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CE 330 Final Sp 08 Name_______________________________________________
AIR
OIL, s = 0.
Pgauge = 2 psig
3’
2’
4.5’
9’
4.8’
6.6’
L = 12’
water
water
3D View
1.8’
Tank B
Tank A
12’
a b
c e d
f
Problem 1 Two tanks are filled with fluids and connected as shown. (a) Determine the pressure at the top and bottom of Tank A and Tank B. (b) Draw and label the pressure distribution on the side abcd of Tank B. (c) Determine the resultant force the water exerts on side abcd and determine the moment it makes about the line cd.
, ,
,
top A bot A air oil w
bot A air
a p psfg psig p
p
γ γ γ
γ
= + (^) ( )( ) ( )
( )
( )
( )
, ,
, ,
,
2
245 .5 9^ 4.
3.93 ; arctan 17.
90 72.3 3.93 / sin 72.3 4.
bot A bot B
top B bot B w
top B o eq w o o
p psfg psig p
p p psfg psig
p
b h ft
ft
L
γ
θ γ
β θ
( ) ( ) ( )( )
( )( )
( )
2
3
sin 72.3 62.4 7.60sin 72.3 6.93*
o o w abcd
cp abcd
cd
ft y ft
F y A
F lb
I
y y f
yA
M arm F ft lb
γ
t
100’ Pgauge = - 12.5 psig
80’ 40
120’
50’
AIR 70’
Tank B WATER
Tank A WATER
ks = 0.0001’ ν = 10 -5^ ft^2 /s D = 2 inches
K = 0. K = 0.
K = 0.
K = 1
=+12.5 psig Problem 2 Water flows by gravity from Tank A to Tank B. (a) Determine the discharge in cfs. Do NOT assume fully rough turbulent flow. (b) Determine the drop in the water surface elevation of Tank A when the flow is zero. Assume that the pressure at the top of Tank B remains 12.5 psig. Also assume that the cross sectional area of both tanks is the same.
( )
( )
( )
2 2 2
2 (^1 2 2 )
2
5 4
12.5* 70 98.
8
1.154 32.6 1260 3.
32.6 1260 3.
. .0006 0. (2 /12)
4 7.6410 2.83
.0006 0.
w
losses
s r
s
p a H z f
fL H H h K Q gD D
f Q
Q f
k f D
Q cfs
Q Re Q D k f D
γ
π
π ν
= + = + =
⎛ ⎞ = + → = (^) ⎜ + Σ ⎟ ⎝ ⎠
= +
=
= = → =
=
= = =
= → =
( ) ( )
( )
( )
4
1 2
6
0.0313 2.39*10 0.
12.5 144 100 70
1 12.5 144 100 70 0. 2 62.
Q cfs Re f
Q cfs DONE
b H H
z z
z f
= → = → =
= →
=
− Δ = + Δ +
⎛ ⎞ Δ = (^) ⎜ − − (^) ⎟= ⎝ ⎠
t
35 o
V = 120 fps
D = 3 inches 30 fps
( )
( )
2
(^2 2 2 ) 1, 2,
(1 cos 35) 139
rel rel
abs x rel rel
x
rel abs^ abs
a V fps Q cfs
Q cfs
F Q V lb
b P F u ft lb sec
HP P horsepower
Q V^ V
HP
g g
HP
⎝ ⎠ ⎝^ ⎠
( )( ) 2 2 2,
2,
2,
2 2 1, 2,
90 30 2 90 30 cos145 115.
sin sin
90 cos 35 90sin 35 73.7 51.
o abs
o abs
rel
rel abs^ abs
horsepower
c V fps
V i
V i j i j
Alternate b
Q V^ V
HP
g
uuuuv v v
uuuuv v v v v
( )
u = 30
V^ 2,rel
= 90
V^ 2,abs
β^145 o^35 o
j fps
62.4 4.42 1202 115.86^2
g
HP horsepower
Problem 4 Darius and Gretta had a mishap when their dad took a corner too fast in their water jet car. The kids were tossed from the car and landed in a fish pond. Fortunately they were able to climb onto a 20-ft long timber that has a square cross section of 1.5 ft x 1.5 ft. The specific gravity of the timber is 0.5. (a) If Gretta weighs 150 lb and Darius weights 300 lb, determine the portion of the timber that is above water before and after they climbed aboard. Express the answer in percent. (b) Assuming that they are each sitting 1 ft from the ends of the timber, determine the dimension, m, on Gretta’s end. Draw a FBD on the distorted figure (b) below.
