Mechanics of Fluids

572 Unsteady flow
Fig. 12.8
The pressure diagrams, Fig. 12.8, show that for the next step – the
reduction of B from 0.10 to 0.08 – matters are slightly complicated
by the return of the first wave, that of magnitude 11.82 m. This rise
of pressure is cancelled as the wave returns (its sign was changed on
reflection from the reservoir) and the wave is then reflected from the
valve with the same (i.e. negative) sign. Consequently the +11.82 m is
dropped from the h column and a −11.82 m takes its place. The next
line of the table is therefore
B (m 1/2 · s −1 ) Head at valve, h (m) u (m · s −1 ) �h (m)
0.08 100 − 11.82 + 13.61 + 15.73 1.094 30.41
+18.18 + 21.01 + 30.41
= 187.12
Since the closure of the valve is not yet completed, the returning
11.82 m wave does not meet an absolutely closed end. Its reflection is
therefore not complete; that is, its amplitude after reflection is somewhat
less than 11.82 m. But the degree of completeness of reflection
happily does not concern us. The amplitude of the new wave (here
30.41 m) must be such as to give a total head (187.12 m) which satisfies
the equation u = 0.08 √ h. The new wave thus effectively incorporates
the loss by reflection of the wave that has just arrived and reversed.
(If the 11.82 m wave had not returned to the valve, �h would have
been 24.29 m. The extra 6.12 m accounts for the fact that only 5.70 m
of the 11.82 m is actually reflected. The total h is the same.) This is
why the interval between successive steps must be either 2l/c or a submultiple
of 2l/c: an ‘old’ wave is reflected from the valve at the same
moment as a new one sets off, and so the new wave can incorporate
the reflection loss of the old.