From Fox-Suggested Extra Problems for Midterm-2
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Transcript of From Fox-Suggested Extra Problems for Midterm-2
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
1/18
Problem 4.66 [Difficulty: 2]
Given: Nozzle hitting stationary cart
Find: Value of M to hold stationary; plot M versu
Solution:
Basic equation: Momentum flux in x direction for the tank
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow 5) Exit velocity is V
Hence Rx M g= V V A( ) V cos ( ) V A( )+= V2
A cos ( ) 1( )= M V
2 A
g1 cos ( )( )=
When = 40o Ms2
9.81 m1000
kg
m3
10m
s
2
0.1 m2
1 cos 40 deg( )( )= M 238 kg=
0 45 90 135 180
1000
2000
3000
Angle (deg)
M(kg)
This graph can be plotted in Excel
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
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Problem 4.67 [Difficulty: 2]
Given: Large tank with nozzle and wire
Find: Tension in wire; plot for range of water depths
Solution:
Basic equation: Momentum flux in x direction for the tank
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow
Hence Rx T= V V A( )= V2
A= 2 g y( ) d
2
4= T
1
2 g y d
2= T is linear with y!
When y = 0.9 m T
21000
kg
m3
9.81m
s2
0.9 m 0.015 m( )2
N s2
kg m= T 3.12N=
0 0.3 0.6 0.9
1
2
3
4
y (m)
T(N)
This graph can be plotted in Excel
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
3/18
Problem 4.71 [Difficulty: 3]
Given: Water tank attached to mass
Find: Whether tank starts moving; Mass to just hold in place
Solution:
Basic equation: Momentum flux in x direction for the tank
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure at exit 4) Uniform flow
Hence Rx V cos ( ) V A( )= V2
D
2
4 cos ( )=
We need to find V. We could use the Bernoulli equation, but here it is known that V 2 g h= where h = 2 m is theheight of fluid in the tank
V 2 9.81m
s2
2 m= V 6.26m
s=
Hence Rx 1000kg
m3
6.26m
s
2
40.05 m( )2 cos 60 deg( )= Rx 38.5 N=
This force is equal to the tension T in the wire T Rx= T 38.5 N=
For the block, the maximum friction force a mass of M = 10 kg can generate is Fmax M g = where is static friction
Fmax 10 kg 9.81m
s2
0.55N s
2
kg m= Fmax 54.0N=
Hence the tension T created by the water jet is less than the maximum friction Fmax; the tank is at rest
The mass that is just sufficient is given by M g Rx=
MRx
g = M 38.5 N
1
9.81
s2
m
1
0.55
kg m
N s2
= M 7.14kg=
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
4/18
Problem 4.78 [Difficulty: 2]
Rx
y
xCS
Given: Water flow through elbow
Find: Force to hold elbow
Solution:
Basic equation: Momentum flux in x direction for the elbow
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure at exit 4) Uniform flow
Hence Rx p1g A1+ V1 V1 A1( ) V2 V2 A2( )= Rx p1g A1 V12
A1 V22
A2+=
From continuity V2 A2 V1 A1= so V2 V1
A1
A2
= V2 10ft
s
4
1= V2 40
ft
s=
Hence Rx 15lbf
in2
4 in2
1.94slug
ft3
10ft
s
2
4 in2
40ft
s
2
1 in2
+
1 ft
12 in
2
lbf s
2
slug ft= Rx 86.9 lbf=
The force is to the left: It is needed to hold the elbow on against the high pressure, plus it generates the large change in x momentum
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
5/18
Problem 4.79 [Difficulty: 2]
Given: Water flow through nozzle
Find: Force to hold nozzle
Solution:
Basic equation: Momentum flux in x direction for the elbow
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Hence Rx p1g A1+ p2g A2+ V1 V1 A1( ) V2 cos ( ) V2 A2( )+= Rx p1g A1 V22
A2 cos ( ) V12
A1+=
From continuity V2 A2 V1 A1= so
V2 V1
A1
A2
= V1
D1
D2
2
= V2 1.5m
s
30
15
2
= V2 6m
s=
Hence Rx 15 103
N
m2
0.3 m( )
2
4 1000
kg
m3
6m
s
2 0.15 m( )
2
4 cos 30 deg( ) 1.5
m
