1047 pág.

# Fenômentos de Transporte

DisciplinaFenômenos de Transporte I12.466 materiais111.646 seguidores
Pré-visualização50 páginas
```rapidly) are the velocity, a+, of the fluid at shutoff, the density
of the fluid, p, and the bulk modulus of the fluid, p = p(dp/dp)~
Find a complete set of dimensionless groups for this problem.
Solution
The dimensions of the bulk modulus are
(5.2.2-6)
Chapter 5: Application of Dimensional Analysis
pmu
p
(v)
p
233
1 -1 -2
0 1 -1
1 - 3 0
1 -1 -2
The dimensional matrix in an MLt system is
M L t
(5.2.2-7)
We have four rows that can be combined in various ways to make 3x3
determinants. From the properties of determinants, the only effect on a
determinant of switching rows is to change the sign. Hence, only combinations
of rows, not permutations of rows, are significant. We can therefore f ~ n n ~ ~
(5.2.2-8)
different 3rd-order determinants from our dimensional matrix. We must see if any
of these are not equal to zero.
From the properties of determinants we know that if any two rows of a
determinant are identical, the determinant is zero, so it immediately follows that
any determinant that contains both the first and fourth rows of our dimensional
matrix will be zero. This eliminates two of our four combinations - the first and
fourth rows combined with 1) the second row and 2) the third row.
This leaves us with two determinants to investigate further. One contains
the first row in combination witb the second and third rows, and the other
contains the fourth row in combination with the second and third rows.
However, since the first and fourth rows are identical, these last two determinants
are identical, leaving only a single determinant to investigate
l 9 The notation for combinations is read \u201cn choose r,\u201d and the general formula in
terms of permutations of r things from a set of n is
= P: = n!
r! r! (n - r )
Fraser, D. A. S. (1958). Statistics: An Introduction. New York, NY, John Wiley and
Sons, Inc.
234 Chapter 5: Application of Dimensional Analysis
and so we have a matrix of rank less than three since we have shown all possible
third-order determinants to be zem.
We proceed to check second-order determinants. It is easy to find by
inspection a 2x2 determinant that is not equal to zero, e.g.
[ ;-;I = 1 (5.2.2- 10)
so the rank of our dimensional matrix is two, and for this example, the rank of
the dimensional matrix, 2, is not equal to the number of fundamental
dimensions, 3. We will obtain, therefore, (4 - 2) = 2 dimensionless groups, not
( 4 - 3 ) = 1.
An acceptable set can be seen to be
(5.2.2-1 1)
5.3 Systematic Analysis of Variables
The proof of Buckingham\u2019s theorem is based on a set of algebraic theorems
concerned with the class of functions known as dimensional ly
homogeneous.2o A systematic approach to finding a set of independent
dimensionless products can be based on this algebraic theory.
We now develop a systematic way to convert a functional relationship
among n independent quantities Q1, Q2, ... Qn to one involving (n-r)
dimensionless products lcl, 1c2, ~ 3 , ... z ~ - ~ , where r is the rank of the
dimensional matrix of the Q\u2019s
*O Brand, L. (1957). The Pi Theorem of Dimensional Analysis. Arch. Rd. Mech. A n d .
1: 35.
Chapter 5: Application of Dimensional Analysis 235
In general, the number of dimensionless products in a complete set is equal to
the total number of variables minus the maximum number of these variables
that will not form a dimensionless product (the rank of the dimensional matrix).
The algorithm to find tbe R'S as follows:
1) Select the number of fundamental dimensions.
2) List the Q variables and their dimensions in terms of the
fundamental dimensions.
3) Select a subset of the Q variables equal in number to the
number of fundamental dimensions such that
none of the selected quantities is dimensionless,
the set includes all the fundamental dimensions and
no two variables of the subset have the same dimensions.
4) The dimensionless products (p's) are found one at a time by
forming the product of the subset variables each raised to an
unknown power times one of the remaining Q variables to a
known power. (This process is repeated n-k times, using all
the remaining Q variables.)
5 ) Apply dimensional homogeneity to each of the products
obtained to determine the unknown exponents.
