Chapter_2_solutions.pdf

Problem 2.1 Problem 2.1

[Difficulty: 1]

2.1

Given:

Velocity fields

Find:

Whether flows are 1, 2 or 3D, steady or unsteady.

Solution: (1) (2) (3) (4) (5) (6) (7) (8)

→ → V = V ( x , y) → → V = V ( x , y) → → V = V ( x) → → V = V ( x) → → V = V ( x) → → V = V ( x , y) → → V = V ( x , y) → → V = V ( x , y , z)

2D 2D 1D 1D 1D 2D 2D 3D

→ → 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 Steady Steady Steady Unsteady Steady Unsteady Steady

Problem 2.2 Problem 2.2

[Difficulty: 1]

2.2

Given:

Velocity fields

Find:

Whether flows are 1, 2 or 3D, steady or unsteady.

Solution: (1) (2) (3) (4) (5) (6) (7) (8)

→ → V = V ( y) → → V = V ( x) → → V = V ( x , y) → → V = V ( x , y) → → V = V ( x) → → V = V ( x , y , z) → → V = V ( x , y) → → V = V ( x , y , z)

1D 1D 2D 2D 1D 3D 2D 3D

→ → 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 Steady Unsteady Unsteady Unsteady Steady Unsteady Steady

Problem 2.3 Problem 2.3 2.3

Given:

Viscous liquid sheared between parallel disks. Upper disk rotates, lower fixed. Velocity field is:

r rω z V = eˆθ h

Find: a.

Dimensions of velocity field.

b.

Satisfy physical boundary conditions.

r

r

To find dimensions, compare to V = V ( x, y , z ) form.

Solution:

r

r

The given field is V = V (r , z ) . Two space coordinates are included, so the field is 2-D. Flow must satisfy the no-slip condition: 1.

r

At lower disk, V = 0 since stationary.

r

z = 0, so V = eˆθ

2.

rω 0 = 0 , so satisfied. h

r

At upper disk, V = eˆθ rω since it rotates as a solid body.

r

z = h, so V = eˆθ

rω h = eˆθ rω , so satisfied. h

[Difficulty: 2]

Problem 2.4 Problem 2.4

[Difficulty: 1]

2.4

Given:

Velocity field

Find:

Equation for streamlines

Streamline Plots

Solution: v u So, separating variables

dy

=

dy y

dx

=

=

B⋅ x⋅ y 2

2

A⋅ x ⋅ y

=

C=1 C=2 C=3 C=4

B⋅ y 4

A⋅ x

B dx ⋅ A x

y (m)

For streamlines

5

3 2

Integrating

The solution is

ln( y ) =

y=

B A

1

⋅ ln( x ) + c = − ⋅ ln( x ) + c 2

1

C x

0

1

2

3

x (m) The plot can be easily done in Excel.

4

5

Problem 2.5 (Difficulty: 2) 2.5 A fluid flow has the following velocity components: 𝑢 = 1 𝑚⁄𝑠 and 𝑣 = 2𝑥 𝑚⁄𝑠. Find an equation for and sketch the streamlines of this flow. Given: The velocity components: 𝑢 = 1 𝑚⁄𝑠 and 𝑣 = 2𝑥 𝑚⁄𝑠. Find: The equation for streamlines and sketch the streamlines. Assumption: The flow is steady and incompressible Solution: Use the definition of streamlines in terms of velocity to determine the equation for the streamlines. By definition, we have: 𝑑𝑑 𝑑𝑑 = 𝑣 𝑢

Or

𝑑𝑑 =

𝑢 𝑑𝑑 𝑣

𝑑𝑑 =

1 𝑑𝑑 2𝑥

Substituting in the velocity components in we obtain:

Integrating both sides, we get:

2𝑥𝑥𝑥 = 𝑑𝑑

𝑥2 = 𝑦 + 𝑐

Where 𝑐 is a constant that can be found for each specific problem. To plot the streamlines, we write: 𝑦 = 𝑥 2 − 𝑐. The plot is shown in the figure.

Problem 2.6 (Difficulty: 2)

2.6 When an incompressible, non-viscous fluid flows against a plate (two-dimensional) flow, an exact solution for the equations of motion for this flow is 𝑢 = 𝐴𝐴, 𝑣 = −𝐴𝐴, with 𝐴 > 0. The coordinate origin is located at the stagnation point 0, where the flow divides and the local velocity is zero. Find the streamlines. Given: The velocity components: 𝑢 = 𝐴𝐴, 𝑣 = −𝐴𝐴

Find: The equation for streamlines and sketch the streamlines. Assumption: The flow is steady and incompressible Solution: Use the definition of streamlines in terms of velocity to determine the equation for the streamlines. By definition, we have: 𝑑𝑑 𝑑𝑑 = 𝑣 𝑢

Substituting in for the velocity components we obtain:

𝑑𝑑 𝑑𝑑 = 𝐴𝐴 −𝐴𝐴 𝑑𝑑 𝑑𝑑 = −𝑦 𝑥

Integrating both sides, we get: �

𝑑𝑑 𝑑𝑑 = −� 𝑥 𝑦

ln 𝑥 = − ln 𝑦 + 𝑐

where 𝑐 is a constant that can be found for each problem Using the relation for logarithms, the streamline equation is:

Or we can rewrite as:

ln 𝑥 𝑦 = +𝑐

𝑥𝑥 = 𝑐1

Where 𝑐1 is a constant.

The plot of the streamlines is shown in the figure as an example:

Problem 2.7 (Difficulty: 2)

5 𝑟

2.7 For the free vortex flow the velocities are 𝑉𝑡 = and 𝑉𝑟 = 0. Assume that lengths are in feet or meters and times are in seconds. Plot the streamlines of this flow. How does the velocity vary with distance from the origin? What is the velocity at the origin (0,0)? 5 𝑟

Given: The velocity components: 𝑉𝑡 = , 𝑉𝑟 = 0

Find: The streamline, how the velocity varies with distance from the origin, and the velocity at the origin (0,0).

Assumption: The flow is steady and incompressible

Solution: Use the definition of streamlines in terms of velocity to determine the equation for the streamlines. By definition, in radial coordinates we have: 1 𝑑𝑑 𝑉𝑟 = 𝑟 𝑑𝑑 𝑉𝑡

Substituting the velocity components we obtain:

Integrating both sides, we get:

𝑑𝑑 =0 𝑑𝑑 𝑟=𝑐

So the streamline can be plotted as:

𝑐 is a constant

The velocity will decrease as the distance to the origin 𝑟 increases as shown in the figure.

vt

The velocity at the origin (0,0) is

r

𝑉𝑟 = 0

𝑉𝑡 = ∞

Problem 2.8 (Difficulty: 2)

2.8 For the forced vortex flow the velocities are 𝑉𝑡 = 𝜔𝜔 and 𝑉𝑟 = 0. Plot the streamlines of this flow. How does the velocity vary with distance from the origin? What is the velocity at the origin(0,0)? Given: The velocity components: 𝑉𝑡 = 𝜔𝜔, 𝑉𝑟 = 0

Assumption: The flow is steady and incompressible Find: The equation for the streamlines, how the velocity varies with distance from origin, the velocity at origin (0,0).

Solution: Use the definition of streamlines in terms of velocity to determine the equation for the streamlines. By definition, in radial coordinates we have: 1 𝑑𝑑 𝑉𝑟 = 𝑟 𝑑𝑑 𝑉𝑡

Substituting the velocity components in we obtain:

Integrating both sides, we get:

𝑑𝑑 =0 𝑑𝑑 𝑟=𝑐

So the streamline can be plotted as:

𝑐 is a constant

The velocity will increase as the distance to the origin 𝑟 increases. For example ω = 1.

𝑉𝑡

The velocity at the origin (0,0) is

r

𝑉𝑟 = 0

𝑉𝑡 = 0

Problem 2.9 Problem 2.6

[Difficulty: 1]

2.9

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. u = a⋅ x ⋅ y = 2 ⋅

At point (2,1/2), the velocity components are

1 m⋅ s 1

2

v = b ⋅ y = −6 ⋅

v

For streamlines

=

u dy

So, separating variables

y

=

dy

=

dx

× 2⋅ m ×

b⋅ y

×

m⋅ s 2

a⋅ x ⋅ y

1 2

⎛ 1 ⋅ m⎞ ⎜ ⎝2 ⎠

⋅m

2

u = 2⋅

m s

3 m v=− ⋅ 2 s

b⋅ y

=

a⋅ x

b dx ⋅ a x b

b

ln( y ) =

Integrating

a

⋅ ln( x) + c

y = C⋅ x

a

−3

y = C⋅ x

The solution is

The streamline passing through point (2,1/2) is given by

1 2

−3

= C⋅ 2

C =

1 3 ⋅2 2

C= 4

y=

4 3

x

20

Streamline for C Streamline for 2C Streamline for 3C Streamline for 4C

16 12 8 4

1

This can be plotted in Excel.

1.3

1.7

2

Problem 2.10 2.10

a= b= 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 1 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

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

Streamline Plot 1.2 c=0

1.0

c=2 c=4

0.8

c=6

y 0.6

0.4 0.2 0.0 0.0

0.5

1.0 x

1.5

2.0

Problem 2.11 Problem 2.10

[Difficulty: 2]

2.11

Given:

Velocity field

Find:

Equation for streamline through (1,3)

Solution: For streamlines

v u

So, separating variables

y

A⋅

dy y

= =

dy dx

x

=

2

A

=

y x

x

dx 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=

dx dt

=

A x

2 ⋅ A⋅ t + c

t=

x

2

2⋅ A

−

c 2⋅ A

Hence the time for a particle to go from x = 1 to x = 2 m is 2

∆t = t( x = 2 ) − t( x = 1 )

∆t =

( 2 ⋅ m) − c 2⋅ A

2

−

( 1 ⋅ m) − c 2⋅ A

2

=

2

4⋅ m − 1⋅ m 2

2 × 2⋅

m s

∆t = 0.75⋅ s

[Difficulty: 3]

Problem 2.12

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 ⋅ π⋅ x + y

)

2

Plotting

=0

K⋅ x

v=

(2

2 ⋅ π⋅ x + y

)

2

=

K 2 ⋅ π⋅ x

160

v( m/s)

80

−1

− 0.5

0

0.5

1

− 80 − 160

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. u=−

On the y axis, x = 0, so

K⋅ y

2 2 2 ⋅ π⋅ ( x + y )

Plotting

=−

K 2 ⋅ π⋅ y

v=

K⋅ x

2 2 2 ⋅ π⋅ ( x + y )

=0

160

v( m/s)

80

−1

− 0.5

0 − 80 − 160

y (km)

0.5

1

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⋅ 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

u v

2

x +y

Then the magnitude of the velocity along y = x is

V=

2

2

K

2

4⋅ π

Plotting

)

2

2

K 4 ⋅ π⋅ x

= −1 r=

then along y = x

u +v =

2

=

1

Slope of trajectory of motion:

r=

(2

2 ⋅ π⋅ x + x

Slope of line y = x:

If we define the radial position:

K⋅ x

v=

1

⋅

x

2

+

1 x

2

=

K 2 ⋅ π⋅ 2 ⋅ x

=

x +x =

2⋅ x

K 2 ⋅ π⋅ r

160

v( m/s)

80

−1

− 0.5

0

0.5

1

− 80 − 160

x (km) This can also be plotted in Excel. K⋅ x

For streamlines

v

=

u

dy dx

(

2

2

2⋅ π⋅ x + y

=

)

K⋅ y

−

(2

2 ⋅ π⋅ x + y So, separating variables

Integrating

)

x y

2

y ⋅ dy = −x ⋅ dx

y

2

2

The solution is

=−

2

x

=−

2

2

2

+c

x +y =C

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 2.13 (Difficulty: 2)

�⃗ = 𝐴𝐴𝚤⃗ − 𝐴𝐴𝚥⃗, where 𝐴 = 2 𝑠 −1 , which can be interpreted to represent flow 2.13 For the velocity field 𝑉 in a corner, show that the parametric equations for particle motion are given by 𝑥𝑝 = 𝑐1 𝑒 𝐴𝐴 and 𝑦𝑝 = 𝑐2 𝑒 −𝐴𝐴 . Obtain the equation for the pathline of the particle located at the point (𝑥, 𝑦) = (2,2) at the instant 𝑡 = 0. Compare this pathline with streamline through the same point. Find: The pathlines and streamlines . Assumption: The flow is steady and incompressible Solution: Use the definitions of pathlines and streamlines in terms of velocity. We relate the velocities to the change in position with time. For the particle motion we have: 𝑑𝑥 = 𝑢 = 𝐴𝐴 𝑑𝑑

Or

𝑑𝑦 = 𝑣 = −𝐴𝐴 𝑑𝑑 𝑑𝑑 = 𝐴 𝑑𝑑 𝑥

Integrating both sides of the equation, we get:

𝑑𝑑 = −𝐴 𝑑𝑑 𝑦 ln 𝑥 = 𝐴𝐴 + 𝑐

ln 𝑦 = −𝐴𝐴 + 𝑐

So the parametric equations for particle motion are given by:

𝑥𝑝 = 𝑒 (𝐴𝐴+𝑐) = 𝑐1 𝑒 𝐴𝐴

With 𝐴 = 2 𝑠 −1 :

𝑦𝑝 = 𝑒 (−𝐴𝐴+𝑐) = 𝑐2 𝑒 −𝐴𝐴 𝑥𝑝 = 𝑒 (𝐴𝐴+𝑐) = 𝑐1 𝑒 2𝑡

𝑦𝑝 = 𝑒 (−𝐴𝐴+𝑐) = 𝑐2 𝑒 −2𝑡

For the pathline: At 𝑡 = 0, 𝑥𝑝 = 𝑥0 = 2, 𝑦𝑝 = 𝑦0 = 2. So the equation for the pathline is For the streamline:

𝑥𝑝 𝑦𝑝 = 𝑥0 𝑦0 = 4

𝑑𝑑 𝑣 −𝐴𝐴 −𝑦 = = = 𝐴𝐴 𝑥 𝑑𝑑 𝑢 Integrating both sides of the equation we get:

𝑑𝑑 −𝑑𝑑 = 𝑥 𝑦

ln 𝑦 = − ln 𝑥 + 𝑐 𝑥𝑥 = 𝑐

For points (𝑥, 𝑦) = (2,2), the constant c = 4 and the equation for the streamline is: 𝑥𝑥 = 4

Comparing the pathline and streamline, it is seen that for steady flow the pathline and streamline coincide as expected.

Problem 2.14 (Difficulty: 2)

2.14 A velocity field in polar coordinates is given with the radial velocity as 𝑉𝑟 = − 𝐴 𝑟

𝐴 𝑟

and the tangential

velocity as 𝑉𝜃 = , where 𝑟 is in 𝑚 and 𝐴 = 10 𝑚2 . Plot the streamlines passing through the 𝜃 = 0 and 𝑟 = 1 𝑚, 2𝑚, 𝑎𝑎𝑎 3𝑚. What does the flow field model?

Assumption: The flow is steady and incompressible

Solution: Use the definition of streamlines in terms of velocity. The definition of a streamline in radial coordinates is: 1 𝑑𝑑 𝑑𝑑 = 𝑉𝜃 𝑟 𝑉𝑟

With the velocity components

Or

𝑑𝑑 1 𝑑𝑑 � �= 𝐴 𝑟 −𝐴 𝑟 𝑟

Integrating both sides:

For the case of 𝜃 = 0 and 𝑟 = 1 𝑚, we have: So the streamline is:

−

𝑑𝑑 = 𝑑𝑑 𝑟

− ln 𝑟 = 𝜃 + 𝑐 𝑐=0 ln 𝑟 = −𝜃

𝑟 = 𝑒𝑒𝑒(−𝜃) The plot of the streamline is

90

1

120

60 0.8 0.6

150

30 0.4 0.2

180

0

330

210

300

240 270

For the case 𝜃 = 0 and 𝑟 = 2 𝑚, we have:

𝑐 = − ln 2

So the streamline is:

ln 𝑟 = −𝜃 − 𝑐 = −𝜃 + ln 2 ln 𝑟 − ln 2 = −𝜃 𝑟 ln = −𝜃 2

𝑟 = 2𝑒𝑒𝑒(−𝜃) 90

2

120

60 1.5 1

150

30

0.5

180

0

210

330

240

300 270

For the case 𝜃 = 0 and 𝑟 = 3 𝑚, we have: So the streamline is:

𝑐 = − ln 3

ln 𝑟 = −𝜃 − 𝑐 = −𝜃 + ln 3 ln 𝑟 − ln 3 = −𝜃

𝑟 ln = −𝜃 3

𝑟 = 3𝑒𝑒𝑒(−𝜃)

The flow field models the circular flow from the center at the origin. 90

3

120

60 2

150

30 1

180

0

330

210

300

240 270

Problem 2.15 (Difficulty: 2)

2.15 The flow of air near the earth’s surface is affected both by the wind and thermal currents. In certain �⃗ = 𝑎𝚤⃗ + 𝑏 �1 − 𝑦� 𝚥⃗ for 𝑦 < ℎ and by 𝑉 �⃗ = 𝑎𝚤⃗ circumstances the velocity field can be represented by 𝑉 for 𝑦 > ℎ. Plot the streamlines for the flow for

𝑏 𝑎

= 0.01, 0.1 𝑎𝑎𝑎 1.