s
2 .3 m( )
2
4
N s2
kg+=
Rx 668 N= The joint is in tension: It is needed to hold the elbow on against the high pressure, plus it generates the largechange in x momentum
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
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Problem 4.82 [Difficulty: 2]
Rx
y
x
CS
Given: Water flow through orifice plate
Find: Force to hold plate
Solution:
Basic equation: Momentum flux in x direction for the elbow
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Hence Rx p1g A1+ p2g A2 V1 V1 A1( ) V2 V2 A2( )+= Rx p1g A1 V22
A2 V12
A1+=
From continuity Q V1 A1= V2 A2=
so V1Q
A1
= 20ft
3
s
4
1
3ft
2
= 229ft
s= and V2 V1
A1
A2
= V1D
d
2
= 229ft
s
4
1.5
2
= 1628ft
s=
NOTE: problem has an error: Flow rate should be 2 ft3/s not 20 ft3/s! We will provide answers to both
Hence Rx 200lbf
in2
4 in( )
2
4 1.94
slug
ft3
1628ft
s
2 1.5 in( )
2
4 229
ft
s
2 4 in( )
2
4
1 ft
12 in
2
lbf s
2
slug ft+=
Rx 51707 lbf=
With more realistic velocities
Hence Rx 200
lbf
in2
4 in( )2
4 1.94slug
ft3 163
ft
s
2 1.5 in( )
2
4 22.9ft
s
2 4 in( )
2
4
1 ft
12 in
2
lbf s2
slug ft+=
Rx 1970 lbf=
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
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Problem *4.91Problem *4.110 [Difficulty: 4]
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
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Problem *4.111 [Difficulty: 4]
CS
Given: Air jet striking disk
Find: Manometer deflection; Force to hold disk
Solution:Basic equations: Hydrostatic pressure, Bernoulli, and momentum flux in x direction
p
V2
2+ g z+ constant=
Assumptions: 1) Steady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (gx = 0)
Applying Bernoulli between jet exit and stagnation point
p
air
V2
2+
p0
air
0+= p0 p1
2air V
2=
But from hydrostatics p0 p SG g h= so h
1
2air V
2
SG g=
air V2
2 SG g=
h 0.002377slug
ft3
225ft
s
2
1
2 1.75
ft3
1.94 slug
s2
32.2 ft= h 0.55 ft= h 6.6 in=
For x momentum Rx V air A V( )= air V2
D
2
4=
Rx 0.002377slug
ft3
225ft
s
2
0.5
12 ft
2
4
lbf s2
slug ft= Rx 0.164 lbf=
The force of the jet on the plate is then F Rx= F 0.164 lbf=
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
10/18
Problem *4.114 [Difficulty: 3]
CS
Given: Water jet striking disk
Find: Expression for speed of jet as function of height; Height for stationary disk
Solution:
Basic equations: Bernoulli; Momentum flux in z direction
p
V2
2+ g z+ constant=
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow
The Bernoulli equation becomesV0
2
2g 0+
V2
2g h+= V
2V0
22 g h= V V0
22 g h=
Hence M g w1 w1 A1( )= V2
A=
But from continuity V0
A0
V
A= so
V A
V0 A0=
Hence we get M g V V A= V0 A0 V02
2 g h=
Solving for h h1
2 gV0
2 M g
V0 A0
2
=
h1
2
s2
9.81 m 10
m
s
2
2 kg9.81 m
s2
m
3
1000 kg
s
10 m
4
25
1000 m
2
2
=
h 4.28 m=
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
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Problem 4.128 [Difficulty: 3]
CS (movesat speed U)
y
x
RxRy
Given: Water jet striking moving vane
Find: Force needed to hold vane to speed U = 10 m/s
Solution:
Basic equations: Momentum flux in x and y directions
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is
constant
Then Rx u1 V1 A1( ) u2 V2 A2( )+= V U( ) V U( ) A[ ] V U( ) cos ( ) V U( ) A[ ]+=
Rx V U( )2
A cos ( ) 1( )=
Using given data
Rx 1000kg
m3
30 10( )m
s
2
0.004 m2
cos 120 deg( ) 1( )N s
2
kg m= Rx 2400 N=
Then Ry v1 V1 A1( ) v2 V2 A2( )+= 0 V U( ) sin ( ) V U( ) A[ ]+=
Ry V U( )2
A sin ( )= Ry 1000kg
m3
30 10( )m
s
2
0.004 m2
sin 120 deg( )N s
2
kg m= Ry 1386N=
Hence the force required is 2400 N to the left and 1390 N upwards to maintain motion at 10 m/s
