458 HYDRAULICS AND ITS APPLICATIONS 



energy ; the whole affords an unique example of the possibilities of 

 engineering science. 



Design of Pelton Wheel. 



EXAMPLE. 



To design a Pelton wheel to work under an effective head of 500 feet 

 and to develop 800 H.P. at 360 revolutions per minute. 



Assuming a coefficient of velocity = '985, the velocity of efflux of 

 the jet = '985 X V 500 X 64'4 



= 177 feet per second. 



Taking the velocity of the pitch circle of the wheel as *46 times that of 

 the jet, we have 



Peripheral velocity of wheel = 81*3 feet per second. 



Q i . o vx ar\ 



.'. Eadius of pitch circle = QAr . = 2'158 feet. 



Zi TT X OOU 



/. Diameter of pitch circle = 4 feet 3| inches. 



Next assuming an efficiency of 85 per cent., we have the energy passing 



the nozzle per second given by -- ^ -- ft. Ibs. = 518,000 ft. Ibs., and 



62*4 X (177) 2 

 since each cubic foot of water contains - - ft. Ibs. = 30,380 



nnn 

 ft. Ibs. in the form of kinetic energy, this requires Q.. 1^ = 17*06 cubic 



feet per second. 



1 7'Ofi 



The required area of the nozzle is thus ^(^r = '0964 sq. ft. 



= 13*89 sq. ins. 

 giving a jet diameter of 4 '20 inches. 



/7 

 Taking n = 7*5 v , this gives the number of buckets as equal to 



I 



7'5 'v *nn = 18*6, or say 20 for convenience in balancing. 



Next applying the formula n = - = a we get, on 



substituting for n, r and t, on reduction s = 2'5 inches, giving the 

 amount by which the buckets must project beyond the pitch circle for 

 continuous impact. For safety it is usual to increase this slightly, say to 

 2'75 inches, giving an extreme wheel diameter of 4 feet 9J inches. 

 The buckets would in this case be about 21 inches wide. 