ℎ

Assumption: The flow is steady and incompressible

Solution: Use the definition of streamlines in terms of velocity to determine the equation for the streamlines. By definition, we have: 𝑑𝑦 𝑣 = 𝑑𝑑 𝑢

Or, substituting the equations for the velocities:

𝑑𝑑 𝑑𝑑 = 𝑣 𝑢

𝑑𝑑 𝑑𝑑 = 𝑦 𝑎 𝑏 �1 − � ℎ

And when y < h

𝑥= And when 𝑦 > ℎ

𝑑𝑑 =0 𝑑𝑑

Integrating both sides of the equation: when 𝑦 < ℎ. when 𝑦 > ℎ.

𝑑𝑑 𝑦 𝑏 �1 − � 𝑎 ℎ

𝑥=−

For the critical point 𝑦 = ℎ, we have 𝑐2 = ℎ

𝑎ℎ ln(𝑏ℎ − 𝑏𝑏) + 𝑐1 𝑏 𝑦 = 𝑐2

For example, the streamline passing through (0,0)：

𝑥=−

when 𝑦 < ℎ.

𝑐1 =

𝑎ℎ ln(𝑏ℎ) 𝑏

𝑎ℎ 𝑎ℎ ln(𝑏ℎ − 𝑏𝑏) + ln(𝑏ℎ) 𝑏 𝑏

Assume ℎ = 𝑏 = 1,. For the first case 𝑎 = 100, the streamline is shown as: The streamline is shown: 1 0.9 0.8 0.7

y

0.6 0.5 0.4 0.3 0.2 0.1 0

0

200

400

600 x

800

1000

1200

80

100

120

For the second case 𝑎 = 10, the streamline is shown as: 1

0.9 0.8 0.7

y

0.6 0.5 0.4 0.3 0.2 0.1 0

0

20

40

60 x

For the third case 𝑎 = 10, the streamline is shown as:

1 0.9 0.8 0.7

y

0.6 0.5 0.4 0.3 0.2 0.1 0

0

2

4

6 x

8

10

12

Problem 2.16 2.16

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

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

t = 20 s 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

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

Streamline Plot (t = 0) 3.5

c=1 c=2 c=3

3.0 2.5

y

2.0 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

c=1 c=2 c=3

1.8 1.6 1.4

y

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

x

Streamline Plot (t = 20s) 2.0

c=1 c=2 c=3

1.8 1.6 1.4

y

1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.0

0.5

1.0

1.5 x

2.0

2.5

Problem 2.17 Problem 2.18

[Difficulty: 2]

2.17

Given:

Time-varying velocity field

Find:

Streamlines at t = 0 s; Streamline through (3,3); velocity vector; will streamlines change with time

Solution: v

For streamlines

u

=

dy

At t = 0 (actually all times!)

dx dy

So, separating variables

y

dy dx

=−

=−

y

=−

dx

a⋅ y ⋅ ( 2 + cos( ω⋅ t) ) a⋅ x ⋅ ( 2 + cos( ω⋅ t) )

y x

x x

Integrating

ln( y ) = −ln( x ) + c

The solution is

y=

C

C =

3

For the streamline through point (3,3)

=−

which is the equation of a hyperbola.

x 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

m

u = 45⋅

3

v = −a⋅ y ⋅ ( 2 + cos( ω⋅ t) ) = 5 ⋅

y

4

2

s

v = −45⋅

1

× 3⋅ m × 3

s

1 s

× 3⋅ m × 3

m s

The velocity vector is tangent to the curve; 0

1

2

3

x

4

5

Tangent of curve at (3,3) is

dx Direction of velocity at (3,3) is

This curve can be plotted in Excel.

dy v u

=− = −1

y x

= −1

Problem 2.18 Problem 2.20

[Difficulty: 3]

2.18

Given:

Velocity field

Find:

Plot of pathline traced out by particle that passes through point (1,1) at t = 0; compare to streamlines through same point at the instants t = 0, 1 and 2s

Solution: up =

dx dt

vp =

= B⋅ x ⋅ ( 1 + A⋅ t)

A = 0.5⋅

For pathlines

Governing equations:

dy

v

For streamlines

dt

u

=

dy dx

Assumption: 2D flow

up =

Hence for pathlines

dx

So, separating variables

x

dx dt

⎛

B⋅ ⎜t+ A⋅

x=e ⎝

2⎞

t

2⎠

+ C1

2

⎠

B⋅ ⎜t+ A⋅

t

⎛

C1

= e ⋅e ⎝

x = c1 ⋅ e ⎝

⎛

x=e ⎝

v u

So, separating variables

Integrating

=

2⎞

⎝

B⋅ ⎜t+ A⋅

For streamlines

vp =

s

dy dx

=

( 1 + A⋅ t) ⋅

y

t

⎛

Using given data

1

dy

⎛

B⋅ ⎜t+ A ⋅

The pathlines are

s

B = 1⋅

= B⋅ ( 1 + A⋅ t) ⋅ dt

ln( x ) = B⋅ ⎜ t + A⋅

Integrating

1

+ C1

2⎠

dt

= C⋅ y

C = 1⋅

1 s

= C⋅ dt

ln( y ) = C⋅ t + C2

⎛

2⎞

dy

B⋅ ⎜t+ A⋅

= c1 ⋅ e ⎝

2⎞

t

2⎠

y=e

C⋅ t+ C2

=e

C2 C⋅ t

⋅e

= c2 ⋅ e

C⋅ t

2⎞

t

2⎠

y = c2 ⋅ e

C⋅ t

2⎞

t

2⎠

y=e

C⋅ t

C⋅ y B⋅ x ⋅ ( 1 + A⋅ t) dy y

=

C dx ⋅ B x

( 1 + A⋅ t) ⋅ ln( y ) =

C B

⋅ ln( x ) + c

which we can integrate for any given t (t is treated as a constant)

C

The solution is

y

1+ A ⋅ t

= const ⋅ x

B

y = const ⋅ x

or

C

y=x

For particles at (1,1) at t = 0, 1, and 2s

C

B

y=x

C

( 1+ A )B

y=x

Streamline and Pathline Plots 5

Streamline (t=0) Streamline (t=1) Streamline (t=2) Pathline

4

y (m)

3

2

1

0

1

2

3

x (m)

4

5

( 1+ 2⋅ A )B

Problem 2.19 Problem 2.22

[Difficulty: 3]

2.19

Given:

Velocity field

Find:

Plot of pathline of particle for t = 0 to 1.5 s that was at point (1,1) at t = 0; compare to streamlines through same point at the instants t = 0, 1 and 1.5 s

Solution: Governing equations:

up =

For pathlines

dx dt

vp =

dy

v

For streamlines

dt

u

=

dy dx

Assumption: 2D flow

Hence for pathlines

So, separating variables

up = dx

dt

= ax

ln⎛⎜

⎞ = a⋅ t x0 ⎝ ⎠ x

x ( t) = x 0⋅ e

Using given data

x ( t) = e

v u

So, separating variables

dy y

Hence

s

=

=

dy dx

x0 = 1 m

a⋅ t

2⋅ t

dy

= b ⋅ y ⋅ ( 1 + c⋅ t )

dt

=

⎞ = b ⋅ ⎛ t + 1 ⋅ c⋅ t2⎞ ⎜ ⎝ 2 ⎠ ⎝ y0 ⎠

ln⎛⎜

b = 2

1 2

c = 0.4

s

y

dy y

= b ⋅ ( 1 + c⋅ t) ⋅ dt

y0 = 1 m

⎛ 1 2⎞ b⋅ ⎜t+ ⋅ c⋅ t ⎝ 2 ⎠

y ( t) = e

2

2⋅ t+ 0.4⋅ t

b ⋅ y ⋅ ( 1 + c⋅ t ) a⋅ x

b ⋅ ( 1 + c⋅ t ) a⋅ x

⋅ dx

which we can integrate for any given t (t is treated as a constant)

⎞ = b ⋅ ( 1 + c⋅ t) ⋅ ln⎛ x ⎞ ⎜x ⎝ y0 ⎠ a ⎝ 0⎠

ln⎛⎜

vp =

y ( t) = e

y

b

The solution is

1

dy = b ⋅ y ⋅ ( 1 + c⋅ t) ⋅ dt

Hence

For streamlines

a = 2

= a⋅ dt

x Integrating

dx

x y = y 0 ⋅ ⎛⎜ ⎞ ⎝ x0 ⎠

a

⋅ ( 1+ c⋅ t)

1 s

b

t = 0

x y = y 0 ⋅ ⎛⎜ ⎞ ⎝ x0 ⎠

a

= x

x t = 1 y = y 0 ⋅ ⎛⎜ ⎞ ⎝ x0 ⎠

b

⋅ ( 1+ c⋅ t)

a

= x

1.4

t = 1.5

x y = y 0 ⋅ ⎛⎜ ⎞ ⎝ x0 ⎠

Streamline and Pathline Plots 10

Streamline (t=0) Streamline (t=1) Streamline (t=1.5) Pathline

8

6

y (m)

For

b

⋅ ( 1+ c⋅ t)

4

2

0

2

4

6

x (m)

8

10

⋅ ( 1+ c⋅ t)

a

= x

1.6

Problem 2.20 Problem 2.23

[Difficulty: 3]

2.20

Given:

Velocity field

Find:

Plot of pathline of particle for t = 0 to 1.5 s that was at point (1,1) at t = 0; compare to streamlines through same point at the instants t = 0, 1 and 1.5 s

Solution: Governing equations:

For pathlines

up =

dx

a =

1 1

vp =

dt

dy

v

For streamlines

dt

u

Assumption: 2D flow

Hence for pathlines

So, separating variables

up = dx

= a⋅ x

dt

vp =

5 s

= a⋅ dt

x Integrating

dx

ln⎛⎜

dy dt

= b⋅ y⋅ t

⎞ = a⋅ t x0 ⎝ ⎠ x

ln⎛⎜

x ( t) = x 0⋅ e

For streamlines

x ( t) = e

5

v

=

u So, separating variables

dy y

Hence

=

=

ln⎛⎜

dy dx

b⋅ t a⋅ x

a⋅ t

y

y ( t) = y 0⋅ e

1 25

1 2

s

= b ⋅ t⋅ dt

y0 = 1 m

2

⋅ b⋅ t

2

2

t

y ( t) = e

50

b⋅ y⋅ t a⋅ x

⋅ dx

which we can integrate for any given t (t is treated as a constant)

⎞ = b ⋅ t⋅ ln⎛ x ⎞ ⎜x ⎝ y0 ⎠ a ⎝ 0⎠ y

b

The solution is

y

⎞ = b ⋅ 1 ⋅ t2 2 ⎝ y0 ⎠

x0 = 1 m

t

Using given data

dy

dy = b ⋅ y ⋅ t⋅ dt

1

Hence

b =

x y = y 0 ⋅ ⎛⎜ ⎞ ⎝ x0 ⎠

a

⋅t

b a

= 0.2

x0 = 1

y0 = 1

=

dy dx

b

x y = y 0 ⋅ ⎛⎜ ⎞ ⎝ x0 ⎠

t = 0

= 1

b

y = y 0 ⋅ ⎛⎜ x

⎞ ⎝ 0⎠ x

t = 5

y = y 0 ⋅ ⎛⎜ x

⎞ ⎝ 0⎠

t = 10

b

= x b

x

⋅t

a

a

⋅t = 1

⋅t

a

= x

2

b a

⋅t = 2

Streamline and Pathline Plots 10

8

6

y (m)

For

⋅t

a

4

2

Streamline (t=0) Streamline (t=1) Streamline (t=1.5) Pathline 0

2

4

6

x (m)

8

10

Problem 2.21 2.21

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 2.22 Problem 2.26

[Difficulty: 4]

2.22

Given:

Velocity field

Find:

Plot streamlines that are at origin at various times and pathlines that left origin at these times

Solution: v

For streamlines

u

=

dy dx

v 0 ⋅ sin⎡⎢ω⋅ ⎛⎜ t −

⎣ ⎝

=

u0

v 0 ⋅ sin⎡⎢ω⋅ ⎛⎜ t − So, separating variables (t=const)

x u0

⎣ ⎝

dy =

u0 v 0 ⋅ cos⎡⎢ω⋅ ⎛⎜ t −

⎞⎤ u0 ⎥ ⎠⎦ + c

ω v 0 ⋅ ⎡⎢cos⎡⎢ω⋅ ⎛⎜ t −

Using condition y = 0 when x = 0

For particle paths, first find x(t)

y= dx dt

⎞⎤ ⎥ ⎠⎦ ⋅ dx

x

⎣ ⎝

y=

Integrating

⎞⎤ u0 ⎥ ⎠⎦ x

⎞⎤ − cos( ω⋅ t)⎤ ⎥ u0 ⎥ ⎠⎦ ⎦ x

⎣ ⎣ ⎝

ω = u = u0

Separating variables and integrating

dx = u 0 ⋅ dt

Using initial condition x = 0 at t = τ

c1 = −u 0 ⋅ τ

x = u 0 ⋅ t + c1

o r

x = u 0 ⋅ ( t − τ)

x ⎞⎤ = v = v 0 ⋅ sin⎡⎢ω⋅ ⎛⎜ t − ⎥ dt ⎣ ⎝ u 0 ⎠⎦

dy

For y(t) we have

and

dy dt

This gives streamlines y(x) at each time t

so

dy

⎡ ⎡

= v = v 0 ⋅ sin⎢ω⋅ ⎢t − dt

⎣ ⎣

u 0 ⋅ ( t − τ) ⎤⎤ u0

⎥⎥ ⎦⎦

= v = v 0 ⋅ sin( ω⋅ τ)

Separating variables and integrating

dy = v 0 ⋅ sin( ω⋅ τ) ⋅ dt

y = v 0 ⋅ sin( ω⋅ τ) ⋅ t + c2

Using initial condition y = 0 at t = τ

c2 = −v 0 ⋅ sin( ω⋅ τ) ⋅ τ

y = v 0 ⋅ sin( ω⋅ τ) ⋅ ( t − τ)

The pathline is then x ( t , τ) = u 0 ⋅ ( t − τ)

y ( t , τ) = v 0 ⋅ 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 Problem 2.28 2.23

[Difficulty: 4]

2.18

Given:

Velocity field

Find:

Plot of streakline for t = 0 to 3 s at point (1,1); compare to streamlines through same point at the instants t = 0, 1 and 2 s

Solution: Governing equations:

For pathlines

up =

dx

vp =

dt

dy

v

For streamlines

dt

u

=

dy dx

Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form

(

x p( t) = x t , x 0 , y 0 , t0

( )

(

)

x st t0 = x t , x 0 , y 0 , t0

)

(

)

and

y p( t) = y t , x 0 , y 0 , t0

and

y st t0 = y t , x 0 , y 0 , t0

( )

(

)

which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) Assumption: 2D flow

For pathlines

So, separating variables

up = dx x

Integrating

dx dt

= B⋅ x ⋅ ( 1 + A⋅ t)

