568 HYDRAULICS AND ITS APPLICATIONS 



the speed of such a wheel is reduced from <*>i to o> 2 , the store of energy 



given out = g I (>i 2 <o 2 2 ) = ^ I ( w i + "2) Oi - ^2) = S E, 



and if co is the mean angular velocity, this may be written : 



1 BE 



2 Ico 8aj = ~2 



Putting I w 2 = -E, this becomes 



i.e., the proportional change in the store of energy in the wheel is twice 

 as great as the proportional change in velocity. 



Suppose, for instance, in tbe numerical example just considered (p. 565), 

 the turbine, rotating at a mean speed of 240 revolutions per minute 

 (co = 8 TT), to be fitted with a flywheel weighing 5 tons and having an 



effective radius of 2'5 feet. The value of E = ^ I 2 now becomes 



1_ 5 x 2,240 25 , 2 

 2 ' 32-2 ' 4 



= 687,000 foot Ibs. 

 If the maximum speed variation on throwing on the excess load 



CV 



is to be 4 per cent, of the mean, so that - - = '04, we have & E, the 



energy given out by the wheel during its retardation, given by 

 8 E = 2 X '04 X 687,000 foot Ibs. 



= 55,000 foot Ibs. 



If the same acceleration in the penstock be assumed, the amount of 



energy required from the stand pipe during the transition may be reduced 



by this amount, and since each square foot of stand pipe area gives up 



2,830 foot Ibs. of energy, this area may be reduced by - r^r- square 



feet = 19'4 square feet. 



This gives an area = 53'6 19*4 = 34'2 square feet, and a diameter 

 of 6 feet 6 inches, as against 8 feet 6 inches without the wheel. 



Evidently with a sufficiently heavy flywheel it would be possible to 

 eliminate the upper cistern altogether. 



For further information on this subject, Papers by E. D. Johnson 1 

 and by Professor Irving P. Church 2 should be consulted. 



1 Am. Soc. C.E., June ; 1908, p. 443. 



a Cornell Civil Engineer, December, 1911, p. 114. 



