2.1
For Problem the velocity fields given below, determine: (a) whether the flow field is one-, two-, or three-dimensional, and why. (b) whether the flow is steady or unsteady, and why. (The quantities a and b are constants.)
Solution (1) V = V ( x) V = V ( x , y) V = V ( x) V = V ( x , z) V = V ( x) V = V ( x , y , z) V = V ( x , y , z) V = V ( x , y) 1D V = V ( t) V V ( t) V V ( t) V V ( t) V V ( t) V = V ( t) V V ( t) V = V ( t) Unsteady
(2)
2D
Steady
(3)
1D
Steady
(4)
2D
Steady
(5)
1D
Steady
(6)
3D
Unsteady
(7)
3D
Steady
(8)
2D
Unsteady
48
49
50
Problem 2.4
A velocity field is given by
r ^ V = axi - bty^ j
where a = 1 s-1 and b = 1 s-2. Find the equation of the streamlines at any time t. Plot several streamlines in the first quadrant at t = 0 s, t = 1 s, and t = 20 s.
Solution For streamlines -b t y v dy = = u dx a x -b t dx dy = y a x -b t a
So, separating variables
Integrating
ln ( y) =
ln ( x)
-b
The solution is
y = c x
a
t
For t = 0 s
y=c
For t = 1 s
y=
c x
For t = 20 s
y = c x
- 20
See the plots in the corresponding Excel workbook
51
Problem 2.4 (In Excel)
A velocity field is given by
r ^ V = axi - bty^ j
-1 -2 where a = 1 s and b = 1 s . Find the equation of the streamlines at any time t . Plot several streamlines in the first quadrant at t = 0 s, t =1 s, and t =20 s.
Solution
-b
The solution is
y = c x
a
t
For t = 0 s
y=c
For t = 1 s
y=
c x
For t = 20 s t=0 c=1 y 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 c=2 y 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 c=3 y 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00
y = c x
- 20
x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00
t =1 s (### means too large to view) c=1 c=2 c=3 x y y y 0.05 20.00 40.00 60.00 0.10 10.00 20.00 30.00 0.20 5.00 10.00 15.00 0.30 3.33 6.67 10.00 0.40 2.50 5.00 7.50 0.50 2.00 4.00 6.00 0.60 1.67 3.33 5.00 0.70 1.43 2.86 4.29 0.80 1.25 2.50 3.75 0.90 1.11 2.22 3.33 1.00 1.00 2.00 3.00 1.10 0.91 1.82 2.73 1.20 0.83 1.67 2.50 1.30 0.77 1.54 2.31 1.40 0.71 1.43 2.14 1.50 0.67 1.33 2.00 1.60 0.63 1.25 1.88 1.70 0.59 1.18 1.76 1.80 0.56 1.11 1.67 1.90 0.53 1.05 1.58 2.00 0.50 1.00 1.50
t = 20 s c=1 y ##### ##### ##### ##### ##### ##### ##### ##### 86.74 8.23 1.00 0.15 0.03 0.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 c=2 y ##### ##### ##### ##### ##### ##### ##### ##### ##### 16.45 2.00 0.30 0.05 0.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 c=3 y ##### ##### ##### ##### ##### ##### ##### ##### ##### 24.68 3.00 0.45 0.08 0.02 0.00 0.00 0.00 0.00 0.00 0.00 0.00
x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00
52
Streamline Plot (t = 0)
3.50 3.00 2.50 2.00
c=1 c=2 c=3
y
1.50 1.00 0.50 0.00 0.00
0.50
1.00
1.50
2.00
x
Streamline Plot (t = 1 s)
70 60 50 40
c=1 c=2 c=3
y
30 20 10 0 0.00
0.50
1.00
1.50
2.00
x
Streamline Plot (t = 20 s)
20 18 16 14 12 10 8 6 4 2 0 -0.15 0.05 0.25 0.45 0.65 0.85 1.05 1.25
c=1 c=2 c=3
y
x
53
54
Problem 2.6
A velocity field is specified as
r ^ V = ax 2 i + bxy ^ j
