1919.j 
Parry.—The Surge-chamber Problem. 
81 
in the surge-chamber below static level. If now the flow be suddenly- 
interrupted by the closing a valve or sluice at the outlet from the surge- 
chamber the energy is absorbed in two ways—viz., in overcoming the 
inertia of the mass, and in friction. The decay of energy at any instant 
is then 
/1 dv 
\p di 
This is balanced by an increase of potential energy in the surge-chamber, 
manifested by the rise of level h above the value —H, at which the level 
stood when the flow had a constant value V. We have then an equation 
and also an identity 
where A is the area of the surge-chamber and dln/dt the velocity of flow 
in the surge-chamber. Differentiating and substituting, we obtain a 
differential equation 
d 2 v q dv a q 
-P K . - .-1— . - 
dP ^ I dt^ A l 
v = o 
the solution of which is 
where 
v = Ae m -}- Be m - 2 f 
m 
, = _ + v k a - - • 9 - 
1 2/ v 4/2 A / 
and 
m. 
Kff _ ^ / KA / 2 a g 
2 1 V IF “ A ’ 1 
In the majority of cases ag/Al is greater than K 2 g 2 /il 2 , and in conse¬ 
quence the expression under the root is negative, and the solution becomes 
v = Ae (a + ^ -p Be (a - 
where a = — ^ and fi = \/~ • % — 
ZL v A l 
KV 
IF 
Transforming, we have 
v — e at {{A + B) cos fit -P i (A —B) sin fit }. 
dv 
Prom the consideration that when t = o, v = V, and — = o, we find that 
at 
A + B 
Consequently 
V and i (A — B) = — - V. 
a 
v ----- Ye at (cos fit — — sin fit) 
,( 2 ) 
which is the equation of an oscillatory discharge having a frequency of 
2 rr/fi and subject to an attenuation or damping governed by e at . 
The level of the water in the surge-chamber in terms of the velocity at 
the instant of closure and of other quantities may be derived by differenti¬ 
ation of equation (2) and substitution in (1), whence we obtain 
h = - * V • e at -f fij sin fit — K • Y ■ e at (cos fit — ^ sin fit) 
6 —Science. 
