466 APPLIED MECHANICS 
(2) Required the diameter of a pipe to connect two reservoirs whiel 
are 10 miles apart. The discharge is to be at the rate of 4,000, ) 0 
gallons per day, and the available head is 350 feet. 
4000000 x 10 1000000 a ; 
62°3 x 24x 60x 60 623x216 
Let d=diameter of pipe in feet, and v= velocity of water through the 
pipe in feet per second. 
Discharge in cubic feet per second = 
™ 72. 1000000 . 14 ,,. 1000000 
4°” 623 x 216’ 623 x 54rd? Ed 
Neglecting all losses except that due to friction in the pipe, then ee 
4i wv? . 4x 10x 5280 1000000? 
cc a med d * 623? x 542n 2d! x 2g’ 
from which d5= 839/. 
1 kf ; 
If be «pat equal to 0-005(1+ 55), then} dita 4195(d-+ 35) 
iat 
which is an awkward equation to solve, and it is simpler to assume a 
value for /, say in this case 0-006, then d= 839 x 0°006 = 5-034, and by 
logarithms d= 1°38. : 
Using this approximate value of d, a more approximate value of 7 is 
1 . 
12 x rh sale 
A more approximate value of dis then d= 3/839 x 0°0053 = 1°35 feet. — 
(3) Reservoirs A and B (Fig. 762) discharge into a reservoir C- 
through pipes AD, BD, and DC, as shown. The lengths of the Pipes 
AD, BD, and DC are 10, 000 feet, P 
6000 feet, and 8000 feet respec- 
tively, and their diameters are 
18 inches, 12 inches, and 21 inches 
respectively. The water levels of 
B and C are 40 feet and 100 feet 
respectively below the water level 
of A. It is required to find the 
rates of flow from A and B. 
Let Q,, Q,, and Q, be the rae, 708 
rates of flow through AD, BD, and DC respectively, in cubic feet per . 
second. 
If h is the loss of head in feet in a pipe of diameter d fect and length — 
1 feet, v the velocity of flow in feet per second, and Q the rate of discharge — 
in cubic feet per second, then 
4b had 
h=f> 7" 99° onVlZ Wee 
hd? . 
= d%y=_al % 
= de wins . 
An average value of / for the pipes in this example may be ere at 
IQ 
about 0-0054, then Q= 43%, and h= a3, 
determined, namely, f= 0-005(1 + 