A = 0.5

1 s

B = 1

1 s

dy

= B⋅ ( 1 + A⋅ t) ⋅ dt

y

2 2 ⎛⎜ t − t0 ⎞ x ⎞ ⎛ ln⎜ = B⋅ ⎜ t − t0 + A⋅ 2 x0 ⎝ ⎠ ⎝ ⎠

ln⎛⎜

⎝

2 2 ⎛⎜ t − t0 ⎞ B⋅ ⎜t− t0+ A⋅ 2 ⎠ x = x0⋅ e ⎝

The pathlines are

vp =

dy dt

= C⋅ y

C = 1

= C⋅ dt

y y0

⎞ = C⋅ t − t ( 0) ⎠

y = y0⋅ e

2 2 ⎛⎜ t − t0 ⎞ B⋅ ⎜t− t0+ A ⋅ 2 ⎠ x p( t) = x 0⋅ e ⎝

( )

C⋅ t− t0

y p( t) = y 0⋅ e

( )

C⋅ t− t0

where x 0, y 0 is the position of the particle at t = t0. Re-interpreting the results as streaklines:

The streaklines are then

2 2 ⎛⎜ t − t0 ⎞ B⋅ ⎜t− t0+ A⋅ 2 ⎠ x st( t0 ) = x 0 ⋅ e ⎝

( )

y st t0 = y 0 ⋅ e

( )

C⋅ t− t0

where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t)

1 s

v

For streamlines

u So, separating variables

=

dy dx

=

( 1 + A⋅ t) ⋅

C⋅ y B⋅ x ⋅ ( 1 + A⋅ t) dy y

=

C dx ⋅ B x C

( 1 + A⋅ t) ⋅ ln( y ) =

Integrating

B

which we can integrate for any given t (t is treated as a constant)

⋅ ln( x ) + const

C

The solution is

y

1+ A ⋅ t

= const ⋅ x

B

2

For particles at (1,1) at t = 0, 1, and 2s

y=x

y=x

1

3

y=x

2

Streamline and Pathline Plots 10

Streamline (t=0) Streamline (t=1) Streamline (t=2) Streakline

8

y (m)

6

4

2

0

2

4

6

x (m)

8

10

Problem 2.24 Problem 2.29

[Difficulty: 4]

2.24

Given:

Velocity field

Find:

Plot of streakline for t = 0 to 3 s at point (1,1); compare to streamlines through same point at the instants t = 0, 1 and 2 s

Solution: Governing equations:

For pathlines

up =

dx

vp =

dt

dy

v

For streamlines

dt

u

=

dy dx

Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form

(

x p( t) = x t , x 0 , y 0 , t0

( )

(

)

x st t0 = x t , x 0 , y 0 , t0

)

(

)

and

y p( t) = y t , x 0 , y 0 , t0

and

y st t0 = y t , x 0 , y 0 , t0

( )

(

)

which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) Assumption: 2D flow

For pathlines

So, separating variables

up = dx x

Integrating

dx dt

= a⋅ x ⋅ ( 1 + b ⋅ t )

a = 1

= a⋅ ( 1 + b ⋅ t) ⋅ dt

2 2 ⎛⎜ t − t0 ⎞ x ⎞ ⎛ ln⎜ = a⋅ ⎜ t − t 0 + b ⋅ 2 ⎝ ⎠ ⎝ x0 ⎠ 2 2 ⎛⎜ t − t0 ⎞ a⋅ ⎜t− t0+ b⋅ 2 ⎠ x = x0⋅ e ⎝

1 s

b =

1

1

5

s

vp = dy y ln⎛⎜

⎝

dy dt

= c⋅ y

c = 1

= c⋅ dt

y y0

⎞ = c⋅ t − t ( 0) ⎠

y = y0⋅ e

( )

c⋅ t− t0

1 s

2 2 ⎛⎜ t − t0 ⎞ a⋅ ⎜t− t0+ b⋅ 2 ⎠ x p( t) = x 0⋅ e ⎝

The pathlines are

y p( t) = y 0⋅ e

( )

c⋅ t− t0

where x 0, y 0 is the position of the particle at t = t0. Re-interpreting the results as streaklines:

The streaklines are then

2 2 ⎛⎜ t − t0 ⎞ a⋅ ⎜t− t0+ b⋅ 2 ⎠ x st( t0 ) = x 0 ⋅ e ⎝

( )

y st t0 = y 0 ⋅ e

( )

c⋅ t− t0

where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) v

For streamlines

u So, separating variables

=

dy dx

=

( 1 + b ⋅ t) ⋅

c⋅ y a⋅ x ⋅ ( 1 + b ⋅ t )

dy y

=

c dx ⋅ a x

( 1 + b ⋅ t) ⋅ ln( y ) =

Integrating

c a

which we can integrate for any given t (t is treated as a constant)

⋅ ln( x ) + const

c

The solution is

y

1+ b⋅ t

= const ⋅ x

a

2

y=x

For particles at (1,1) at t = 0, 1, and 2s

y=x

3

1

y=x

2

Streamline and Pathline Plots 5

Streamline (t=0) Streamline (t=1) Streamline (t=2) Streakline

4

y (m)

3

2

1

0

1

2

3

x (m)

4

5

Problem 2.25 Problem 2.30

[Difficulty: 4]

2.25

Given:

Velocity field

Find:

Plot of pathline for t = 0 to 3 s for particle that started at point (1,2) at t = 0; compare to streakline through same point at the instant t = 3

Solution: Governing equations:

up =

For pathlines

dx

vp =

dt

dy dt

Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form

(

x p( t) = x t , x 0 , y 0 , t0

( )

(

)

x st t0 = x t , x 0 , y 0 , t0

)

(

)

and

y p( t) = y t , x 0 , y 0 , t0

and

y st t0 = y t , x 0 , y 0 , t0

( )

(

)

which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) Assumption: 2D flow

For pathlines

So, separating variables

up = dx

dt

= a⋅ x ⋅ t

ln⎛⎜

⎞ = a ⋅ ⎛ t2 − t 2⎞ 0 ⎠ ⎝ ⎝ x0 ⎠ 2

x = x0⋅ e

2

⋅ ⎛t − t0 ⎝ 2

a

x p( t) = x 0⋅ e

1 4

1 2

s

b =

1

m

3

s

vp =

dy dt

=b

dy = b ⋅ dt

x

a

The pathlines are

a =

= a⋅ t⋅ dt

x Integrating

dx

2

2⎞

⎠

)

(

)

y = y0 + b⋅ t − t0

⋅ ⎛t − t0 ⎝

2⎞

2

(

y − y0 = b⋅ t − t0

⎠

(

y p( t) = y 0 + b ⋅ t − t0

)

where x 0, y 0 is the position of the particle at t = t0. Re-interpreting the results as streaklines: a

The pathlines are then

( )

x st t0 = x 0 ⋅ e

2

⋅ ⎛t − t0 ⎝ 2

2⎞

⎠

( )

(

y st t0 = y 0 + b ⋅ t − t0

where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t)

)

Streakline and Pathline Plots 2

Streakline Pathline

y (m)

1.5

1

0.5

0

1

2

x (m)

3

4

Problem 2.26 Problem 2.31

[Difficulty: 4]

2.26

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

For streamlines

u

=

a⋅ y

Integrating

3

dy dx

b

=

⌠ ⌠ ⎮ 2 ⎮ a ⋅ y dy = ⎮ b dx ⌡ ⌡

or 2

a⋅ y

3

= b⋅ x + c

For the streamline through point (6,6)

c = 60 and

For particle that passed through (1,4) at t = 0

u=

dx

v=

dy

dt

dt

= a⋅ y

3

y = 6 ⋅ x + 180 ⌠ ⌠ ⎮ 2 ⎮ 1 dx = x − x 0 = ⎮ a ⋅ y dt ⌡ ⌡

2

⌠ ⌠ ⎮ 1 dy = ⎮ b dt ⌡ ⌡

=b t

⎛

⌠ 2 x − x 0 = ⎮ a ⋅ y 0 + b ⋅ t dt ⌡

Then

(

)

x0 = 1

y0 = 4

2

x = 1 + 16⋅ t + 8 ⋅ t +

y = y0 + b⋅ t = y0 + 2⋅ t

x = x 0 + a⋅ ⎜ y 0 ⋅ t + b ⋅ y 0 ⋅ t + 2

2

4 3 ⋅t 3

t

⌠ 2 x − x 0 = ⎮ a ⋅ y 0 + b ⋅ t dt ⎮ ⌡t

(

)

y = 6⋅ m

⌠ ⌠ ⎮ 1 dy = ⎮ b dt ⌡ ⌡

⎡

(

y = y0 + b⋅ t − t0

x = x 0 + a⋅ ⎢y 0 ⋅ t − t0 + b ⋅ y 0 ⋅ ⎛ t − t0 ⎝

⎣

2

(

)

2

2⎞

⎠

+

2

)

⋅ ⎛ t − t0 3 ⎝

b

3

3⎞⎤

0

3 ( ) 3

4

Hence, with x 0 = -3, y 0 = 0 at t0 = 1

x = −3 +

Evaluating at t = 3

x = 31.7⋅ m

⋅ t −1 =

( 3 1

3

⋅ 4 ⋅ t − 13

This is a steady flow, so pathlines, streamlines and streaklines always coincide

)

3

x = 26.3⋅ m

At t = 1 s

y = 4 + 2⋅ t

For particle that passed through (-3,0) at t = 1

2 3⎞

b ⋅t

⎝

0

Hence, with

We need y(t)

y = 2⋅ ( t − 1) y = 4⋅ m

⎠⎥⎦

⎠

Problem 2.27 2.32 Problem

[Difficulty: 3]

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

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

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

2.0

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 Problem 2.34

[Difficulty: 3]

2.28

Given:

Velocity field

Find:

Equation for streamline through point (2.5); coordinates of particle at t = 2 s that was at (0,4) at t = 0; coordinates of particle at t = 3 s that was at (1,4.25) at t = 1 s; compare pathline, streamline, streakline

Solution: Governing equations:

v

For streamlines

u

=

dy

dx

up =

For pathlines

dx

dt

vp =

dy dt

Assumption: 2D flow Given data

For streamlines

a = 2 v u

So, separating variables

a b

Integrating

=

m

b = 1

s

dy dx

1 s

x0 = 2

y0 = 5

x = 1

x = x

b⋅ x

=

a

⋅ dy = x ⋅ dx

1 2 2 ⋅ y − y0 = ⋅ ⎛ x − x0 ⎞ ⎝ ⎠ 2 b a

(

)

2

The solution is then

x 2 2 y = y0 + ⋅ ⎛ x − x0 ⎞ = +4 ⎝ ⎠ 4 2⋅ a

Hence for pathlines

up =

b

dx dt

=a

Hence

dx = a⋅ dt

Integrating

x − x 0 = a⋅ t − t 0

vp =

dy dt

= b⋅ x

dy = b ⋅ x ⋅ dt

(

)

(

)

dy = b ⋅ ⎡x 0 + a⋅ t − t0 ⎤ ⋅ dt ⎣ ⎦ a 2 2 y − y 0 = b ⋅ ⎡⎢x 0 ⋅ t − t0 + ⋅ ⎛ ⎛ t − t0 ⎞ ⎞ − a⋅ t0 ⋅ t − t0 ⎥⎤ ⎝ ⎝ ⎠⎠ 2 ⎣ ⎦

(

The pathlines are

(

x = x 0 + a⋅ t − t 0

)

)

(

)

a 2 2 y = y 0 + b ⋅ ⎡⎢x 0 ⋅ t − t0 + ⋅ ⎛ ⎛ t − t0 ⎞ ⎞ − a⋅ t0 ⋅ t − t0 ⎥⎤ ⎝ ⎝ ⎠⎠ 2 ⎣ ⎦

(

)

(

)

For a particle that was at x 0 = 0 m, y 0 = 4 m at t0 = 0s, at time t = 2 s we find the position is

(

)

x = x 0 + a⋅ t − t 0 = 4 m

a 2 2 y = y 0 + b ⋅ ⎡⎢x 0 ⋅ t − t0 + ⋅ ⎛ ⎛ t − t0 ⎞ ⎞ − a⋅ t0 ⋅ t − t0 ⎤⎥ = 8m ⎝ ⎝ ⎠⎠ 2 ⎣ ⎦

(

)

(

)

For a particle that was at x 0 = 1 m, y 0 = 4.25 m at t0 = 1 s, at time t = 3 s we find the position is

(

)

x = x 0 + a⋅ t − t 0 = 5 m

a 2 2 y = y 0 + b ⋅ ⎡⎢x 0 ⋅ t − t0 + ⋅ ⎛ ⎛ t − t0 ⎞ ⎞ − a⋅ t0 ⋅ t − t0 ⎤⎥ = 10.25 m ⎝ ⎝ ⎠⎠ 2 ⎣ ⎦

(

)

(

)

For this steady flow streamlines, streaklines and pathlines coincide; the particles refered to are the same particle!

Streamline and Position Plots 15

Streamline Position at t = 1 s Position at t = 5 s Position at t = 10 s 12

y (m)

9

6

3

0

1.2

2.4

3.6

x (m)

4.8

6

Problem 2.29 Problem 2.35

[Difficulty: 4]

2.29

Given:

Velocity field

Find:

Coordinates of particle at t = 2 s that was at (1,2) at t = 0; coordinates of particle at t = 3 s that was at (1,2) at t = 2 s; plot pathline and streakline through point (1,2) and compare with streamlines through same point at t = 0, 1 and 2 s

Solution : Governing equations:

For pathlines

up =

dx

dy

vp =

dt

For streamlines

dt

v u

=

dy dx

Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form

(

x p( t) = x t , x 0 , y 0 , t0

( )

)

(

x st t0 = x t , x 0 , y 0 , t0

)

(

)

and

y p( t) = y t , x 0 , y 0 , t0

and

y st t0 = y t , x 0 , y 0 , t0

( )

(

)

which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) Assumption: 2D flow Given data

Hence for pathlines

a = 0.2

up =

dx dt

1 s

b = 0.4

m 2

s

= a⋅ y

vp =

dy dt

= b⋅ t

Hence

dx = a⋅ y ⋅ dt

dy = b ⋅ t⋅ dt

For x

b 2 2 dx = ⎡⎢a⋅ y 0 + a⋅ ⋅ ⎛ t − t0 ⎞⎤⎥ ⋅ dt ⎝ ⎠⎦ 2 ⎣

Integrating

⎡⎢ 3 t 3 ⎤⎥ 0 t 2 x − x 0 = a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ − − t0 ⋅ ( t − t0 )⎥ 3 2 ⎣3 ⎦

The pathlines are

⎡⎢ 3 t 3 ⎤⎥ 0 t 2 x ( t ) = x 0 + a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ − − t0 ⋅ ( t − t0 )⎥ 3 2 ⎣3 ⎦

b 2 2 y − y0 = ⋅ ⎛ t − t0 ⎞ ⎠ 2 ⎝

b

b

These give the position (x,y) at any time t of a particle that was at (x 0,y 0) at time t0

Note that streaklines are obtained using the logic of the Governing equations, above

b 2 2 y ( t) = y0 + ⋅ ⎛ t − t0 ⎞ ⎠ 2 ⎝

The streaklines are

⎡⎢ 3 t 3 ⎤⎥ 0 t 2 x ( t 0 ) = x 0 + a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ − − t0 ⋅ ( t − t0 )⎥ 3 2 ⎣3 ⎦

b 2 2 y t0 = y 0 + ⋅ ⎛ t − t0 ⎞ ⎝ ⎠ 2

( )

b

These gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) For a particle that was at x 0 = 1 m, y 0 = 2 m at t0 = 0s, at time t = 2 s we find the position is (from pathline equations)

⎡⎢ 3 t 3 ⎤⎥ 0 t 2 x = x 0 + a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ − − t0 ⋅ ( t − t0 )⎥ = 1.91m 3 2 ⎣3 ⎦

b 2 2 y = y 0 + ⋅ ⎛ t − t0 ⎞ = 2.8 m ⎝ ⎠ 2

b

For a particle that was at x 0 = 1 m, y 0 = 2 m at t0 = 2 s, at time t = 3 s we find the position is

⎡⎢ 3 t 3 ⎤⎥ 0 t 2 x = x 0 + a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ − − t0 ⋅ ( t − t0 )⎥ = 1.49m 3 2 ⎣3 ⎦

b 2 2 y = y 0 + ⋅ ⎛ t − t0 ⎞ = 3.0 ⎠ 2 ⎝

b

For streamlines

v

=

u So, separating variables

dy dx

y ⋅ dy =

2

Integrating

=

b a

y − y0

y =

a⋅ y

⋅ t⋅ dx

2

=

2

The streamlines are then

b⋅ t

2

y0 +

where we treat t as a constant

b⋅ t a

(

⋅ x − x0

2⋅ b⋅ t a

(

)

and we have

)

