Instructor Solutions Manual Fox and McDonald's Introduction to Fluid Mechanics 10th Edition by John W. Mitchell PDF

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If you are interested in the instructor solution manual and / or PDF ebook Order it Now! Contact email: markrainsun(@)gmail(.)com Please send your request via e-mail.. It is NOT free Problem 8.1 (Difficulty: 2) 8.1 Consider For incompressible flow incompressible in in flow a circular a circulartube,...

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If you are interested in the instructor solution manual and / or PDF ebook Order it Now! Contact email: markrainsun(@)gmail(.)com Please send your request via e-mail.. Problem 8.1 It is NOT free (Difficulty: 2)

8.1 Consider For incompressible flow in in a circular derive expressions for Reynolds terms incompressible flow a circulartube, channel. Derive general expressions for number Reynoldsinnumber of (a) volume flow rate and tube diameter and (b) mass flow rate and tube diameter. in terms of (a) volume flow rate and tube diameter and (b) mass flow rate and tube diameter. The Reynolds number is 1800 in a section where the tube diameter is 10 𝑚𝑚. Find the Reynolds number for the same flow rate in a section where the tube diameter is 6 𝑚𝑚.

Assumptions: steady, incompressible flow Solution: Use the continuity equation and the basic definitions: 𝑅𝑅 =

�⃗ 𝜌𝜌𝑉 𝜇

�⃗ 𝑄 = 𝐴𝑉

�⃗ 𝑚̇ = 𝜌𝜌𝑉 𝐴=

Then

Also

𝑅𝑅 =

𝜋𝐷 2 4

�⃗ 𝜌𝜌 𝑄 4𝜌𝜌 𝑄 4𝜌𝜌 4𝑄 𝜌𝜌𝑉 = = = = 2 𝜇 𝐴 𝜇 𝜋𝐷 𝜇𝜇𝜇 𝜐𝜐𝜐 𝜇 𝑅𝑅 =

�⃗ 𝐴 4𝐷 𝑚̇ 4𝑚̇ 𝐷 𝜌𝑉 = = 2 𝜇 𝜋𝐷 𝜋𝜋𝜋 𝜇 𝐴

We have the following equation from above:

𝑄=

𝜐𝜐𝜐𝜐𝜐 4

For the same flow rate in section with different channel diameter, 𝐷1 𝑅𝑅1 = 𝐷2 𝑅𝑅2 𝑅𝑅2 =

𝐷1 10 𝑚𝑚 𝑅𝑅1 = × 1800 = 3000 6 𝑚𝑚 𝐷2

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8.2 maximum ofthat air at laminar conditions in a 4-in.-diameter pipe atpipe an at 8.2Determine What is thethe maximum flowflow raterate of air may occur at laminar condition in a 4 in diameter absolute pressure of 30 psia and 100oF. Determine the maximum flow rate when (a) the pressure an absolute pressure of 30 𝑝𝑝𝑝𝑝 and 100 ℉ ? If the pressure is raised to 60 𝑝𝑝𝑝𝑝, what is the maximum isflow raised psia, and (b) theis temperature is raised 200oF. Describe the reason rateto? 60 If the temperature raise to 200 ℉, what istothe maximum flow rate? Explainfor thethe differences differences in flow rates in terms of the physical mechanisms involved. in answers in terms of the physical mechanisms involved. Find: The maximum flow rate for laminar flow. Assumption: Air behaves as an ideal gas Solution: The basic equations are the definition of Reynolds number, the continuity expression, and the ideal gas law 𝑅𝑅 =

𝑉𝑉𝑉 𝑉𝑉 = 𝜈 𝜇

𝑚 = 𝜌̇ 𝐴 𝑉 𝜌=

We have

𝑝 𝑅𝑅

𝑝 = 30 𝑝𝑝𝑝𝑝 = 4320 𝑅 = 1716

𝑙𝑙𝑙 ∙ 𝑓𝑓 𝑠𝑠𝑠𝑠 ∙ °𝑅

𝑇 = 100 ℉ = 560 °𝑅

Thus the density is 𝑝 𝜌= = 𝑅𝑅

𝑙𝑙𝑙 𝑓𝑓 2

𝑙𝑙𝑙 𝑠𝑠𝑠𝑠 𝑙𝑙𝑙 ∙ 𝑠2 𝑓𝑓 2 = 0.0045 = 0.0045 𝑙𝑙𝑙 ∙ 𝑓𝑓 𝑓𝑓 3 𝑓𝑓 4 1716 × 560 °𝑅 𝑠𝑠𝑠𝑠 ∙ °𝑅 4320

