Unformatted text preview: Problem 2.1 Given: Velocity fields Find: [1] Whether flows are 1, 2 or 3D, steady or unsteady. Solution:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8) →
V
→
V
→
V
→
V
→
V
→
V
→
V
→
V →
=V
→
=V
→
=V
→
=V
→
=V
→
=V
→
=V
→
=V ( y) 1D ( x) 1D ( x , y) 2D ( x , y) 2D ( x) 1D ( x , y , z) 3D ( x , y) 2D ( x , y , z) 3D →
V
→
V
→
V
→
V
→
V
→
V
→
V
→
V →
=V
→
≠V
→
=V
→
=V
→
=V
→
≠V
→
=V
→
≠V ( t) Unsteady ( t) Steady ( t) Unsteady ( t) Unsteady ( t) Unsteady ( t) Steady ( t) Unsteady ( t) Steady Problem 2.2 [2] Problem 2.3 Given: Velocity field Find: [1] Equation for streamlines Streamline Plots Solution: 5 So, separating variables dy
B dx
=⋅
y
Ax 3 Integrating ln ( y) = C=1
C=2
C=3
C=4 4 The solution is y= C B
1
⋅ ln ( x) + c = − ⋅ ln ( x) + c
A
2 y (m) For streamlines v
dy
B⋅ x⋅ y
B⋅ y
=
=
=
2
u
dx
A⋅ x
A⋅ x 2
1 x
0 The plot can be easily done in Excel. 1 2 3 x (m) 4 5 Problem 2.4 [2] t=0 x
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(### means too large to view)
c=1 c=2 c=3
x
y
y
y
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y
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86.74
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y
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173.47
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y
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260.21
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0.0 0.5 1.0 1.5 2.0 x Streamline Plot (t = 1 s)
70 c=1 60 c=2
50 c=3 y 40
30
20
10
0
0.0 0.5 1.0 1.5 2.0 x Streamline Plot (t = 20 s)
20
18 c=1 16 c=2 14 c=3 y 12
10
8
6
4
2
0
0.0 0.2 0.4 0.6 x 0.8 1.0 1.2 Problem 2.6 [1] Given: Velocity field Find: Whether field is 1D, 2D or 3D; Velocity components at (2,1/2); Equation for streamlines; Plot Solution:
The velocity field is a function of x and y. It is therefore 2D.
At point (2,1/2), the velocity components are u = a ⋅ x⋅ y = 2⋅
2 1
1
× 2⋅ m × ⋅ m
m⋅ s
2 v = b ⋅ y = −6⋅ 1
1⎞
× ⎛ ⋅m⎟
⎜
m⋅ s ⎝ 2 ⎠ 2 u = 2⋅ m
s 3m
v=− ⋅
2s 2 For streamlines v
dy
b⋅y
b⋅y
=
=
=
u
dx
a ⋅ x⋅ y
a⋅x So, separating variables dy
b dx
=⋅
y
ax Integrating ln ( y) = The solution is y = C⋅ x b
⋅ ln ( x) + c
a y = C⋅ x b
a −3 The streamline passing through point (2,1/2) is given by 1
−3
= C⋅ 2
2 C= 13
⋅2
2 C=4 Streamline for C
Streamline for 2C
Streamline for 3C
Streamline for 4C 12
8
4
1 This can be plotted in Excel. 1.3 1.7 4
3 x 20
16 y= 2 a= 1
b= 1
C=
x
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y
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y
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y
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1.2
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c=6 y 0.6 0.4
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x 1.5 2.0 A = 10
B = 20
C=
1
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y
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c=1 3.0 c=2
c=4 2.5 c = 6 ((x,y) = (1.2) 2.0
y x
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x 1.5 2.0 Problem 2.9 Given: Velocity field Find: [2] Equation for streamline through (1,3) Solution: A⋅ y
2 For streamlines v
dy
=
=
u
dx So, separating variables dy
dx
=
y
x Integrating ln ( y) = ln ( x) + c The solution is y = C⋅ x which is the equation of a straight line. For the streamline through point (1,3) 3 = C⋅ 1 C=3 and y = 3⋅ x For a particle up = or x⋅ dx = A⋅ dt x= x
A
x dx
A
=
dt
x = y
x 2 2⋅ A ⋅ t + c t= c
x
−
2⋅ A 2⋅ A Hence the time for a particle to go from x = 1 to x = 2 m is
2 Δt = t ( x = 2) − t ( x = 1) Δt = 2 2 2 ( 2⋅ m) − c ( 1⋅ m) − c
4⋅ m − 1⋅ m
−
=
2
2⋅ A
2⋅ A
m
2 × 2⋅
s Δt = 0.75⋅ s [3] Problem 2.10 Given: Flow field Find: Plot of velocity magnitude along axes, and y = x; Equation of streamlines Solution:
K⋅ y u=− On the x axis, y = 0, so ( 2 2 2⋅ π⋅ x + y ) =0 K⋅ x v= ( ) 2 2 2⋅ π⋅ x + y Plotting = K
2⋅ π⋅ x 100 v( m/s) 50 − 10 −5 0 5 10 − 50
− 100 x (km)
The velocity is perpendicular to the axis, is very high close to the origin, and falls off to zero.
This can also be plotted in Excel.
K⋅ y u=− On the y axis, x = 0, so ( 2 2⋅ π⋅ x + y ) 2 =− K
2⋅ π⋅ y K⋅ x v= (2 ) 2 2⋅ π⋅ x + y =0 100 Plotting u ( m/s) 50 − 10 −5 0
− 50
− 100 y (km)
The velocity is perpendicular to the axis, is very high close to the origin, and falls off to zero. 5 10 This can also be plotted in Excel.
K⋅ x u=− On the y = x axis ( 2 2⋅ π⋅ x + x The flow is perpendicular to line y = x: =− ) 2 K
4⋅ π⋅ x Slope of line y = x: ( ) 2 2 2 2 2⋅ π⋅ x + x = K
4⋅ π⋅ x 1 Slope of trajectory of motion: u
= −1
v 2 r= If we define the radial position: K⋅ x v= 2 x +y
2 Then the magnitude of the velocity along y = x is V = then along y = x 2 u +v = r= x +x = 2⋅ x 1
1
K
K
K
⋅
+
=
=
4⋅ π x2 x2
2⋅ π⋅ r
2⋅ π⋅ 2⋅ x Plotting
100 V(m/s) 50
− 10 −5 0 5 10 − 50
− 100 r (km)
This can also be plotted in Excel.
K⋅ x For streamlines v
dy
=
=
u
dx ( 2 2) 2⋅ π⋅ x + y K⋅ y − (2 y⋅ dy = −x⋅ dx Integrating y
x
= − +c
2
2 The solution is x +y = C ) x
y 2 2⋅ π⋅ x + y
So, separating variables =− 2 2 2 2 which is the equation of a circle. Streamlines form a set of concentric circles.
This flow models a vortex flow. See Example 5.6 for streamline plots. Streamlines are circular, and the velocity approaches infinity as we
approach the center. In Problem 2.11, we see that the streamlines are also circular. In a real tornado, at large distances from the center, the
velocities behave as in this problem; close to the center, they behave as in Problem 2.11. Problem 2.11 [3] Given: Flow field Find: Plot of velocity magnitude along axes, and y = x; Equation for streamlines Solution:
u=− On the x axis, y = 0, so M⋅ y
=0
2⋅ π Plotting v= M⋅ x
2⋅ π 1000 v (m/s) 500
− 10 −5 0 5 10 − 500
− 1000 x (km)
The velocity is perpendicular to the axis and increases linearly with distance x.
This can also be plotted in Excel.
u=− On the y axis, x = 0, so M⋅ y
2⋅ π Plotting v= M⋅ x
=0
2⋅ π 1000 u (m/s) 500
− 10 −5 0
− 500
− 1000 y (km)
The velocity is perpendicular to the axis and increases linearly with distance y.
This can also be plotted in Excel. 5 10 u=− On the y = x axis The flow is perpendicular to line y = x: M⋅ y
M⋅ x
=−
2⋅ π
2⋅ π v= Slope of line y = x: 1 Slope of trajectory of motion: u
= −1
v 2 r= If we define the radial position: M⋅ x
2⋅ π 2 x +y
2 Then the magnitude of the velocity along y = x is V = 2 then along y = x u +v = 2 r= 2 x +x = 2⋅ x M
M⋅ r
2
2 M⋅ 2⋅ x
⋅ x +x =
=
2⋅ π
2⋅ π
2⋅ π Plotting 1000 V(m/s) 500
− 10 −5 0 5 10 − 500
− 1000 r (km)
This can also be plotted in Excel.
M⋅ x
2⋅ π For streamlines v
dy
=
=
u
dx So, separating variables y⋅ dy = −x⋅ dx Integrating y
x
= − +c
2
2 The solution is x +y = C 2 2 M⋅ y
−
2⋅ π =− x
y 2 2 which is the equation of a circle. The streamlines form a set of concentric circles.
This flow models a rigid body vortex flow. See Example 5.6 for streamline plots. Streamlines are circular, and the velocity approaches zer
as we approach the center. In Problem 2.10, we see that the streamlines are also circular. In a real tornado, at large distances from the
center, the velocities behave as in Problem 2.10; close to the center, they behave as in this problem. Problem 2.12 [3] Given: Flow field Find: Plot of velocity magnitude along axes, and y = x; Equations of streamlines Solution:
q⋅x u=− On the x axis, y = 0, so ( 2 2 2⋅ π⋅ x + y ) =− u (m/s) Plotting − 10 q
2⋅ π⋅ x q⋅ y v=− (2 35
25
15
5
−5 0
− 15
− 25
− 35 −5 ) 2 2⋅ π⋅ x + y =0 5 10 x (km)
The velocity is very high close to the origin, and falls off to zero. It is also along the axis. This can be plotted in Excel.
q⋅ x u=− On the y axis, x = 0, so ( 2 2⋅ π⋅ x + y v (m/s) Plotting − 10 −5 ) 2 =0 q⋅ y v=− 35
25
15
5
−5 0
− 15
− 25
− 35 The velocity is again very high close to the origin, and falls off to zero. It is also along the axis. ) 2 2⋅ π⋅ x + y y (km) This can also be plotted in Excel. ( 2 5 =− q
2⋅ π⋅ y 10 q⋅ x u=− On the y = x axis ( 2 ) 2⋅ π⋅ x + x
The flow is parallel to line y = x: =− 2 q
4⋅ π⋅ x 2 r= 2 x +y
2 then along y = x 2 u +v = 2⋅ π⋅ x + x =− q
4⋅ π⋅ x − 10 r= 2 2 x +x = 35
25
15
5
−5 0
− 15
− 25
− 35 −5 5 r (km)
This can also be plotted in Excel.
