THEORETICAL AND MEASURED PUMPING POWER OF WINDMILLS. QQ 



Losses of Power. 



I. THAT DUE TO THE AXIAL COxMPONEN'L OF WIND PRESSURE. 



Equation (5 ) gives the value of this component on any /^s. 



Multiply this pressure by f, the coefficient of friction, and by 



2Qep* where e equals the radius of friction and p the number of 



revolutions of wheel per second, we have 



, r . . 



/^Lj=^fK — As(c sin X — V cos x)'^sin x 2i_|ep . . ( 17) 



Dividing (17) by (4) after multiplying the latter by 2[^e we 

 have for the ratio of the work lost per revolution in this axis fric- 

 tion to the work per revolution of the mill 



fK '- 



ALj g l\^{c sin X — V cos x)"^sin x 2(_|ep f|_Jtanx 



/^^L^ r /\^s(c sin x — v cos x)2cos x 2[jRp R ' 



g 

 Example: — 12 ft. mill with f=:.o7, e==3/( ft., R:=.4. 56 ft. 



A L .07 X— X • 8 

 From ( 18 ) ^ ---" — * — ~^=.ooc^-=\-, per cent of power of 

 ' aLo 4-56X.6 ^ - i^ 



mill. This friction loss increases with e and x, and decreases as 



R increases. 



II. THAT DUE TO SKIN FRICTION. 



From mechanics we have for this loss of power expressed in 

 head 



32f'hQ2 



"''^cm- ■■■■■■■ ('^^ 



in which h is the height to which the water is raised, Q the number 

 of cubic feet raised per second, d the diameter of the pipe and f 

 the coefficient of skin friction. 



III. THAT DUE TO AXLIC FRICTION. 



For this loss of work per revolution we have 



AL3=rie'f[G + Fr(h+h2)] (20) 



in which G is the weight of the wheel, piston and pump-rod, F the 

 cross-section of the water column, r the heaviness of water, and e' 

 the radius of axle. 



Example: — 12 ft. mill with G — 950 lbs., double-acting pump 

 having 4^ -inch cylinder and 12-inch stroke, 30 strokes per minute, 

 h:=25 ft., back geared 2:1, e' i inch, and f-=.o7. 



*A] = 3.Ul.-i9. 



