Page 134 - Bird R.B. Transport phenomena

§4.1 Time-Dependent Flow of Newtonian Fluids 119
We now apply the boundary and initial conditions in the following way:

B.C. 1: Because $ f = 0 at 77 = 0, the function / must be zero at 77 = 0. Hence С must be zero.
B.C. 2: Because ф = 0 at 77 = 1, the function / must be zero at 17 = 1. This will be true if В = 0
(
or if sin с is zero. The first choice would lead to / = 0 for all 17, and that would be physically
unacceptable. Therefore we make the second choice, which leads to the requirement that с =
0, ±тг, ±2тг, ±3тг, • • • • We label these various admissible values of с (called "eigenvalues'') as
c and write
n
c n = птт, with n = 0, ±1, ±2, ±3, • • • (4.1-32)

There are thus many admissible functions /„ (called "eigenfunctions") that satisfy Eq. 4.1-29
and the boundary conditions; namely,
/„ = B sin niTT), with w = 0, ±1, ±2, ±3, • • • (4.1-33)
n
The corresponding functions satisfying Eq. 4.1-28 are called g n and are given by
2
2
g n = Л ехр(-и тг т), with n = 0, ±1, ±2, ±3, • • • (4.1-34)
;
I.C.: The combinations f g satisfy the partial differential equation for <j> in Eq. 4.1-25, and
t
n n
so will any superposition of such products. Therefore we write for the solution of Eq.
4.1-25
2
2
ф = ^ А» ехр(-и тг т) sin HTTT j (4.1-35)
{
in which the expansion coefficients D tl = A B n have yet to be determined. In the sum, the term
n
n = 0 does not contribute; also since sin(-7?)7ri7 = —sin(+п)тгг1, we may omit all the terms
with negative values of n. Hence, Eq. 4.1-35 becomes
2
2
<f> t = 2 А, ехр(-и тг т) sin птгт) (4.1-36)
H = l
According to the initial condition, <f> t = 1 - 17 at т = 0, so that
s
1 - V = Ё A, ^ И7Г77 (4.1-37)
We now have to determine all the D n from this one equation! This is done by multiplying
both sides of the equation by sin тттт\, where m is an integer, and then integrating over the
physically pertinent range from 17 = 0 to 17 = 1, thus:

I (1-17) sin min]dr\ = 2 A I s m П7ГГ 1 s m WTTTJ^TJ (4.1-38)
Jo »=i Jo
The left side gives l/ттг; the integrals on the right side are zero when n Ф m and \ when n =
m. Hence the initial condition leads to

D,,,=4 (4.1-39)
The final expression for the dimensionless velocity profile is obtained from Eqs. 4.1-24, 36,
and 39 as
9
2
2
, т) = (1 - 77) - 2 (ш) ехр(-и тг т) sin птгг) (4.1-40)
The solution thus consists of a steady-state-limit term minus a transient term, which fades out
with increasing time.