1. A nozzle discharges 2cfs with a
1. A nozzle discharges 2cfs with a velocity of 50ft/sec. The nozzle ivelocity of 50ft/sec. The nozzle is inclined downward so that as the jet s inclined downward so that as the jet strikesstrikes a fixed curved vane it is directed 30 degrees from the horizontal. The jet is deflected upward 90 degrees, a fixed curved vane it is directed 30 degrees from the horizontal. The jet is deflected upward 90 degrees, making an angle of 60 degrees with the horizontal as
making an angle of 60 degrees with the horizontal as it leaves the vane. Determineit leaves the vane. DetermineXXandandYYcomponents of thecomponents of the force exerted.
force exerted.
Sol’n. Sol’n. V
V11= = 50ft/s 50ft/s VV1x1x= 50cos30= 50cos3000 VV1y1y= - 50sin30= - 50sin3000 Q
Q = = 2cfs 2cfs VV2x2x=50cos60=50cos6000 VV2y2y=50sin60=50sin6000
F
FXX==
(V(V1x1x– V– V2x2x))=
=
$&{$
$&{$
%$$
%$$
(50cos30(50cos3000- 50cos60- 50cos6000))F
FXX= = 71lb 71lb to to the the rightright
F
FYY==
(V(V1y1y– V– V2y2y))=
=
$&{$
$&{$
%$$
%$$
(-50sin30(-50sin3000– 50sin0– 50sin000))F
FYY= = 265lb 265lb downwardsdownwards
2. A jet having a diameter of 2 inches and a velocity of 50ft/sec is deflected by a vane which is curved trough 2. A jet having a diameter of 2 inches and a velocity of 50ft/sec is deflected by a vane which is curved trough an angle of 60 degrees and which is moving with a velocity of 20ft/sec in the same direction as the jet. an angle of 60 degrees and which is moving with a velocity of 20ft/sec in the same direction as the jet. Determine
DetermineXXandandYYcomponents of the force exerted and direction and velocity of water leaving the components of the force exerted and direction and velocity of water leaving the vane.vane.
Sol’n. Sol’n. u = V u = V11– V’– V’ u = 50 – 20 = 3 u = 50 – 20 = 30ft/s0ft/s Q’ = Au Q’ = Au = =
&&
ӘӘ
#$
#$
$$
әә
$$
(30)(30) Q’ = 0.6545cfs Q’ = 0.6545cfs V V1x1x= V= V11= 50ft/s= 50ft/s V V1y1y= 0= 0 VV2y2y= u sin60= u sin6000= 30sin60= 30sin6000 = 25.98 ft/s
= 25.98 ft/s V
V2x2x=V’ + u cos60=V’ + u cos6000= 20 + 30cos60= 20 + 30cos6000
= 35ft/s = 35ft/s F FXX==
(V(V1x1x– V– V2x2x)) = =$&{"'&'
$&{"'&'
%$$
%$$
(50 - 35)(50 - 35) F FXX= 19lb= 19lb F FYY==
(V(V1y1y– V– V2y2y)) = =$&{"'&'
$&{"'&'
%$$
%$$
(0 – 25.98)(0 – 25.98) F FYY= 33lb= 33lb V V22=={
{ ˢˢ
$
$$$
++ˢˢ
$
$$$
)) V V22=={
{ ŷŹ
ŷŹ
$$
++ŶŹ
ŶŹ
$$
)) V V22= 43.59 ft/s= 43.59 ft/s Tanβ = V Tanβ = V2y2y/ V/ V2x2x β= 25.98 / 35 β= 25.98 / 35 β = 36.58 β = 36.58004. A submarine when running submerged in sea water has a drag coefficient of 0.15 and a projected area of 65 sq. ft. Determine the drag force for a velocity of 15miles per hour and the horsepower required to overcome this force.
Sol’n.
Specific gravity of Sea Water = 1.03 V = 15mi/hr. = 22ft/s FD= CDγ A
$
FD= 0.15(62.2 x 1.03) (65) [$$
${%$$
]
FD= 4860lb3. A jet having a diameter of 11/2 inch and a velocity of 60ft/sec strikes a series of vanes so arranged
in the periphery of a wheel that the entire jet strikes the vanes. The deflection angle of the vane is 170 degrees. Determine the maximum amount of work that can be done and the direction and absolute velocity of water leaving the vanes
1. An earth canal carries a depth of water of 6ft. The canal is 20ft wide on the bottom and has side’s slopes of 1.5 horizontal to 1 vertical. S = 0.0002. Using a value of n of 0.025, compute the discharge
by the Manning formula and with this discharge determine: a. the value of n in the Kutter formula. b.
the value on m in the Basin formula.
Sol’n: Θ = 56.30 y = 6sec56.300 y = 10.81ft P = b + 2y P = 20 +2(10.81) P = 41.62ft A = 20(6) + 2(
#
$
(6) (9) A = 174ft2 R =˓ ˜9
=ŵŻŸ ŸŵźŶ
9
R = 4.18 Q = A (#&
)
R2/3 S1/2 Q = 174 (#&
""$'
)
(4.18)2/3 (0.0002)1/2 Q = 379cfs. Q = AC{˞˟
C =
=#&{&#{""""$
%
a.) By Kutter Formula
C =
&#'
#
{&#'
75.33 =
&#'
#
{&#'
n = 0.0253 b.) by Bazin Formula C =#'
#
75.33 =#'
#
m = 2.23 C = 75.332. What should be the slope of a planed timber flume rectangular in cross-section, to carry 30cfs if the width is 4ft and the depth of flow is to be 1.5ft? Is the flow at upper or lower stage? Determine the critical depth and critical slope for the given discharge and width.
