THE STAND PIPE 565 



sensibly uniform, the energy entering the wheel from the supply reservoir 

 during the change may also be calculated, so that the total energy sup- 

 plied to the wheel during the period of transition may be obtained. 

 Equating this to the energy required to develop the required horse 

 power, the time necessary to produce the required change in the velocity 

 of flow, and hence the acceleration may be calculated in terms of the 

 stand pipe area. Having obtained this, it only remains to equate to the 

 maximum acceleration consistent with good speed regulation. 



The value of the latter factor depends largely on the exasfc requirements 

 of the plant as regards speed regulation. Where this is to be very close, 

 as in a plant for electric driving, the acceleration should not exceed that 

 given by the formula 



075 g h fl 

 a = j -- ft. per sec. per sec. 



= 2'4 j ft. per sec. per sec. 



EXAMPLE. 



Consider a turbine, supplied under a head of 60 feet, through a pen- 

 stock 4 feet diameter and 200 feet long, and working under a normal 

 load of 300 B.H.P. Assuming the efficiency of the turbine to be *80, 



this necessitates a supply of energy = - ^ -- = 206,000 foot Ibs. 



per second. 



And since the energy entering the wheel ) AO A ( 7 ^ fl } 

 casing per second j = v \ n 2 g m } 



Where a = area of penstock = 12*57 square feet. 



h = supply head = 60 feet. 



/ = coefficient of friction = *005 (say). 



This gives on substitution and reduction 



v = 4'4 feet per second 

 as the velocity of flow along the penstock. 



Similarly, if the maximum increase in load is one of 50 per cent, up to 

 450 H.P., the new velocity, when steady flow is once more attained, will 

 be 6*64 feet per second. 



Now, if hj is the head at the entrance to the turbine casing, and 

 therefore at the stand pipe, we have, in the first case 



(I 



= 60 - : 3 (1 + 1) 

 = 59-4 feet. 