Algebraic manipulation of the dimensionless products that does not change
their number will not destroy the completeness of the set. Thus, a II may be
replaced by any power of itself, by its product of any other ~t raised to any
power, or by its product with a numerical constant.
d e 5.3-1 Drqg force on a S D ~
The drag force, F, on a smooth sphere of diameter D suspended in a stream
of flowing fluid depends on the freestream velocity, v,, the fluid density, p, and
the fluid viscosity, p.
(5.3-2)
236 Chapter 5: Application of Dimensional Analysis
Using the FLt system,
(a) Write the dimensional matrix.
(b) Show that the rank of the dimensional matrix is 3.
(c) Using (I?, D, p) as the fundamental set, determine the
dimensionless group incorporating v,.
Solution
a) The dimensional matrix in the indicated system is
b) We can easily fmd a non-zero third-order determinant, e.g.,
1 0 : :-:I 1 1 = (l)[(l)(-1) -(1)(0)] = - 1
(5.3 -3)
(5.3-4)
c) Choosing Q1= F, Q2 = D, and Q3 = p as the subset, (n - k) = (6 - 3) = 3
F a + c = O 3 a = - 1
L: l + b - 2 ~ = 0 3 b = l
t: - l + c = O 3 c = l
(5.3-5)
Example 5.3-2 Dimensionless prouus for -flow over a -flat ulate
For flow over a flat plate, the pertinent variables in predicting the velocity
in the xdirection are
Chapter 5: Application of Dimensional Analysis
v, = f (x, y, v,, p, p) (5.3-6)
Using the MLt system
(a) Write tbe dimensional matrix.
(b) Show that the rank of the dimensional matrix is 3.
(c) Using (x, p, p) as the fundamental set, determine the
dimensionless group incorporating v,.
Solution
&quot;x x Y V - P P
M I O 0 0 0 1 1
L 1 1 1 1 - 3 - 1
4-1 0 0-1 0 - 1
0 1 1
1 - 3 - 1
- 1 0 -1
(5.3-7)
= 0 - 1 - 3 = - 4 + 0 * r a n k = 3
(5.3-8)
c) Choosing Q1= x, Q2 = p, and Q3 = p as the subset
7c = v, xa p b p c [4] [LIa [3Ib [B]'
M: b + c = O =+ a = l
L: l + a - 3 b - c = 0 * b = l
t: - 1 - c = o * c = - 1
237
(5.3-9)
238 Chapter 5: Application of Dimensional Analysis
W(Lt2)
L
L
L/t
M/(Lt)
M/L3
L
U D l e 5.3-3 Consistencv 0 f d imension less pro U D s across
pvstem of di-
F/L2 F/L2 F/(Ft2/M)2 F/L2
= M2/(Ft4)
L L Ft2/M L I
L L Ft2/M L
L J 0 ° * 5 L/t Ft2/(Mt) L/t - - (LF/M)O.~ = Fth4
M / [ L 0 ° * 5 J (Ft2/L)/(Lt) M/[(Ft2/M)t M/(Lt)
M/L3 (Ft2/L)/L3 M/(Ft2/M)3 M/L3
1
= M2/(Ft3)
- - ( ~ ~ ) 0 . 5 ~ - 1 . 5 = F a 2
= Ft2/L4 = M4/(F3t6)
L L Ft2/M L
ML/(Ft2)
The following table shows a list of possible variables for flow in a pipe.
Apply the algorithm for obtaining dimensionless groups in each of the five
fundamental systems of units MLt, FML, FLt, FMt, and FMLt, and show that
the results obtained are equivalent. Details of the application of the algorithm for
the FML, FLt, and FMt systems are omitted and left as an exercise in Problem
5.20. The rank of the dimensional matrix may be shown in each case to be equal
to the number of fundamental dimensions. The algorithm presented will work for
systems of four fundamental dimensions as well as three if the dimensional
conversion factor & is included in the set of Q\u2019s purely as a formal procedure.
For the MLt system
(5.3- 10)
21 From Newton\u2019s law, F = M U 2 .
22 From Newton\u2019s law, t = (ML/F)0*5.
23 From Newton\u2019s law, M = Ft2/L.
24 From Newton\u2019s law, L = Ft2/M.
Chapter 5: Application of Dimensional Analysis 239```