q⋅ y −
v
dy
=
=
u
dx (2 q⋅ x (2 ln ( y) = ln ( x) + c The solution is y = C⋅ x y
x dy
dx
=
y
x Integrating ) = 2 2⋅ π⋅ x + y
So, separating variables ) 2 2⋅ π⋅ x + y
− This flow field corresponds to a sink (discussed in Chapter 6). 2⋅ x 1
1
q
q
q
⋅
+
=
=
2
2
4⋅ π x
2⋅ π⋅ r
2⋅ π⋅ 2⋅ x
x Plotting For streamlines ) 2 v
=1
u Then the magnitude of the velocity along y = x is V = V(m/s) ( 2 1 Slope of trajectory of motion:
If we define the radial position: Slope of line y = x: q⋅ x v=− which is the equation of a straight line. 10 Problem 2.13 [2] t=0
x
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C=1
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C=1
y
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y
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1.48 Streamline Plot (t = 0)
3.5 c=1
c=2
c=3 3.0
2.5 y 2.0
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1.0
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0.0
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 x Streamline Plot (t = 1s)
2.0 c=1
c=2
c=3 1.8
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1.4 y 1.2
1.0
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0.0
0.0 0.1 0.2 0.3 0.4 0.5 0.6 x Streamline Plot (t = 20s)
2.0 c=1
c=2
c=3 1.8
1.6
1.4 y 1.2
1.0
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0.0
0.0 0.5 1.0 1.5
x 2.0 2.5 Problem 2.15 [4] Given: Pathlines of particles Find: Conditions that make them satisfy Problem 2.10 flow field; Also Problem 2.11 flow field; Plot pathlines Solution:
The given pathlines are xp = −a⋅ sin ( ω⋅ t) The velocity field of Problem 2.10 is u=− K⋅ y yp = a⋅ cos ( ω⋅ t) ( 2 2⋅ π⋅ x + y K⋅ x v= ) 2 (2 ) 2 2⋅ π⋅ x + y If the pathlines are correct we should be able to substitute xp and yp into the velocity field to find the velocity as a function of time:
K⋅ y u=− ( 2 2 2⋅ π⋅ x + y
K⋅ x v= ( 2 2 2⋅ π⋅ x + y ) ) =− K⋅ a ⋅ cos ( ω⋅ t ) ( 2 2 2 2⋅ π⋅ a ⋅ sin( ω⋅ t ) + a ⋅ cos ( ω⋅ t )
=− ( K⋅ ( −a ⋅ sin( ω⋅ t ) )
2 2 2 2⋅ π⋅ a ⋅ sin( ω⋅ t ) + a ⋅ cos ( ω⋅ t ) 2 2 =− ) =− ) K⋅ cos ( ω⋅ t )
2⋅ π⋅ a (1) K⋅ sin( ω⋅ t )
2⋅ π⋅ a (2) We should also be able to find the velocity field as a function of time from the pathline equations (Eq. 2.9):
dxp
=u
dt
u= dxp
dt dxp
dt
= −a ⋅ ω⋅ cos ( ω⋅ t ) Comparing Eqs. 1, 2 and 3 u = −a ⋅ ω⋅ cos ( ω⋅ t ) = − Hence we see that a⋅ ω = K
2⋅ π⋅ a v= K ⋅ cos ( ω⋅ t )
2⋅ π⋅ a
or (2.9) =v
dyp
dt = −a⋅ ω⋅ sin ( ω⋅ t) v = −a⋅ ω⋅ sin ( ω⋅ t) = − ω= K
2⋅ π⋅ a 2 (3) K ⋅ sin ( ω⋅ t)
2⋅ π⋅ a for the pathlines to be correct. The pathlines are a = 300 m
a = 400 m
a = 500 m 400 To plot this in Excel, compute xp and yp for t
ranging from 0 to 60 s, with ω given by the
above formula. Plot yp versus xp. Note that
outer particles travel much slower! 200 − 400 − 200 0 200 This is the free vortex flow discussed in
Example 5.6 400 − 200 − 400 u=− The velocity field of Problem 2.11 is M⋅ y
2⋅ π v= M⋅ x
2⋅ π If the pathlines are correct we should be able to substitute xp and yp into the velocity field to find the velocity as a function of time:
u=− M⋅ y
M⋅ ( a⋅ cos ( ω⋅ t) )
M⋅ a⋅ cos ( ω⋅ t)
=−
=−
2⋅ π
2⋅ π
2⋅ π v= M⋅ x
M⋅ ( −a⋅ sin ( ω⋅ t) )
M⋅ a⋅ sin ( ω⋅ t)
=
=−
2⋅ π
2⋅ π
2⋅ π Recall that u= dxp
= −a⋅ ω⋅ cos ( ω⋅ t)
dt Comparing Eqs. 1, 4 and 5 u = −a⋅ ω⋅ cos ( ω⋅ t) = − Hence we see that ω= (4) M
2⋅ π M⋅ a⋅ cos ( ω⋅ t)
2⋅ π for the pathlines to be correct. (5) v= dyp
dt = −a⋅ ω⋅ sin ( ω⋅ t) v = −a⋅ ω⋅ sin ( ω⋅ t) = − M⋅ a⋅ sin ( ω⋅ t)
2⋅ π (3) The pathlines
To plot this in Excel, compute xp and yp for t
ranging from 0 to 75 s, with ω given by the
above formula. Plot yp versus xp. Note that
outer particles travel faster! 400 200 − 400 − 200 0 200 400 − 200 − 400 Note that this is rigid body rotation! − 600 a = 300 m
a = 400 m
a = 500 m This is the forced vortex flow discussed in
Example 5.6 Problem 2.16 [2] Given: Timevarying velocity field Find: Streamlines at t = 0 s; Streamline through (3,3); velocity vector; will streamlines change with time Solution:
For streamlines v
dy
a ⋅ y⋅ ( 2 + cos ( ω⋅ t ) )
y
=
=−
=−
u
dx
a ⋅ x⋅ ( 2 + cos ( ω⋅ t ) )
x At t = 0 (actually all times!) dy
y
=−
dx
x So, separating variables dy
dx
=−
y
x Integrating ln ( y) = −ln ( x) + c The solution is y= For the streamline through point (3,3) C= C
x which is the equation of a hyperbola. 3
3 C=1 y= and 1
x The streamlines will not change with time since dy/dx does not change with time.
At t = 0 5 u = a⋅ x⋅ ( 2 + cos ( ω⋅ t) ) = 5⋅ 1
× 3⋅ m × 3
s m
s u = 45⋅ 3 v = −a⋅ y⋅ ( 2 + cos ( ω⋅ t) ) = 5⋅ y 4 2 v = −45⋅ 1 1
s × 3⋅ m × 3 m
s The velocity vector is tangent to the curve;
0 1 2 3 4 5 This curve can be plotted in Excel. Tangent of curve at (3,3) is dy
y
= − = −1
dx
x Direction of velocity at (3,3) is x v
= −1
u Problem 2.17 [3] Problem 2.18 [3] Pathline
t
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00 x
1.00
1.00
1.01
1.03
1.05
1.08
1.12
1.17
1.22
1.29
1.37
1.46
1.57
1.70
1.85
2.02
2.23
2.47
2.75
3.09
3.49 Streamlines
t=0
x
y
1.00
1.00
1.00
0.78
1.00
0.61
1.00
0.47
1.00
0.37
1.00
0.29
1.00
0.22
1.00
0.17
1.00
0.14
1.00
0.11
1.00
0.08
1.00
0.06
1.00
0.05
1.00
0.04
1.00
0.03
1.00
0.02
1.00
0.02
1.00
0.01
1.00
0.01
1.00
0.01
1.00
0.01 y
1.00
0.78
0.61
0.47
0.37
0.29
0.22
0.17
0.14
0.11
0.08
0.06
0.05
0.04
0.03
0.02
0.02
0.01
0.01
0.01
0.01 t=1s
x
1.00
1.00
1.01
1.03
1.05
1.08
1.12
1.17
1.22
1.29
1.37
1.46
1.57
1.70
1.85
2.02
2.23
2.47
2.75
3.09
3.49 t=2s
x
1.00
1.00
1.01
1.03
1.05
1.08
1.12
1.17
1.22
1.29
1.37
1.46
1.57
1.70
1.85
2.02
2.23
2.47
2.75
3.09
3.49 y
1.00
0.97
0.88
0.75
0.61
0.46
0.32
0.22
0.14
0.08
0.04
0.02
0.01
0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00 y
1.00
0.98
0.94
0.87
0.78
0.68
0.57
0.47
0.37
0.28
0.21
0.15
0.11
0.07
0.05
0.03
0.02
0.01
0.01
0.00
0.00 Pathline and Streamline Plots
1.0 Pathline
Streamline (t = 0)
Streamline (t = 1 s)
Streamline (t = 2 s) 0.8 y 0.6
0.4
0.2
0.0
0.0 0.5 1.0 1.5 2.0
x 2.5 3.0 3.5 Problem 2.20 [3] Problem 2.21 [3] Given: Flow field Find: Pathline for particle starting at (3,1); Streamlines through same point at t = 1, 2, and 3 s Solution:
For particle paths dx
= u = a ⋅ x⋅ t
dt and dy
=v=b
dt Separating variables and integrating dx
= a ⋅ t ⋅ dt
x or ln ( x) = dy = b ⋅ dt or y = b⋅ t + c2 12
⋅ a⋅ t + c1
2 Using initial condition (x,y) = (3,1) and the given values for a and b
c1 = ln ( 3⋅ m )
0.05⋅ t and c2 = 1⋅ m and y = 4⋅ t + 1 2 The pathline is then x = 3⋅ e For streamlines (at any time t) v
dy
b
=
=
u
dx
a ⋅ x⋅ t So, separating variables dy = Integrating y= b dx
⋅
a⋅t x
b
⋅ ln ( x) + c
a⋅ t We are interested in instantaneous streamlines at various times that always pass through point (3,1). Using a and b values:
c = y− y = 1+ The streamline equation is b
4
⋅ ln ( x) = 1 −
⋅ ln ( 3)
a⋅ t
0.1⋅ t
40
t x⎞
⋅ ln ⎛ ⎟
⎜
⎝ 3⎠ 30 Pathline
Streamline (t=1)
Streamline (t=2)
Streamline (t=3) 20 y 10 0 1 2 3 4 5 − 10
− 20 x These curves can be plotted in Excel. Problem 2.22 [4] Given: Velocity field Find: Plot streamlines that are at origin at various times and pathlines that left origin at these times Solution:
⎡⎛ v
dy
=
=
u
dx For streamlines v0⋅ sin⎢ω⋅ ⎜ t − ⎣⎝ u0 v0⋅ sin⎡ω⋅ ⎛ t −
⎢⎜
dy = So, separating variables (t=const) x ⎞⎤
u0 ⎟⎥
⎠⎦ ⎣⎝ x ⎞⎤
u0 ⎟⎥ u0 ⎠⎦ ⋅ dx x ⎞⎤
v0⋅ cos ⎡ω⋅ ⎛ t −
⎢ ⎜ u ⎟⎥
0 ⎠⎦
⎣⎝
y=
+c
ω Integrating x ⎞⎤
⎤
v0⋅ ⎡cos ⎡ω⋅ ⎛ t −
⎢ ⎢ ⎜ u ⎟⎥ − cos ( ω⋅ t)⎥
0 ⎠⎦
⎣ ⎣⎝
⎦
y=
ω Using condition y = 0 when x = 0 For particle paths, first find x(t) dx
= u = u0
dt Separating variables and integrating dx = u0⋅ dt Using initial condition x = 0 at t = τ c1 = −u0⋅ τ For y(t) we have dy
x ⎞⎤
= v = v0⋅ sin ⎡ω⋅ ⎛ t − ⎟⎥
⎢⎜ u
dt
0 ⎠⎦
⎝ This gives streamlines y(x) at each time t x = u0⋅ t + c1 or x = u0⋅ ( t − τ)
so ⎡ ⎡ u0⋅ ( t − τ)⎤⎤
dy
= v = v0⋅ sin ⎢ω⋅ ⎢t −
⎥⎥
dt
u0
⎣⎣
⎦⎦ dy
= v = v0⋅ sin ( ω⋅ τ)
dt and
Separating variables and integrating dy = v0⋅ sin ( ω⋅ τ) ⋅ dt y = v0⋅ sin ( ω⋅ τ) ⋅ t + c2 Using initial condition y = 0 at t = τ c2 = −v0⋅ sin ( ω⋅ τ) ⋅ τ y = v0⋅ sin ( ω⋅ τ) ⋅ ( t − τ) The pathline is then
x ( t , τ) = u0⋅ ( t − τ) y ( t , τ) = v0⋅ sin ( ω⋅ τ) ⋅ ( t − τ) These terms give the path of a particle (x(t),y(t)) that started at t = τ. 0.5
0.25
0 1 2 − 0.25
− 0.5 Streamline t = 0s
Streamline t = 0.05s
Streamline t = 0.1s
Streamline t = 0.15s
Pathline starting t = 0s
Pathline starting t = 0.05s
Pathline starting t = 0.1s
Pathline starting t = 0.15s
The streamlines are sinusoids; the pathlines are straight (once a water particle is fired it travels in a straight line).
