Jökull - 01.12.1977, Side 88

of images and observe diat because of the small
amplitude condition, the pressure field in F for
t -» oo will be
Ps (P) = (-fo/^C) [(l/rPQ) + (1 /rPQ’)]» (18)
and the associated surface amplitude is given by
hS (S) = -f„/277/>gCrSQ (19)
where Q' = (0,0,—d) and
rt>Q = [(x~x')2 + (y-y')2 + (z-d)2]^> (20)
rPQ’= [(x-x')2 + (y-y')2 + (z+dm (2i)
rSQ = Kx-x')2 + (y-y')2 + d2]%- (22)
Equation (18) gives the stationary flow pres-
sure due to the sink at Q. To obtain the correct
initial condition at t = 0, we have only to add
to pH the pressure field due to a surface ampli-
tude distribution of —hs imposed at t = 0 on F
in equilibrium. The response to this initial
condition is given by (15) and has then to be
added to (18). The integral in (15) with h0 =
—hs can be evaluated with the help of the
potential theoretical identity
f /rrQ’ = (z/2tt) (l/r3pu) (f/fuq) dau> (23)
2
where U = (x", y"), dal: = dx"dy" and
rpu = [(x-x")2 + (y-y'O + z2]^ 24)
ruQ = [(x"-x02 + (y"-yO + d2]% (25)
Applying (23) we find that the solution to our
problem is
P(P>t) =
Ho/4nC) [(I/rPQ) + (l/rPQ>) - (2/rpq,t)], (26)
wliere t > 0 and
rPQ’t = [(x-x02 + (y-y')2 + (z+at+d)2]>/2 (27)
The elevation of the fluid surface is obtained
from (26) by taking z = 0 and h = p/gp, viz.,
h (S,t) = - (v0/27rgC) [(1 /rBQ) - (1 /rSQ,t)] (28)
where v0 = fn/p is the volume rate of the sink,
S = (x,y) is a point on £ and
rSQ’t = [(x-x02 + (y-y')2 + (at+d)2]+ (29)
DISCUSSION
Equation (27) reveals that the effect of the
free fluid surface on the pressure drawdown
due to the concentrated constant sink of strength
f0 starting at time t = 0 can be represented by
the pressure field due to a stationary image
sink of strength f0 located at Q' = (x', y', —d)
and a moving image source of strength 2f0
located at Q't = [x', y', —(at+d)]. At time t =
0+ the image sink and i/2 of the image source
cancel resulting in an initial pressure field of
p (P,0+) = - (f0/47rC) [(1 /rPQ) - (l/rPQ,)]. (30)
At very large times, that is at t >> d/a, when
the image source has retreated far into tlie
negative half space, the third term in (26) be-
comes negligible and the pressure field reaches
its stationary value ps given by (18).
The fluid surface is then at
h» (s) = - vo/27rgCrSQ, (31)
and the final stationary position vertically above
the sink is consequently
h8(S0) = -v0/27rgCd, (32)
where the point S0 = (x', y', 0).
Moreover, we find tliat the rate of drawdown
vertically above the sink is
r(So>1) = -dh(S0,t)/dt
= (v0/2tt</>) (at+d)-2. ^
It is interesting to note that the initial rate of
drawdown is
r (So>0+) = vc/2tT(f>V, (34)
and the time required to reach l/2 of the sta-
tionary drawdown vertically above the sink
given by (32) is
t,A = d/a. (35)
We will also consider the recovery of the
fluid surface following a period of withdrawal
by the concentrated sink located at Q =
(x', y', d). We assume that the fluid surface is in
equilibrium, at t = 0, that is, h (S,0) = 0. Let
fluid withdrawal by the sink at Q start at
t = 0+ and continue for a period of time t0.
The withdrawal is then discontinued and the
86 JÖKULL 27. ÁR