§ 53. _Flow from a Vessel when the Effective Head varies with the
Time._--Various useful problems arise relating to the time of emptying
and filling vessels, reservoirs, lock chambers, &c., where the flow is
dependent on a head which increases or diminishes during the
operation. The simplest of these problems is the case of filling or
emptying a vessel of constant horizontal section.
[Illustration: FIG. 64.]
_Time of Emptying or Filling a Vertical-sided Lock Chamber._--Suppose
the lock chamber, which has a water surface of [Omega] square ft., is
emptied through a sluice in the tail gates, of area [omega], placed
below the tail-water level. Then the effective head producing flow
through the sluice is the difference of level in the chamber and tail
bay. Let H (fig. 64) be the initial difference of level, h the
difference of level after t seconds. Let -dh be the fall of level in
the chamber during an interval dt. Then in the time dt the volume in
the chamber is altered by the amount -[Omega]dh, and the outflow from
the sluice in the same time is c[omega][root](2gh)dt. Hence the
differential equation connecting h and t is
c[omega] [root](2gh) dt + [Omega]h = 0.
For the time t, during which the initial head H diminishes to any
other value h,
_ _
/h /t
-{[Omega]/(c[omega] [root]2g)} | dh/[root]h = | dt.
_/H _/0
.: t = 2[Omega]([root]H - [root]h) / {c[omega] [root](2g)}
= ([Omega]/c[omega]){[root](2H/g) - [root](2h/g)}.
For the whole time of emptying, during which h diminishes from H to 0,
T = ([Omega]/c[omega]) [root](2H/g).
Comparing this with the equation for flow under a constant head, it
will be seen that the time is double that required for the discharge
of an equal volume under a constant head.
The time of filling the lock through a sluice in the head gates is
exactly the same, if the sluice is below the tail-water level. But if
the sluice is above the tail-water level, then the head is constant
till the level of the sluice is reached, and afterwards it diminishes
with the time.