These curves can be plotted in Excel. 3 Problem 2.23 Given: Velocity field Find: [5] Plot streakline for first second of flow Solution:
Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form ( xp( t ) = x t , x0 , y0 , t0 ) and ( yp ( t) = y t , x0 , y0 , t0 ) where x0, y0 is the position of the particle at t = t0, and reinterprete the results as streaklines () ( xst t0 = x t , x0 , y0 , t0 ) and () ( ) yst t0 = y t , x0 , y0 , t0 which gives the streakline at t, where x0, y0 is the point at which dye is released (t0 is varied from 0 to t)
For particle paths, first find x(t) dx
= u = u0
dt Separating variables and integrating dx = u0⋅ dt For y(t) we have and Separating variables and integrating ( x = x0 + u0⋅ t − t0 or dy
x ⎞⎤
= v = v0⋅ sin⎡ω⋅ ⎛ t −
⎢ ⎜ u ⎟⎥
dt
0 ⎠⎦
⎣⎝
x0 ⎞⎤
⎡⎛
dy
= v = v0⋅ sin⎢ω⋅ ⎜ t0 −
⎟⎥
dt
u0
⎣⎝
⎠⎦
x0 ⎞⎤
⎡⎛
dy = v0⋅ sin ⎢ω⋅ ⎜ t0 − ⎟⎥ ⋅ dt
u0
⎣⎝
⎠⎦ () ( The streakline is then xst t0 = x0 + u0 t − t0 With ) so ) ( ⎣⎣ ⎦⎦ x0 ⎞⎤
⎡⎛
y = y0 + v0⋅ sin ⎢ω⋅ ⎜ t0 − ⎟⎥ ⋅ t − t0
u0
⎣⎝
⎠⎦
x0 ⎞⎤
⎡⎛
yst t0 = y0 + v0⋅ sin ⎢ω⋅ ⎜ t0 − ⎟⎥ ⋅ t − t0
u0
⎣⎝
⎠⎦ ( () () ( ) () ()( yst t0 = v0⋅ sin ⎡ω⋅ t0 ⎤ ⋅ t − t0
⎣
⎦ Streakline for First Second
2 y (m) 1
0 2 4 6 −1
−2 x (m)
This curve can be plotted in Excel. For t = 1, t0 ranges from 0 to t. 8 ) ( x0 = y0 = 0
xst t0 = u0⋅ t − t0 ) ⎡ ⎡ x0 + u0⋅ t − t0 ⎤⎤
dy
= v = v0⋅ sin⎢ω⋅ ⎢t −
⎥⎥
dt
u0 10 ) ) Problem 2.24 [3] Part 1/2 Problem 2.24 [3] Part 2/2 Problem 2.25 [3] Part 1/2 Problem 2.25 [3] Part 2/2 Problem 2.26 [4] Part 1/2 Problem 2.26 [4] Part 2/2 Problem 2.27 Solution Pathlines:
t
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
2.20
2.40
2.60
2.80
3.00
3.20
3.40
3.60
3.80
4.00 [3] The particle starting at t = 3 s follows the particle starting at t = 2 s;
The particle starting at t = 4 s doesn't move!
Starting at t = 0
x
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00 Starting at t = 1 s y
0.00
0.40
0.80
1.20
1.60
2.00
2.40
2.80
3.20
3.60
4.00
3.80
3.60
3.40
3.20
3.00
2.80
2.60
2.40
2.20
2.00 Starting at t = 2 s x y x y 0.00
0.20
0.40
0.60
0.80
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00 0.00
0.40
0.80
1.20
1.60
2.00
1.80
1.60
1.40
1.20
1.00
0.80
0.60
0.40
0.20
0.00 0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00 0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00 Streakline at t = 4 s
x
2.00
1.80
1.60
1.40
1.20
1.00
0.80
0.60
0.40
0.20
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00 Pathline and Streakline Plots
4
3
2
1 y
0
0.5 0.0 0.5 1.0 1.5 2.0 Pathline starting at t = 0
Pathline starting at t = 1 s
Pathline starting at t = 2 s
Streakline at t = 4 s 1
2
3 x 2.5 y
2.00
1.60
1.20
0.80
0.40
0.00
0.40
0.80
1.20
1.60
2.00
1.80
1.60
1.40
1.20
1.00
0.80
0.60
0.40
0.20
0.00 Problem 2.28 [4] Given: 2D velocity field Find: Streamlines passing through (6,6); Coordinates of particle starting at (1,4); that pathlines, streamlines and
streaklines coincide Solution:
v
dy
=
=
u
dx For streamlines b ⌠
⌠
2
⎮
⎮
a ⋅ y dy = ⎮ b dx
⎮
⌡
⌡ or 2 a⋅y 3 Integrating a⋅y
= b⋅x + C
3 For the streamline through point (6,6) C = 60 For particle that passed through (1,4) at t = 0 u= dx
2
= a⋅y
dt ⌠
⌠
2
⎮
⎮
1 dx = x − x0 = ⎮ a ⋅ y dt
⎮
⌡
⌡ v= dy
=b
dt ⌠
⌠
⎮
⎮
1 dy = ⎮ b dt
⎮
⌡
⌡ t Then
Hence, with x0 = 1 y0 = 4 3 y = 6⋅ x + 180 and ⎛ ⌠
2
x − x0 = ⎮ a ⋅ y0 + b ⋅ t dt
⌡0
2 43
x = 1 + 16⋅ t + 8⋅ t + ⋅ t
3 ( ) 2 y = y0 + b⋅ t = y0 + 2⋅ t
2 x = x0 + a⋅ ⎜ y0 ⋅ t + b⋅ y0⋅ t + ⎝ t ⌠
2
x − x0 = ⎮ a⋅ y0 + b⋅ t dt
⌡t ( ) 0 y = 6⋅ m ⌠
⌠
⎮
⎮
1 dy =
b dt
⎮
⎮
⌡
⌡ ( y = y0 + b ⋅ t − t0 ) ⎡2
b ⎛3
2
2
3⎤
x = x0 + a ⋅ ⎢y0 ⋅ t − t0 + b ⋅ y0⋅ ⎛ t − t0 ⎞ +
⋅ t − t0 ⎞⎥
⎝
⎠ 3⎝
⎠⎦
⎣ ( 4 Hence, with x0 = 3, y0 = 0 at t0 = 1 x = −3 + Evaluating at t = 3 (3 ) ⋅ t −1 = 2 ) 1 ( x = 31.7⋅ m 3 23
b ⋅t ⎞
⎟
3⎠ x = 26.3⋅ m At t = 1 s y = 4 + 2⋅ t For particle that passed through (3,0) at t = 1 3 3 ) ⋅ 4⋅ t − 13 This is a steady flow, so pathlines, streamlines and streaklines always coincide but we need y(t) y = 2⋅ ( t − 1)
y = 4⋅ m Problem 2.29 [4] Part 1/2 Problem 2.29 [4] Part 2/2 Problem 2.30 [4] Part 1/2 Problem 2.30 [4] Part 2/2 Problem 2.31 [4] Part 1/2 Problem 2.31 [4] Part 2/2 Problem 2.32 [2] Problem 2.33 [2] Data: Using procedure of Appendix A.3:
T (oC)
0
100
200
300
400 μ(x105)
1.86E05
2.31E05
2.72E05
3.11E05
3.46E05 T (K)
273
373
473
573
673 T (K)
273
373
473
573
673 T3/2/μ
2.43E+08
3.12E+08
3.78E+08
4.41E+08
5.05E+08 The equation to solve for coefficients
S and b is
32 T μ S
⎛1⎞
= ⎜ ⎟T +
b
⎝b⎠ From the builtin Excel
Linear Regression functions: Hence:
b = 1.531E06
S = 101.9 Slope = 6.534E+05
Intercept = 6.660E+07
R2 = 0.9996 kg/m.s.K1/2
K Plot of Basic Data and Trend Line
6.E+08
Data Plot 5.E+08 Least Squares Fit 4.E+08
T3/2/μ 3.E+08 2.E+08
1.E+08
0.E+00
0 100 200 300 400
T 500 600 700 800 Problem 2.35 [2] Given: Velocity distribution between flat plates Find: Shear stress on upper plate; Sketch stress distribution Solution:
τyx = μ⋅ Basic equation τyx = − 2
8⋅ umax⋅ y
⎡
du
d
2⋅ y ⎞ ⎤
⎛ 4⎞
=
umax⋅ ⎢1 − ⎛
⎜ ⎟ ⎥ = umax⋅ ⎜ − 2 ⎟ ⋅ 2⋅ y = −
2
dy
dy
⎣ ⎝ h ⎠⎦
h
⎝h⎠ du
dy 8⋅ μ⋅ umax⋅ y
2 h
At the upper surface y= h
2 h = 0.1⋅ mm and umax = 0.1⋅ m
s − 3 N ⋅s
⋅
2 μ = 1.14 × 10 1000⋅ mm ⎞
m 0.1
1⋅ m
1
− 3 N ⋅s
⋅
× 0.1⋅ ×
⋅ mm ×
×⎛
×
⎜
⎟
2
1⋅ m ⎠
s
2
1000⋅ mm ⎝ 0.1⋅ mm m 2 τyx = −8 × 1.14 × 10 Hence m τyx = −4.56⋅ N
2 m The upper plate is a minus y surface. Since τyx < 0, the shear stress on the upper plate must act in the plus x direction. ⎞
⎛ 8⋅ μ⋅ umax ⎟
⋅y
2
⎜h
⎟
⎝
⎠ τyx ( y) = −⎜ The shear stress varies linearly with y 0.05
0.04
0.03 y (mm) 0.02
0.01
−5 −4 −3 −2 −1
0
− 0.01 1 − 0.02
− 0.03
− 0.04
− 0.05 Shear Stress (Pa) 2 3 4 5 (Table A.8) Problem 2.36 [2] Given: Velocity distribution between parallel plates Find: Force on lower plate Solution:
Basic equations F = τyx⋅ A τyx = μ⋅ du
dy 2
8⋅ umax⋅ y
⎡
du
d
2⋅ y ⎞ ⎤
⎛ 4⎞
=
umax⋅ ⎢1 − ⎛
⎜ ⎟ ⎥ = umax⋅ ⎜ − 2 ⎟ ⋅ 2⋅ y = −
2
dy
dy
⎣ ⎝ h ⎠⎦
h
⎝h⎠ so τyx = − 8⋅ μ⋅ umax⋅ y
h At the lower surface y=− h
2 Hence 2 m
s
2 μ = 1.14 × 10 A = 1⋅ m
− 3 N⋅ s
⋅
2 2 2 (Table A.8) m m −0.1
1⋅ m
11
1000⋅ mm ⎞
− 3 N⋅ s
⋅
× 0.05 ×
⋅ mm ×
×⎛
×
⎜⋅
⎟
2
s
2
1000⋅ mm ⎝ 0.1 mm
1⋅ m ⎠ F = −8 × 1⋅ m × 1.14 × 10 m F = 2.28⋅ N 8⋅ A⋅ μ⋅ umax⋅ y
h h = 0.1⋅ mm and umax = 0.05⋅ F=− and (to the right) 2 Problem 2.37 [2] Explain how an ice skate interacts with the ice surface. What mechanism acts to reduce
sliding friction between skate and ice?