⋅ x − x0 =

x0 = 1 m

4 ⋅ t⋅ ( x − 1) + 4

y0 = 2

m

m

Pathline Plots

Streamline Plots

5

15

Pathline (t0=0) Pathline (t0=2) Streakline

12

3

y (m)

y (m)

4

9

2

6

1

3

0

0.6

1.2

x (m)

Streamline (t=0) Streamline (t=1) Streamline (t=2) Streamline (t=3)

1.8

2.4

3

0

2

4

6

x (m)

8

10

Problem 2.30 Problem 2.36

[Difficulty: 4]

2.30

Given:

Velocity field

Find:

Coordinates of particle at t = 2 s that was at (2,1) at t = 0; coordinates of particle at t = 3 s that was at (2,1) at t = 2 s; plot pathline and streakline through point (2,1) and compare with streamlines through same point at t = 0, 1 and 2 s

Solution: Governing equations:

For pathlines

up =

dx

vp =

dt

dy

v

For streamlines

dt

u

=

dy dx

Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form

(

x p( t) = x t , x 0 , y 0 , t0

( )

)

(

x st t0 = x t , x 0 , y 0 , t0

)

(

)

and

y p( t) = y t , x 0 , y 0 , t0

and

y st t0 = y t , x 0 , y 0 , t0

( )

(

)

which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) Assumption: 2D flow Given data

m

a = 0.4

2

b = 2

s Hence for pathlines

up =

dx dt

m 2

s

= a⋅ t

vp =

dy dt

=b

Hence

dx = a⋅ t⋅ dt

dy = b ⋅ dt

Integrating

a 2 2 x − x0 = ⋅ ⎛ t − t0 ⎞ ⎝ ⎠ 2

y − y0 = b⋅ t − t0

The pathlines are

a 2 2 x ( t) = x0 + ⋅ ⎛ t − t0 ⎞ ⎠ 2 ⎝

y ( t) = y0 + b⋅ t − t0

(

)

(

)

(

)

These give the position (x,y) at any time t of a particle that was at (x 0,y 0) at time t0 Note that streaklines are obtained using the logic of the Governing equations, above The streaklines are

a 2 2 x t0 = x 0 + ⋅ ⎛ t − t0 ⎞ ⎠ 2 ⎝

( )

( )

y t0 = y 0 + b ⋅ t − t0

These gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) For a particle that was at x 0 = 2 m, y 0 = 1 m at t0 = 0s, at time t = 2 s we find the position is (from pathline equations) a 2 2 x = x 0 + ⋅ ⎛ t − t0 ⎞ = 2.8 m ⎠ 2 ⎝

(

)

y = y 0 + b ⋅ t − t0 = 5 m

For a particle that was at x 0 = 2 m, y 0 = 1 m at t0 = 2 s, at time t = 3 s we find the position is a 2 2 x = x 0 + ⋅ ⎛ t − t0 ⎞ = 3 m ⎝ ⎠ 2

v

(

b

=

dy

So, separating variables

dy =

b

Integrating

b y − y0 = ⋅ x − x0 a⋅ t

The streamlines are then

b 5⋅ ( x − 2) y = y0 + ⋅ x − x0 = +1 t a⋅ t

For streamlines

u

dx

a⋅ t

=

)

y = y 0 + b ⋅ t − t0 = 3 m

a⋅ t

⋅ dx

where we treat t as a constant

(

(

)

and we have

x0 = 2 m

m

)

Pathline Plots

Streamline Plots

8

8

Pathline (t0=0) Pathline (t0=2) Streakline

Streamline (t=0) Streamline (t=1) Streamline (t=2)

6

y (m)

6

y (m)

y0 = 1

4

2

4

2

0

1

2

3

x (m)

4

5

0

1

2

3

x (m)

4

5

Problem 2.31 Problem 2.37

[Difficulty: 2]

2.31

A

Given:

Sutherland equation

Find:

Corresponding equation for kinematic viscosity

1

Solution: Governing equation:

μ=

b⋅ T

2

1+

S

p = ρ⋅ R⋅ T

Sutherland equation

Ideal gas equation

T

Assumptions: Sutherland equation is valid; air is an ideal gas

The given data is

−6

b = 1.458 × 10

kg

⋅

1

m⋅ s⋅ K

The kinematic viscosity is

where

ν=

b' =

μ ρ

=

μ⋅ R⋅ T

=

p

S = 110.4 ⋅ K

R = 286.9 ⋅

1

3

3

2

2

2

R⋅ T b ⋅ T R⋅ b T b'⋅ T ⋅ = ⋅ = S S S p p 1+ 1+ 1+ T T T

b' = 4.129 × 10

p

2

−9

m

1.5

K

N⋅ m kg⋅ K

× 1.458 × 10

−6

⋅

⋅s

2

kg 1

m⋅ s⋅ K

m

×

3

= 4.129 × 10

ν=

b'⋅ T

2

1+

S

with T

b' = 4.129 × 10

2

−9

⋅

m

3

101.3 × 10 ⋅ N

2

s⋅ K

3

Hence

p = 101.3 ⋅ kPa

kg⋅ K

2

R⋅ b

b' = 286.9 ⋅

J

2

−9

⋅

m

3

s⋅ K

2

S = 110.4 K

2

Check with Appendix A, Table A.10. At T = 0 °C we find

2 −5 m

T = 273.1 K

ν = 1.33 × 10

⋅

s

3 2 −9 m

4.129 × 10

3

s⋅ K

ν =

1+

× ( 273.1 ⋅ K)

2 2 −5 m

2

ν = 1.33 × 10

110.4

⋅

Check!

s

273.1

At T = 100 °C we find

2 −5 m

T = 373.1 K

ν = 2.29 × 10

⋅

s

3 2

−9 m 4.129 × 10

s⋅ K

ν =

1+

3

× ( 373.1 ⋅ K)

2 2 −5 m

2

ν = 2.30 × 10

110.4

⋅

Check!

s

373.1

Viscosity as a Function of Temperature

−5

2.5× 10

Kinematic Viscosity (m2/s)

Calculated Table A.10

−5

2× 10

−5

1.5× 10

0

20

40

60

Temperature (C)

80

100

Problem 2.32 Problem 2.38

[Difficulty: 2]

2.32

Given:

Sutherland equation with SI units

Find:

Corresponding equation in BG units

1

Solution: Governing equation:

μ=

b⋅ T

2

1+

S

Sutherland equation T

Assumption: Sutherland equation is valid

The given data is

−6

b = 1.458 × 10

kg

⋅

S = 110.4 ⋅ K

1

m⋅ s⋅ K

2 1

Converting constants

−6

b = 1.458 × 10

kg

⋅

1

m⋅ s⋅ K

Alternatively

b = 2.27 × 10

−8

Also

S = 110.4 ⋅ K ×

lbm 0.454 ⋅ kg

×

slug 32.2⋅ lbm

μ=

b⋅ T

2

1+

S

ft

×

⎛ 5⋅ K ⎞ ⎜ ⎝ 9⋅ R ⎠

2

−8

b = 2.27 × 10

1

⋅

×

slug 1

ft⋅ s⋅ R

2

− 8 lbf ⋅ s

lbf ⋅ s

b = 2.27 × 10

slug⋅ ft

⋅

1 2

2

ft ⋅ R

9⋅ R

S = 198.7 ⋅ R

5⋅ K

with T in Rankine, µ in T

0.3048⋅ m

2

1

and

×

2

slug ft⋅ s⋅ R

×

lbf ⋅ s ft

2

2

Check with Appendix A, Table A.9. At T = 68 °F we find

T = 527.7 ⋅ R

μ = 3.79 × 10

− 7 lbf ⋅ s

⋅

ft

1 − 8 lbf ⋅ s

2.27 × 10

1 2

ft ⋅ R

μ =

1+

× ( 527.7 ⋅ R)

2

2

2

μ = 3.79 × 10

198.7

− 7 lbf ⋅ s

⋅

ft

Check!

2

527.7

At T = 200 °F we find

T = 659.7 ⋅ R

μ = 4.48 × 10

− 7 lbf ⋅ s

⋅

ft

2

1 − 8 lbf ⋅ s

2.27 × 10

1 2

μ =

ft ⋅ R 1+

× ( 659.7 ⋅ R)

2

2

198.7 659.7

μ = 4.48 × 10

− 7 lbf ⋅ s

⋅

ft

2

Check!

Problem 2.33 2.33

Data:

Using procedure of Appendix A.3: T (oC) 0 100 200 300 400

µ(x105) 1.86E-05 2.31E-05 2.72E-05 3.11E-05 3.46E-05

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

T

3 2

µ

S ⎛ 1 ⎞ = ⎜ ⎟T + b b ⎝ ⎠

From the built-in Excel Linear Regression functions:

Hence: b = 1.531E-06 S = 101.9

Slope = 6.534E+05 Intercept = 6.660E+07

. .

1/2

kg/m s K K

2 R = 0.9996

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 2.34 Problem 2.40

[Difficulty: 2]

2.34

Given:

Velocity distribution between flat plates

Find:

Shear stress on upper plate; Sketch stress distribution

Solution: Basic equation

du τyx = μ⋅ dy τyx = −

At the upper surface

Hence

y=

du

=

dy

d dy

⎡

u max⋅ ⎢1 −

⎣

2 ⎛ 2 ⋅ y ⎞ ⎤⎥ = u ⋅ ⎛ − 4 ⎞ ⋅ 2⋅ y = − 8 ⋅ umax⋅ y ⎜ max ⎜ 2 2 ⎝ h ⎠⎦ h ⎝ h ⎠

8 ⋅ μ⋅ u max⋅ y h

h

2

and

2

τyx = −8 × 1.14 × 10

− 3 N⋅ s

⋅

2

h = 0.1⋅ mm

× 0.1⋅

m

m s

×

0.1 2

u max = 0.1⋅

⋅ mm ×

1⋅ m 1000⋅ mm

×

m s

− 3 N⋅ s

μ = 1.14 × 10

⋅

2

(Table A.8)

m

2 ⎛ 1 × 1000⋅ mm ⎞ ⎜ 1⋅ m ⎠ ⎝ 0.1⋅ mm

N τyx = −4.56⋅ 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 ⎜ h2 ⎝ ⎠

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

Problem 2.35 (Difficulty: 1)

2.35 What is the ratio between the viscosities of air and water at 10℃ ? What is the ratio between their kinematic viscosities at this temperature and standard barometric pressure ? Given: The temperature 10℃ .

Find: Ratio between the viscosities of air and water at 10℃ . Ratio of kinematic viscosities at this temperature and pressure. Assumption: The standard barometric pressure is sea level pressure. Air can be treated as an ideal gas Solution: At 10℃, for the viscosities:

𝜇𝑎𝑎𝑎 = 0.018 × 10−3 𝑃𝑃

For the densities at STP:

𝑎𝑎𝑎

𝜇𝐻2 𝑜 = 1.4 × 10−3 𝑃𝑃 ∙ 𝑠

𝜇𝑎𝑎𝑎 0.018 × 10−3 𝑃𝑃 = = 0.013 1.4 × 10−3 𝑃𝑃 𝜇𝐻2 𝑜

𝑘𝑘 𝑚3 𝑘𝑘 𝜌𝑎𝑎𝑎 = 1.225 3 𝑚 1 at 15℃, using the ideal gas relation where ρ ∝ at constant pressure 𝜌𝐻2 𝑜 = 1000

𝜌𝑎𝑎𝑎 = 1.225 ×

The ration of kinematic viscosities at 10℃. 𝑣𝑎𝑎𝑎 𝑣𝐻2 𝑜

𝑇

(15 + 273) 𝑘𝑘 𝑘𝑘 = 1.247 (10 + 273) 𝑚3 𝑚3

0.018 × 10−3 𝑃𝑃 𝑘𝑘 1.247 3 𝑚 = = 10.3 1.4 × 10−3 𝑃𝑃 𝑘𝑘 1000 3 𝑚

The dynamic viscosity of air is much less than that of water but the kinematic viscosity is greater.

Problem 2.36 (Difficulty: 2)

2.36 Calculate the velocity gradients and shear stress for 𝑦 = 0, 0.2, 0.4 and 0.6 𝑚, if the velocity profile is a quarter-circle center having its center 0.6 𝑚 from the boundary. The fluid viscosity is 7.5 × 10−4

Given: The fluid viscosity 𝜇 = 7.5 × 10−4 Find: The velocity gradient Solution:

𝑑𝑑 𝑑𝑑

𝑁𝑁 . 𝑚2

The equation for a quarter-circle with y measured up from the surface of the plate is:

Or, expanding the expression:

The velocity gradient is:

�

2 𝑢 2 𝑦 � +� − 1� = 1 10 0.6

𝑢2 = 278(1.2𝑦 − 𝑦 2 ) 2u

And the shear stress is 𝜏=𝜇

𝑑𝑑 = 278(1.2 − 2𝑦) 𝑑𝑑

1.2 − 2𝑦 𝑑𝑑 = 139 � � 𝑢 𝑑𝑑

𝑑𝑑 𝑁𝑁 1.2 − 2𝑦 1.2 − 2𝑦 = 7.5 × 10−4 2 ∙ 139 � � = 0.104 � � 𝑑𝑑 𝑚 𝑢 𝑢

When 𝑦 = 0, from the equation for the velocity we have 𝑢=0

𝑚 𝑠

𝑁𝑁 . 𝑚2

And for the gradient we have

And

When 𝑦 = 0.2

1 𝑑𝑑 =∞ 𝑠 𝑑𝑑 𝜏=∞

𝑁 𝑚2 𝑚 𝑠

𝑢 = 7.46

1 𝑑𝑑 = 14.9 𝑠 𝑑𝑑

When 𝑦 = 0.4

𝜏 = 0.0111

𝑁 𝑚2

𝑚 𝑠

𝑢 = 9.43

1 𝑑𝑑 = 5.90 𝑠 𝑑𝑑

When 𝑦 = 0.6

𝜏 = 0.0044 𝑢 = 10

𝑁 𝑚2

𝑚 𝑠

𝑑𝑑 1 =0 𝑑𝑑 𝑠

𝜏=0

𝑁 𝑚2

Problem 2.37 (Difficulty: 2)

2.37 A very large thin plate is centered in a gap of width 0.06 𝑚 with different oils of unknown viscosities above and below; one viscosity is twice the other. When the plate is pulled at a velocity of 0.3

𝑚 , 𝑠

the resulting force on one square meter of plate due to the viscous shear on both sides is 29 𝑁..

Assuming viscous flow and neglecting all end effects, calculate the viscosities of the oils.

Given: Viscosity: 𝜇2 = 2𝜇1 . Width of gap: ℎ = 0.06 𝑚. Velocity:𝑉 = 0.3

𝐹 = 29

𝑁 𝑚2

𝑚 . 𝑠

Force per square meter:

Find: 𝜇1 and 𝜇2

Assumption: Viscous flow with linear velocity profiles, negligible end effects.