where a = 2 m-1s-1 and b = - 6 m-1s-1, and the coordinates are measured in meters. Is the flow field one-, two-, or three-dimensional? Why? Calculate the velocity components at the point (2, 1/2). Develop an equation for the streamline passing through this point. Plot several streamlines in the first quadrant including the one that passes through the point (2, 1/2). Solution The velocity field is a function of x and y. It is therefore At point (2,1/2), the velocity components are u = a x = 2
2
2D
1 2 ( 2 m) m s 1 1 2 m m 2 m s b x y b y v dy = = = 2 a x u dx a x dy b dx = y a x b ln ( y) = ln ( x) a
u = 8
m s m s
v = b x y = -6
v = -6
For streamlines
So, separating variables
Integrating
y = c x
b a
= c x
-3
The solution is
y=
c x
3
See the plot in the corresponding Excel workbook
55
Problem 2.6 (In Excel)
A velocity field is specified as
r ^ V = ax 2 i + bxy ^ j
where a = 2 m-1s-1, b = - 6 m-1s-1, and the coordinates are measured in meters. Is the flow field one-, two-, or three-dimensional? Why? Calculate the velocity components at the point (2, 1/2). Develop an equation for the streamline passing through this point. Plot several streamlines in the first quadrant including the one that passes through the point (2, 1/2). Solution
The solution is
c= x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 1 2 3 4 y y y y 8000 16000 24000 32000 1000 2000 3000 4000 125 250 375 500 37.0 74.1 111.1 148.1 15.6 31.3 46.9 62.5 8.0 16.0 24.0 32.0 4.63 9.26 13.89 18.52 2.92 5.83 8.75 11.66 1.95 3.91 5.86 7.81 1.37 2.74 4.12 5.49 1.00 2.00 3.00 4.00 0.75 1.50 2.25 3.01 0.58 1.16 1.74 2.31 0.46 0.91 1.37 1.82 0.36 0.73 1.09 1.46 0.30 0.59 0.89 1.19 0.24 0.49 0.73 0.98 0.20 0.41 0.61 0.81 0.17 0.34 0.51 0.69 0.15 0.29 0.44 0.58 0.13 0.25 0.38 0.50
y=
c x
3
Streamline Plot
4.0
c=1
3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 0.0 0.5 1.0 1.5 2.0
c=2 c=3 c = 4 ((x,y) = (2,1/2)
y
x
56
Problem 2.7
Solution Streamlines are given by -A y v dy = = u dx A x + B dy dx = -A y A x + B - 1 1 B ln ( y) = ln x + A A A
So, separating variables
Integrating
The solution is
y=
C x+ B A
For the streamline that passes through point (x,y) = (1,2)
C = y x +
B 20 = 2 1 + = 6 A 10
y=
6 x+ 20 10
y=
6 x+2
See the plot in the corresponding Excel workbook
57
Problem 2.7 (In Excel)
Solution
The solution is
y=
C x+ B A
6 y 3.00 2.86 2.73 2.61 2.50 2.40 2.31 2.22 2.14 2.07 2.00 1.94 1.88 1.82 1.76 1.71 1.67 1.62 1.58 1.54 1.50 A = 10 B = 20
C= x 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 1 y 0.50 0.48 0.45 0.43 0.42 0.40 0.38 0.37 0.36 0.34 0.33 0.32 0.31 0.30 0.29 0.29 0.28 0.27 0.26 0.26 0.25 2 y 1.00 0.95 0.91 0.87 0.83 0.80 0.77 0.74 0.71 0.69 0.67 0.65 0.63 0.61 0.59 0.57 0.56 0.54 0.53 0.51 0.50 4 y 2.00 1.90 1.82 1.74 1.67 1.60 1.54 1.48 1.43 1.38 1.33 1.29 1.25 1.21 1.18 1.14 1.11 1.08 1.05 1.03 1.00
Streamline Plot
3.5
c=1
3.0 2.5 2.0
c=2 c=4 c = 6 ((x,y) = (1.2)
y
1.5 1.0 0.5 0.0 0.0 0.5 1.0 1.5 2.0
x
58
Problem 2.8
Solution Streamlines are given by v dy b x y = = 3 u dx a x
3
So, separating variables
dy y
3
=
b dx a x
2
Integrating
-
1 2 y
2
=
b 1 - + C a x
The solution is
y=
1
b + C 2 a x
Note: For convenience the sign of C is changed.