OpenEnded Problem Statement: Explain how an ice skate interacts with the ice
surface. What mechanism acts to reduce sliding friction between skate and ice?
Discussion: The normal freezing and melting temperature of ice is 0°C (32°F) at
atmospheric pressure. The melting temperature of ice decreases as pressure is increased.
Therefore ice can be caused to melt at a temperature below the normal melting
temperature when the ice is subjected to increased pressure.
A skater is supported by relatively narrow blades with a short contact against the ice. The
blade of a typical skate is less than 3 mm wide. The length of blade in contact with the ice
may be just ten or so millimeters. With a 3 mm by 10 mm contact patch, a 75 kg skater is
supported by a pressure between skate blade and ice on the order of tens of megaPascals
(hundreds of atmospheres). Such a pressure is enough to cause ice to melt rapidly.
When pressure is applied to the ice surface by the skater, a thin surface layer of ice melts
to become liquid water and the skate glides on this thin liquid film. Viscous friction is
quite small, so the effective friction coefficient is much smaller than for sliding friction.
The magnitude of the viscous drag force acting on each skate blade depends on the speed
of the skater, the area of contact, and the thickness of the water layer on top of the ice.
The phenomenon of static friction giving way to viscous friction is similar to the
hydroplaning of a pneumatic tire caused by a layer of water on the road surface. Problem 2.38 Given: Velocity profile Find: [2] Plot of velocity profile; shear stress on surface Solution:
ρ⋅ g ⎛
y⎞
u=
⋅ ⎜ h⋅ y − ⎟ ⋅ sin ( θ)
2⎠
μ⎝
2 The velocity profile is u Hence we can plot umax ⎡y = 2⋅ ⎢ ⎣h − 2 ρ⋅ g h
so the maximum velocity is at y = h umax =
⋅ ⋅ sin ( θ)
μ2 2
1 ⎛ y⎞ ⎤
⋅⎜ ⎟ ⎥
2 ⎝h⎠ ⎦ 1 y/h 0.8
0.6
0.4
0.2 0 0.2 0.4 0.6 0.8 1 u/umax
This graph can be plotted in Excel
The given data is h = 0.1⋅ in μ = 2.15 × 10 − 3 lbf ⋅ s
⋅
2 θ = 45⋅ deg ft du Basic equation τyx = μ⋅ At the surface y = 0 τyx = 0.85 × 1.94⋅ du dy slug
ft 3 × 32.2⋅ ft
2 s dy = μ⋅ 2
d ρ⋅ g ⎛
y⎞
⋅ ⎜ h⋅ y − ⎟ ⋅ sin ( θ) = ρ⋅ g⋅ ( h − y) ⋅ sin ( θ)
2⎠
dy μ ⎝ τyx = ρ⋅ g⋅ h⋅ sin ( θ) Hence τyx = μ⋅ × 0.1⋅ in × 1⋅ ft
12⋅ in 2 × sin ( 45⋅ deg) × lbf ⋅ s
slug⋅ ft The surface is a positive y surface. Since τyx > 0, the shear stress on the surface must act in the plus x direction. τyx = 0.313⋅ lbf
ft 2 Problem 2.39 [2] Problem 2.40 [2] Problem 2.41 [2] Given: Data on tape mechanism Find: Maximum gap region that can be pulled without breaking tape Solution:
Basic equation τyx = μ⋅ du
dy and Here F is the force on each side of the tape; the total force is then
The velocity gradient is linear as shown
The area of contact is F = τyx⋅ A
FT = 2⋅ F = 2⋅ τyx⋅ A du
V−0
V
=
=
dy
c
c y A = w⋅ L c
t
F,V Combining these results x
c V
FT = 2⋅ μ⋅ ⋅ w⋅ L
c
FT ⋅ c Solving for L L= The given data is FT = 25⋅ lbf Hence L = 25⋅ lbf × 0.012⋅ in × L 2⋅ μ⋅ V⋅ w
c = 0.012⋅ in μ = 0.02⋅ slug
ft⋅ s V = 3⋅ ft
s w = 1⋅ in 1⋅ ft
1
1 ft⋅ s 1 s 1 1 12⋅ in slug⋅ ft
××
⋅
×⋅×
×
×
2
1⋅ ft
12⋅ in 2 0.02 slug 3 ft 1 in
s ⋅ lbf L = 2.5 ft Problem 2.42 Given: Flow data on apparatus Find: [2] The terminal velocity of mass m Solution:
Given data: Dpiston = 73⋅ mm Reference data: ρwater = 1000⋅ kg
3 Dtube = 75⋅ mm Mass = 2⋅ kg L = 100⋅ mm SGAl = 2.64 (maximum density of water) m μ = 0.13⋅ From Fig. A.2:, the dynamic viscosity of SAE 10W30 oil at 25oC is: N ⋅s
m 2 The terminal velocity of the mass m is equivalent to the terminal velocity of the piston. At that terminal speed, the acceleration of the
piston is zero. Therefore, all forces acting on the piston must be balanced. This means that the force driving the motion
(i.e. the weight of mass m and the piston) balances the viscous forces acting on the surface of the piston. Thus, at r = Rpiston:
2 ⎞⎤
⎡
⎛ π⋅ D
⎢
⎜
piston ⋅ L ⎟⎥
⎛d ⎞
⎢Mass + SGAl⋅ ρwater⋅ ⎜
⎟⎥ ⋅ g = τrz⋅ A = ⎜ μ⋅ dr Vz ⎟ ⋅ (π⋅ Dpiston⋅ L)
4
⎣
⎝
⎠⎦
⎝
⎠ The velocity profile within the oil film is linear ...
Therefore V
d
Vz =
dr
⎛ Dtube − Dpiston ⎞
⎜
⎟
2
⎝
⎠ Thus, the terminal velocity of the piston, V, is: g⋅ ⎛ SGAl⋅ ρwater⋅ π⋅ Dpiston ⋅ L + 4⋅ Mass⎞ ⋅ Dtube − Dpiston
⎠
V= ⎝
8⋅ μ⋅ π⋅ Dpiston⋅ L
2 or V = 10.2 m
s ( ) Problem 2.43 [3] Given: Flow data on apparatus Find: Sketch of piston speed vs time; the time needed for the piston to reach 99% of its new terminal speed. Solution:
Given data: Dpiston = 73⋅ mm Reference data: ρwater = 1000⋅ kg
3 Dtube = 75⋅ mm L = 100⋅ mm SGAl = 2.64 V0 = 10.2⋅ (maximum density of water) (From Problem 2.42) m From Fig. A.2, the dynamic viscosity of SAE 10W30 oil at 25oC is: μ = 0.13⋅ N⋅ s
2 m
The free body diagram of the piston after the cord is cut is: 2⎞
⎛ π⋅ D
⎜
piston ⎟
Wpiston = SGAl⋅ ρwater⋅ g⋅ ⎜
⎟⋅L
4
⎝
⎠ Piston weight: Viscous force: Fviscous ( V) = τrz⋅ A V
⎤ ⋅ π⋅ D
Fviscous( V) = μ⋅ ⎡
piston⋅ L
⎢1
⎥
⎢ ⋅ Dtube − Dpiston ⎥
⎣2
⎦ or ( dV
mpiston⋅
= Wpiston − Fviscous( V)
dt Applying Newton's second law: 8⋅ μ
SGAl⋅ ρwater⋅ Dpiston⋅ Dtube − Dpiston Therefore dV
= g − a⋅ V where
dt a= If V = g − a⋅ V dX
dV
= −a ⋅
dt
dt The differential equation becomes then dX
= −a ⋅ X
dt ( where m
s ) X ( 0) = g − a ⋅ V 0 ) ( ) − a⋅ t X ( t) = X0⋅ e The solution to this differential equation is: or ( ) − a⋅ t g − a⋅ V ( t) = g − a⋅ V0 ⋅ e g ( − a⋅ t) g
