CE 3500 Fluid Mechanics – Fall 2014
Test #2
Name:_____________________________
1 Parabolic Gate (10pts + 10pts)
Find the a) pressure force (10pts) and b) the center of the pressure (10pts) on a parabolic gate
shown in the figure.
1
b
1
b
p ( y ) =− y γ , w=a y , dA=w dy → dA=a y dy
h
Fp
∫
=
h
p ( y ) dA
yR F p
=
0
0
h
=
=
=
a γ∫ y
h
b+1
b
=
dy
0
[
b
y
2b+1
aγ
b
h
2b+1
aγ
∫ y p ( y ) dA
]
2 b+1 h
b
=
0
2 b+1
b
=
a γ∫ y
aγ
[
2 b+1
b
dy
0
b
y
3 b+1
aγ
]
3 b+1 h
b
b
h
3b+1
y R=
y R F p 2b+1
=
h
Fp
3b+1
0
3 b+1
b
1
CE 3500 Fluid Mechanics – Fall 2014
Test #2
Name:_____________________________
2 Triangular Gate (10pts)
A gate having the cross-section shown in figure below is w [ L] wide and is hinged at C . The
gate weighs F w [ MLT −2 ] , and it mass center is d [ L] to the right of the plane BC . Determine
the vertical reaction at A on the gate when the water level is h [ L] above the base. All contact
surfaces are smooth.
1
2
1
Forces acting on the gate are F h = γ h2 w at z + h ; F v =γ h b w at b . F w at d
2
3
2
and F A at b therefore
(
)
2
1
F A b= F w d − F h z + h −F v b
3
2
(
)
1
2
1
F A b= F w d − γ h 2 w z + h −γ h b w b
2
3
2
F A b= F w d −γ h w
F w d −γ h w
F A=
[
[ ( ) ]
[
1
2
1
1
1
1
h z + h + b 2 = F w d −γ h w h z+ h 2 + b 2
2
3
2
2
3
2
1
1
1
h z + h 2 + b2
2
3
2
b
]
]
2
CE 3500 Fluid Mechanics – Fall 2014
Test #2
Name:_____________________________
3 Concrete steps (10pts)
Concrete is poured into the forms to produce a set of steps. Determine the weight of the sandbag
needed to keep the bottomless forms from lifting off the ground. The weight of the forms is
85 lb, and the specific weight of the concrete is 150 lb/ft3.
A=w d =3 [ ft ] 10 [ .in ] 1
[ ft ]
=2.5 [ ft 2 ]
12 [ .in ]
p 1=2h γ concrete=150 [ lb/ ft 3 ] 2×8 [ in. ] =200 [ lb/ ft 2 ]
p 2=h γ concrete=150 [ lb/ ft 3 ] 8 [ in. ]=100 [ lb/ ft 2 ]
p 1 A+ p2 A=W bag+W form
W bag=( p1 + p 2 ) A−W form =( 200 [ lb/ ft ] +100 [ lb/ ft
2
2
] ) 2.5 [ ft 2 ] −85 [ lb ]=665 [ lb ]
3
CE 3500 Fluid Mechanics – Fall 2014
Test #2
Name:_____________________________
4 Cylindrical Bulge (10pts + 10pts)
Determine a) the resultant hydrostatic force (10pts) and center of pressure (10pts) on a L length
(into the page as shown in side view) of the semi cylindrical bulge.
