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The solutions to the final exam problems for ce 330, focusing on hydraulics and open channel flow. The problems include determining resultant forces, velocities, discharges, energy losses, and forces on sluice gates and vane series. Students can use this document as a study resource for understanding hydraulics concepts and solving similar problems.
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CE 330, Final Exam, Sp 05 Name ___________________________________________ (Problems 1-4 = 65%; problems 5-7= 35%)
Problem 1 The body is submerged in a water tank in four different positions. Its dimension into the tank is 10 feet and the depth of water is 16 feet. Determine the
on the body for each of the 4 positions. Draw FBDs for each case on the picture below.
3-D View
10’10’
3-D View
16’ a b (^) c
d
16’ a b (^) c
d
Problem 2 The velocity in the pipe shown below is 10 fps. Determine the discharge and the elevation of the lower tank. Then plot the EGL and HGL on the next page after filling in the values in the table for sections a-f as shown below.
Z (^) a=100’
n = 10-5^ ft^2 /sec ks = 0.0001’
a b c^ d^ ef
V = 10 fps
Z (^) a=100’
n = 10-5^ ft^2 /sec ks = 0.0001’
a b c^ d^ ef
V = 10 fps
Problem 3 A 0.5-inch diameter free jet enters the vane series as shown. Determine the following (a) The magnitude of V (^) 1,abs.
30 o
45 o
u = 30 fps
20 o
V1,abs
30 o
45 o
u = 30 fps
20 o
V1,abs
20 o
V1,abs
(b) The absolute velocity vector of the outgoing jet. (c) The horsepower the jet loses going through the vane. (d) The force vector exerted by the vane series on the jet. (x and y components)
Problem 4 Water flows as a free jet from the 4-inch diameter pipe shown. Neglect friction and local losses and determine (a) The discharge. (b) The maximum height of the jet relative to the large tank water surface elevation. (c) The velocity vector just before the jet strikes the water in the 4’x4’x4’ water tank. (d) The increases in the readings of Scales A and B when the jet impinges on t water in the tank. Draw FBDs here since you will be using the momentum equation. The reading of Scale B is zero before the jet impinges.
he
(e) What is the total reading on Scale A with the impinging jet if the tank weighs 400 lb when empty?
55 o
D = 4”
1
2
3
4
4’x4’x4’ tank of water
Large tank (water)
Scale A
4’ Scale B
55 o
D = 4”
1
2
3
4
4’x4’x4’ tank of water
Large tank (water)
Scale A
4’ Scale B
Problem 6 The channel on the right has a slope of 0.001, n = .030.
(a) Determine discharge if the uniform depth of flow is 9 ft as shown. (b) Determine the Froude number for the uniform flow. (c) Determine the critical depth for the channel.
Problem 7 A long trapezoidal channel with a 15-ft base width and 1.5 to 1 side slopes flows at a uniform depth of 10 ft. If n = .020 and So = 0.0008 determine the following.
10’
y (^2)
flow
(a) The discharge and the specific energy, (b) The minimum width of an RCB required to pass the flow without creating a choke if the entrance loss coefficient is 0.5. Assume inlet control. (c) The amount by which the culvert inlet must be dropped if an 18-ft wide RCB is used and a choke is to be avoided. (d) The depth in the trapezoidal channel just upstream from the culvert if an 18-ft RCB is used with no dropped inlet. (e) For case (d), determine the distance upstream from the culvert where the depth equals 11 ft. Use the step method with 2 steps.
G C
M
G C
M
Where G = center of gravity C = center of buoyancy M = metacenter Ioo = second area moment about plane where water surface cuts body
If GM positive, stable If GM negative, unstable
water = 62.4 lb/ft 3
Forces on an inclined plane
F = pcA = γ ybar(sin β ) A acts at (xcp, ycp) where
ycp = ybar + I/ybarA and I is the second moment about the centroidal x-axis.