⎛
V ( t) = ⎜ V0 − ⎞ ⋅ e
+
⎟
a⎠
a
⎝ Therefore Plotting piston speed vs. time (which can be done in Excel) Piston speed vs. time
12
10
8
V ( t) 6
4
2 0 1 2 3 t The terminal speed of the piston, Vt, is evaluated as t approaches infinity
Vt = g
a or Vt = 3.63 The time needed for the piston to slow down to within 1% of its terminal velocity is: ⎛ V −g
0a
1⎜
t = ⋅ ln ⎜
a
⎜ 1.01⋅ Vt −
⎝ ⎞
⎟
⎟
g
⎟
a⎠ or t = 1.93 s m
s Problem 2.44 [3] Part 1/2 Problem 2.44 [3] Part 2/2 Problem 2.45 [4] Ff = τ ⋅ A
x, V, a M⋅ g Given: Data on the block and incline Find: Initial acceleration; formula for speed of block; plot; find speed after 0.1 s. Find oil viscosity if speed is 0.3 m/s after 0.1 s Solution:
Given data M = 5⋅ kg From Fig. A.2 μ = 0.4⋅ A = ( 0.1⋅ m ) 2 d = 0.2⋅ mm θ = 30⋅ deg N⋅ s
2 m Applying Newton's 2nd law to initial instant (no friction)⋅ a = M⋅ g⋅ sin( θ) − Ff = M⋅ g⋅ sin( θ)
M
so M⋅ a = M⋅ g⋅ sin( θ) − Ff Applying Newton's 2nd law at any instant
so M⋅ a = M⋅ μ⋅ A
⋅V
g⋅ sin ( θ) −
M⋅ d
− Integrating and using limits or V = 5⋅ kg × 9.81⋅ m
2 s du
V
⋅ A = μ⋅ ⋅ A
dy
d M⋅ d ⎛
μ⋅ A
⎞
⋅ ln ⎜ 1 −
⋅ V⎟ = t
μ⋅ A ⎝
M⋅ g⋅ d⋅ sin ( θ) ⎠ − μ⋅ A ⎞
⎛
⋅t
⎜
M⋅ g⋅ d⋅ sin ( θ)
M⋅ d ⎟
V ( t) =
⋅ ⎝1 − e
⎠ μ⋅ A × 0.0002⋅ m⋅ sin ( 30⋅ deg) × m Ff = τ ⋅ A = μ⋅ = dt 2 s
V ( 0.1⋅ s) = 0.404⋅ and m
ainit = 4.9
2
s dV
μ⋅ A
= M⋅ g⋅ sin( θ) −
⋅V
dt
d dV Separating variables At t = 0.1 s m
ainit = g⋅ sin( θ) = 9.81⋅ × sin( 30⋅ deg)
2
s m 0.4⋅ N⋅ s⋅ ( 0.1⋅ m) ⎡ ⎛ 0.4⋅ 0.01 ⎞⎤ 2
−⎜
⋅ 0.1⎟
⎢
N⋅ s
5⋅ 0.0002
⎠⎥
×
× ⎣1 − e ⎝
⎦
2 kg⋅ m The plot looks like V (m/s) 1.5 1 0.5 0 0.2 0.4 0.6 0.8 t (s) To find the viscosity for which V(0.1 s) = 0.3 m/s, we must solve ⎡ − μ⋅ A M⋅ g⋅ d⋅ sin ( θ) ⎢
M⋅ d
V ( t = 0.1⋅ s) =
⋅ ⎣1 − e
μ⋅ A ⋅ ( t=0.1⋅ s)⎤ ⎥
⎦ The viscosity μ is implicit in this equation, so solution must be found by manual iteration, or by any of a number of classic
rootfinding numerical methods, or by using Excel's Goal Seek
Using Excel: μ = 1.08⋅ N⋅ s
2 m 1 Problem 2.46 [3] Problem 2.47 [2] Problem 2.48 [3] NOTE: Figure is wrong  length is 0.85 m Given: Data on double pipe heat exchanger Find: Whether noslip is satisfied; net viscous force on inner pipe Solution:
For the oil, the velocity profile is Check the noslip condition. When For the water, the velocity profile is Check the noslip condition. When 2
⎡
r ⎞⎤
uz( r ) = umax⋅ ⎢1 − ⎛ ⎟ ⎥
⎢ ⎜ Rii ⎥
⎣ ⎝ ⎠⎦ 2 where umax = Rii ⋅ Δp
4⋅ μ⋅ L ⎡ ⎛ R ⎞ 2⎤
ii ⎥
⎢
r = Rii
uz( Rii) = umax⋅ 1 − ⎜ ⎟ = 0
⎢
Rii ⎥
⎣ ⎝ ⎠⎦
2
2
⎞
⎛
1 Δp ⎜ 2 2 Roi − Rio
r⎞
uz( r ) =
⋅
⋅ R −r −
⋅ ln ⎛
⎜ ⎟⎟
4⋅ μ L ⎜ io
Rio ⎠ ⎟
Rio ⎞
⎛
⎝
ln ⎜
⎟
⎟
⎜
⎝ Roi ⎠
⎠
⎝
2
2
⎛
⎛ Roi ⎞ ⎞
1 Δp ⎜ 2
2 Roi − Rio
r = Roi
uz (Roi) =
⋅
⋅ Rio − Roi −
⋅ ln ⎜
⎟⎟
4⋅ μ L ⎜
Rio ⎠ ⎟
Rio ⎞
⎛
⎝
ln ⎜
⎟
⎟
⎜
⎝
⎝ Roi ⎠
⎠
1 Δp ⎡ 2
2
2
2
uz Roi =
⋅
⋅ R − Roi + ⎛ Roi − Rio ⎞⎤ = 0
⎝
⎠⎦
4⋅ μ L ⎣ io () ⎛
⎛ Rio ⎞ ⎞
1 Δp ⎜ 2
2 Roi − Rio
uz Rio =
⋅
⋅ Rio − Rio −
⋅ ln ⎜
⎟⎟ = 0
4⋅ μ L ⎜
Rio ⎟
Rio ⎞
⎛
⎝⎠
ln ⎜
⎜
⎟
⎟
Roi
2 When 2 () r = Rio ⎝ ⎝ ⎠ ⎠ The noslip condition holds on all three surfaces.
The given data is Rii = 7.5⋅ cm
− 3⋅ mm Rii = 3.45⋅ cm
2 Δpw = 2.5⋅ Pa Rio = 7.5⋅ cm
2 Δpoil = 8⋅ Pa 11⋅ cm
− 3⋅ mm Roi = 5.2⋅ cm
2 L = 0.85⋅ m μw = 1.25 × 10 The viscosity of water at 10oC is (Fig. A.2) Rio = 3.75⋅ cm Roi = − 3 N⋅ s
⋅
2 m − 2 N⋅ s
⋅
2 μoil = 1 × 10 The viscosity of SAE 1030 oil at 100oC is (Fig. A.2) m For each, shear stress is given by For water τrx = μ⋅
τrx = μ⋅ du
dr ⎞⎤
⎡
⎛
d
1 Δpw ⎜ 2 2 Roi − Rio
r ⎞ ⎟⎥
= μw ⋅ ⎢
⋅
⋅ Rio − r −
⋅ ln ⎛
⎜⎟
dr ⎢ 4⋅ μw L ⎜
⎛ Rio ⎞
⎝ Rio ⎠ ⎟⎥
ln ⎜
⎢
⎜
⎟
⎟⎥
Roi
2 duz ( r)
dr ⎣ ⎝ ⎝ 2 ⎠ ⎠⎦ 2⎞ ⎛
Roi − Rio
1 Δpw ⎜
τrx = ⋅
⋅ − 2⋅ r −
4L⎜
⎛ Rio ⎞
ln ⎜
⎜
⎟⋅r
Roi
2 ⎝ ⎝ ⎟
⎟
⎟
⎠⎠ ⎛
Roi − Rio ⎞
1 Δpw ⎜
⎟ ⋅ 2⋅ π ⋅ R ⋅ L
Fw = τrx⋅ A = ⋅
⋅ −2⋅ Rio −
io
4L⎜
Rio ⎞
⎟
⎛
ln ⎜
⎜
⎟ ⋅ Rio ⎟
Roi
2 so on the pipe surface ⎝ Fw Hence Fw ⎝ 2 ⎠ ⎠ 2
2
⎛
R − Rio ⎞
⎜ −R 2 − oi
⎟
= Δpw⋅ π⋅
⎜ io
⎛ Rio ⎞ ⎟
2⋅ ln ⎜
⎜
⎟⎟
⎝
⎝ Roi ⎠ ⎠
2⎤
⎡
⎢
⎡( 5.2⋅ cm) 2 − ( 3.75⋅ cm) 2⎤ × ⎛ 1⋅ m ⎞ ⎥
2⎣
⎦⎜
⎟
N
1⋅ m ⎞
⎢
⎝ 100⋅ cm ⎠ ⎥
= 2.5⋅
× π × −⎛ 3.75⋅ cm ×
−
⎟
⎢⎜
⎥
2
100⋅ cm ⎠
3.75 ⎞
⎝
m
2⋅ ln ⎛
⎜
⎟
⎢
⎥
⎣
⎝ 5.2 ⎠
⎦ Fw = 0.00454 N
This is the force on the rnegative surface of the fluid; on the outer pipe itself we also have Fw = 0.00454 N
duz ( r) 2
2⋅ μoil⋅ umax⋅ r
Δpoil⋅ r
⎡
d
r ⎞⎤
⎥
= μoil⋅ umax⋅ ⎢1 − ⎛
=−
⎜R ⎟ ⎥ = −
2
⎢
dr
2⋅ L
Rii
⎣ ⎝ ii ⎠ ⎦ For oil τrx = μ⋅ so on the pipe surface Foil = τrx⋅ A = − dr Δpoil⋅ Rii
2⋅ L ⋅ 2⋅ π⋅ Rii⋅ L = −Δpoil⋅ π⋅ Rii This should not be a surprise: the pressure drop just balances the friction! 2 Hence Foil = −8⋅ N
2 m ⎛
⎝ × π × ⎜ 3.45⋅ cm × 1⋅ m ⎞
⎟
100⋅ cm ⎠ 2 This is the force on the rpositive surface of the fluid; on the pipe it is equal and opposite
The total force is F = Fw + Foil Note we didn't need the viscosities because all quantities depend on the Δp's! Foil = −0.0299 N
Foil = 0.0299 N
F = 0.0345 N Problem 2.49 [3] NOTE: Figure is wrong  length is 0.85 m Given: Data on counterflow heat exchanger Find: Whether noslip is satisfied; net viscous force on inner pipe Solution:
The analysis for Problem 2.48 is repeated, except the oil flows in reverse, so the pressure drop is 2.5 Pa not 2.5 Pa.
For the oil, the velocity profile is Check the noslip condition. When For the water, the velocity profile is Check the noslip condition. When ⎡ ⎛ r ⎞ 2⎤