(
)
(
2
r2 π
2 r π
, F down=γ L hr +r +
F up =γ L h r +r −
4
4
2
(
F vR= F down− F up =γ L hr+ r 2 +
F hR =γ L
C v=
[
)
) ( )
r2 π
r2π
r2π
−hr−r 2 +
=γ L
4
4
2
h+h+2 r
2r =2 γ L ( h+ r ) r =2 γ L ( hr+ r 2 )
2
(
)
4
γ L ( h+r ) 2 rh+ h+ r 2 r 2
3
F hR
]
C h=
4
r
3π
4
CE 3500 Fluid Mechanics – Fall 2014
Test #2
Name:_____________________________
5 Circular Plate (10pts)
A pump supplies water under pressure to a large tank as shown. The circular-plate valve fitted in
the short discharge pipe on the tank pivots about its diameter A–A and is held shut against the
water pressure by a latch at B. Show that the force on the latch is independent of the supply
pressure, p, and the height of the tank, h.
y R=
I xc
+h
hA
F R=h γ A
F B r =F R ( y R−h ) → F B=
γ I xc
ML−2 T −2 L 4
→ F B=
r
L
5
CE 3500 Fluid Mechanics – Fall 2014
Test #2
Name:_____________________________
6 Anchored Timber (10pts + 10pts)
The homogeneous L=10 [ m ] long timber is h=0.15 [ m ] by w=0.35 [ m ] in cross section.
Determine a) the specific weight of the timber (10pts) and the tension in the rope (10pts). The
specific weight of the water is γ water =9.8 [ kN m−3 ] .
V timber= L w h=10 [ m ] 0.15 [ m ] 0.35 [ m ]=0.525 [ m3 ]
V water =L water w h=8 [ m ] 0.15 [ m ] 0.35 [ m ] =0.420 [ m 3 ]
F B=γ water V water =4.116 [ kN ]
FB
L water
L
L
=F G
→ F G =F B water =3.29 [ kN ]
2
2
L
F G =γtimber V timber → γ timber =
FG
−3
=6.27 [ kN m ]
V timber
F rope =F B− F G =0.826 [ kN ]
6
CE 3500 Fluid Mechanics – Fall 2014
Test #2
Name:_____________________________
7 Floating Cylinder Tank (20pts + 10pts)
Thin-walled, d =1 [ m ] diameter cylindrical tank is H =3 [ m ] tall and has a mass of m=90 [ kg ] .
The closed top of the tank has a pressure gauge and the open bottom is lowered into water
(density ρ wat =1000 [ kg m−3 ] ) and held in position ( h=0.6 [ m ] portion sticking out from the
water) by a steel block (density ρ block =7840 [ kg m−3 ] ). Assume that the air in the tank is
p
p
+ p atm
compressed at constant temperature ( ρatm = gauge
=constant ). The atmospheric pressure
ρ
air
comp
is p atm =101.33 [ kPa ] .
Determine:
a) the reading on the pressure gage at the top of the tank (20pts);
b) the volume of the steel block (10pts).
The first part of the problem leads to a second order equation: a x 2 + b x+c=0 that can be solved
by the following formula: x 1,2=
−b± √ b 2−4 ac
2a
You can solve the second part (without completing the first part) by knowing that the height of
the compressed air in the tank is hair =2.581 [ m ] .
7
CE 3500 Fluid Mechanics – Fall 2014
Test #2
Name:_____________________________
p atm + p gauge p atm
ρair
p atm
,
ρcomp = ρair → ρ comp = p
gauge + p atm
ρ
ρ air V cyl =ρcomp V air → hair = ρ air H =
comp
p gauge=ρwat g ( hair −h )=ρwat g
(
p atm
H,
p gauge + p atm
)
p atm
H −h ,
p gauge + p atm
( p gauge +ρwat g h )( p gauge + p atm)− p atm ρ wat g H =0 ,
p 2gauge + p gauge ρwater g h+ p gauge p atm + p atm ρwater g ( h−H )=0
√
2
−ρwater g h− p atm ± (ρ water g h + p atm ) −4 p atm ρ water g ( h−H )
p gauge=
=18.915 [ kPa ]
2
hair =
patm
d2
3
H =2.581 [ m ] , V buoy =( h air −h ) π=1.541 [ m ]
p gauge + p atm
4
ρ wat g ( V buoy +V block )=mg +ρ block g V block → ρwat V buoy −m=( ρblock −ρwat ) V block
ρ V −m
3
V block = ρwat buoy
=0.208 [ m ]
block −ρwat
8