Solution: Use Newton’s law relating shear stress to viscosity and velocity gradient. The relation between the two viscosities is 𝜇2 = 2𝜇1

Because the gaps are equal and the plate velocity is the same for both fluids, the velocity gradient is the same for both sides of the plate: 𝑚 0.3 . 𝑉 1 𝑑𝑑 𝑠 = = = 10 𝑠 𝑑𝑑 0.5ℎ 0.5 × 0.06 𝑚

For a Newtonian fluid with a linear velocity profile, we have 𝜏=𝜇

𝑑𝑑 ∆𝑉 =𝜇 𝑑𝑑 ∆𝑦

The force on the plate due to the top layer of fluid is

𝑚 0.3 ∆𝑉 𝑠 = 𝜇 101 𝜏1 = 𝜇1 = 𝜇1 1 ∆𝑦 0.03 𝑚

Similarly, the force on bottom of the plate is

𝑚 0.3 ∆𝑉 𝑠 = 𝜇 10 1 = 𝜇2 𝜏2 = 𝜇2 2 𝑠 ∆𝑦 0.03 𝑚

The total force per unit area equals the sum of the two shear stresses, where for the 1 m2 plate the shear stress is equal to 29 N/m2. 𝑁 𝐹 = 𝜏1 + 𝜏2 = 29 2 𝑚 𝐴

Or, since µ2 = times µ1

𝜇1 =

1 1 𝑁 𝜇1 101 + 𝜇2 10 = 3𝜇1 10 = 29 2 𝑠 𝑠 𝑚

1 𝑁 𝑠 𝑁∙𝑠 × 29 2 × = 0.967 2 = 0.967 𝑃𝑃 ∙ 𝑠 3 𝑚 10 𝑚 𝜇2 = 2𝜇1 = 1.934 𝑃𝑃 ∙ 𝑠

Problem 2.38 Problem 2.44

[Difficulty: 2]

2.38

Given:

Ice skater and skate geometry

Find:

Deceleration of skater

τ yx = µ

y

Solution: Governing equation:

du τyx = μ⋅ dy

ΣFx = M ⋅ ax

du dy

V = 20 ft/s

h x L

Assumptions: Laminar flow The given data is

W = 100 ⋅ lbf

V = 20⋅ − 5 lbf ⋅ s

μ = 3.68 × 10

⋅

ft Then

ft

L = 11.5⋅ in

s

w = 0.125 ⋅ in

Table A.7 @32oF

2

du V ft 1 12⋅ in − 5 lbf ⋅ s τyx = μ⋅ = μ⋅ = 3.68 × 10 ⋅ × 20⋅ × × 2 dy h s 0.0000575 ⋅ in ft ft lbf τyx = 154 ⋅ 2 ft

Equation of motion

ΣFx = M ⋅ ax

ax = −

τyx⋅ A⋅ g W

ax = −154

lbf ft

ax = −0.495 ⋅

−W τyx⋅ A = ⋅a g x

or

2

ft 2

s

=−

τyx⋅ L⋅ w⋅ g W

× 11.5⋅ in × 0.125 ⋅ in × 32.2⋅

ft 2

s

×

1 100 ⋅ lbf

×

ft

2

( 12⋅ in)

2

h = 0.0000575 ⋅ in

Problem 2.39 Problem 2.46

[Difficulty: 2]

2.39

Given:

Block moving on incline on oil layer

Find:

Speed of block when free, pulled, and pushed

Solution:

y

U

Governing equations:

x

x

du

τyx = μ⋅ dy

f

N W

ΣFx = M ⋅ ax

d

θ

Assumptions: Laminar flow The given data is

M = 10⋅ kg

W = M⋅ g

W = 98.066 N

d = 0.025 ⋅ mm

θ = 30⋅ deg

F = 75⋅ N

− 1 N⋅s

μ = 10

⋅

w = 250 ⋅ mm

Fig. A.2 SAE 10-39 @30oC

2

m Equation of motion

ΣFx = M ⋅ ax = 0

The friction force is

du U 2 f = τyx⋅ A = μ ⋅ ⋅ A = μ ⋅ ⋅ w dy d

Hence for uphill motion

F = f + W ⋅ sin ( θ) = μ ⋅

For no force:

U =

d ⋅ W⋅ sin( θ) 2

F − f − W ⋅ sin ( θ) = 0

so

U d

U =

d ⋅ ( F − W⋅ sin( θ) ) 2

μ⋅ w

U=

d ⋅ ( F − W⋅ sin( θ) )

(For downpush change sign of W)

2

μ⋅ w

U = 0.196

m

U = 0.104

m

μ⋅ w

Pushing up:

2

⋅ w + W ⋅ sin ( θ)

s

s

Pushing down:

U =

d ⋅ ( F + W ⋅ sin ( θ) ) 2

μ⋅w

U = 0.496

m s

Problem 2.40 Problem 2.48

[Difficulty: 2]

2.40

Given:

Flow data on apparatus

Find:

The terminal velocity of mass m

Solution: Given data:

Dpiston = 73⋅ mm

Dtube = 75⋅ mm

Mass = 2 ⋅ kg

Reference data:

kg ρwater = 1000⋅ 3 m

(maximum density of water)

L = 100 ⋅ mm

μ = 0.13⋅

From Fig. A.2:, the dynamic viscosity of SAE 10W-30 oil at 25oC is:

SG Al = 2.64

N⋅ s 2

m

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 ⎥ ⎢Mass + SGAl⋅ ρwater⋅ ⎜ ⎥ ⋅ g = τrz⋅ A = 4 ⎣ ⎝ ⎠⎦

⎛ μ⋅ d V ⎞ ⋅ π⋅ D ⎜ z ( piston⋅ L) ⎝ dr ⎠

The velocity profile within the oil film is linear ... d Vz = dr

Therefore

V

⎛ Dtube − Dpiston ⎞ ⎜ 2 ⎝ ⎠

Thus, the terminal velocity of the piston, V, is:

g ⋅ ⎛ SG Al⋅ ρwater⋅ π⋅ Dpiston ⋅ L + 4 ⋅ Mass⎞ ⋅ Dtube − Dpiston ⎝ ⎠ 2

V =

or

V = 10.2

8 ⋅ μ⋅ π⋅ Dpiston⋅ L m s

(

)

Problem 2.41 (Difficulty: 2)

2.41 A vertical gap 25 mm wide of infinite extent contains oil of specific gravity 0.95 and viscosity 2.4 Pa ∙ s. A metal plate 1.5 m × 1.5 m × 1.6 mm weighting 45 N is to be lifted through the gap at a constant speed of 0.06 𝑚⁄𝑠.Estimate the force required.

Given: Plate size:1.5 m × 1.5 m × 1.6 mm .Width of gap: 25 𝑚𝑚. Specific gravity:0.95. Viscosity: 2.4 Pa ∙ s. Weight: 45 N. Speed: 0.06 𝑚⁄𝑠.

Find: The force required F 𝑇 .

Assumption: Viscous flow. Neglecting all end effects. Linear velocity profile in the gap. Solution: Make a force balance on the plate. Use Newton’s law of viscosity to relate the viscous force on the plate to the viscosity and velocity. We need to calculate all the individual forces. There are the force due to gravity (weight), the buoyancy force, and the drag force. Buoyancy force: 𝐹𝐵 = 𝜌𝜌𝜌 = 𝑆𝑆 ∙ 𝜌𝐻2 𝑜 𝑔𝑔 = 0.95 × 998.2 × 9.81 × 1.5 × 1.5 × 0.0016 = 33.5 𝑁

Drag force: The viscous shear stress is given by

𝜏=𝜇

𝑑𝑑 ∆𝑢 = 𝜇 𝑑𝑑 ∆𝑥

The force on both sides of the plate is 𝑚 0.06 ∆𝑢 𝑠 × (1.5 𝑚)2 = 55.4 𝑁 𝐴 = 2 × 2.4 𝑃𝑃 𝑠 × 𝐹𝜏 = 2 𝜇 ∆𝑥 0.0117 𝑚

The force required to maintain motion is 𝐹𝑇 , by the force balance equation we have: The total force is then

𝐹𝑇 + 𝐹𝐵 = 𝑊 + 𝐹𝜏

𝐹𝑇 = 𝑊 + 𝐹𝜏 − 𝐹𝐵 = 45 𝑁 + 55.4 𝑁 − 33.5 𝑁 = 66.9 𝑁

Problem 2.42 (Difficulty: 2)

2.42 A cylinder 8 in in diameter and 3 ft long is concentric with a pipe of 8.25 in. Between cylinder and pipe there is an oil film. What force is required to move the cylinder along the pipe at a constant velocity of 3 fps? The kinematic viscosity of the oil is 0.006

𝑓𝑓 2 . 𝑠

The specific gravity is 0.92.

Given: Cylinder diameter: 𝐷𝑐 = 8 𝑖𝑖. Cylinder length: L = 3 𝑓𝑓. Pipe diameter: 𝐷𝑝 = 8.25 𝑖𝑖. Cylinder

velocity:V = 3

𝑓𝑓 . 𝑠

Oil viscosity: 𝑣 = 0.006

Find:The force required F 𝑇 .

𝑓𝑓 2 . 𝑠

Specific gravity:𝑆𝑆 = 0.92.

Assumption: Viscous flow with linear velocity profile in oil, negligible end effects. Solution: Use Newton’s law of viscosity to evaluate the viscous force.

The gap ℎ between the cylinder and pipe is: ℎ=

8.25 − 8 𝑖𝑖 = 0.125 𝑖𝑖 = 0.0104 𝑓𝑓 2

The contact area 𝐴 between cylinder and oil is:

1 𝑓𝑓 = 12 𝑖𝑖

The dynamic viscosity:

𝐴 = 𝜋𝐷𝑐 𝐿 = 𝜋 ×

𝜇 = 𝑣𝑣 = 𝑣 ∙ 𝑆𝑆 ∙ 𝜌𝐻2 𝑜 = 0.006

8 × 3 𝑓𝑓 2 = 6.28 𝑓𝑓 2 12

𝑓𝑓 2 𝑙𝑙𝑙 𝑙𝑙𝑙 × 0.92 × 62.4 3 = 0.344 𝑓𝑓 𝑓𝑓 ∙ 𝑠 𝑠

The drag force is, assuming a linear velocity profile in the fluid

𝑓𝑓 3 𝑑𝑑 𝑉 𝑙𝑙𝑙 𝑙𝑙𝑙 ∙ 𝑓𝑓 𝑠 2 F𝐷 = 𝜇𝜇 = 𝜇𝜇 = 0.344 × 6.28 𝑓𝑓 × = 605 = 19 𝑙𝑙𝑙 𝑑𝑑 ℎ 𝑓𝑓 ∙ 𝑠 𝑠2 0.0104 𝑓𝑓

Problem 2.43 (Difficulty: 2)

2.43 Crude oil at 20℃ fills the space between two concentric cylinders 250 𝑚𝑚 high and with diameters of 150 𝑚𝑚 and 156 𝑚𝑚 . What torque is required to rotate the inner cylinder at 12 𝑟𝑟𝑟, the outer cylinder remaining stationary?

Given: Temperature: T = 20 ℃. Cylinder height: H = 250 𝑚𝑚. Outer cylinder diameter: 𝐷𝑜 = 156 𝑚𝑚. Inner cylinder diameter: 𝐷𝐼 = 150 𝑚𝑚. Rotating speed: 12 𝑟𝑟𝑟. Find: The required torque Γ.

Assumption: Linear velocity profile in fluid, viscous flow, neglect all end effects. Solution: Use Newton’s law of viscosity to find the force on the surfaces The torque equals force times radius: 𝑇 = 𝐹𝐷 𝑅𝐼

The velocity of inner cylinder is: 𝑉 = 𝜔𝑅𝐼 = 12

𝑟 1 𝑚𝑚𝑚 1 150 𝑚 × × �2𝜋 � × 𝑚 = 0.0942 𝑚𝑚𝑚 60 𝑠 𝑟 2 × 1000 𝑠

The dynamic viscosity of crude oil at Temperature = 20 ℃.:

𝜇 = 0.00718 𝑃𝑃 ∙ 𝑠

Newton’s law of viscosity with a linear velocity profile is

The drag force is:

𝜏=𝜇 F𝐷 = 𝜇

𝑑𝑑 ∆𝑢 = 𝜇 𝑑𝑑 ∆𝑟

𝑉 𝐴 (𝐷 0.5 × 𝑜 − 𝐷𝐼 )

𝑚 𝑠 F𝐷 = 0.00718 𝑃𝑃 ∙ 𝑠 × × (𝜋 × 0.15 𝑚 × 0.25 𝑚) = 0.0266 𝑁 0.5 × (0.006 𝑚) 0.0942

The torque is then

𝑇 = 𝐹𝐷 𝑅𝐼 = 0.0266 𝑁 × 0.075 𝑚 = 0.002 𝑁 ∙ 𝑚

Problem 2.44 Problem 2.49 2.44

[Difficulty: 3]

2.40

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

Dtube = 75⋅ mm

L = 100 ⋅ mm

Reference data:

kg ρwater = 1000⋅ 3 m

(maximum density of water)

(From Problem 2.40 2.48))

μ = 0.13⋅

From Fig. A.2, the dynamic viscosity of SAE 10W-30 oil at 25oC is:

m V0 = 10.2⋅ s

SG Al = 2.64

N⋅ s 2

m The free body diagram of the piston after the cord is cut is: Piston weight:

2⎞ ⎛⎜ π⋅ D piston Wpiston = SGAl⋅ ρwater⋅ g ⋅ ⎜ ⋅L 4 ⎝ ⎠

Viscous force:

Fviscous( V) = τrz⋅ A

or

Fviscous( V) = μ⋅ ⎡⎢ 1

⎤ ⋅ π⋅ D ⎥ ( piston⋅ L) ⋅ D − D ⎢ ( tube piston)⎥ ⎣2 ⎦ dV mpiston⋅ = Wpiston − Fviscous( V) dt

Applying Newton's second law:

Therefore

dV dt

If

= g − a⋅ V

V = g − a⋅ V

V

where

then

The differential equation becomes

a =

dX dt dX dt

The solution to this differential equation is:

8⋅ μ

(

SGAl⋅ ρwater⋅ Dpiston⋅ Dtube − Dpiston = −a⋅

)

dV dt

= −a⋅ X

X( t) = X0 ⋅ e

− a⋅ t

where

X( 0 ) = g − a⋅ V0

or

g − a⋅ V( t) = g − a⋅ V0 ⋅ e

(

)

− a⋅ t

Therefore

g ( − a⋅ t) g V( t) = ⎛⎜ V0 − ⎞ ⋅ e + a⎠ a ⎝

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 t

The terminal speed of the piston, Vt, is evaluated as t approaches infinity g Vt = a

or

m Vt = 3.63 s

The time needed for the piston to slow down to within 1% of its terminal velocity is:

⎛ V − g ⎞ ⎜ 0 a t = ⋅ ln⎜ ⎟ g a ⎜ 1.01⋅ Vt − a⎠ ⎝ 1

or

t = 1.93 s

3

Problem 2.45

Problem 2.50

[Difficulty: 3]

2.45

Given:

Block on oil layer pulled by hanging weight

Find:

Expression for viscous force at speed V; differential equation for motion; block speed as function of time; oil viscosity

Mg

Solution: Governing equations:

x y

Ft du τyx = μ⋅ dy

ΣFx = M ⋅ ax

Ft

Fv mg

N

Assumptions: Laminar flow; linear velocity profile in oil layer M = 5 ⋅ kg

Equation of motion (block)

ΣFx = M ⋅ ax

so

dV Ft − Fv = M ⋅ dt

( 1)

Equation of motion (block)

ΣFy = m⋅ ay

so

dV m⋅ g − Ft = m⋅ dt

( 2)

Adding Eqs. (1) and (2)

dV m⋅ g − Fv = ( M + m) ⋅ dt

The friction force is

du V Fv = τyx⋅ A = μ⋅ ⋅ A = μ⋅ ⋅ A dy h

Hence

m⋅ g −

To solve separate variables

W = m⋅ g = 9.81⋅ N

μ⋅ A h

M+m

dt =

m⋅ g − t=−

Hence taking antilogarithms

1−

⋅ V = ( M + m) ⋅

μ⋅ A h

( M + m) ⋅ h μ⋅ A

μ⋅ A m⋅ g ⋅ h

A = 25⋅ cm

2

The given data is

h = 0.05⋅ mm

dV dt

⋅ dV ⋅V ⋅ ⎛⎜ ln⎛⎜ m⋅ g −

⎝ ⎝ −

⋅V = e

μ⋅ A

μ⋅ A ( M+ m) ⋅ h

h ⋅t

⋅ V⎞ − ln( m⋅ g ) ⎞ = −

⎠

⎠

( M + m) ⋅ h μ⋅ A

⋅ ln⎛⎜ 1 −

⎝

μ⋅ A m⋅ g ⋅ h

⋅ V⎞

⎠

⎡ − m⋅ g ⋅ h ⎢ V= ⋅ ⎣1 − e

Finally

μ⋅ A ( M + m) ⋅ h

μ⋅ A

⋅ t⎤

⎥ ⎦

The maximum velocity is V =

m⋅ g ⋅ h μ⋅ A

In Excel: The data is

M= m=

5.00 1.00

kg kg

To find the viscosity for which the speed is 1 m/s after 1 s use Goal Seek with the velocity targeted to be 1 m/s by varying

g= 0=

9.81

the viscosity in the set of cell below:

1.30

m/s2 N.s/m2

A= h=

25 0.5

cm 2 mm

t (s) 1.00

V (m/s) 1.000

Speed V of Block vs Time t t (s) 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 2.10 2.20 2.30 2.40 2.50 2.60 2.70 2.80 2.90 3.00