Darius
Gretta
m
(a)
m
150 lb
300 lb
20 ft
18 ft
1.5 ft x 1.5 ft‘
Specific Gravity = 0.
(b)
( )
2
3 3
wood wood wood wood z b b
b w displaced displaced
above
w w
a V ft
W V lb
F F F before
F V V ft
V ft after
h b l l
m
20 ft 1.5 ft x 1.5 ft‘
Specific Gravity = 0.
h l
lb
( ) ( ) ( )
( ) ( )
w w
h
M
h l
h l h l
Combine
h h h
l ft
m l ft
ft
⎣ ⎝^ ⎠ ⎦
Problem 6 Water flows through the rectangular channel as shown. Flow is subcritical at Sections 1 and 5, critical at Section 2 and supercritical at Sections 3 and 4. Determine Q, y 2 , y 3 , b 1 , b 2 and b 3.
y 1 = 4.2’ y 2 y 4 = 2.15’
y 5 = 3.31’
b 1 =? (^) b 2 =?
b 4 = 8’ b 5 = 8’
Δz= 1’
Maximum Constriction
y 3 =?
b 3 = 7’
( )
( )
5 5 4 4 4 5
4 5 2
2 2 (^2 4 4 2 ) 4
2
2 2 2 3 2 3 2 3 (^3 )
( ) 1 1
2
32.2(43.82 18.49) 200 1 1 17.2 26.
200 1 1 2.15 1 5. 2 64.4 17.
/1.5 5.25 /1.5 3.
5.25 1. (^2 2 )
c c
g y A y A M M Q
A A
by where y A and A by
Q cfs
Q E E y gA
y y E ft
Q Q E y y y ft gA (^) g y
q
− = → = −
= =
− = = −
= + = + + = + + =
= = = =
= = + = + → =
ft
( )( )
3 2 2
2 2 2 2 1 2 1 1
37.2 /
/ 5.
200 5.25 4.2 5. 2 32.2 4.
gy cfs ft
b Q q ft
E E b ft b
= =
= =
= = = + → =
Problem 7 Water flowing in a trapezoidal channel at a uniform depth of 13.5 feet will pass through a rectangular culvert (RCB). The culvert is under inlet control thus the depth at the throat of the culvert. y 2 , is critical. (a) Determine the discharge and the critical depth for the trapezoidal channel. Is the channel mild or steep? (b) Determine the minimum width of the RCB required to avoid raising the water surface elevation in the trapezoidal channel. (c) If the culvert width is 10 feet, determine the depth in the trapezoidal channel just upstream from the culvert. Also, what type of gradually varied flow profile will exist in the channel? (d) If the culvert width is 20 feet, determine the depth in the trapezoidal channel just upstream from the culvert. Also, what type of gradually varied flow profile will exist in the channel?
y (^) o = 12 ft
n = 0.025; So = 0.0005; b = 15 ft; m = 1
RCB (box culvert)
Flow
K ent = 0.
y (^1)
y 2 = y (^) c
y (^) o = 13.5 ft
y 2 = y (^) c
y (^1)
Kent = 0.
RCB
Box Culvert
( )
( ) 3 2 2 3 2
1 2
' 1917 ;
13.88 ; 0.
(15 )^1917 15 2(1) 32.
(1.5 ) ( ) 2
o r
c c
c
ent o c
a Manning s Q cfs
E ft F
A Q y y T g y
y ft
K b E E y z z 1
→ =
= =
= → =
=
= = + + −
3 2
2
. 13.88 (1.5 ) 1. 2
127 /
1917
127
c c
c
c
min
y y
y ft
q gy cfs ft
Q b ft q
= + =
= =
= =
= = =
( )
( )
2 2 2 1/ 3 2
1
1
2 2 2 1/ 3 2
1
1917 192 10
1.75 10.44 18.
1917 96 20
1.75 6.58 11.
c
SE
SE
c
SE
SE
Q c q c
1
fs b
q y f g
E
t
ft
y ft
Q d q cfs b
q y ft g
E f
y ft
= = =
⎛ ⎞ = (^) ⎜ ⎟ = ⎝ ⎠
t
= = →
→ =
= = =
⎛ ⎞ = (^) ⎜ ⎟ = ⎝ ⎠
= = →
→ =
