468 APPLIED MECHANICS 
may be varied. Diameter of pipe=d feet, length of pipe =/ feet, total 
head at nozzle=h feet, and v=velocity of water through the pipe 
in feet per second. Neglect the loss of ~ 
energy at the entrance to the pipe and at : 
the nozzle. 
Energy delivered at the nozzle per lb. of 
4i 
water per second =h—/- =a 39° 
Total energy delivered at the nozzle per second 
Al v 
atm 
i” aul - I-43 sa) 
Fie. 763. 
g 
: m7d?vw Al 
€ =H= ei J Vea ee 
Horse-power delivered at nozzle = H = TxbE fas fd ot dy 
To find when H is a maximum 
dH __7d*w wah _ 3f. 4] £). 
dv 4x 550 d 29 
ic > 
Hence H is a maximum when 3/: a om =h, or when v= nee 
d’wh |ghd 
The maximum horse-power is therefore = a 550 se . 
If there were no friction, the horse-power delivered would be 
_ Tavwh | 
14x 550 
H 4l # 
The efficiency is therefore Hi, 1-/f- id’ Ey 
When H is a maximum, the efficiency is 3. 
If P is the pressure of the water in lbs. per square foot just before . 
passing the valve at the nozzle, a the area of the cross section of the pipe, — 
a, the area through the valve or through the contracted nozzle, and v, 
the velocity of the water through the contracted nozzle, then . 
a ot ere ee 
therefore — aki if tay 29” and — » »/ Qghd — 4flv2/° 
1 
coiate ta _  Ighd Oh d 
When H is a maximum, v= af 6fl then aR ef Sil . 
ExampLe.—Let d=0°5 feet, 7= 400 feet, h = 300 feet, and f= 0°006. © 
dw Al v 
Th H= fom ( :. 
en rea a ee = 0:0667r(100 — 0:099402). 
, ; ghd _ 
H is a maximum when v= oe fl 
Maximum horse-power = 0°0667 x 18°31(100 — 0°0994 x 18°312) =81 “4. 
Efficiency per cent. = 100(1 = 7" 5)" 100 — 0-099402, 
= 18°31 feet per ‘ge0ond, . 