FN =
ˢ
˧ˤ
9
; Q = VA 30 = V (6), V = 4.997 FN =ŸŻ
ŷŶŶ{ŵŹ
9
FN = 0.719 < 1 (Upper stage) A = bd = (1.5x6) A = 6ft2 P =
+ 2(d) P =
#'
+ 2 (1.5) P = 7 R =˓ ˜9
=ź Ż9
Q = A (#&
)
R2/3S1/2; n = 0.012 30 = 6 (#&
""#$
)
(ź Ż9
)2/3 S1/2 S = 0.002 dc =
J
Ŷ
˧
; q =˝ I9
= 7.5 dc =
ŻŹ
Ŷ
ŷŶŶ
dc= 1.2ft VC = (#&
)
Rc2/3 Sc1/2; n = 0.012 VC =J VU9
=ŻŹ ŵŶ
9
= 6.25 ft/s RC =˓ U
9
=Ÿ{ŵŶ
:
Ÿ-Ŷ{ŵŶ
=ŷ Ÿ9
6.25 = (#&
)
(ŷ Ÿ9
2/3Sc1/2 Sc = 0.00373. Find the most efficient cross-section and the required grade of a trapezoidal canal in clean earth with good alignment to carry 470cfs at a velocity of 3ft per sec. assuming side slopes of 2 horizontal to 1 vertical Sol’n. Q = 470cfs V = 3ft/s y =
Ź
d Θ = 63.430 Q = AV 470 = A (3) A = 156.67 ft2 x = 2y b + 2 (2d) = 2 ( Ź
d) b + 4d = 2 Ź
d b = d (2 Ź
– 4) A = bd +#
$
(
d) (2d) (2) A = [d (2 Ź
– 4)] d +#
$
(
d) (2d) (2) A = .472d2 + 2d2 A = 2.472 d2; A = 156.67ft2 d = 7.96 ft R =ˤ Ŷ9
R =Żź Ŷ9
= 3.98 b = 7.96 (2 Ź
– 4) b = 3.76ft Q = A (#&
)
R2/3S1/2; n = 0.02 470 = 156.67 (#&
""$
)
(3.98)2/3S1/2; n = 0.02 S = 0.000264. What should be the bottom width and the depth of flow for a concrete-lined canal of most efficient
cross-section with side slopes 3/4 horizontal to 1 vertical to carry 1200cfs on a grade of 5ft per mile?
What is the velocity of flow?
Sol’n: Q = 1200cfs n =
'
'$'
= 0.00095 x = 2y b + 2(0.75d) = 2 (1.25 d) b + 1.5d = 2.5d b = d A = bd +#
$
(
d) (0.75d) (2) A = d2+ 0.75d2 Q = A (#&
)
R2/3S1/2; n = 0.013 1200 = (d2+ 0.75d2) (#&
""#%
)
(
$
)
2/3 (0.00095)1/2 b = d = 8.58ft Q = VA V =#$""
{'
"'{'
V = 9.31 ft/s5. Determine the slope in feet per mile that a circular sewer 5ft in diameter must have when flowing at its maximum capacity if the mean velocity is 8ft per sec.
cos
ß
$
=Ŷŵ
$'
, θ = 57.670 d = 0.938(5) = 4.69 A = πr2-^
Ŷ
õ
%"
+$
ŵ
r sin θ = π (2.5)2-
{
ŶŹ
Ŷ
{ŹŻźŻ
%"
+ŵ
$
(2.5) sin (57.67) A = 19.13ft2 Q = AV = 19.13(8) =153.04 R =˓ ˜9
=#%
b{{!!
= 0.4949 Q = A (#&
)
R2/3S1/2; n = 0.000087 153.04 = 19.13 (#&
"""""
)
(0.4949)2/3S1/2 S = 5.639x10-7(5280) S = 0.002966. A smooth wooden flume with vertical sides is 8ft wide and has a grade of 0.002. The discharge is 190cfs, the depth of water being 3ft. It is proposed to construct a weir in the flume which will raise the water surface 1ft at a section 1000ft upstream. Assuming the water to have a plane surface, what will be the depth of water 500ft upstream from the weir?
7. Water upon leaving the spillway of a dam passes over a level concrete apron 200ft wide. Conditions are such that a hydraulic jump will form on the apron. When the discharge is 50cfs per ft. width of channel and the velocity where the water leaves the spillway is 45 ft/sec and the depth after the jump is 10ft., Determine the distance downstream from the dam to the place where the jump occurs. Sol’n: q = 50ft3/s V1– 45ft/s b = 200ft d3= 10ft q = v1d1 50 = 45d1 d1= 1.11ft
=
{
$
'"
%$$
=
{#"{
$
#"
d2= 1.37ft q = v2d2 50 = v2(1.37) v2= 36.5 ft/s A1= bd1 = 200(1.1) A1= 222 P1= 200 + 2 (1.11) P1= 202.22 R1=˓ ˜9
=ŶŶŶ ŶŴŶŶŶ
9
= 1.098 A2= bd2 = 200(1.37) A2= 274 P2= 200 + 2 (1.37) P2= 202.74 R2=˓ ˜9
=ŶŻŸ ŶŴŶŻŸ
9
= 1.35 Rm=#"#%'
$
=
1.224 Vm=%'&'
$
=
40.75 S =
#&{
È
n = 0.0113 ={""##%
{&"'
#&{#$$&
È
=
0.1089 H1=
$
+ d
1=
{&'
${%$$
+ 1.11 = 32.55 H2=
$
+ d
2=
{%'
${%$$
+ 1.37 = 22.06 L =
=
$$"%$''
""#"
L = 96ft5. A 24-inch cast-iron pipe, 3/4 thick and 2,000ft long discharges water from a reservoir under a head
of 80ft. If a valve at the discharge end is closed in 6 seconds, determine the magnitude of the resulting water hammer.