⎥
uz ( r) = umax⋅ ⎢1 − ⎜
⎢
Rii ⎟ ⎥
⎣ ⎝ ⎠⎦ 2 where umax = Rii ⋅ Δp
4⋅ μ⋅ L ⎡ ⎛ R ⎞ 2⎤
ii ⎥
⎢
r = Rii
uz (Rii) = umax⋅ 1 − ⎜
⎟ =0
⎢
Rii ⎥
⎣ ⎝ ⎠⎦
2
2
⎞
⎛
1 Δp ⎜ 2 2 Roi − Rio
⎛ r ⎞⎟
uz ( r) =
⋅
⋅ R −r −
⋅ ln ⎜
⎟
4⋅ μ L ⎜ io
⎛ Rio ⎞
⎝ Rio ⎠ ⎟
ln ⎜
⎟
⎟
⎜
⎝
⎝ Roi ⎠
⎠
2
2
⎛
⎛ Roi ⎞ ⎞
1 Δp ⎜ 2
2 Roi − Rio
r = Roi
uz (Roi) =
⋅
⋅ R − Roi −
⋅ ln ⎜
⎟⎟
4⋅ μ L ⎜ io
Rio ⎟
⎛ Rio ⎞
⎝⎠
ln ⎜
⎜
⎟
⎟
Roi
⎝
⎝⎠
⎠
1 Δp ⎡ 2
2
2
2
uz Roi =
⋅
⋅ Rio − Roi + ⎛ Roi − Rio ⎞⎤ = 0
⎣
⎝
⎠⎦
4⋅ μ L () ⎛
⎛ Rio ⎞ ⎞
1 Δp ⎜ 2
2 Roi − Rio
uz Rio =
⋅
⋅ R − Rio −
⋅ ln ⎜
⎟⎟ = 0
4⋅ μ L ⎜ io
Rio ⎟
⎛ Rio ⎞
⎝⎠
ln ⎜
⎜
⎟
⎟
Roi
2 When 2 () r = Rio ⎝ ⎝ ⎠ ⎠ The noslip condition holds on all three surfaces.
The given data is Rii = 7.5⋅ cm
− 3⋅ mm Rii = 3.45⋅ cm
2 Δpw = −2.5⋅ Pa Rio = 7.5⋅ cm
2 Δpoil = 8⋅ Pa 11⋅ cm
− 3⋅ mm Roi = 5.2⋅ cm
2 L = 0.85⋅ m μw = 1.25 × 10 The viscosity of water at 10oC is (Fig. A.2) Rio = 3.75⋅ cm Roi = − 3 N⋅ s
⋅
2 m The viscosity of SAE 1030 oil at 100oC − 2 N⋅ s
μoil = 1 × 10 ⋅
2 is (Fig. A.2) m For each, shear stress is given by For water τrx = μ⋅
τrx = μ⋅ du
dr
duz ( r)
dr ⎞⎤
⎡
⎛
d
1 Δpw ⎜ 2 2 Roi − Rio
r ⎞ ⎟⎥
= μw ⋅ ⎢
⋅
⋅ Rio − r −
⋅ ln ⎛
⎜ R ⎟ ⎟⎥
dr ⎢ 4⋅ μw L ⎜
⎛ Rio ⎞
⎝ io ⎠
ln ⎜
⎢
⎜
⎟
⎟⎥
Roi
2 ⎣ ⎝ ⎝ 2 ⎠ ⎠⎦ 2⎞ ⎛
Roi − Rio
1 Δpw ⎜
τrx = ⋅
⋅ − 2⋅ r −
4L⎜
⎛ Rio ⎞
ln ⎜
⎜
⎟⋅r
Roi
2 ⎝ ⎝ ⎟
⎟
⎟
⎠⎠ ⎛
Roi − Rio ⎞
1 Δpw ⎜
⎟ ⋅ 2⋅ π ⋅ R ⋅ L
Fw = τrx⋅ A = ⋅
⋅ −2⋅ Rio −
io
4L⎜
Rio ⎞
⎟
⎛
ln ⎜
⎜
⎟ ⋅ Rio ⎟
Roi
2 so on the pipe surface ⎝ Fw Hence Fw ⎝ 2 ⎠ ⎠ 2
2
⎛
R − Rio ⎞
⎜ −R 2 − oi
⎟
= Δpw⋅ π⋅
⎜ io
⎛ Rio ⎞ ⎟
2⋅ ln ⎜
⎜
⎟⎟
⎝
⎝ Roi ⎠ ⎠
2⎤
⎡
⎢
⎡( 5.2⋅ cm) 2 − ( 3.75⋅ cm) 2⎤ × ⎛ 1⋅ m ⎞ ⎥
2⎣
⎦⎜
⎟
N
1⋅ m ⎤
⎢
⎝ 100⋅ cm ⎠ ⎥
= −2.5⋅
× π × −⎡( 3.75⋅ cm) ×
−
⎥
⎢⎢
⎥
2
100⋅ cm⎦
3.75 ⎞
⎣
m
2⋅ ln ⎛
⎜
⎟
⎢
⎥
⎣
⎝ 5.2 ⎠
⎦ Fw = −0.00454 N
This is the force on the rnegative surface of the fluid; on the outer pipe itself we also have Fw = −0.00454 N
duz ( r) 2
2⋅ μoil⋅ umax⋅ r
Δpoil⋅ r
⎡
d
r ⎞⎤
⎥
= μoil⋅ umax⋅ ⎢1 − ⎛
=−
⎜R ⎟ ⎥ = −
2
⎢
dr
2⋅ L
Rii
⎣ ⎝ ii ⎠ ⎦ For oil τrx = μ⋅ so on the pipe surface Foil = τrx⋅ A = − dr Δpoil⋅ Rii
2⋅ L ⋅ 2⋅ π⋅ Rii⋅ L = −Δpoil⋅ π⋅ Rii This should not be a surprise: the pressure drop just balances the friction! 2 Hence Foil = −8⋅ N
2 m ⎛
⎝ × π × ⎜ 3.45⋅ cm × 1⋅ m ⎞
⎟
100⋅ cm ⎠ 2 This is the force on the rpositive surface of the fluid; on the pipe it is equal and opposite
The total force is F = Fw + Foil Note we didn't need the viscosities because all quantities depend on the Δp's! Foil = −0.0299 N
Foil = 0.0299 N
F = 0.0254 N Problem 2.50 Given: Flow between two plates Find: [2] Force to move upper plate; Interface velocity Solution:
The shear stress is the same throughout (the velocity gradients are linear, and the stresses in the fluid at the interface
must be equal and opposite).
Hence du1
du2
τ = μ1⋅
= μ2⋅
dy
dy Solving for the interface velocity Vi V Vi =
1+ Then the force required is μ1⋅ or μ1 h2
⋅
μ2 h1 1⋅ = Vi
h1 = μ2⋅ ( V − Vi)
h2 where Vi is the interface velocity m s
0.1 0.3
1+
⋅
0.15 0.5 Vi
N ⋅s
m
1
1000⋅ mm
2
F = τ ⋅ A = μ1⋅ ⋅ A = 0.1⋅
× 0.714⋅ ×
×
× 1⋅ m
2
h1
s 0.5⋅ mm
1⋅ m
m Vi = 0.714 F = 143 N m
s Problem 2.51 [2] Problem 2.52 [2] Problem 2.53 [2] Problem 2.54 [2] Problem 2.55 Given: Data on the viscometer Find: [4] Time for viscometer to lose 99% of speed Solution:
The given data is R = 50⋅ mm H = 80⋅ mm a = 0.20⋅ mm I = 0.0273⋅ kg⋅ m 2 μ = 0.1⋅ N ⋅s
m 2 I⋅ α = Torque = −τ ⋅ A⋅ R The equation of motion for the slowing viscometer is where α is the angular acceleration and τ is the viscous stress, and A is the surface area of the viscometer
The stress is given by τ = μ⋅ du
V−0
μ⋅ V
μ⋅ R ⋅ ω
= μ⋅
=
=
dy
a
a
a where V and ω are the instantaneous linear and angular velocities.
2 dω
μ⋅ R ⋅ ω
μ⋅ R ⋅ A
=−
⋅ A⋅ R =
⋅ω
a
dt
a Hence I⋅ α = I⋅ Separating variables dω
μ⋅ R ⋅ A
=−
⋅ dt
ω
a⋅I 2 2 − Integrating and using IC ω = ω0 ω ( t ) = ω0⋅ e μ⋅ R ⋅ A
⋅t
a⋅I
2 − The time to slow down by 99% is obtained from solving Note that A = 2⋅ π ⋅ R ⋅ H 0.01⋅ ω0 = ω0⋅ e μ⋅ R ⋅ A
⋅t
a⋅ I t=− a⋅ I
2 ⋅ ln ( 0.01) μ⋅ R ⋅ A
t=− so so a⋅ I
3 ⋅ ln ( 0.01) 2⋅ π⋅ μ⋅ R ⋅ H
2 2 0.0002⋅ m⋅ 0.0273⋅ kg⋅ m
m
t=−
⋅
⋅
2⋅ π
0.1⋅ N⋅ s 2 N⋅ s
⋅
⋅
⋅ ln ( 0.01)
3 0.08⋅ m kg⋅ m
( 0.05⋅ m)
1 1 t = 4.00 s Problem 2.56 [4] Problem 2.57 [4] Part 1/2 Problem 2.57 [4] Part 2/2 Problem 2.58 Given: Shockfree coupling assembly Find: [3] Required viscosity Solution:
Basic equation τrθ = μ⋅ du
dr Shear force F = τ⋅ A Assumptions: Newtonian fluid, linear velocity profile
τrθ = μ⋅ V2 = ω2(R + δ) δ τrθ = μ⋅ V1 = ω1R Then P = T⋅ ω2 = F⋅ R⋅ ω2 = τ ⋅ A2⋅ R⋅ ω2 = ( ) ( Torque T = F⋅ R Power P = T⋅ ω ⎡ω1⋅ R − ω2⋅ ( R + δ)⎤
du
ΔV
⎦
= μ⋅
= μ⋅ ⎣
δ
dr
Δr ( ω1 − ω2 )⋅ R
δ Because δ << R ) μ⋅ ω1 − ω2 ⋅ R
⋅ 2⋅ π⋅ R⋅ L⋅ R⋅ ω2
δ 3 2⋅ π⋅ μ⋅ ω2⋅ ω1 − ω2 ⋅ R ⋅ L
P=
δ
Hence P⋅ δ μ= ( ) 3 2⋅ π⋅ ω2⋅ ω1 − ω2 ⋅ R ⋅ L
−4 μ= 10⋅ W × 2.5 × 10
2⋅ π μ = 0.202⋅ N⋅ s
2 m ⋅m 2 × 1 min
1 min
1
1
N ⋅ m ⎛ rev ⎞
⎛ 60⋅ s ⎞
⋅
×
⋅
×
×
×
×⎜
⎟ ×⎜
⎟
3 0.02⋅ m
9000 rev 1000 rev
s⋅ W ⎝ 2⋅ π⋅ rad ⎠
⎝ min ⎠
( .01⋅ m ) μ = 2.02 poise which corresponds to SAE 30 oil at 30oC. 2 Problem 2.59 [4] Problem 2.60 [4] The data is 2
N (rpm) μ (N·s/m )
10
0.121
20
0.139
30
0.153
40
0.159
50
0.172
60
0.172
70
0.183
80
0.185 The computed data is
ω (rad/s) ω/θ (1/s) η (N·s/m2x103)
1.047
2.094
3.142
4.189
5.236
6.283
7.330
8.378 120
240
360
480
600
720
840
960 121
139
153
159
172
172
183
185 From the Trendline analysis
k = 0.0449
n  1 = 0.2068
n = 1.21 The fluid is dilatant The apparent viscosities at 90 and 100 rpm can now be computed
N (rpm) ω (rad/s)
90
9.42
100
10.47 ω/θ (1/s) η (N·s/m2x103) 1080
1200 191
195 Viscosity vs Shear Rate 2
3
η (N.s/m x10 ) 1000
Data
Power Trendline 100 η = 44.94(ω/θ)0.2068
R2 = 0.9925
10
100 1000
Shear Rate ω/θ (1/s) Problem 2.62 (In Excel) Given: Viscometer data
Find: Value of k and n in Eq. 2.17
Solution:
τ (Pa)
du/dy (s1)
The data is
0.0457
5
0.119
10
0.241
25
0.375
50
0.634
100
1.06
200
1.46
300
1.78
400 [3] Shear Stress vs Shear Strain
10 Data τ (Pa) Power Trendline
1
1 10 100 τ = 0.0162(du/dy) 0.7934
R2 = 0.9902 0.1 0.01 du/dy (1/s) k = 0.0162
n = 0.7934 Hence we have Blood is pseudoplastic (shear thinning) We can compute the apparent viscosity from
du/dy (s1) η (N·s/m2)
5
10
25
50
100
200
300
400 0.0116
0.0101
0.0083
0.0072
0.0063
0.0054
0.0050
0.0047 η= k (du/dy )n 1 2
o
μ water = 0.001 N·s/m at 20 C Hence, blood is "thicker" than water! 1000 Problem 2.63 (In Excel) Given:
Find:
Solution: [4] Data on insulation material
Type of material; replacement material The velocity gradient is
du/dy = U/ δ Data and
computations where δ = τ (Pa)
50
100
150
163
171
170
202
246
349
444 U (m/s)
0.000
0.000
0.000
0.005
0.01
0.03
0.05
0.1
0.2
0.3 0.001 m du/dy (s1)
0
0
0
5
10
25
50
100
200
300 τy =
μp = Hence we have a Bingham plastic, with 154 Pa
N·s/m2 0.963 At τ = 450 Pa, based on the linear fit du/dy = 307 s1 For a fluid with τy = 250 Pa we can use the Bingham plastic formula to solve for μ p given τ , τ y and du/dy from above μp = N·s/m2 0.652 Shear Stress vs Shear Strain
500
450 τ (Pa) 400
350
300 Linear data fit:
τ = 0.9632(du/dy ) + 154.34
R2 = 0.9977 250
200
150
100
50
0
0 50 100 150 200 du/dy (1/s) 250 300 350 Problem 2.64 [5] Problem 2.65 [5] Problem 2.66 Given: Conical bearing geometry Find: [4] Expression for shear stress; Viscous torque on shaft Solution: ds τ = μ⋅ Basic equation du
dy dT = r ⋅ τ ⋅ dA dz
z AA Infinitesimal shear torque
r Assumptions: Newtonian fluid, linear velocity profile (in narrow clearance gap), no slip condition
tan( θ) =
τ = μ⋅ Then r
z Section AA r = z⋅ tan( θ) so du
Δu
( ω⋅ r − 0)
μ⋅ ω⋅ z⋅ tan( θ)
= μ⋅
= μ⋅
=
dy
Δy
( a − 0)
a U = ωr a As we move up the device, shear stress increases linearly (because rate of shear strain does)
But from the sketch dz = ds⋅ cos ( θ) dA = 2⋅ π⋅ r ⋅ ds = 2⋅ π⋅ r ⋅ The viscous torque on the element of area is dT = r⋅ τ⋅ dA = r⋅ Integrating and using limits z = H and z = 0 T= μ⋅ ω⋅ z⋅ tan( θ)
dz
⋅ 2⋅ π⋅ r⋅
a
cos ( θ)
3 μ = 0.2⋅ Using given data, and N⋅ s
2 dz
cos ( θ) π⋅ μ⋅ ω⋅ tan( θ) ⋅ H
2⋅ a⋅ cos ( θ) 3 dT = 3 2⋅ π⋅ μ⋅ ω⋅ z ⋅ tan( θ)
⋅ dz
a⋅ cos ( θ) 4 from Fig. A.2 m
T= π
2 × 0.2⋅ N⋅ s
2 m × 75⋅ rev
s 3 4 × tan( 30⋅ deg) × ( 0.025⋅ m) × 1
0.2 × 10 −3 ⋅m × 1
cos ( 30⋅ deg) × 2⋅ π⋅ rad
rev T = 0.0643⋅ N⋅ m Problem 2.67 [5] Problem 2.68 [5] Problem 2.69 Given: [5] Geometry of rotating bearing Find: Expression for shear stress; Maximum shear stress; Expression for total torque; Total torque Solution:
