mv 



Mr. J. L. Woodbridge on Turbines. 317 



— -, as in No. 6 ; hence the total change will be the sum 



r 



of these, and the rate of change will be the sum divided by dt, 

 which result multiplied by the momentum, mv, will give the 

 reaction, the components of which will be 



along the vane, 



. -fi — T-'-r\ normal to the vane. 



L p dt dp dt J 



This completes the reactions. Next consider the pressure 

 in the wheel. The intensity of the pressure on the two sides 



ab and cd differs by an amount dp = -^- dp. The area of the 



face is dc X x=xpdd sin. y, and the force due to the difference 

 of pressures will be 



xpdd sin y-r- dp. 



If dp is positive, which will he the case when the pressure 

 on dc exceeds that on ab, the force acts backwards, and the 

 preceding expression will be minus along the vane. 



In regard to the pressure normal to the vane, if a uniform 

 pressure p existed from one end of the vane VW to the other, 

 the resultant effect would be zero, since the pressure in one 

 direction on YW would equal the opposite pressure on XY. 

 If, however, the pressure at a exceeds that at d by an amount 

 — dp, since Va is longer than X6, the pressure on Ya, due to 

 this — dp, will exceed that on X6 by an amount 



—dp. xx ah=— dp. x.pdO cos y= — xp cos 7 dd -J- dp. 



Collecting these several reactions, we have 





Normal to the vane. 





Along the vane. 



(2) 



+ moo? p cos 7. 





-J- moo? p sin 7. 



(*) 



—mcov. 





0. 



(5) 



dp 

 —moo sin 7 -f. 

 dt 





d ? 

 + m&> cosy -/-. 



' dt 



(6) 



d P 

 —mco cot y cos 7 — . 





dp 

 — mo) cos y -~» 



' dt 



(7) 



0. 





dp dv 

 dt dp' 



(8) 



Tcoty dp dy dpi 

 L p ' dt dp dtj ' 



0. 



(9) 



— xp cos y~ dpdd. 





— xp sin 7 -j- dpdd. 