V (m/s) 0.000 0.155 0.294 0.419 0.531 0.632 0.722 0.803 0.876 0.941 1.00 1.05 1.10 1.14 1.18 1.21 1.25 1.27 1.30 1.32 1.34 1.36 1.37 1.39 1.40 1.41 1.42 1.43 1.44 1.45 1.46

1.6 1.4 1.2 1.0

V (m/s) 0.8 0.6 0.4 0.2 0.0 0.0

0.5

1.0

1.5

t (s)

2.0

2.5

3.0

Problem 2.46 Problem 2.51

[Difficulty: 4]

2.46

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)

so

M ⋅ a = M ⋅ g ⋅ sin( θ) − Ff = M ⋅ g ⋅ sin( θ) ainit = g ⋅ sin( θ) = 9.81⋅

m 2

× sin( 30⋅ deg)

s M ⋅ a = M ⋅ g ⋅ sin( θ) − Ff

Applying Newton's 2nd law at any instant

so

M⋅ a = M⋅

dV

g ⋅ sin( θ) −

−

Integrating and using limits

or

V = 5 ⋅ kg × 9.81⋅

m 2

s

V( 0.1⋅ s) = 0.404 ⋅

m s

M⋅ d μ⋅ A

μ⋅ A M⋅ d

⋅ ln⎛⎜ 1 −

⎝

m 2

s

du V Ff = τ⋅ A = μ⋅ ⋅ A = μ⋅ ⋅ A dy d

and μ⋅ A

= M ⋅ g ⋅ sin( θ) −

dV

Separating variables

At t = 0.1 s

dt

ainit = 4.9

d

⋅V

= dt ⋅V

μ⋅ A M ⋅ g ⋅ d ⋅ sin( θ)

⋅ V⎞ = t

⎠

− μ⋅ A ⎞ ⎛ ⋅t ⎜ M ⋅ g ⋅ d ⋅ sin( θ) M⋅ d V( t) = ⋅⎝1 − e ⎠

μ⋅ A

2

× 0.0002⋅ m⋅ sin( 30⋅ deg) ×

2

m

0.4⋅ N⋅ s⋅ ( 0.1⋅ m)

2

×

N⋅ s

kg⋅ m

⎛ 0.4⋅ 0.01 ⋅ 0.1⎞⎤ ⎡ −⎜ ⎢ 5⋅ 0.0002 ⎠⎥ × ⎣1 − e ⎝ ⎦

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

V( t = 0.1⋅ s) =

M ⋅ g ⋅ d ⋅ sin( θ) μ⋅ A

− μ⋅ A ⎡ ⋅ ( t= 0.1⋅ s )⎤ ⎢ ⎥ M⋅ d ⋅ ⎣1 − e ⎦

The viscosity µ is implicit in this equation, so solution must be found by manual iteration, or by any of a number of classic root-finding numerical methods, or by using Excel's Goal Seek

Using Excel:

μ = 1.08⋅

N⋅ s 2

m

1

Problem 2.47 (Difficulty: 1)

2.47 A torque of 4 𝑁 ∙ 𝑚 is required to rotate the intermediate cylinder at 30

of the oil. All the cylinders are 450 𝑚𝑚 long. Neglect end effects.

Given: Cylinder height: H = 450 𝑚𝑚. Rotation speed: 30

𝑟 . 𝑚𝑚𝑚

𝑟 . 𝑚𝑚𝑚

Gap between cylinder: ℎ = 0.003 𝑚.

Radius of intermediate cylinder: 𝑅 = 0.15 𝑚. Torque: T = 4 𝑁 ∙ 𝑚. Find:The oil viscosity 𝜇.

Assumption: Linear velocity profile in fluid, viscous flow, negligible end effects.

Solution: Use Newtons’s viscosity law to evaluate the force on the cylinder Newton’s law of viscosity for a linear velocity profile is

The velocity of intermediate cylinder: 𝑉 = 𝜔𝜔 = 30

𝜏=𝜇

𝑑𝑑 ∆𝑢 = 𝜇 𝑑𝑑 ∆𝑟

𝑟 1 𝑚𝑚𝑚 1 𝑚 × × 2𝜋 × 𝑅 𝑚 = 0.471 𝑚𝑚𝑚 60 𝑠 𝑟 𝑠

The drag force on both sides of the cylinder is:

The torque is given by:

The area is:

𝐹𝐷 = 2𝜇𝜇

𝑉 ℎ

𝑇 = 𝐹𝐷 ∙ 𝑅 = 2𝜇𝜇 A = 2πRH

Calculate the viscosity

𝑉 ∙𝑅 ℎ

The viscosity is then: 𝜇=

𝑇ℎ 4 𝑁 ∙ 𝑚 × 0.003 𝑚 𝑁∙𝑆 = = 0.2 = 0.2 𝑃𝑃 ∙ 𝑠 𝑚 2𝐴𝐴𝐴 2 × 2𝜋 × 0.15 𝑚 × 0.45 𝑚 × 0.471 × 0.15 𝑚 𝑚2 𝑠

Problem 2.48 (Difficulty: 2)

2.48 A circular disk of diameter d is slowly rotated in a liquid of large viscosity µ at a small distance h from a fixed surface. Derive an expression for the torque T necessary to maintain an angular velocity ω. Neglect centrifugal effects.

Given: Disk diameter: d. Distance to fixed surface: ℎ. Viscosity: 𝜇. Angular velocity: ω. Radius of Find: Torque: T.

Assumption: Linear velocity profile in gap between the two disks, viscous flow, negligible end effects, negligible centrifugal force effects.

Solution: Use Newton’s law of viscosity with a linear velocity profile to find the forces 𝜏=𝜇

∆𝑢 𝑑𝑑 = 𝜇 ∆𝑦 𝑑𝑑

The velocity at the interface of the fluid and disk varies with radius 𝑉=𝜔𝑟

The expression for shear stress is then

𝜏= 𝜇

𝜔𝑟 ℎ

The incremental torque the product of the radius and the force per unit area and the area from the center to the outer radius. The total torque is the integral from the centerline to the outer radius. 𝑑/2

𝑇=�

0

𝑟 �𝜇

𝜔𝑟 � 2𝜋𝜋 𝑑𝑑 ℎ

Or 𝑇=

𝑑/2 𝜇𝜔 2𝜋 � 𝑟 3 𝑑𝑑 ℎ 0

𝑇=

𝜋 𝜇 𝜔 𝑑4 32 ℎ

Problem 2.49 (Difficulty: 2)

2.49 The fluid drive shown transmits a torque T for steady-state conditions (𝜔1 and 𝜔2 constant). Derive an expression for the slip (𝜔1 − 𝜔2 ) in terms of 𝑇, 𝜇, 𝑑 𝑎𝑎𝑎 ℎ. For values 𝑑 = 6 𝑖𝑖, ℎ = 0.2 𝑖𝑖., SAE 30 oil at 75 𝐹, a shaft rotation of 120 𝑟𝑟𝑟, and a torque of 0.003 ft-lbf, determine the slip.

Given: 𝑑 = 6 𝑖𝑖, ℎ = 0.2 𝑖𝑖 , rotation: 120 𝑟𝑟𝑟. SAE 30 oil at 75 𝐹

Find: The slip (𝜔1 − 𝜔2 ).

Assume: Linear velocity profile in the viscous fluid Solution: Use Newton’s law of viscosity to relate the viscous forces to the torque and slip From the force balance equation we have: 𝑑 2

𝑑 2

𝑇 = � 𝜏 ∙ 𝑑𝑑 ∙ 𝑟 = � 𝜏 ∙ 2𝜋𝜋𝜋𝜋 ∙ 𝑟 0

0

Assuming a linear velocity profile in the space between the two disks, Newton’s law of viscosity is 𝑑𝑑 ∆𝑢 = 𝜇 𝑑𝑑 ∆𝑥

The velocity difference varies with radius

𝜏=𝜇

So the shear stress is

∆𝑢 = (𝜔1 − 𝜔2 )𝑟

The torque is then:

𝜏=𝜇 𝑑 2

𝑇 = � 2𝜋𝜋 0

(𝜔1 − 𝜔2 )𝑟 ℎ

(𝜔1 − 𝜔2 ) 3 (𝜔1 − 𝜔2 ) 4 𝑟 𝑑𝑑 = 𝜋𝜋 𝑑 ℎ 32ℎ

Solving for the slip:

The viscosity for SAE 30 oil at 75 𝐹 is: The slip is then:

𝜇 = 0.008

(𝜔1 − 𝜔2 ) =

(𝜔1 − 𝜔2 ) =

32 𝑇ℎ 𝜋 𝜇 𝑑4

𝑙𝑙𝑙 ∙ 𝑠2 1 𝑙𝑙𝑙 ∙ 𝑠 𝑠𝑠𝑠𝑠 = 0.008 = 0.008 𝑓𝑓 2 𝑓𝑓 ∙ 𝑠 𝑓𝑓 𝑓𝑓 ∙ 𝑠

32 × 0.003 𝑓𝑓 − 𝑙𝑙𝑙 × 0.0167 𝑓𝑓 𝑟𝑟𝑟 = 1.019 = 9.73 𝑟𝑟𝑟 𝑙𝑙𝑙 ∙ 𝑠 𝑠 4 (0.5 𝜋 × 0.008 × 𝑓𝑓) 𝑓𝑓 2

Problem 2.50 Problem 2.52

[Difficulty: 3]

2.50

Given:

Block sliding on oil layer

Find:

Direction of friction on bottom of block and on plate; expression for speed U versus time; time required to lose 95% of initial speed

Solution:

U

Governing equations:

du τyx = μ⋅ dy

ΣFx = M ⋅ ax

Fv y

h

Assumptions: Laminar flow; linear velocity profile in oil layer

x The bottom of the block is a -y surface, so τyx acts to the left; The plate is a +y surface, so τyx acts to the right Equation of motion

ΣFx = M ⋅ ax

The friction force is

du U 2 Fv = τyx⋅ A = μ⋅ ⋅ A = μ⋅ ⋅ a dy h

Hence

−

2

1 U

dU

⋅ U = M⋅

h

⋅ dU = −

μ⋅ a

dt

2

⋅ dt

M⋅ h

2

⎞ = − μ⋅ a ⋅ t U0 M⋅ h ⎝ ⎠

ln⎛⎜

dU Fv = M ⋅ dt

U

U

To solve separate variables

μ⋅ a

so

2

−

Hence taking antilogarithms

U = U0 ⋅ e

t=−

Solving for t

M⋅ h μ⋅ a

Hence for

U U0

= 0.05

t = 3.0⋅

2

μ⋅ a

M⋅ h

⋅t

t ⋅ ln⎛⎜

⎞

⎝ U0 ⎠

M⋅ h μ⋅ a

U

2

Problem 2.54 Problem 2.51

[Difficulty: 3]

2.51

Given:

Data on annular tube

Find:

Whether no-slip is satisfied; location of zeroshear stress; viscous forces

Solution: The velocity profile is

Check the no-slip condition. When

2 2 ⎛ ⎞ r 2 2 Ro − Ri ⎜ u z( r) = ⋅ ⋅ Ri − r − ⋅ ln⎛⎜ ⎞ 4⋅ μ L ⎜ ⎝ Ri ⎠ ⎟ ⎛ Ri ⎞ ln⎜ ⎜ ⎝ ⎝ Ro ⎠ ⎠

∆p

1

2 2 ⎛ Ro − Ri ⎛ Ro ⎞ ⎞ 2 2 ⎜ u z( R o ) = ⋅ ⋅ Ri − Ro − ⋅ ln⎜ 4⋅ μ L ⎜ ⎝ Ri ⎠ ⎟ ⎛ Ri ⎞ ln⎜ ⎜ ⎝ ⎝ Ro ⎠ ⎠

1

r = Ro

∆p

1 ∆p ⎡ 2 2 2 2 u z Ro = ⋅ ⋅ R − Ro + ⎛ Ro − Ri ⎞⎤ = 0 ⎝ ⎠⎦ 4⋅ μ L ⎣ i

( )

When

r = Ri

2 2 ⎛ Ro − Ri ⎛ Ri ⎞ ⎞ 2 2 ⎜ u z( R i ) = ⋅ ⋅ Ri − Ri − ⋅ ln⎜ =0 Ri ⎟ 4⋅ μ L ⎜ ⎛ Ri ⎞ ⎝ ⎠ ln⎜ ⎜ Ro ⎝ ⎝ ⎠ ⎠

1

∆p

The no-slip condition is satisfied.

The given data is

The viscosity of the honey is

Ri = 5 ⋅ mm

Ro = 25⋅ mm

μ = 5⋅

N⋅ s 2

m

∆p = 125 ⋅ kPa

L = 2⋅ m

The plot looks like

Radial Position (mm)

25 20 15 10 5

0

0.25

0.5

0.75

Velocity (m/s) For each, shear stress is given by

du τrx = μ⋅ dr

τrx = μ⋅

duz( r) dr

2 2 ⎡ ⎛ ⎞⎤ 1 ∆p ⎜ 2 r 2 Ro − Ri ⎢ = μ⋅ ⋅ ⋅ Ri − r − ⋅ ln⎛⎜ ⎞ ⎥ dr ⎢ 4 ⋅ μ L ⎜ ⎝ Ri ⎠ ⎟⎥ ⎛ Ri ⎞ ln⎜ ⎢ ⎜ ⎥ ⎣ ⎝ ⎝ Ro ⎠ ⎠⎦

d

⎛ Ro − Ri 1 ∆p ⎜ τrx = ⋅ ⋅ −2 ⋅ r − 4 L ⎜ ⎛ Ri ⎞ ln⎜ ⋅r ⎜ ⎝ ⎝ Ro ⎠ 2

Hence

2

For zero stress

−2 ⋅ r −

Ro − Ri

2⎞

⎟ ⎠

2

⎛ Ri ⎞ ln⎜ ⋅r ⎝ Ro ⎠

2

=0

r =

or

2 2⎞ ⎛ 2 Ro − Ri ⎜ Fo = ∆p⋅ π⋅ −Ro − ⎜ ⎛ Ri ⎞ ⎟ 2 ⋅ ln⎜ ⎜ ⎝ ⎝ Ro ⎠ ⎠

2

2

⎛ Ri ⎞ 2 ⋅ ln⎜ ⎝ Ro ⎠

⎛ Ro − Ri ⎞ 1 ∆p ⎜ Fo = τrx⋅ A = ⋅ ⋅ −2 ⋅ Ro − ⋅ 2 ⋅ π⋅ R o ⋅ L 4 L ⎜ ⎛ Ri ⎞ ⎟ ln⎜ ⋅ Ro ⎜ ⎝ ⎝ Ro ⎠ ⎠ 2

On the outer surface

Ri − Ro

r = 13.7⋅ mm

⎡ ⎢ ⎡⎣( 25⋅ mm) 2 − ( 5⋅ mm) 2⎤⎦ × ⎛⎜ 1 ⋅ m ⎞ 2 N 1 ⋅ m ⎢ 3 ⎝ 1000⋅ mm ⎠ ⎞ − Fo = 125 × 10 ⋅ × π × −⎛⎜ 25⋅ mm × ⎢⎝ 5 2 1000⋅ mm ⎠ m 2 ⋅ ln⎛⎜ ⎞ ⎢ ⎣ ⎝ 25 ⎠ Fo = −172 N

⎛ Ro − Ri ⎞ 1 ∆p ⎜ Fi = τrx⋅ A = ⋅ ⋅ −2 ⋅ Ri − ⋅ 2 ⋅ π⋅ R i ⋅ L 4 L ⎜ ⎛ Ri ⎞ ⎟ ln⎜ ⋅ Ri ⎜ ⎝ ⎝ Ro ⎠ ⎠ 2

On the inner surface

2

2 2⎞ ⎛ 2 Ro − Ri ⎜ Fi = ∆p⋅ π⋅ −Ri − ⎜ ⎛ Ri ⎞ ⎟ 2 ⋅ ln⎜ ⎜ ⎝ ⎝ Ro ⎠ ⎠

Hence

2 ⎡ 2 2⎤ ⎛ 1 ⋅ m ⎞ ⎢ ⎡ ( 25 ⋅ mm ) − ( 5 ⋅ mm ) × ⎣ ⎦ ⎜ 2 1⋅ m ⎞ ⎢ 3 N ⎝ 1000⋅ mm ⎠ Fi = 125 × 10 ⋅ × π × −⎛⎜ 5 ⋅ mm × − ⎢⎝ 5 2 1000⋅ mm ⎠ m 2 ⋅ ln⎛⎜ ⎞ ⎢ ⎣ ⎝ 25 ⎠

Fi = 63.4 N Note that

Fo − Fi = −236 N

and

∆p⋅ π⋅ ⎛ Ro − Ri ⎝ 2

2⎞

⎠ = 236 N

The net pressure force just balances the net viscous force!