Forces on a curved surface and bouyant force (hor.) Fx = Pc,x Ax; Fy = Pc,y Ay acts at center of pressure of projected area
acts through centroid of V*
bouyant
acts through centroid of Vdisplaced
y (^) cp
b/2 b/
b
y (^) cp
b/2 b/
b
3 12
1 CG where I bh V
I GM = oo^ − oo =^3
3
I bh
bh
tri
rect
I = h b h b
( )
( )
ρ water =
3
min
1 2
2
550 ( )
550 ( )
2
slug
ft
lb g
P P P
h h h
Qeff
horsepower h
Q
horsepower eff whereh
H h H h h
g
V z
P H
water water
absollute gauge atmospheric
losses friction or
t
p
p t losses
= =
= +
= +
=
=
= + +
γ ρ
γ
γ
γ
CONTINUITY EQUATION
For one-dimensional flows
∑ρV^ i
. A i =^ d/dt (∫ρdV)
where V (^) i and A (^) i are vectors and dV is the differential volume.
For steady flow
∑ρV^ i
. A I = 0
or for incompressible, constant density fluids
Q in = Q out
Note for vanes the relative velocity vector is constant in magnitude and it is tangent to the vane everywhere.
Energy Equation H 1 = H 2 + hlosses where H = z (^) bed + y + V^2 /2g = z (^) bed+ E E = y + V^2 /2g
Mom. Eq. M 1 = M 2 + Fobs/ γ^ Friction is neglected for vanes. where M = y^2 barA + Q^ /gA where ybar is the distance Use Q* = V^ absA for a vane series from the water surface to the centroid of^ and Q* = V^ A for a single vane the cross section.
rel 2 2
z o oz i i z
y o o y i i y
x o ox i ix
P = Fxu = γ Q(V 1 -V 2 ) 2g
V (^) abs = Vrel +u (vector addition!!!)
a^2 = b^2 + c^2 - 2bc cos θ 1
a
b c
θ 2 θ 1
θ (^3)
sin θ 1 = sin θ 2 = sin θ 3 a b c
.
xu^ =^ γ Q(V 1 -V 2 ) (eff) (eff) 2g
Open Channel Flow Equations Definitions Energy Equation H = H + h
o c o c o c
w o f
w
MildSlope y y Critical Slopey y SteepSlopey y
nQ P S bed slope S frictionslope
B surfacewidth A area P wetted perimeter
gA
Fr
10 / 3
(^24) / 3
3
1 2 losses^2
z E 2g
H z y bed
2 where = bed + + = +
2
2 2
2gA
y 2g
E = y+ = +
γ
Momentum Eq. M 1 =M 2 +
where gA
M y A
2 = +
Critical flow
1 gA
Fr (^3) c
c
2 (^2) = =
2 c
2 c c
c c c 2gA
y 2B
E =y + = +
Rect culverts
2
2
2
b
q Vy gy
q = = E = y +
yc E c 3
g
q ; y ( gy
q Fr 1/
2 3 c
2 (^2) = = =
Uniform flow (^) 2/3 1/ w
5/ So P
n
(1-Fr ) x x (y -y) o f,,ave
ave 2 = 1 + 2 1
2
where (^) 10/ ave
4/ 2 w,ave f,ave A
nQ S =(
gA
F ave (^3) ave
ave
2 r^2 =
( ) 2
y y^1 y^2 ave
= ; Aave = byave + myave^2 ; Bave = b + 2 myave ; Pw , ave = b + 2 yave ( m^2 + 1 )
b m
y 1
b m
y 1
y dist fromwsto centroid
by my yA
P b y m
B b my
A by my
w
2 3
2
2
(z -z ) 2g
E y (1 K ) 2 1
2 2 1 = 2 + + ent +
If section 2 is critical, like a culvert under inlet control, then
2g 2
V (^22) yc rect culverts =
y (^1) y 2 = y (^) c
y (^1) 2 )y (z -z^ ) y^2 = y^ c
i.e. E 1 =(1.5+ ent c+ 2 1
c
c B
2g 2
non-rect culverts =
(z -z ) 2B
E y (1 K ) 2 1 c
c i.e. 1 = 2 + + ent +
Steep Channel yc > yo
y (^) c
S 2 curve
Mild Channel y (^) c < yo
y (^) o
y (^) o y (^) c
CDL
CDL
NDL
NDL
Steep Channel yc > yo
y (^) c
S 2 curve
Mild Channel y (^) c < yo
y (^) o
y (^) o y (^) c
CDL
CDL
NDL
NDL
Eq. 1 ( ) lake c
ent c c E B
k A y =
Eq. 2 Mild channel
( ) lake wo
ent o o o (^) P E
n
k S y = ⎥
3
4
,
If slope is unknown
Alternate 4-