See the plot in the corresponding Excel workbook
59
Problem 2.8 (In Excel)
Solution
The solution is
y=
1 2
C= x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 0 y 0.16 0.22 0.32 0.39 0.45 0.50 0.55 0.59 0.63 0.67 0.71 0.74 0.77 0.81 0.84 0.87 0.89 0.92 0.95 0.97 1.00 2 y 0.15 0.20 0.27 0.31 0.33 0.35 0.37 0.38 0.39 0.40 0.41 0.41 0.42 0.42 0.43 0.43 0.44 0.44 0.44 0.44 0.45 4 y 0.14 0.19 0.24 0.26 0.28 0.29 0.30 0.30 0.31 0.31 0.32 0.32 0.32 0.32 0.33 0.33 0.33 0.33 0.33 0.33 0.33
b + C a x
6 y 0.14 0.18 0.21 0.23 0.24 0.25 0.26 0.26 0.26 0.27 0.27 0.27 0.27 0.27 0.27 0.27 0.27 0.28 0.28 0.28 0.28
a= 1 b= 1
Streamline Plot
1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
c=0 c=2 c=4 c=6
y
x
60
Problem 2.9
61
Problem 2.10
Problem 2.11
Solution Streamlines are given by -b x v dy = = u dx a y t
So, separating variables
a t y dy = -b x dx
Integrating
1 1 2 2 a t y = - b x + C 2 2
The solution is
y=
C-
b x a t
2
For t = 0 s
x=c
For t = 1 s
y=
C - 4 x
2
For t = 20 s
y=
C-
x 5
2
See the plots in the corresponding Excel workbook
63
Problem 2.11 (In Excel)
Solution The solution is y= C- b x
2
a t
For t = 0 s
x=c
For t = 1 s
y=
C - 4 x
2
For t = 20 s t=0 x 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00
y=
C-
x 5
2
t =1 s C=1 y 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 C=2 y 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 C=3 y 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 x 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 0.275 0.300 0.325 0.350 0.375 0.400 0.425 0.450 0.475 0.500 C=1 y 1.00 1.00 0.99 0.99 0.98 0.97 0.95 0.94 0.92 0.89 0.87 0.84 0.80 0.76 0.71 0.66 0.60 0.53 0.44 0.31 0.00 C=2 y 1.41 1.41 1.41 1.41 1.40 1.39 1.38 1.37 1.36 1.34 1.32 1.30 1.28 1.26 1.23 1.20 1.17 1.13 1.09 1.05 1.00 C=3 y 1.73 1.73 1.73 1.73 1.72 1.71 1.71 1.70 1.69 1.67 1.66 1.64 1.62 1.61 1.58 1.56 1.54 1.51 1.48 1.45 1.41
t = 20 s x 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 C=1 y 1.00 1.00 1.00 0.99 0.98 0.97 0.96 0.95 0.93 0.92 0.89 0.87 0.84 0.81 0.78 0.74 0.70 0.65 0.59 0.53 0.45 C=2 y 1.41 1.41 1.41 1.41 1.40 1.40 1.39 1.38 1.37 1.36 1.34 1.33 1.31 1.29 1.27 1.24 1.22 1.19 1.16 1.13 1.10 C=3 y 1.73 1.73 1.73 1.73 1.72 1.72 1.71 1.70 1.69 1.68 1.67 1.66 1.65 1.63 1.61 1.60 1.58 1.56 1.53 1.51 1.48
64
Streamline Plot (t = 0)
3.5 3.0 2.5 2.0
c=1 c=2 c=3
y
1.5 1.0 0.5 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 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6
c=1 c=2 c=3
y
x
Streamline Plot (t = 20s)
2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.5 1.0 1.5 2.0 2.5
c=1 c=2 c=3
y
x
65
66
67
68
Problem 2.15
Solution Pathlines are given by dx = u = a x t dt dy = v = -b y dt
So, separating variables
dx = a t dt x
dy = -b dt y
Integrating
ln ( x) =
1 2 a t + c1 2
ln ( y) = -b t + c2
For initial position (x0,y0)
x = x0 e
a 2 t 2
y = y0 e
- b t
Using the given data, and IC (x0,y0) = (1,1) at t = 0
x=e
0.05 t
2
y=e
-t
69
Problem 2.15 (In Excel)
Solution Using the given data, and IC (x0,y0) = (1,1) at t = 0, the pathline is The streamline at (1,1) at t = 0 s is x= 1
