Hw1 Soln.pdf

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 2.6 (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:

[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 2.31 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 2.33 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 3.2 Problem 3.3

[Difficulty: 2]

3.2

Given:

Data on flight of airplane

Find:

Pressure change in mm Hg for ears to "pop"; descent distance from 8000 m to cause ears to "pop."

Solution: Assume the air density is approximately constant constant from 3000 m to 2900 m. From table A.3 ρSL = 1.225⋅

kg

ρair = 0.7423 ⋅ ρSL

3

m

ρair = 0.909

kg 3

m

We also have from the manometer equation, Eq. 3.7 Δp = −ρair ⋅ g ⋅ Δz Combining

ΔhHg =

ρair ρHg

ΔhHg =

⋅ Δz =

0.909 13.55 × 999

Δp = −ρHg ⋅ g ⋅ ΔhHg

and also ρair SGHg ⋅ ρH2O

SGHg = 13.55 from Table A.2

⋅ Δz

× 100 ⋅ m

ΔhHg = 6.72⋅ mm

For the ear popping descending from 8000 m, again assume the air density is approximately constant constant, this time at 8000 m. From table A.3

ρair = 0.4292 ⋅ ρSL

ρair = 0.526

kg 3

m We also have from the manometer equation ρair8000 ⋅ g ⋅ Δz8000 = ρair3000 ⋅ g ⋅ Δz3000 where the numerical subscripts refer to conditions at 3000m and 8000m. Hence Δz8000 =

ρair3000 ⋅ g ρair8000 ⋅ g

⋅ Δz3000 =

ρair3000 ρair8000

⋅ Δz3000

Δz8000 =

0.909 × 100 ⋅ m 0.526

Δz8000 = 173 m

Problem 3.8 (Difficulty: 1)

3.8 An open vessel contains carbon tetrachloride to a depth of 6 𝑓𝑓 and water on the carbon tetrachloride to a depth of 5 𝑓𝑓 . What is the pressure at the bottom of the vessel?

Given: Depth of carbon tetrachloride: ℎ𝑐 = 6 𝑓𝑓. Depth of water: ℎ𝑤 = 5 𝑓𝑓. Find: The gage pressure 𝑝 at the bottom of the vessel.

Assumption: The gage pressure for the liquid surface is zero. The fluid is incompressible. Solution: Use the hydrostatic pressure relation to detmine pressures in a fluid.

Governing equation: Hydrostatic pressure in a liquid, with z measured upward:

The density for the carbon tetrachloride is:

The density for the water is:

𝑑𝑑 = −𝜌 𝑔 = −𝛾 𝑑𝑑

𝜌𝑐 = 1.59 × 103 𝜌𝑤 = 1.0 × 103

𝑘𝑘 𝑠𝑠𝑠𝑠 = 3.09 3 𝑚 𝑓𝑓 3

𝑘𝑘 𝑠𝑠𝑠𝑠 = 1.940 3 𝑚 𝑓𝑓 3

Using the hydrostatic relation, the gage pressure 𝑝 at the bottom of the vessel is: 𝑝 = 3.09

𝑝 = 𝜌𝑐 𝑔ℎ𝑐 + 𝜌𝑤 𝑔ℎ𝑤

𝑠𝑠𝑠𝑠 𝑓𝑓 𝑠𝑠𝑠𝑠 𝑓𝑓 𝑙𝑙𝑙 × 32.2 2 × 6 𝑓𝑓 + 1.940 × 32.2 2 × 5 𝑓𝑓 = 909 2 = 6.25 𝑝𝑝𝑝 3 3 𝑓𝑓 𝑠 𝑓𝑓 𝑠 𝑓𝑓

Problem 3.20 (Difficulty: 1)

3.20 With the manometer reading as shown, calculate 𝑝𝑥 .

Given: Oil specific gravity: 𝑆𝑆𝑜𝑜𝑜 = 0.85 Depth: ℎ1 = 60 𝑖𝑖𝑖ℎ. ℎ2 = 30 𝑖𝑖𝑖ℎ.

Find: The pressure 𝑝𝑥 .