τ = μ⋅ Basic equation du
dy dT = r⋅ τ⋅ dA Assumptions: Newtonian fluid, narrow clearance gap, laminar motion
du
u−0
u
=
=
dy
h
h r = R⋅ sin ( θ) u = ω⋅ r = ω⋅ R⋅ sin ( θ) h = a + R⋅ ( 1 − cos ( θ) ) From the figure dA = 2⋅ π⋅ r⋅ dr = 2⋅ π R⋅ sin ( θ) ⋅ R⋅ cos ( θ) ⋅ dθ du
μ⋅ ω⋅ R⋅ sin ( θ)
=
dy
a + R⋅ ( 1 − cos ( θ) ) Then τ = μ⋅ To find the maximum τ set d ⎡ μ⋅ ω⋅ R⋅ sin ( θ) ⎤
⎢
⎥=0
dθ ⎣ a + R⋅ ( 1 − cos ( θ) )⎦ so R⋅ cos ( θ) − R + a⋅ cos ( θ) = 0 θ = acos ⎛
⎜ R⋅ μ⋅ ω⋅ ( R⋅ cos ( θ) − R + a⋅ cos ( θ) )
( R + a − R⋅ cos ( θ) ) 2 R⎞
⎛ 75 ⎞
⎟ = acos⎜
⎟
R + a⎠
⎝
⎝ 75 + 0.5 ⎠ kg
m⋅ s =0 θ = 6.6⋅ deg
2 70 rad
1
N⋅ s
τ = 12.5⋅ poise × 0.1⋅
× 2⋅ π ⋅ ⋅
× 0.075⋅ m × sin ( 6.6⋅ deg) ×
×
poise
60 s
[ 0.0005 + 0.075⋅ ( 1 − cos ( 6.6⋅ deg) ) ] ⋅ m m⋅ kg
τ = 79.2⋅ N
2 m θmax The torque is ⌠
⌠
⎮
⎮
T=
r⋅ τ⋅ A dθ =
⎮
⎮
⌡
⌡0 4 2 μ⋅ ω⋅ R ⋅ sin ( θ) ⋅ cos ( θ)
dθ
a + R⋅ ( 1 − cos ( θ) ) where This integral is best evaluated numerically using Excel, Mathcad, or a good calculator ⎛ R0 ⎞
θmax = asin ⎜ ⎟
⎝R⎠
−3 T = 1.02 × 10 ⋅ N⋅ m θmax = 15.5⋅ deg Problem 2.70 [2] Problem 2.71 [2] Slowly fill a glass with water to the maximum possible level. Observe the water level
closely. Explain how it can be higher than the rim of the glass.
OpenEnded Problem Statement: Slowly fill a glass with water to the maximum
possible level before it overflows. Observe the water level closely. Explain how it can be
higher than the rim of the glass.
Discussion: Surface tension can cause the maximum water level in a glass to be higher
than the rim of the glass. The same phenomenon causes an isolated drop of water to
“bead up” on a smooth surface.
Surface tension between the water/air interface and the glass acts as an invisible
membrane that allows trapped water to rise above the level of the rim of the glass. The
mechanism can be envisioned as forces that act in the surface of the liquid above the rim
of the glass. Thus the water appears to defy gravity by attaining a level higher than the
rim of the glass.
To experimentally demonstrate that this phenomenon is the result of surface tension, set
the liquid level nearly as far above the glass rim as you can get it, using plain water. Add
a drop of liquid detergent (the detergent contains additives that reduce the surface tension
of water). Watch as the excess water runs over the side of the glass. Problem 2.72 Given: Data on size of various needles Find: [2] Which needles, if any, will float Solution:
For a steel needle of length L, diameter D, density ρs, to float in water with surface tension σ and contact angle θ, the
vertical force due to surface tension must equal or exceed the weight
2 2⋅ L⋅ σ⋅ cos ( θ) ≥ W = m ⋅ g = −3 N σ = 72.8 × 10 From Table A.4
From Table A.1, for steel Hence π⋅ D
⋅ ρs⋅ L⋅ g
4 8⋅ σ⋅ cos ( θ)
=
π⋅ SG⋅ ρ⋅ g ⋅ m D≤ or θ = 0⋅ deg 8⋅ σ⋅ cos ( θ)
π⋅ ρs⋅ g and for water ρ = 1000⋅ kg
3 m SG = 7.83
3 2 8
m
s
kg⋅ m
−3 N
−3
× 72.8 × 10 ⋅ ×
×
×
= 1.55 × 10 ⋅ m = 1.55⋅ mm
2
π⋅ 7.83
m 999⋅ kg 9.81⋅ m N ⋅ s Hence D < 1.55 mm. Only the 1 mm needles float (needle length is irrelevant) Problem 2.73 [5] Plan an experiment to measure the surface tension of a liquid similar to water. If
necessary, review the NCFMF video Surface Tension for ideas. Which method would be
most suitable for use in an undergraduate laboratory? What experimental precision could
be expected?
OpenEnded Problem Statement: Plan an experiment to measure the surface tension of
a liquid similar to water. If necessary, review the NCFMF video Surface Tension for
ideas. Which method would be most suitable for use in an undergraduate laboratory?
What experimental precision could be expected?
Discussion: Two basic kinds of experiment are possible for an undergraduate laboratory:
1. Using a clear smalldiameter tube, compare the capillary rise of the unknown liquid with that of a
known liquid (compare with water, because it is similar to the unknown liquid).
This method would be simple to set up and should give fairly accurate results. A vertical
traversing optical microscope could be used to increase the precision of measuring the liquid
height in each tube.
A drawback to this method is that the specific gravity and co ntact angle of the two liquids must be
the same to allow the capillary rises to be compared.
The capillary rise would be largest and therefore easiest to measure accurately in a tube with the
smallest practical diameter. Tubes of several diameters could be used if desired. 2. Dip an object into a pool of test liquid and measure the vertical force required to pull the object
from the liquid surface.
The object might be made rectangular (e.g., a sheet of plastic material) or circular (e.g., a metal
ring). The net force needed to pull the same object from each liquid should be proportional to the
surface tension of each liquid.
This method would be simple to set up. However, the force magnitudes to be measured would be
quite small.
A drawback to this method is that the contact angles of the two liquids must be the same. The first method is probably best for undergraduate laboratory use. A quantitative
estimate of experimental measurement uncertainty is impossible without knowing details
of the test setup. It might be reasonable to expect results accurate to within ± 10% of the
true surface tension. *Net force is the total vertical force minus the weight of the object. A buoyancy correction would be
necessary if part of the object were submerged in the test liquid. Problem 2.74 [2] Problem 2.75 [2] Given: Boundary layer velocity profile in terms of constants a, b and c Find: Constants a, b and c Solution:
⎛y
⎛y
u = a + b ⋅ ⎜ ⎞ + c⋅ ⎜ ⎞
⎟
⎟
⎝δ⎠
⎝δ⎠ Basic equation 2 Assumptions: No slip, at outer edge u = U and τ = 0
At y = 0 0=a a=0 At y = δ U = a+b+c b +c = U (1) At y = δ τ = μ⋅ b + 2⋅ c = 0 (2) du
=0
dy
2 0= d
y
y
b
y
b
c
a + b ⋅ ⎛ ⎞ + c⋅ ⎛ ⎞ =
+ 2⋅ c⋅
=
+ 2⋅
⎜⎟
⎜⎟
2
dy
δ⎠
δ⎠
δ
δ
δ
⎝
⎝
δ From 1 and 2 c = −U b = 2⋅ U Hence y⎞
y⎞
u = 2⋅ U ⋅ ⎛ ⎟ − U ⋅ ⎛ ⎟
⎜
⎜
δ⎠
⎝
⎝ δ⎠ 2 u
y⎞
y⎞
= 2⋅ ⎛ ⎟ − ⎛ ⎟
⎜
⎜
U
δ⎠ ⎝ δ⎠
⎝ 2 Dimensionless Height 1
0.75
0.5
0.25 0 0.25 0.5 Dimensionless Velocity 0.75 1 Problem 2.76 [2] Given: Boundary layer velocity profile in terms of constants a, b and c Find: Constants a, b and c Solution:
Basic equation ⎛y
⎛y
u = a + b ⋅ ⎜ ⎞ + c⋅ ⎜ ⎞
⎟
⎟
⎝δ⎠
⎝δ⎠ 3 Assumptions: No slip, at outer edge u = U and τ = 0
At y = 0 0=a a=0 At y = δ U = a+b+c b +c = U (1) At y = δ τ = μ⋅ b + 3⋅ c = 0 (2) du
=0
dy
3 0= 2 d
y
y
b
y
b
c
a + b ⋅ ⎛ ⎞ + c⋅ ⎛ ⎞ =
+ 3⋅ c⋅
=
+ 3⋅
⎜⎟
⎜⎟
3
dy
δ⎠
δ⎠
δ
δ
δ
⎝
⎝
δ From 1 and 2 c=− Hence u= U
2 b= 3
⋅U
2 3⋅ U ⎛ y ⎞ U ⎛ y ⎞
⋅⎜ ⎟ − ⋅⎜ ⎟
2 ⎝ δ⎠ 2 ⎝ δ⎠ 3 u
3 y⎞ 1 y⎞
= ⋅⎛ ⎟ − ⋅⎛ ⎟
⎜
⎜
U
2 ⎝ δ⎠ 2 ⎝ δ⎠ 3 Dimensionless Height 1
0.75
0.5
0.25 0 0.25 0.5 Dimensionless Velocity 0.75 1 Problem 2.77 Given: Local temperature Find: [1] Minimum speed for compressibility effects Solution:
Basic equation V = M⋅ c
c= Hence and k⋅ R⋅ T M = 0.3 for compressibility effects For air at STP, k = 1.40 and R = 286.9J/kg.K (53.33 ft.lbf/lbmoR). V = M⋅ c = M⋅ k ⋅ R⋅ T
V = 0.3 × ⎡1.4 × 53.33⋅
⎢ ⎣ 1
2 ft⋅ lbf
32.2⋅ lbm⋅ ft
⎤ 60⋅ mph
×
× ( 60 + 460) ⋅ R⎥ ⋅
2
lbm⋅ R
ft
lbf⋅ s
⎦ 88⋅
s V = 229⋅ mph Problem 2.78 [2] 3 NOTE: Flow rate should be Given: ft
min Geometry of and flow rate through garden hose Find: 0.75⋅ At which point becomes turbulent Solution:
For pipe flow (Section 26) Re = Also flow rate Q is given by Basic equation Q= ρ⋅ V ⋅ D
= 2300
μ for transition to turbulence 2 π⋅ D
⋅V
4 We can combine these equations and eliminate V to obtain an expression for Re in terms of D
Re = Hence D= ρ⋅ V ⋅ D
ρ⋅ D 4⋅ Q
4⋅ Q⋅ ρ
=
⋅
=
= 2300
2
μ
μ π⋅ D
π⋅ μ ⋅ D
4⋅ Q⋅ ρ
2300⋅ π⋅ μ From Appendix A:
0.209⋅ μ = 1.25 × 10 − 3 N⋅ s
⋅
×
2 m 1⋅ ρ = 1.94⋅ ft lbf ⋅ s ft
N⋅ s slug
3 (Approximately) 2 (Approximately, from
Fig. A.2) μ = 2.61 × 10 − 4 lbf ⋅ s
⋅
2 ft 2 m
3 Hence D= 2 2 0.75⋅ ft
4
1⋅ min 1.94⋅ slug
ft
lbf ⋅ s
12⋅ in
×
×
×
×
×
×
3
−4
min
slug⋅ ft
2300⋅ π
60⋅ s
1⋅ ft
2.61⋅ 10 ⋅ lbf ⋅ s
ft The nozzle is tapered: Din = 1⋅ in Dout = Din
4 Dout = 0.5⋅ in Linear ratios leads to the distance from Din at which D = 0.617 in
Lturb = L⋅ D − Din
Dout − Din Lturb
L L = 5⋅ in = D = 0.617⋅ in
NOTE: For wrong flow
rate, will be 1/10th of
this! D − Din
Dout − Din Lturb = 3.83⋅ in NOTE: For wrong flow
rate, this does not apply!
