389
Chapter 5 x Dimensional Analysis and Similarity
Solution: Establish the variables and their dimensions:
M
fcn( R , :, P , T )
{ML2/T2}
{L} {1/T} {M/LT} {1}
Then n 5 and j 3, hence we expect n j
arrangement, as follows:
M
P :R 3
5 3
2 Pi groups, capable of only one reasonable
M
P:T R 3
fcn(T ); if M v T , then
constant
Ans.
See Prob. 1.56 of this Manual, for an analytical solution.
5.36 The rate of heat loss, Qloss through a window is a function of the temperature difference
'T, the surface area A, and the R resistance value of the window (in units of ft2˜hr˜qF/Btu): Qloss
fcn('T, A, R). (a) Rewrite in dimensionless form. (b) If the temperature difference doubles, how
does the heat loss change?
Solution: First figure out the dimensions of R: {R} {T34/M}. Then note that n
and j 3 dimensions, hence we expect only 4 3 one Pi group, and it is:
31
Qloss R
A 'T
Const , or:
Qloss
Const
A 'T
R
4 variables
Ans. (a)
(b) Clearly (to me), Q v 'T: if 't doubles, Qloss also doubles. Ans. (b)
P5.37
The volume flow Q through an orifice plate is a function of pipe diameter D, pressure
drop 'p across the orifice, fluid density U and viscosity P, and orifice diameter d. Using D, U,
and 'p as repeating variables, express this relationship in dimensionless form.
Solution: There are 6 variables and 3 primary dimensions (MLT), and we already know that
j = 3, because the problem thoughtfully gave the repeating variables. Use the pi theorem to find the
three pi’s:
31
D a U b 'p c Q ; Solve for a
32
D a U b 'p c d ; Solve for a
33
D a U b 'p c P ; Solve for a
2, b 1 / 2, c 1 / 2. Thus
1 b 0 c
0. Thus
1, b 1 / 2, c 1 / 2. Thus
31
31
31
Q U 1/ 2
D 2 'p1 / 2
d
D
P
D U 1 / 2 'p1 / 2