homework solutions.week 10
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8/13/2019 Homework Solutions.week 10
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5.80: PROBLEM DEFINITION
Situation:A tank is draining through an ori ce.h1 = 3 m , h = 0.5 m.D
T = 0 .6 m, D 2 = 3 cm
Find:Time required for the water surface to drop the speci ed distance (3 to 0.5 m).
SOLUTION
From Example 5-6 the time to decrease the elevation from h1 to h is
t = 2AT 2gA2 (h1 / 21 h1 / 2 )=
2
/ 4 (0.6 m)2
3
0.5
m1 / 2
p 2 9.81 m/ s2 (/ 4) (0.03m)
2
t = 185 s
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5.108: PROBLEM DEFINITION
Situation:A sphere moves below the surface in water.D = 1 ft , h = 12 ft.
Find:Speed at which cavitation occurs.
Properties:Water ( 50 F), Table A.5: = 1.94 slug/ ft3 .
PLAN
Apply the Bernoulli equation between the free stream and the maximum width.
SOLUTION
Let po be the pressure on the streamline upstream of the sphere. The minimumpressure will occur at the maximum width of the sphere where the velocity is 1.5times the free stream velocity.Bernoulli equation
po + 12
V 2o + h o = p + 12
(1.5V o )2 + (h o + 0 .5)
Solving for the pressure p gives
p = po 0.625V 2o 0.5
The pressure at a depth of 12 ft is 749 lbf/ft 2 . The density of water is 1.94 slugs/ft 3
and the speci c weight is 62.4 lbf/ft 3 . At a temperature of 50 o F, the vapor pressureis 0.178 psia or 25.6 psfa. Substituting into the above equation
25.6 psfa = 749 psfa (0.625)(1.94) V 2o (0.5)(62.4)692.2 = 1.21V 2o
Solving for V o gives
V o = 23.9 ft/s
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5.111: PROBLEM DEFINITION
Situation:A hydrofoil is tested in water.h = 4 ft , V = 25ft / s.
Find:Speed that cavitation begins.Properties:
p0 = 2 .5 psi vacuum.Water ( 50 F) Table A.5, pv = 0.178 psia.
PLAN
Consider a point ahead of the foil (at same depth as the foil) and the point of minimumpressure on the foil, and apply the pressure coe ffi cient de nition between these twopoints.
SOLUTION
pmin = 2.5 144 = 360 psf gage p0 = 4 = 4 62.4 = 249.6 psf
Then
C p = ( pmin p0 )
V 20 / 2 =
( 360 249.6)(1.94 slug/ ft3 / 2) (25ft/ s)2
C p = 1.005
Now let pmin = pvapor
= 0 .178 psia = 14.52 psia = 2, 091 psfgThen
1.005 = 249.6 psf + 2 , 091(1.94 slug/ ft3 / 2)V 20
V 0 = 49.0 ft/s
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6.9: PROBLEM DEFINITION
Situation:A water jet is lling a tank.m = 25kg, V = 25 L.
d = 30 mm, v = 25 m/s.
Find:Force on the bottom of the tank (N).Force acting on the stop block (N).
Assumptions:Steady ow.
Properties:Water (15 C), Table A.5: = 999kg/ m3 , = 9800N/ m3 .
PLANApply the momentum equation in the x-direction and in the y-direction.
SOLUTION
Force and momentum diagrams
Momentum equation ( x-direction)
XF x = Xcs movox Xcs m i vixF = ( mv cos70o)
= Av2 cos70o
Calculations
Av 2 = 999kg/ m3
(0.03m)2
4 !(25m/ s)2= 441.3 N
F = (441.3 N)(cos70o)= 150.9 N
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F = 150.9 N pushing to the left on the stop block
y-direction
XF y =
Xcs
movoy
Xcs
m i viy
N W = ( mv sin70o)N = W + Av 2 sin70o
Calculations:
W = W tank + W water= (25 kg) 9.81 m/ s
2
+ (0 .025m3 )(9800N/ m3 )
= 490.3 N
N = W + Av2
sin70o
= (490 .3 N) + (441 .3 N)sin70o
N = 905 N acting upward
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8/13/2019 Homework Solutions.week 10
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6.12: PROBLEM DEFINITION
Situation:Horizontal round jet strikes a plate.Q = 2 cfs, F x = 200 lbf.
Find:Speed of water jet (ft/s).Sketch:
Properties:Water (70 o F) , Table A.5: = 1.94 slug/ft 3 .
PLAN
Apply the momentum equation to a control volume surrounding the plate.
SOLUTION
Force and momentum diagrams
Momentum equation ( x-direction)
X F x = mv1 xF x = ( mv1 ) = Qv1
v1 = F x
Q
= 200lbf
1.94 slug/ ft3 2 ft3 / s
v1 = 51.5 ft/ s
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6.15: PROBLEM DEFINITION
Situation:Water jet from a re hose on a boat.d = 4 in, V = 60 mph = 88.0 ft/s.
Find:Tension in cable (lbf).
Sketch:
Properties:Water ( 50 F), Table A.5: = 1.94 slug/ ft3 .
PLAN
Apply the momentum equation.
SOLUTION
Force and momentum diagrams
Flow rate
m = AV = 1.94 slug/ ft3 (2/ 12ft)2 (88.0 ft/ s)= 14 .90 slug/ s
Momentum equation ( x-direction)
X F = m (vo ) xT = mV cos60o
T = (14 .90 slug/ s)(88.0 ft/ s)cos60o
= 656 lbf
T = 656 lbf
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