Solution:
You have to determine a mathematical expression with parameters (C1, C2, C3,
C4), based on a set of experimental data.
If you should have 4 sets of experimental data with various values of (s, v, Re),
you should have to solve an algebraic system with unknown (C1, C2, C3, C4).
The system is not linear, but could be reduced to a linear one by applying the
logarithm function:
Ra  C1 sC2  v C3  ReC 4

ln Ra  lnC1 C2ln s  C3lnv  C4lnRe
(1)
(2)
By writing the above relation for every set of data, we will get a linear algebraic
system from where we can determine the unknown (Ci), i = 1,…4.
Note that we can denote
C1’ = lnC1
(3)
However, when experimental conditions are available, we need more than 4 sets
of data in order to cover as much as possible the measuring errors.
So, the problem becomes more interesting when we have (N) sets of data, N > 4.
In this case, the algebraic system becomes overdetermined (that is, there are
more equations than unknown) and the way to “solve” such a system is the socalled “least squares method”.
In short, this method determines the parameters (C1, C2, C3, C4) so that the
deviations between the values of (Rak) given by (1) for each set of (sk, vk, Rek)
and the actual measured values of (Rak) are minimum.
In other words, the sum of all squares of these deviations has to be minimum.
We express that in a mathematical way as follows:
 Rak  C1 skC 2  v kC 3  RekC 4 
N
2
 Min.
(4)
k 1
where N = number of sets of experimental data (N > 4)
Remark: The method uses squares of distances, and not simple distances in
order to avoid the misresults due to some negative values.
Using (2), we can write a simpler expression than the above one:
N
 ln Rak  C1'C2  ln sk  C3  ln v k  C 4  ln Rek 2  Min.
(5)
k 1
In order to solve this problem, we need to differentiate with respect to each
parameter (Ck), like in calculus:
 N
2
 ln Rak  C1'C 2  ln sk  C3  ln v k  C 4  ln Rek    0 , i  1, 4
Ci k 1

(6)
By performing the derivatives, we will have:
N

     2ln Rak  C1'C 2  ln sk  C3  ln v k  C 4  ln Rek   0
C1'
k 1

N  C1'  C2  ln sk  C3  ln v k  C4  ln Rek  ln Rak
(7)
where
N
C1'  NC1'
k 1

     2 ln sk ln Rak  C1'C 2  ln sk  C3  ln v k  C 4  ln Rek   0
C 2
k 1
N
 C1' ln sk  C 2   ln sk   C3   ln v k ln sk  C 4   ln Rek ln sk   ln Rak ln sk
2
(8)
N

     2 ln v k ln Rak  C1'C 2  ln sk  C3  ln v k  C 4  ln Rek   0
C3
k 1
 C1' ln v k  C 2   ln v k ln sk  C 3   ln v k   C 4   ln Rek ln v k 
2
 ln Rak ln v k
(9)

 
C 4
N
  2 ln Rek ln Rak  C1'C 2  ln sk  C3  ln v k  C 4  ln Rek   0
k 1
 C1' ln Rek  C 2   ln Rek ln sk  C 3   ln Rek ln v k  C 4   ln Rek  
2
 ln Rak ln Rek
(10)
The equations (7), (8), (9) and (10) build the system, which determines the
constants that we are looking for.
For a better organization of work, we will write this system under a matriceal form
below:

N

  ln sk

  ln v k
  ln Re
k

 ln sk
 ln v k
 ln Rek   C1'    ln Rak 
2
 ln sk 
 ln sk ln v k  ln sk ln Rek    C 2     ln Rak ln sk 
 ln sk ln v k
 ln v k 2  ln v k ln Rek   C3    ln Rak ln v k 
 ln sk ln Rek  ln v k ln Rek  ln Rek 2  C 4    ln Rak ln Rek 
( 11)
(all sums are taken from k = 1 to N)
The above system can be solved using an appropriate computation tool, like
Excel, MatCad or MatLab.
My opinion is that, in using Excel, one can compute easily the sums in (11) and
by using the Solver, one can solve the system and find the constants C1’, C2,
C3, C4.
At the end, don’t forget to compute C1 by exponential, from (3):
C1  eC1'
( 12)
In this way, you will find a formula (1)-parallel to one that matches best to all your
experimental data.
Final remark: If, after finding the constants, you try to test the formula (1), don’t
expect to get exactly the values of (Rak) by experiment; you will get instead the
closest results to “all” your experimental data. In fact, this is the main idea of the
least squares method.
I hope this is the answer that you are looking for.
If additional explanations are needed, let me know. Good luck!