Problem 2.55 Problem 2.52

[Difficulty: 3]

2.52

Given:

Data on flow through a tube with a filament

Find:

Whether no-slip is satisfied; location of zero stress;stress on tube and filament

Solution: V( r) =

The velocity profile is

Check the no-slip condition. When

r=

r=

d 2

⋅

∆p

2

ln⎛⎜

V⎛⎜

D⎞

=

⎝2⎠

1

⋅

∆p

16⋅ μ L

⋅

∆p

d⎞

⋅ ln⎛⎜

2

2

(

2

2

D −d

⋅ d −d −

2

2

2⋅ r ⎞ ⎞

⎝ d ⎠⎟ ⎠

⎛⎜ 16⋅ μ L ⎜ ⎜⎝ 1

⋅ ⎡⎣d − D + D − d

⎛⎜ 16⋅ μ L ⎜ ⎜⎝ 1

2

⎝ D⎠

2

V( d ) =

2

D −d

2

⋅ d − 4⋅ r −

D

V( D) =

When

⎛⎜ 16⋅ μ L ⎜ ⎜⎝ 1

∆p

2

2

⋅ d −D −

2

D −d ln⎛⎜

2

d⎞

⎝ D⎠

⋅ ln⎛⎜

D ⎞⎞

⎝ d ⎠⎟ ⎠

)⎦ = 0

2⎤

2

d ln⎛⎜ ⎞ ⎝ D⎠

⋅

⋅ ln⎛⎜

d ⎞⎞

⎝ d ⎠⎟ ⎠

=0

The no-slip condition is satisfied. The given data is

d = 1 ⋅ μm

The viscosity of SAE 10-30 oil at 100 oC is (Fig. A.2)

D = 20⋅ mm

∆p = 5 ⋅ kPa

− 2 N⋅ s

μ = 1 × 10

⋅

2

m

L = 10⋅ m

The plot looks like

Radial Position (mm)

10 8 6 4 2

0

0.25

0.5

0.75

1

Velocity (m/s)

du τrx = μ⋅ dr

For each, shear stress is given by

dV( r) d τrx = μ⋅ = μ⋅ dr dr

2 2 ⎡⎢ 1 ∆p ⎛⎜ 2 2 ⋅ r ⎞ ⎞⎤⎥ 2 D − d ⋅ ⋅ d − 4⋅ r − ⋅ ln⎛⎜ ⎢ 16⋅ μ L ⎜ ⎟⎥ d ⎝ Di ⎠ ⎥ ln⎛⎜ ⎞ ⎢⎣ ⎜⎝ ⎝ D⎠ ⎠⎦

1 ∆p ⎛⎜ D −d ⎞ τrx( r) = ⋅ ⋅ −8 ⋅ r − ⎟ d 16 L ⎜ ln⎛⎜ ⎞ ⋅ r ⎜⎝ ⎝ D⎠ ⎠ 2

2

−8 ⋅ r −

For the zero-stress point

D −d

2

2

d ln⎛⎜ ⎞ ⋅ r ⎝ D⎠

2

=0

or

r =

2

d −D

d 8 ⋅ ln⎛⎜ ⎞ ⎝ D⎠

r = 2.25⋅ mm

Radial Position (mm)

10

7.5

5

2.5

−3

−2

−1

0

1

2

3

4

Stress (Pa)

Using the stress formula

D τrx⎛⎜ ⎞ = −2.374 Pa 2

⎝ ⎠

d τrx⎛⎜ ⎞ = 2.524 ⋅ kPa 2

⎝ ⎠

Problem 2.53 (Difficulty: 2)

2.53 The lubricant has a kinematic viscosity of 2.8 × 10−5 piston is 6

𝑚 , 𝑠

𝑉=6

and 𝑆𝑆 of 0.92. If the mean velocity of the

approximately what is the power dissipated in the friction?

Given: The kinematic viscosity: 𝑣 = 2.8 × 10−5 𝑚 . 𝑠

𝑚2 𝑠

The configuration is shown in the figure.

𝑚2 .Specific 𝑠

gravity: 𝑆𝑆 = 0.92 .

Mean velocity:

Assumption: Linear velocity profile in the lubricant, negligible end effects. Find: Power 𝑃𝑓 dissipated in the friction.

Solution: Use Newton’s law of viscosity to relate the viscous shear stress to the velocities

The shear stress is given by Newton’s law of viscosity 𝜏=𝜇

The density of the lubricant is:

𝑑𝑑 ∆𝑢 = 𝜇 𝑑𝑑 ∆𝑟

𝜌 = 𝑆𝑆 ∙ 𝜌𝐻2 0 = 0.92 × 998

The dynamic viscosity of the lubricant is: 𝜇 = 𝑣𝑣 = 2.8 × 10−5

The drag force:

𝑚2 𝑘𝑘 𝑘𝑘 × 918 3 = 2.57 × 10−2 = 2.57 × 10−2 𝑃𝑃 ∙ 𝑠 𝑚 𝑚∙𝑠 𝑠 𝐹𝐷 = 𝜇𝜇

𝐹𝐷 = 2.57

× 10−2

𝑘𝑘 𝑘𝑘 = 918 3 3 𝑚 𝑚

∆𝑢 ∆𝑟

𝑚 𝑠 𝑃𝑃 ∙ 𝑠 × (𝜋 × 0.15 𝑚 × 0.3 𝑚) × = 218 𝑁 0.0001 𝑚 6

The power 𝑃𝑓 dissipated in the friction is the product of the force and velocity: 𝑃𝑓 = 𝐹𝐷 𝑉 = 218 𝑁 × 6

𝑚 = 1308 𝑊 𝑠

Problem 2.54 (Difficulty: 1)

2.54 Calculate the approximate viscosity of the oil.

Given: Velocity: 𝑉 = 0.6 Slope: 𝑠𝑠𝑠𝑠 =

5 . 13

𝑓𝑓 . 𝑠

Gravity: 𝑊 = 25 𝑙𝑙𝑙. Area: 2 𝑓𝑓 × 2𝑓𝑓. Gap: ℎ = 0.05 𝑖𝑖.

Assumption: Linear velocity profile in oil, negligible end effects. Find: Viscosity of the oil. Solution: Use Newton’s law of viscosity to find the relation between shear stress and velocity. The force balance equation is that the drag force equals the component of the weight along the surface: 𝐹𝐷 = 𝑊 ∙ 𝑠𝑠𝑠𝑠 =

The drag force is found using Newton’s law of viscosity 𝜏=𝜇

5 𝑊 13

𝑑𝑑 ∆𝑢 = 𝜇𝜇 𝑑𝑑 ∆𝑦

The drag force is then, where the velocity profile is assumed linear:

From the force balance

The viscosity of the oil is:

𝐹𝐷 = 𝜇𝜇 𝜇𝜇

∆𝑢 𝑉 = 𝜇𝜇 ∆𝑦 ℎ

𝑉 5 = 𝑊 ℎ 13

5 ℎ 5 𝜇= 𝑊∙ = × 25 𝑙𝑙𝑙 × 13 𝐴𝐴 13

0.05 𝑓𝑓 12

4 𝑓𝑓 2 × 0.6

𝑓𝑓 𝑠

= 0.0167

𝑙𝑙𝑙 ∙ 𝑠 𝑓𝑓 2

Problem 2.55 (Difficulty: 2)

2.55 Calculate the approximate power lost in friction in this ship propeller shaft bearing.

Given: Rotation speed: 200 Viscosity: 𝜇 = 0.72 𝑃𝑃 ∙ 𝑠.

𝑟 𝑚𝑚𝑚

. Gap: ℎ = 0.23 𝑚𝑚. Length: 𝐿 = 1 𝑚. Shaft diameter: 𝐷 = 0.36 𝑚.

Assumption: Linear velocity profile in fluid, negligible end effects.

Find: Power 𝑃𝑑 lost in friction.

Solution: Use Newton’s law of viscosity to relate the viscous force to the velocity 𝜏=𝜇

𝑑𝑑 ∆𝑢 = 𝜇 𝑑𝑑 ∆𝑟

The drag force is given by the product of the shear stress and area. For the linear velocity in the fluid: 𝐹𝐷 = 𝐴 𝜇

The velocity is given by:

So we have:

𝑉 = 𝜔𝜔 = 200

∆𝑢 𝑉 =𝐴𝜇 ∆𝑟 ℎ

𝑟 1 𝑚𝑚𝑚 𝑟𝑟𝑟 0.36 𝑚 × × �2𝜋 �× 𝑚 = 3.77 𝑚𝑚𝑚 60 𝑠 𝑟 2 𝑠

𝑚 𝑠 = 13340 𝑁 𝐹𝐷 = 0.72 𝑃𝑃 ∙ 𝑠 × (𝜋 × 0.36 𝑚 × 1 𝑚) × 0.00023 𝑚 3.77

The power 𝑃𝑑 lost in friction is the product of force and velocity: 𝑃𝑑 = 𝐹. 𝑉 = 13340 𝑁 × 3.77

𝑚 = 50.3 𝑘𝑘 𝑠

Problem 2.56 Problem 2.56

[Difficulty: 2]

2.56

Given:

Flow between two plates

Find:

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 V i

Then the force required is

(

Vi V − Vi μ1 ⋅ = μ2 ⋅ h1 h2

or

V

Vi = 1+

μ1 h 2 ⋅ μ2 h 1

1⋅ = 1+

)

where V i is the interface velocity

m s

0.1 0.3 ⋅ 0.15 0.5

m Vi = 0.714 s

Vi N⋅ s m 1 1000⋅ mm 2 F = τ⋅ A = μ1 ⋅ ⋅ A = 0.1⋅ × 0.714 ⋅ × × × 1⋅ m h1 2 s 0.5⋅ mm 1⋅ m m

F = 143 N

Problem 2.57 Problem 2.58 2.57

[Difficulty: 2]

Problem 2.58 Problem 2.60

2.58

[Difficulty: 2]

Problem 2.59 Problem 2.62 2.59

Difficulty: [2]

Problem 2.60 Problem 2.64

[Difficulty: 3]

2.60

Given: Shock-free coupling assembly Find:

Required viscosity

Solution: du τrθ = μ⋅ dr

Basic equation

Shear force

F = τ⋅ A

Assumptions: Newtonian fluid, linear velocity profile

τrθ = μ⋅

V1 = ω1R

P = T⋅ ω2 = F⋅ R⋅ ω2 = τ⋅ A2 ⋅ R⋅ ω2 = P=

Hence

(

P = T⋅ ω

Power

⎡⎣ω1⋅ R − ω2 ⋅ ( R + δ)⎤⎦ du ∆V τrθ = μ⋅ = μ⋅ = μ⋅ δ dr ∆r

V2 = ω2(R + δ)

δ

Then

Torque T = F⋅ R

)

(

)

μ⋅ ω1 − ω2 ⋅ R δ

(ω1 − ω2)⋅ R

Because δ << R

δ

⋅ 2 ⋅ π⋅ R⋅ L⋅ R⋅ ω2

3

2 ⋅ π⋅ μ⋅ ω2 ⋅ ω1 − ω2 ⋅ R ⋅ L δ P⋅ δ

μ=

(

)

3

2 ⋅ π⋅ ω2 ⋅ ω1 − ω2 ⋅ R ⋅ L μ =

10⋅ W × 2.5 × 10

μ = 0.202 ⋅

2⋅ π N⋅ s 2

m

−4

⋅m

×

1

⋅

min

9000 rev

μ = 2.02⋅ poise

×

1

⋅

min

1000 rev

×

1 ( .01⋅ m)

3

×

1 0.02⋅ m

×

N⋅ m s⋅ W

2

×

⎛ rev ⎞ × ⎛ 60⋅ s ⎞ ⎜ ⎜ ⎝ 2 ⋅ π⋅ rad ⎠ ⎝ min ⎠

which corresponds to SAE 30 oil at 30oC.

2

Problem 2.61 Problem 2.66 2.61

[Difficulty: 4]

Problem 2.62

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/m x10 ) 1.047 120 121 2.094 240 139 3.142 360 153 4.189 480 159 5.236 600 172 6.283 720 172 7.330 840 183 8.378 960 185 2

3

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) 1080 1200

η (N·s/m2x103) 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 2.70 Problem 2.63

[Difficulty: 3]

2.63

Given: Viscometer data Find:

Value of k and n in Eq. 2.17

Solution:

τ (Pa)

du/dy (s-1)

0.0457 0.119 0.241 0.375 0.634 1.06 1.46 1.78

5 10 25 50 100 200 300 400

Shear Stress vs Shear Strain 10

Data Power Trendline

τ (Pa)

The data is

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

The apparent viscosity from

Blood is pseudoplastic (shear thinning)

η =

du/dy (s-1) 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

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 2.64 Problem 2.72 2.64

[Difficulty: 5]

Problem 2.65 Problem 2.74 2.65

[Difficulty: 5]

Problem 2.76 Problem 2.66

[Difficulty: 5]

2.66

Geometry of rotating bearing

Given:

Expression for shear stress; Maximum shear stress; Expression for total torque; Total torque

Find: Solution:

τ = μ⋅

Basic equation

du

dT = r⋅ τ⋅ dA

dy

Assumptions: Newtonian fluid, narrow clearance gap, laminar motion From the figure

h = a + R⋅ ( 1 − cos( θ) )

dA = 2 ⋅ π⋅ r⋅ dr = 2 ⋅ π R⋅ sin( θ) ⋅ R⋅ cos( θ) ⋅ dθ

du

To find the maximum τ set

d ⎡ μ⋅ ω⋅ R⋅ sin( θ) ⎤ ⎢ ⎥=0 dθ ⎣ a + R⋅ ( 1 − cos( θ) ) ⎦

R⋅ μ⋅ ω⋅ ( R⋅ cos( θ) − R + a⋅ cos( θ) )

so

( R + a − R⋅ cos( θ) )

τ = 79.2⋅

2

⎞ = acos⎛ 75 ⎞ ⎜ ⎝ R + a⎠ ⎝ 75 + 0.5 ⎠

θ = acos⎛⎜

kg

poise

h

a + R⋅ ( 1 − cos( θ) )

R⋅ cos( θ) − R + a⋅ cos( θ) = 0 m⋅ s

h

u

=

μ⋅ ω⋅ R⋅ sin( θ)

τ = μ⋅

dy

=

dy

=

u−0

u = ω⋅ r = ω⋅ R⋅ sin( θ)

Then

τ = 12.5⋅ poise × 0.1⋅

du

r = R⋅ sin( θ)

R

=0

θ = 6.6⋅ deg 2

× 2 ⋅ π⋅

70 rad 1 N⋅ s ⋅ × 0.075 ⋅ m × sin( 6.6⋅ deg) × × 60 s [ 0.0005 + 0.075 ⋅ ( 1 − cos( 6.6⋅ deg) ) ] ⋅ m m⋅ kg

N 2

m

The torque is

⌠ T = ⎮ r⋅ τ⋅ A dθ = ⌡

θ ⌠ max 4 2 μ⋅ ω⋅ R ⋅ sin( θ) ⋅ cos( θ) ⎮ dθ ⎮ a + R⋅ ( 1 − cos( θ) ) ⌡ 0

wher e

This integral is best evaluated numerically using Excel, Mathcad, or a good calculator

⎛ R0 ⎞ θmax = asin⎜ ⎝R⎠

T = 1.02 × 10

−3

⋅ N⋅ m

θmax = 15.5⋅ deg

Problem 2.67 Problem 2.77 2.67

[Difficulty: 2]

Problem 2.68 Problem 2.78

[Difficulty: 2]

2.68

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 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 =

4

⋅ ρs⋅ L⋅ g

π⋅ SG ⋅ ρ⋅ g

θ = 0 ⋅ deg

m

8 ⋅ σ⋅ cos( θ) π⋅ ρs⋅ g

and for water

ρ = 1000⋅

kg 3

m

SG = 7.83

From Table A.1, for steel 8 ⋅ σ⋅ cos( θ)

⋅

D≤

or

−3 N

σ = 72.8 × 10

From Table A.4

Hence

π⋅ D

=

8 π⋅ 7.83

× 72.8 × 10

−3 N

⋅

m

3

×

m

999 ⋅ kg

2

×

s

9.81⋅ m

Hence D < 1.55 mm. Only the 1 mm needles float (needle length is irrelevant)

×

kg⋅ m 2

N⋅ s

−3

= 1.55 × 10

⋅ m = 1.55⋅ mm

Problem 2.69 Problem 2.79

[Difficulty: 3]

2.69

Given:

Caplillary rise data

Find:

Values of A and b

Solution: D (in.) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1

∆h (in.) 0.232 0.183 0.090 0.059 0.052 0.033 0.017 0.010 0.006 0.004 0.003

A = 0.403 b = 4.530 The fit is a good one (R2 = 0.9919)

Capillary Rise vs. Tube Diameter

∆h (in.)