Assumption: Fluids are incompressible Solution: Use the hydrostatic relation to find the pressures in the fluid

Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 = −𝜌 𝑔 = −𝛾 𝑑𝑑

Integrating with respect to z for an incompressible fluid, we have the relation for the pressure difference over a difference in elevation (h):

Repeated application of this relation yields

The specific weight for mercury is:

The pressure at the desired location is 𝑝𝑥 = 0.85 × 62.4

∆𝑝 = 𝜌𝜌ℎ

𝑝𝑥 = 𝑆𝑆𝑜𝑜𝑜 𝛾𝑤𝑤𝑤𝑤𝑤 ℎ1 + 𝛾𝑀 ℎ2 𝛾𝑀 = 845

𝑙𝑙𝑙 𝑓𝑓 3

𝑙𝑙𝑙 60 𝑙𝑙𝑙 30 𝑙𝑙𝑙 × � � 𝑓𝑓 + 845 3 × � � 𝑓𝑓 = 2380 2 = 16.5 𝑝𝑝𝑝 3 𝑓𝑓 12 𝑓𝑓 12 𝑓𝑓

Problem 3.42 (Difficulty: 2)

3.42 A circular gate 3 𝑚 in diameter has its center 2.5 𝑚 below a water surface and lies in a plane sloping at 60°. Calculate magnitude, direction and location of total force on the gate. Find: The direction, magnitude of the total force 𝐹. Assumptions: Fluid is static and incompressible

Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the surface of the liquid: 𝑑𝑑 = 𝜌𝑔=𝛾 𝑑𝑑 𝐹𝑅 = � 𝑝 𝑑𝑑

For the magnitude of the force we have:

A free body diagram of the gate is

𝑦 ′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑 𝐹 = � 𝑝𝑝𝑝 𝐴

The pressure on the gate is the pressure at the centroid, which is yc = 2.5 m. So the force can be calculated as: 𝐹 = 𝜌𝜌ℎ𝑐 𝐴 = 999

𝑘𝑘 𝑚 𝜋 × 9.81 2 × 2.5 𝑚 × × (3 𝑚)2 = 173200 𝑁 = 173.2 𝑘𝑘 3 𝑚 𝑠 4

The direction is perpendicular to the gate.

For the location of the force we have: 𝑦 ′ = 𝑦𝑐 +

𝐼𝑥�𝑥� 𝐴𝑦𝑐

The y axis is along the plate so the distance to the centroid is: 𝑦𝑐 =

The area moment of inertia is

The area is

𝐼𝑥�𝑥� = 𝐴=

So 𝑦 ′ = 2.89 𝑚 +

2.5 𝑚 = 2.89 𝑚 sin 60°

𝜋𝐷 4 𝜋 = × (3 𝑚)4 = 3.976 𝑚4 64 64 𝜋 2 𝜋 𝐷 = × (3 𝑚)2 = 7.07 𝑚2 4 4

3.976 𝑚4 = 2.89 𝑚 + 0.1946 𝑚 = 3.08 𝑚 7.07 𝑚2 × 2.89 𝑚

The vertical location on the plate is

ℎ′ = 𝑦 ′ sin 60° = 3.08 𝑚 ×

The force acts on the point which has the depth of 2.67 𝑚.

√3 = 2.67 𝑚 2

Problem 3.67 Problem 3.83

[Difficulty: 2]

3.67

Given:

Canoe floating in a pond

Find:

What happens when an anchor with too short of a line is thrown from canoe

Solution: Governing equation:

FB = ρ w gVdisp = W Before the anchor is thrown from the canoe the buoyant force on the canoe balances out the weight of the canoe and anchor:

FB1 = Wcanoe + Wanchor = ρ w gVcanoe1 The anchor weight can be expressed as

Wanchor = ρ a gVa

so the initial volume displaced by the canoe can be written as

Vcanoe1 =

Wcanoe ρ a + Va ρw g ρw

After throwing the anchor out of the canoe there will be buoyant forces acting on the canoe and the anchor. Combined, these buoyant forces balance the canoe weight and anchor weight:

FB2 = Wcanoe + Wanchor = ρ w gVcanoe2 + ρ w gVa

Vcanoe 2 =

Wcanoe Wa + − Va ρw g ρw g

Vcanoe 2 =

Wcanoe ρ a Va − Va + ρw g ρw

Using the anchor weight,

Hence the volume displaced by the canoe after throwing the anchor in is less than when the anchor was in the canoe, meaning that the canoe is floating higher.