- y = x 10
x=e
0.05 t
2
y=e
-t
The streamline at (1,1) at t = 1 s is
The streamline at (1,1) at t = 2 s is Pathline t 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.75 2.50 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 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
- y= x 5
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
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
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
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 0.98 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 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Pathline Streamline (t = 0) Streamline (t = 1 s) Streamline (t = 2 s)
y
x
70
71
72
73
74
75
76
Problem 2.20 (In Excel)
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 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 y 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00 2.40 2.80 3.20 3.60 4.00 4.40 4.80 5.20 5.60 6.00 Starting at t = 1 s x y Starting at t = 2 s x y Streakline at t = 4 s x 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 y 0.00 0.40 0.80 1.20 1.60 2.00 2.40 2.80 3.20 3.60 4.00 4.20 4.40 4.60 4.80 5.00 5.20 5.40 5.60 5.80 6.00
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 0.20 0.40 0.60 0.80 1.00 1.40 1.80 2.20 2.60 3.00 3.40 3.80 4.20 4.60 5.00
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
0.00 0.40 0.80 1.20 1.60 2.00 2.40 2.80 3.20 3.60 4.00
Pathline and Streamline Plots
6
5
4
y
Pathline starting at t = 0 Pathline starting at t = 1 s Pathline starting at t = 2 s Streakline at t = 4 s
3
2
1
0 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
x
79
80
Problem 2.22
81
Problem 2.22 (cont'd)
82
Problem 2.23
84
85
86
87
Problem 2.26
88
Problem 2.27
89
Problem 2.28 (In Excel)
Solution Pathlines: T (oC) 0 100 200 300 400 Data: T (K) 273 373 473 573 673 (x105) 1.86E-05 2.31E-05 2.72E-05 3.11E-05 3.46E-05 Using procedure of Appendix A.3: T (K) 273 373 473 573 673 T3/2/ 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
S T32 1 = T + b b
From the built-in Excel Linear Regression functions: Slope = 6.534E+05 Intercept = 6.660E+07 R2 = 0.9996 Plot of Basic Data and Trend Line
6.E+08
Hence: b = 1.53E-06 S = 101.9 kg/m.s.K1/2 K
5.E+08
Data Plot Least Squares Fit
4.E+08
T3/2/ 3.E+08
2.E+08
1.E+08
0.E+00 0 100 200 300 400 500 600 700 800
T
91
92
93
Problem 2.31
94
95
96
97
Problem 2.35
98
99
100
101
Problem 2.38
A block 0.2 m square, with 5 kg mass, slides down a smooth incline, 30 below the horizontal, a film of SAE 30 oil at 20C that is 0.20 mm thick. If the block is released from rest at t = 0, wh is its initial acceleration? Derive an expression for the speed of the block as a function of time. the curve for V(t). Find the speed after 0.1 s. If we want the mass to instead reach a speed of 0.3 m/s at this time, find the viscosity of the oil we would have to use. Ff = A
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 A = ( 0.2 m)
2
x, V, a
M g
d = 0.2 mm