0.3

∆h = 0.403e-4.5296D R2 = 0.9919

0.2

0.1

0.0 0.0

0.2

0.4

0.6 D (in.)

0.8

1.0

1.2

Problem 2.70 (Difficulty: 3)

2.70 Calculate and plot the maximum capillary rise of water at 20 C to be expected in a vertical glass tube as a function of tube diameters from 0.5 to 2.5 mm.

Given: Temperature: 𝑇 = 20 ℃. Diameter: 𝐷 𝑓𝑓𝑓𝑓 0.5 𝑡𝑡 2.5 𝑚𝑚.

Find: Maximum capillary rise ∆ℎ.

Solution: Use the relation for capillary force to find the rise. For the force balance on the water we have the capillary force and the weight of the volume of water: � 𝐹𝑧 = 𝜎𝐻2 𝑜 𝜋𝜋 cos 𝜃 − 𝜌𝐻2 𝑜 𝑔∆𝑉 = 0 1 ∆𝑉 = 𝜋𝐷 2 ∆ℎ 4

So we have: ∆ℎ =

4𝜎𝐻2 𝑜 𝜋𝜋 cos 𝜃 4𝜎𝐻2 𝑜 cos 𝜃 = 𝜌𝐻2𝑜 𝑔𝑔𝐷 2 𝜌𝐻2 𝑜 𝑔𝑔

When 𝜃 = 0, we have the maximum ∆ℎ for specific tube diameter . The surface tension for water is given by

The rise is then

For the range

𝜎𝐻2 𝑜 = 0.0728

𝑁 𝑚

𝑁 4 × 0.0728 × 1 4𝜎𝐻2 𝑜 cos 𝜃 2.97 × 10−5 𝑚 = = 𝑚 ∆ℎ = 𝑘𝑘 𝑚 𝐷 𝜌𝐻2 𝑜 𝑔𝐷 998 3 × 9.81 2 × 𝐷 𝑚 𝑠 0.0005 𝑚 ≤ 𝐷 ≤ 0.0025 𝑚

The plot of rise versus diameter is

Problem 2.71 (Difficulty: 2)

2.71 Calculate the maximum capillary rise of water at 20 ℃ to be expected between two vertical, clean glass plates spaced 1𝑚𝑚 apart.

Given: Temperature: 𝑇 = 20 ℃ . Distance between two plate: 𝐷 = 1 𝑚𝑚. Find: Maximum capillary rise ∆ℎ.

Solution: Use the relation for capillary force to find the rise The force balance equation equates the capillary force to the weight of the water: � 𝐹𝑧 = 𝜎𝐻2 𝑜 ∙ 2𝐿 cos 𝜃 − 𝜌𝐻2 𝑜 𝑔𝑔𝑔∆ℎ = 0

where L is the width of the plate. Solving for ∆h:

For the maximum capillary rise:

∆ℎ =

The surface tension for water is given by

𝜎𝐻2 𝑜 ∙ 2𝐿 cos 𝜃 2𝜎𝐻2 𝑜 cos 𝜃 = 𝜌𝐻2 𝑜 𝑔𝑔𝑔 𝜌𝐻2 𝑜 𝑔𝑔 𝜃=0 𝜎𝐻2 𝑜 = 0.0728

𝑁 𝑚

𝑁 2 × 0.0728 2𝜎𝐻2 𝑜 cos 𝜃 𝑚 ∆ℎ = = = 0.0149 𝑚 = 14.9 𝑚𝑚 𝑘𝑘 𝑚 𝜌𝐻2 𝑜 𝑔𝑔 998 3 × 9.8 2 × 0.001 𝑚 𝑚 𝑠

Problem 2.72 (Difficulty: 2)

2.72 Calculate the maximum capillary depression of mercury to be expected in the vertical glass tube 1 𝑚𝑚 in diameter at 15.5 ℃.

Given: Temperature: 𝑇 = 15.5 ℃ 𝑜𝑜 60℉ . Distance between two plate: 𝐷 = 1𝑚𝑚 𝑜𝑜 0.04 𝑖𝑖.

Find: Maximum capillary depression ∆ℎ.

Solution: Use the relation for capillary force to find the rise The force balance equation per width of the plate equates the capillary force to the weight of the water: � 𝐹𝑧 = 𝜎𝜎𝜎 cos 𝜃 − 𝜌𝜌∆𝑉 = 0

Where the volume is

1 ∆𝑉 = 𝜋𝐷 2 ∆ℎ 4

Solving for the depression:

For mercury, the surface tension is

∆ℎ =

4𝜎𝜎𝜎 cos 𝜃 4𝜎 cos 𝜃 = 𝜌𝜌𝜌𝐷 2 𝜌𝜌𝜌 𝜎 = 0.51

And the density is

For the maximum capillary depression:

𝑁 𝑚

𝛾 = 𝜌𝜌 = 133

𝑘𝑘 𝑚3

𝜃 = 130 ° for mercury.

The depression is ∆ℎ =

4𝜎 cos 𝜃 4 × 0.51 × cos(130 °) = 𝑚 = −9.86 𝑚𝑚 𝛾𝛾 133 × 1000 × 0.001

Problem 2.73 Problem 2.82 2.73

[Difficulty: 2]

Problem 2.74 Problem 2.84

[Difficulty: 2]

2.74

Given:

Boundary layer velocity profile in terms of constants a, b and c

Find:

Constants a, b and c

Solution: Basic equation

u = a + b ⋅ ⎛⎜

y⎞

+ c⋅ ⎛⎜

⎝δ⎠

y⎞

3

⎝δ⎠

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)

0=

dy

d dy

From 1 and 2

c=−

Hence

u=

Dimensionless Height

du

=0

a + b ⋅ ⎛⎜

y⎞

⎝δ⎠

U

2

⋅ ⎛⎜

y⎞

⎝δ⎠

−

U 2

y⎞

⎝δ⎠

b=

2

3⋅ U

+ c⋅ ⎛⎜ 3 2

⋅ ⎛⎜

3

=

b δ

+ 3 ⋅ c⋅

y

2

3

=

δ

b δ

+ 3⋅

c δ

⋅U

y⎞

⎝δ⎠

3

u U

=

3 2

⋅ ⎛⎜

y⎞

⎝δ⎠

−

1 2

⋅ ⎛⎜

y⎞

3

⎝δ⎠

1 0.75 0.5 0.25

0

0.25

0.5

Dimensionless Velocity

0.75

1

Problem 2.75 Problem 2.86

[Difficulty: 3]

2.75

Given:

Geometry of and flow rate through tapered nozzle

Find:

At which point becomes turbulent

Solution: Basic equation

Re =

For pipe flow (Section 2-6)

ρ⋅ V⋅ D μ

= 2300

for transition to turbulence

2

π⋅ D

Q=

Also flow rate Q is given by

4

⋅V

We can combine these equations and eliminate V to obtain an expression for Re in terms of D and Q Re =

ρ⋅ V⋅ D μ

=

ρ⋅ D 4 ⋅ Q 4 ⋅ Q⋅ ρ ⋅ = 2 μ π⋅ μ⋅ D π⋅ D

Re =

4 ⋅ Q⋅ ρ π⋅ μ⋅ D

For a given flow rate Q, as the diameter is reduced the Reynolds number increases (due to the velocity increasing with A -1 or D -2). Hence for turbulence (Re = 2300), solving for D

The nozzle is tapered:

Carbon tetrachloride:

Din = 50⋅ mm

μCT = 10

D=

4 ⋅ Q⋅ ρ 2300⋅ π⋅ μ

Dout =

− 3 N⋅ s

⋅

Din

Dout = 22.4⋅ mm

5

(Fig A.2)

For water

2

ρ = 1000⋅

3

m

m SG = 1.595

kg

ρCT = SG⋅ ρ

(Table A.2)

ρCT = 1595

kg 3

m For the given flow rate

Q = 2⋅

L

4 ⋅ Q⋅ ρCT

min

π⋅ μCT⋅ Din

For the diameter at which we reach turbulence

But

L = 250 ⋅ mm

D =

= 1354

4 ⋅ Q⋅ ρCT 2300⋅ π⋅ μCT

LAMINAR

4 ⋅ Q⋅ ρCT π⋅ μCT⋅ Dout

D − Din Dout − Din

Lturb = 186 ⋅ mm

TURBULENT

D = 29.4⋅ mm

and linear ratios leads to the distance from D in at which D = 29.4⋅ mm Lturb = L⋅

= 3027

Lturb L

=

D − Din Dout − Din

Problem 2.76 Problem 2.87

[Difficulty: 2]

2.76

Given:

Data on water tube

Find:

Reynolds number of flow; Temperature at which flow becomes turbulent

Solution: Basic equation

At 20oC, from Fig. A.3 ν = 9 × 10

For the heated pipe

Hence

Re =

For pipe flow (Section 2-6)

Re = ν=

V⋅ D ν V⋅ D

2300

2 −7 m

⋅

and so

s

= 2300 =

1 2300

ρ⋅ V⋅ D μ

Re = 0.25⋅

=

m s

V⋅ D ν

× 0.005 ⋅ m ×

9 × 10

for transition to turbulence

× 0.25⋅

m s

× 0.005 ⋅ m

From Fig. A.3, the temperature of water at this viscosity is approximately

ν = 5.435 × 10 T = 52⋅ C

1

2 −7m

s

−7

⋅

s 2

m

Re = 1389

Problem 2.77 Problem 2.88

[Difficulty: 3]

2.77

Given:

Data on supersonic aircraft

Find:

Mach number; Point at which boundary layer becomes turbulent

Solution: Basic equation

V = M⋅ c

Hence

M=

V c

c=

and

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

At 27 km the temperature is approximately (from Table A.3)

T = 223.5 ⋅ K 1 2

2 ⎞ ⋅ ⎛⎜ 1 × 1 ⋅ kg⋅ K × 1⋅ N⋅ s × 1 ⋅ 1 ⎞ M = 2.5 M = ⎛⎜ 2700 × 10 ⋅ × hr 3600⋅ s ⎠ ⎝ 1.4 286.9 N⋅ m 223.5 K ⎠ kg⋅ m ⎝ 3 m

For boundary layer transition, from Section 2-6 Then

Retrans =

ρ⋅ V⋅ x trans

1 ⋅ hr

Retrans = 500000 μ ⋅ Retrans

x trans =

so

μ

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

μ=

b ⋅T

2

1+

S

where

3

−6

b = 1.458 × 10

1

m⋅ s ⋅ K

3

S = 110.4 ⋅ K

2

− 5 N⋅s

− 5 kg

μ = 1.459 × 10

m⋅ s

⋅

2

m

− 5 kg

x trans = 1.459 × 10

kg m

kg

⋅

T

μ = 1.459 × 10

Hence

ρ = 0.0297

m

1

For µ

kg

ρ = 0.02422× 1.225⋅

At this altitude the density is (Table A.3)

⋅

m⋅ s

× 500000×

3

m 1 1 hr 3600⋅ s ⋅ × × ⋅ × 3 m 0.0297 kg 2700 1⋅ hr 10 1

x trans = 0.327m

Problem 2.78 Problem 2.89

[Difficulty: 2]

2.78

Given:

Type of oil, flow rate, and tube geometry

Find:

Whether flow is laminar or turbulent

Solution: ν=

Data on SAE 30 oil SG or density is limited in the Appendix. We can Google it or use the following

At 100 oC, from Figs. A.2 and A.3

− 3 N⋅ s

μ = 9 × 10

⋅

ν = 1 × 10

2

− 3 N⋅ s

⋅

2

1

×

1 × 10

m Hence

The specific weight is

SG =

ρ

−5

⋅

s 2

×

⋅

kg⋅ m

ρ = 900

2

γ = ρ⋅ g

γ = 900 ⋅

kg 3

2

Q=

π⋅ D 4

⋅V

V=

so

Then

Hence

V = Re =

4 π

10

3

⋅m

1 ⋅ mL

× 1.11 × 10

m 2

2

N⋅ s

×

3 N

γ = 8.829 × 10 ⋅

kg⋅ m

s

4⋅ Q 2

×

1 1 ⋅ 9 s

Q = 1.111 × 10

3 −5 m

2 1 1 1000⋅ mm ⎞ ⎛ ⋅ ×⎜ ⋅ × s 1⋅ m ⎠ ⎝ 12 mm

V = 0.0981

ρ⋅ V⋅ D μ

Re = 900 ⋅

kg 3

m

Flow is laminar

3

m

π⋅ D

−6

Q = 100 ⋅ mL ×

3

SG = 0.9

× 9.81⋅

m For pipe flow (Section 2-6)

kg m

kg ρwater = 1000⋅ 3 m

ρwater

ρ

s

s ⋅N

m

ρ=

so

2 −5 m

m ρ = 9 × 10

μ

× 0.0981⋅

m s

× 0.012 ⋅ m ×

1 9 × 10

2

⋅

m

− 3 N⋅ s

2

×

N⋅ s

kg⋅ m

Re = 118

m s

3 −5m

s

μ ν

Problem 2.79 Problem 2.90

[Difficulty: 2]

2.79

Given:

Data on seaplane

Find:

Transition point of boundary layer

Solution: For boundary layer transition, from Section 2-6

Retrans = 500000

Then

Retrans =

At 45oF = 7.2 oC (Fig A.3)

ρ⋅ V⋅ x trans μ

2 −5 m

ν = 0.8 × 10

⋅

s

V⋅ x trans

=

ν 10.8⋅

×

− 5 ft

⋅

ft

V

− 5 ft

s

ν = 8.64 × 10

m

⋅

2

s

s

2

⋅ 500000 ×

s

ν⋅ Retrans

2

2

1⋅

x trans = 8.64 × 10

x trans =

so

1 100 ⋅ mph

×

60⋅ mph 88⋅

x trans = 0.295 ⋅ ft

ft s

As the seaplane touches down:

At 45oF = 7.2 oC (Fig A.3)

2 −5 m

ν = 1.5 × 10

⋅

s

10.8⋅ ×

− 4 ft

⋅

2 − 4 ft

s

ν = 1.62 × 10

2

1⋅

x trans = 1.62 × 10

ft

m

2

s

s

2

s

⋅

⋅ 500000 ×

1 100 ⋅ mph

×

60⋅ mph 88⋅

ft s

x trans = 0.552 ⋅ ft

Problem 2.80 Problem 2.91

[Difficulty: 3]

2.80

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 Table appropriate At 5.5 km the temperature is approximately

252

c=

The speed of sound is obtained from where

k = 1.4 R = 286.9

J/kg·K

c = 318

m/s

V = 700

km/hr

V = 194

m/s

K

k ⋅ R ⋅T

(Table A.6)

We also have

or

Hence M = V/c or M = 0.611 V = M · c = 0.611·c

To compute V for constant M , we use

V = 677 At a height of 8 km: km/hr NOTE: Realistically, the aiplane will fly to a maximum height of about 10 km! T (K)

4

262

5

259

5

256

6

249

7

243

8

236

9

230

10

223

11

217

12

217

13

217

14

217

15

217

16

217

17

217

18

217

19

217

20

217

22

219

24

221

26

223

28

225

30

227

40

250

50

271

60

256

70

220

80

181

90

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

Speed vs. Altitude

713 709 750

704 695 686 677 668 658

700

649 649 649 649 649 649 649 649

Speed V (km/hr)

z (km)

650

649 649 651 654

600

657 660 663 697 725 705 653 592 592

550 0

20

40

60

Altitude z (km)

80

100