Flow will not become
turbulent. Problem 2.79 [3] Given: Data on supersonic aircraft Find: Mach number; Point at which boundary layer becomes turbulent Solution:
Basic equation V = M⋅ c Hence M= c= and V
=
c k⋅ R⋅ T For air at STP, k = 1.40 and R = 286.9J/kg.K (53.33 ft.lbf/lbmoR). V
k⋅ R⋅ T
T = 223.5⋅ K At 27 km the temperature is approximately (from Table A.3) M = ⎛ 2700 × 10 ⋅
⎜ 3m ⎝ hr 1 kg⋅ K 1⋅ N⋅ s
1⋅ hr ⎞ ⎛ 1
1
1⎞
⋅
×
×
⋅⎟
⎟⋅⎜ ×
kg⋅ m
3600⋅ s ⎠ ⎝ 1.4 286.9 N⋅ m
223.5 K ⎠
2 × For boundary layer transition, from Section 26
Then Retrans = ρ⋅ V⋅ xtrans
μ 1
2 M = 2.5 Retrans = 500000
μ⋅ Retrans xtrans = so ρ⋅ V We need to find the viscosity and density at this altitude and pressure. The viscosity depends on temperature only, but at 223.5 K =  50oC,
it is off scale of Fig. A.3. Instead we need to use formulas as in Appendix A
ρ = 0.02422 × 1.225⋅ At this altitude the density is (Table A.3) For μ μ= b⋅ T
1+ μ = 1.459 × 10 Hence ρ = 0.0297 3 m 1
2 S
T kg where b = 1.458 × 10 −6 m⋅ s⋅ K
− 5 kg μ = 1.459 × 10 m⋅ s 1
2 S = 110.4⋅ K − 5 N⋅ s
⋅
2 m − 5 kg xtrans = 1.459 × 10 3 m kg ⋅ kg ⋅ m⋅ s × 500000 × 3 m
1
1 hr 3600⋅ s
×
×
⋅×
0.0297 kg 2700 103 m
1⋅ hr
1 ⋅ xtrans = 0.327 m Problem 2.80 [2] Given: Data on water tube Find: Reynolds number of flow; Temperature at which flow becomes turbulent Solution:
Basic equation Re = For pipe flow (Section 26)
2
−7 m At 20oC, from Fig. A.3 ν = 9 × 10
For the heated pipe Re = Hence ν= ⋅ s V⋅ D
= 2300
ν and so ρ⋅ V ⋅ D
V⋅ D
=
μ
ν Re = 0.25⋅ m
1
s
× 0.005⋅ m ×
⋅
−7 2
s
9 × 10
m for transition to turbulence V⋅ D
1
m
=
× 0.25⋅ × 0.005⋅ m
2300
2300
s ν = 5.435 × 10 From Fig. A.3, the temperature of water at this viscosity is approximately T = 52⋅ C 2
−7m s Re = 1389 Problem 2.81 Given: Type of oil, flow rate, and tube geometry Find: [2] Whether flow is laminar or turbulent Solution:
Data on SAE 30 oil SG or density is limited in the Appendix. We can Google it or use the following At 100oC, from Figs. A.2 and A.3 μ = 9 × 10 ν = 1 × 10 − 3 N⋅ s
⋅
×
2 1 ρ = 9 × 10 1 × 10 m The specific weight is SG = ρ
ρwater −5 s ⋅ 2 m ⋅ × ρ = 900 2 s ⋅N kg γ = 900⋅ V= so
−6 Q = 100⋅ mL × Hence kg 3 m
2 s N⋅ s
kg⋅ m 3N
3 γ = 8.829 × 10 ⋅ m 4⋅ Q
π⋅ D 2 3 10 ⋅ m
11
×⋅
1⋅ mL
9s Q = 1.111 × 10 4
11
1000⋅ mm ⎞
−5 m
× 1.11 × 10 ⋅
×⎛ ⋅
×
⎜
⎟
s
π
1⋅ m ⎠
⎝ 12 mm
ρ⋅ V ⋅ D
Re =
μ 2 V = 0.0981 2 2 m
N⋅ s
Re = 900⋅
× 0.0981⋅ × 0.012⋅ m ×
⋅
×
3
− 3 N⋅ s kg⋅ m
s
m
9 × 10
Flow is laminar 3 2 × V= kg kg SG = 0.9 3 × 9.81⋅ 3 Then μ
ν m m
π⋅ D
⋅V
4 ρ= m γ = ρ⋅ g Q= so s kg⋅ m ρwater = 1000⋅ 2 For pipe flow (Section 26) μ
ρ 2
−5 m − 3 N⋅ s
⋅
2 m Hence ν= m 1 Re = 118 m
s 3
−5m s Problem 2.82 Given: Data on seaplane Find: [2] Transition point of boundary layer Solution:
For boundary layer transition, from Section 26 Retrans = 500000 Then Retrans = ρ⋅ V⋅ xtrans V⋅ xtrans = μ ν so xtrans = ν⋅ Retrans
V 2 At 45oF = 7.2oC (Fig A.3) −5 m ν = 0.8 × 10 ⋅ 10.8⋅ 2 s × 2 1⋅ − 5 ft xtrans = 8.64 × 10 ⋅ ft
s ⋅ 500000 × As the seaplane touches down:
At 45oF = 7.2oC (Fig A.3) − 5 ft ⋅ m
s 2 s ν = 8.64 × 10 2 s 1
60⋅ mph
×
100⋅ mph
ft
88⋅
s xtrans = 0.295⋅ ft 2 −5 m ν = 1.5 × 10 ⋅ 10.8⋅ 2 s × 2 1⋅ − 4 ft xtrans = 1.62 × 10 ⋅ ft
s m
s 2 s ν = 1.62 × 10 ⋅ 500000 × 1
60⋅ mph
×
100⋅ mph
ft
88⋅
s − 4 ft ⋅ 2 s xtrans = 0.552⋅ ft Problem 2.83 (In Excel) [3] Given:
Data on airliner
Find:
Sketch of speed versus altitude (M = const)
Solution:
Data on temperature versus height can be obtained from Table A.3
At 5.5 km the temperature is approximatel
The speed of sound is obtained from c =
k = 1.4
R = 286.9 J/kg·K c = 318 k ⋅ R ⋅T km/hr V = 194 K m/s V = 700 where 252 m/s (Table A.6) We also have or Hence M = V/c or
M = 0.611
To compute V for constant M , we use V = M · c = 0.611 ·c V = 677
km/hr
At a height of 8 km
NOTE: Realistically, the aiplane will fly to a maximum height of about 10 km!
T (K) 4
5
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
26
28
30
40
50
60
70
80
90 262
259
256
249
243
236
230
223
217
217
217
217
217
217
217
217
217
217
219
221
223
225
227
250
271
256
220
181
181 c (m/s) V (km/hr)
325
322
320
316
312
308
304
299
295
295
295
295
295
295
295
295
295
295
296
298
299
300
302
317
330
321
297
269
269 713
709
704
695
686
677
668
658
649
649
649
649
649
649
649
649
649
649
651
654
657
660
663
697
725
705
653
592
592 Speed vs. Altitude
750 700
Speed V (km/hr) z (km) 650 600 550
0 20 40 60 Altitude z (km) 80 100 Problem 2.84 [4] How does an airplane wing develop lift?
OpenEnded Problem Statement: How does an airplane wing develop lift?
Discussion: The sketch shows the crosssection of a typical airplane wing. The airfoil
section is rounded at the front, curved across the top, reaches maximum thickness about a
third of the way back, and then tapers slowly to a fine trailing edge. The bottom of the
airfoil section is relatively flat. (The discussion below also applies to a symmetric airfoil
at an angle of incidence that produces lift.) NACA 2412 Wing Section It is both a popular expectation and an experimental fact that air flows more rapidly over
the curved top surface of the airfoil section than along the relatively flat bottom. In the
NCFMF video Flow Visualization, timelines placed in front of the airfoil indicate that
fluid flows more rapidly along the top of the section than along the bottom.
In the absence of viscous effects (this is a valid assumption outside the boundary layers
on the airfoil) pressure falls when flow speed increases. Thus the pressures on the top
surface of the airfoil where flow speed is higher are lower than the pressures on the
bottom surface where flow speed does not increase. (Actual pressure profiles measured
for a lifting section are shown in the NCFMF video Boundary Layer Control.) The
unbalanced pressures on the top and bottom surfaces of the airfoil section create a net
force that tends to develop lift on the profile. ...
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 Spring '11
 TUZLA
 Shear Stress, Velocity

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