= 30 deg
From Fig. A.2
= 0.4
N s m
2
Applying Newton's 2nd law to initial instant (no friction) M a = M g sin ( ) - Ff = M g sin ( ) m ainit = g sin ( ) = 9.81 sin ( 30) 2 s m ainit = 4.9 2 s
so
Applying Newton's 2nd law at any instant M a = M g sin ( ) - Ff
102
and
Ff = A =
du V A = A dy d
so
M a = M
A dV = M g sin ( ) - V dt d
Separating variables
dV A V g sin ( ) - M d
= dt
Integrating and using limits - M d A ln 1 -
M g d sin ( )
A
V = t
or
t M g d sin ( ) M d 1 - e V ( t) = A
- A
0.4 0.3 V (m/s) 0.2 0.1 0 0.05 0.1 0.15 0.2 t (s) 0.25 0.3 0.35
103
At t = 0.1 s
V = 5 kg 9.81
m s
2
0.0002 m sin ( 30)
m
2
0.4 N s ( 0.2 m)
0.4 0.04 0.1 - N s 5 0.002 1 - e 2 2
kg m
V = 0.245
m s
To find the viscosity for which V(0.1 s) = 0.3 m/s, we must solve
V ( t = 0.1 s) =
M g d sin ( ) A
1 - e
- A ( t=0.1 s) M d
The viscosity is implicit in this equation, so solution must be found by manual iteration, or by of a number of classic root-finding numerical methods, or by using Excel's Goal Seek From the Excel workbook for this problem the solution is = 0.27 N s m
2
Excel workbook
104
Problem 2.38 (In Excel)
A block 0.2 m square, with 5 kg mass, slides down a smooth incline, 30 below the horizontal, on a film of SAE 30 oil at 20C that is 0.20 mm thick. If the block is released from rest at t = 0, what is its initial acceleration? Derive an expression for the speed of the block as a function of time. Plot the curve for V (t ). Find the speed after 0.1 s. If we want the mass to instead reach a speed of 0.3 m/s at this time, find the viscosity of the oil we would have to use. Solution Ff = A x, V, a
The solution is
V( t) =
t M gd sin( ) M d 1 - e A
- A
M g
The data is
M= = = A= d=
5.00 30 0.40 0.04 0.2
kg deg N.s/m2 m2 mm
Speed V of Block vs Time t t (s) 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.30 V (m/s) 0.000 0.045 0.084 0.117 0.145 0.169 0.189 0.207 0.221 0.234 0.245 0.254 0.262 0.268 0.274 0.279 0.283 0.286 0.289 0.292 0.294 0.296 0.297 0.299 0.300 0.301 0.302 0.302 0.303 0.304 0.304
0.35 0.3 0.25 0.2
V (m/s)
0.15 0.1 0.05 0 0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
t (s)
To find the viscosity for which the speed is 0.3 m/s after 0.1 s use Goal Seek with the velocity targeted to be 0.3 by varying the viscosity in the set of cell below: t (s) 0.10 V (m/s) 0.300 for
2 = 0.270 N.s/m
105
106
107
Problem 2.41
108
Problem 2.42
109
110
Problem 2.44
The viscometer of Problem 2.43 is being used to verify that the viscosity of a particular fluid is = 0.1 N.s/m2. Unfortunately the cord snaps during the experiment. How long will it take the cylinder to lose 99% of its speed? The moment of inertia of the cylinder/pulley system is 0.0273 kg.m2.
Given: Data on the viscometer Find: 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
= 0.1
N s m
2
The equation of motion for the slowing viscometer is I = Torque = - A R where is the angular acceleration and viscometer The stress is given by = V R du V-0 = = = dy a a a
where V and are the instantaneous linear and angular velocities.
