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GENERAL PRINCIPLES OF HYDRAULICS 475 
29. A channel, with a bottom width of 30 feet, and side slopes of 2 hori- 
zontal to 1 vertical, flows full of water to a depth of 7 feet. Find the velocity 
flow in, and the discharge of, the channel, the inclination being 1 in 5000, 
of 
and ¢ in the Chézy formula 100. [Inst.C.E.] 
30. A channel with cement sides is 4 feet wide at the bottom, and its sides 
ag at 2 vertical to 1 horizontal. The slope of the channel is 1 in 500. 
will be the depth of the water for a discharge of 12,000 gallons per 
minute. Take the coefficient of friction as 0-006. (U.L.] 
$1. A channel in which the water is to run 3 feet deep has to discharge 
60 cubic feet per second, the velocity of flow being 2°5 feet per second. The 
sides slope at 1°5 vertically to 1 horizontally, and the value of ¢ in the formula 
v=c,/mi may be taken as 110, feet and seconds being units. Find the width of 
the el at the bottom, and the hydraulic inclination necessary. (U.L.] 
2. Apply Kutter’s formula to find the rate of discharge in cubic feet per 
second of a channel having a bed width of 20 feet, side slopes of 14 horizontal 
to 1 vertical, depth 6 feet, and longitudinal slope 1 in 5000. Take n, the co- 
efficient of roughness, = 0°02. 
33. Water flows in a pipe without filling it, Show that the velocity of flow 
for a given slope is a maximum when the wetted perimeter subtends an angle 0 
at the centre given by the equation @=tan 0, and that @=2574 degrees nearly. 
$4. The cross section of a closed channel is a square with a diagonal vertical. 
is the side of the square, and y is the depth of the water line below the apex. 
obra for maximum discharge y=0'127s, and that for maximum velocity 
=0° 
, 35. A cast-iron pipe 18 inches in diameter is laid with a slope of 1 in 1000. 
Water flows through this pipe with a depth of 135 inches, Taking ¢ in the 
formula v=c,/mi as 125, find the discharge in gallons.per hour. 
414. Impact of a Jet on a Flat Vane.—Casz I. Direction of jet 
perpendicular to vane. Vane at rest (Fig. 770).—A =sectional area of 
jet. v=velocity of jet before impact. W-=weight of liquid reaching 
vane per second. w=weight of unit of volume of liquid. P= total 
normal pressure on vane due to impact of jet. 
Since the motion of the liquid in the direction 
in which the jet is moving is entirely destroyed, the 
loss of momentum per second in that direction is v—=S====4qe-P 
ae and therefore 
2 . 
pa We_ war? _ ows. v  QwAh, Fra. 770. 
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where hf is the-head due to the velocity v. But wAh is the static 
pressure on an area A due to a head h. Therefore the total dynamic 
pressure due to the impact of the jet on the vane is equal to twice the 
static pressure on an area A due toa head h. In 
other words, the total dynamic pressure of the jet 
issuing under a head h will balance a_ static 
peas on an area due to a head 2k. This may 
demonstrated by the apparatus shown in Fig. 
771, where B is a tank containing water to a 
constant height h above the axis of a tube pro- 
jecting from the side of the tank. C is another 
tank containing water to a height 2h above Fic. 771 
the axis of a projecting tube of the same size and eS 
shape as that projecting from B. A flat plate D is suspended loosely 
against the mouth of the tube on C, and is held there by the force of 
the jet from B, as shown. Experimentally, the head in C will be slightly 
