,…
D 2 , D 3 ,… (the independent dimensionless groups , IDGs) MUST NOT
contain v 1 but can contain v 2 , v 3 , v 4 ,…
v 1 depends on { v 2 , v 3 , v 4 ,…}
D 1 depends on { D 2 , D 3 ,…}
- Buckingham’s rule tells you the minimum number of DGs you can form.
Exactly what you do with the information arrived at depends on the form of the
dimensionless relationship found. We will next consider some illustrative examples.
15. FORMS OF DIMENSIONLESS RELATIONSHIPS
15.1 D 1 depends on { - }
Let us consider the simple pendulum. We want to find how its period [ t ] depends on its
length [ l ] and the acceleration due to gravity [ g ].
The dependent variable is clearly t , so we have:
Quantity t depends on { l , g }
Dimensions T L LT
Using the elimination method, we can choose to “target” dimension L or T first. Let’s pick
T. We can eliminate T in the dependent variable by multiplying by g , i.e.:
Quantity t g depends on { l , g }
Dimensions
2 0. 5 T LT =L
! L LT
Now g is the only quantity with dimension T, so we can eliminate it, using the PDC:
Quantity t g depends on { l }
Dimensions
- 5 (^) L L
Now we can target the one remaining dimension [L]. We can eliminate this by dividing the
dependent quantity by l :
Quantity l
t g depends on { l }
Dimensions 1 L
L
5
5 = L
Now l is the only quantity with dimension L, so we can eliminate it:
Quantity l
g t depends on { - }
Dimensions - -
Buckingham’s rule sanity check: N = 3 , M = 2 , K = 1 : K " N! M? Yes!
The correct interpretation of this result is not, as might be thought, that the DDG [ t g l ]
must be zero, but that, as the DDG does not depend on anything, it cannot vary, i.e.:
=const. l
g t
We already happen to know that the value of the (dimensionless) constant is 2π, but, if we
did not know this, we could measure it experimentally. Note that, in an ideal world, we
Quantity Symbol Typical Units Dimensions
heat transfer coefficient h W m
−^2 K −^1 MT −^3 Θ −^1
thermal capacity of fluid cp kJ kg
−^1 K −^1 L
2 T −^2 Θ −^1
thermal conductivity of fluid^ λ^ W m
−^1 K −^1 MLT −^3 Θ −^1
dynamic viscosity of fluid^ μ^ kg m
−^1 s −^1 ML −^1 T −^1
density of fluid^ ρ^ kg m
−^3 ML −^3
velocity of fluid V m s
−^1 LT −^1
diameter of pipe d m L
Hopefully, it is obvious that h is the dependent variable. So our starting point is:
Quantity h depends on { cp , λ, μ, ρ, V , d }
Dimensions !
3 T
M
2
2
T
L
3 T
ML
LT
M
3 L
M
T
L
L
There are seven variables involving four dimensions so Buckingham’s rule shows us that
we should expect at least 7! 4 = 3 dimensionless groups.
Finding the dimensionless groups for this example using the elimination method is quite
time consuming (not to mention very boring when done live in a lecture), so this is included
as a worked example at the end of this section. This shows that the result obtained is:
Quantity !
hd depends on { !
cp μ , μ
! Vd }
Dimensions - - -
We have formed three dimensionless groups, so Buckingham’s rule checks out. In fact, if
you look at page 4 of the Thermofluids Data Book, you will see that all three of these
dimensionless groups have names (because these three groups crop up a lot in dimensional
analysis within heat transfer):
Re
Vd
μ
! Reynolds Number; Pr
c (^) p
!
μ Prandtl Number; Nu
hd
!
Nusselt Number
Thus, a more compact way of writing our final result is:
Quantity Nu depends on { Pr , Re }
If we wanted to find the relationship between Nu , Pr and Re experimentally, we would have
to conduct a series of experiments in which we systematically varied Pr and Re and
measured Nu. The results would then define a surface such as the one shown below.
We could also try to fit an algebraic equation to our data to obtain an empirical
correlation. This has already been done for many commonly occurring situations within
heat transfer. You can find the results of such a process on page 5 of the Thermofluids Data
Book. For instance, for turbulent flow in pipes with a constant wall temperature:
- 8 0. 4 Nu = 0. 023! Re! Pr
Now have a go at Questions 5, 6 and 7 on the Examples Paper.
Reynolds Number
Prandtl Number
Nusselt Number
Quantity !
h depends on { !
cp μ , μ
! V
, V , d }
Dimensions L
L
T
L
L
T
2
T
L
L
V is now the only quantity with dimension T, so we can eliminate it:
Quantity !
h depends on { !
cp μ , μ
! V
, d }
Dimensions L
L
L
Finally, we can target dimension L by multiplying h! and! V μ by d :
Quantity !
hd depends on { !
cp μ , μ
! Vd , d }
Dimensions L 1 L
! = - L 1
L
! = L
d is now the only quantity with dimension L, so we can eliminate it, reaching our final
result:
Quantity !
hd depends on { !
cp μ , μ
! Vd }
Dimensions - - -
16. ALTERNATIVE WAYS TO FORM DIMENSIONLESS RELATIONSHIPS
As the previous example showed the elimination method for finding dimensionless
relationships can be rather laborious. There are other valid methods for finding
dimensionless groups. These can be quicker, but need more care/thought in their execution.
16.1 The indicial method
This is very commonly presented in textbooks on dimensional analysis. However, it is best
suited to problems involving only one dimensionless group, so you will probably not find
this technique very useful in solving complex problems.
Example: Consider again the simple pendulum.
Quantity t depends on { l , g }
Dimensions T L LT
The indicial method assumes the dimensionless relationship is of the form:
Relation t is proportional to
! l ×
! g
Dimensions T
! L
! (LT )
" 2
We can invoke the PDC and compare the dimensions on each side of the equation. This
leads to the following simultaneous equations:
For T: 1 =" 2!
For L: 0 =" +!
These can be easily solved to give: "=! 0. 5 and != 0. 5 so:
g
l t! l g =
- 5 " 0. 5 or =const. l
g t
Note that when the number of indices (= the number of variables) on the RHS is greater
than the number of dimensions involved, the indicial equations cannot be solved simply and
finding the dimensionless groups by this method becomes more difficult. This happens
when more than one dimensionless group is involved.