111
Hence R R A d =- A R = I = I dt a a Separating variables d =- R A dt a I
2 2
Integrating and using IC = 0
R A t a I
2
-
( t) = 0 e
The time to slow down by 99% is obtained from solving
R A t a I
2
-
0.01 0 = 0 e
so
t=-
a I R A
2
ln ( 0.01)
Note that
A = 2 R H
so
t=-
a I 2 R H
3
ln ( 0.01)
t=-
0.0002 m 0.0273 kg m 2
2
m 1 1 N s ln ( 0.01) 0.1 N s ( 0.05 m) 3 0.08 m kg m
2
2
t = 4s
112
Problem 2.45
Problem 2.46
Problem 2.46 (cont'd)
113
Problem 2.47
114
115
Problem 2.49
116
Problem 2.50
Given: Data from viscometer Find: The values of coefficients k and n; determine the kind of non-Newtonial fluid it is; estimate viscosity at 90 and 100 rpm
Solution The velocity gradient at any radius r is du r = dy r tan ( )
where (rad/s) is the angular velocity
=
2 N 60
where N is the speed in rpm
du For small , tan() can be replace with , so = dy
n-1 du du = du k dy dy dy
From Eq 2.11.
du where is the apparent viscosity. Hence = k dy
n-1
= k
n-1
117
The data in the table conform to this equation. The corresponding Excel workbook shows how Excel's Trendline analysis is used to fit the data. From Excel k = 0.0449 n = 1.21
( 90 rpm) = 0.191
N s m
2
( 100 rpm) = 0.195
N s m
2
For n > 1 the fluid is dilatant
118
Problem 2.50 (In Excel)
Solution The data is N (rpm) 10 20 30 40 50 60 70 80 (N.s/m ) 0.121 0.139 0.153 0.159 0.172 0.172 0.183 0.185 Viscosity vs Shear Rate / (1/s) 120 240 360 480 600 720 840 960 (N.s/m2x103) 121 139 153 159 172 172 183 185
1000
2
The computed data is (rad/s) 1.047 2.094 3.142 4.189 5.236 6.283 7.330 8.378
2 3 (N.s/m x10 )
Data Power Trendline
100
y = 44.94x0.2068 R2 = 0.9925
From the Trendline analysis
10
k = 0.0449 n - 1 = 0.2068 n = 1.21
100
1000
Shear Rate / (1/s) The fluid is dilatant
The apparent viscosities at 90 and 100 rpm can now be computed N (rpm) 90 100 (rad/s) 9.42 10.47 / (1/s) 1080 1200 (N.s/m2x103) 191 195
119
Problem 2.51
121
Problem 2.52
Problem 2.53
Problem 2.54
125
126
127
Problem 2.57
128
Problem 2.58
You intend to gently place several steel needles on the free surface of the water in a large tank. The needles come in two lengths: Some are 5 cm long, and some are 10 cm long. Needles of each length are available with diameters of 1 mm, 2.5 mm, and 5 mm. Make a prediction as to which needles, if any, will float. Given: Data on size of various needles Find: Which needles, if any, will float Solution For a steel needle of length L, diameter D, density s, to float in water with surface tension an contact angle , the vertical force due to surface tension must equal or exceed the weight 2 L cos ( ) W = m g = 8 cos ( ) s g mN m kg m
3
D 4
2
s L g
or
D
From Table A.4
= 72.8
= 0 deg
and for water
= 999
From Table A.1, for steel SG = 7.83 Hence 8 cos ( ) SG g
3 2
=
8 7.83
72.8 10
-3 N
m s kg m -3 = 1.55 10 m m 999 kg 9.81 m N s2
Hence D < 1.55 mm. Only the 1 mm needles float (needle length is irrelevant)
129
Problem 2.59
130
Problem 2.60
132
Problem 2.62
133