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
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Problem 4.129 [Difficulty: 2]
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
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Problem 4.134 [Difficulty: 3]
CS (moves to
left at speed Vc)
y
x
Rx
Vj + Vc
Vj + Vc
t
R
Given: Water jet striking moving cone
Find: Thickness of jet sheet; Force needed to move cone
Solution:
Basic equations: Mass conservation; Momentum flux in x direction
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is
constant
Then V1 A1 V2 A2+ 0= Vj Vc+( ) Dj
2
4 Vj Vc+( ) 2 R t+ 0= (Refer to sketch)
Hence t
Dj2
8 R=t
1
84 in( )2
1
9 in=t 0.222 in=
Using relative velocities, x momentum is
Rx u1 V1 A1( ) u2 V2 A2( )+= Vj Vc+( ) Vj Vc+( ) Aj Vj Vc+( ) cos ( ) Vj Vc+( ) Aj+=
Rx Vj Vc+( )2
Aj cos ( ) 1( )=
Using given data
Rx 1.94slug
ft3
100 45+( )ft
s
2
4
12 ft
2
4 cos 60 deg( ) 1( )
lbf s2
slug ft= Rx 1780 lbf=
Hence the force is 1780 lbf to the left; the upwards equals the weight
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
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Problem 4.138 [Difficulty: 2]
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
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Problem 4.142 [Difficulty: 3]
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
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Problem 4.156 [Difficulty: 3]
Given: Data on system
Find: Jet speed to stop cart after 1 s; plot speed & position; maximum x; time to return to origin
Solution: Apply x momentum
Assumptions: 1) All changes wrt CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet area
The given data is 999kg
m3
= M 100 kg= A 0.01 m2
= U0 5m
s=
Then arf M u1 V U+( ) A[ ] u2 m2+ u3 m3+=
where arfdU
dt= u1 V U+( )= and u2 u3= 0=
HencedU
dt M V U+( )
2 A= or
dU
dt
V U+( )2
A
M= which leads to
d V U+( )
V U+( )2
A
Mdt
=
Integrating and using the IC U= U0 at t= 0 U VV U0+
1 A V U0+( )
Mt+
+=
To find the jet speed Vto stop the cart after 1 s, solve the above equation for V, with U= 0 and t= 1 s. (The equation becomes a
quadratic in V). Instead we use Excel's Goal Seekin the associated workbook
From Excel V 5m
s=
For the position x we need to integratedx
dtU= V
V U0+
1 A V U0+( )
M
t+
+=
The result is x V tM
Aln 1
A V U0+( )M
t+
+=
This equation (or the one for Uwith U= 0) can be easily used to find the maximum value ofx by differentiating, as well as the time for x
to be zero again. Instead we use Excel's Goal Seekand Solverin the associated workbook
From Excel xmax 1.93 m= t x 0=( ) 2.51 s=
The complete set of equations is U VV U0+
1 A V U0+( )
Mt+
+= x V tM
Aln 1
A V U0+( )M
t+
+=
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8/7/2019 From Fox-Suggested Extra Problems for Midterm-2
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The plots are presented in the Excel workbook:
(s) x (m) U (m/s) To find V for U = 0 in 1 s, use Goal Seek
0.0 0.00 5.00
0.2 0.82 3.33 (s) U (m/s) V (m/s)
0.4 1.36 2.14 1.0 0.00 5.00
0.6 1.70 1.25
0.8 1.88 0.56 To find the maximum x , use Solver
1.0 1.93 0.00
1.2 1.88 -0.45 s x m
1.4 1.75 -0.83 1.0 1.93
1.6 1.56 -1.15
1.8 1.30 -1.43 To find the time at which x = 0 use Goal Seek
2.0 0.99 -1.67
2.2 0.63 -1.88 (s) x (m)
2.4 0.24 -2.06 2.51 0.00
2.6 -0.19 -2.22
2.8 -0.65 -2.37
3.0 -1.14 -2.50
Cart Speed U vs Time
-3
-2
-1
0
1
2
3
4
5
6
0.0 0.5 1.0 1.5 2.0 2.5 3.0
t (s)
U(m/
s)
Cart Position x vs Time
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
0.0 0.5 1.0 1.5 2.0 2.5 3.0
t (s)
x(m)