For the maximum laminar flow we have the Reynolds number at the critical value:

For this situation

𝑅𝑅𝑐𝑐𝑐𝑐 = 2300 𝐷 = 4 𝑖𝑖

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The maximum velocity is then:

The cross section area is:

−7 𝑙𝑙𝑙 ∙ 𝑠 × 2300 𝜇𝜇𝜇 3.94 × 10 𝑓𝑓 𝑓𝑓 2 𝑉= = = 0.605 2 𝑙𝑙𝑙 ∙ 𝑠 4 𝜌𝜌 𝑠 � 𝑓𝑓� × 0.0045 12 𝑓𝑓 4

𝐴=

The maximum flow rate:

𝑚̇ = 𝜌𝜌𝜌 = 0.0045

𝜋𝐷 2 4

=

𝜋×�

2 4 𝑓𝑓� 12 = 0.0873 𝑓𝑓 2 4

𝑓𝑓 𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠 × 0.605 × 0.0873 𝑓𝑓 2 = 2.38 × 10−4 3 𝑠 𝑠 𝑓𝑓

If the pressure is raised up to 60 𝑝𝑝𝑝 = 8640

𝑙𝑙𝑙 , 𝑓𝑓 2

the density of the air will become:

𝑙𝑙𝑙 𝑠𝑠𝑠𝑠 𝑙𝑏𝑏 ∙ 𝑠2 𝑓𝑓 2 𝜌= = 0.009 = 0.009 𝑙𝑙𝑙 ∙ 𝑓𝑓 𝑓𝑓 3 𝑓𝑓 4 1716 × 560 °𝑅 𝑠𝑠𝑠𝑠 ∙ °𝑅 8640

The maximum velocity in this case is:

−7 𝑙𝑙𝑙 ∙ 𝑠 × 2300 𝜇𝜇𝜇 3.94 × 10 𝑓𝑓 𝑓𝑓 2 𝑉= = = 0.302 2 𝑙𝑙𝑙 ∙ 𝑠 4 𝜌𝜌 𝑠 � 𝑓𝑓� × 0.009 12 𝑓𝑓 4

And the maximum flow rate:

𝑚̇ = 𝜌𝜌𝜌 = 0.009

𝑠𝑠𝑠𝑠 𝑓𝑓 𝑠𝑠𝑠𝑠 × 0.302 × 0.0873 𝑓𝑓 2 = 2.38 × 10−4 3 𝑓𝑓 𝑠 𝑠

The maximum flow rate is the same. The reason is that the density cancels out of the flow rate using Reynolds number: 𝑚̇ = 𝜌𝜌𝜌 =

𝜇𝜇𝜇𝜇 𝜇𝜇𝜇 𝜌𝜌 = 𝐷 𝜌𝜌

The pressure will not change the viscosity 𝜇 (as an assumption)

When the temperature is raised to 200℉, the viscosity decreases. The density also decreases, but we have seen that this has no effect. The viscosity is 𝜇 = 4.49 × 10−7

𝑙𝑙𝑙 ∙ 𝑠 𝑓𝑓 2

If you are interested in the instructor solution manual and / or PDF ebook Order it Now! Contact email: markrainsun(@)gmail(.)com Please send your request via e-mail.. Thus the flow rate is the ratio of the previous flow rate times the ratio of viscosities It is NOT free 𝑚̇ = 2.38 × 10−4

−7 𝑙𝑙𝑙 ∙ 𝑠 𝑠𝑠𝑠𝑠 4.49 × 10 𝑠𝑠𝑠𝑠 𝑓𝑓 2 × = 2.71 × 10−4 𝑙𝑙𝑙 ∙ 𝑠 𝑠 𝑠 3.94 × 10−7 𝑓𝑓 2

The reason for the increase is that the viscosity is a function of temperature.

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8.38.5 An incompressible fluid flows between two infinite stationary parallel plates. The velocity profile is given by 𝑢 = 𝑢𝑚𝑚𝑚 (𝐴𝑦 2 + 𝐵𝐵 + 𝐶), where 𝐴, 𝐵, 𝐶 are constants and 𝑦 is measured upward from the lower plate. The total gap width is ℎ units. Use appropriate boundary conditions to express the magnitude and units of the constants in terms of ℎ. Develop an expression for volume flow rate per unit depth and �⃗ �𝑢𝑚𝑚𝑚 . evaluate the ratio 𝑉

Assumptions: The flow is steady and the fluid id incompressible Solution: Use the continuity expression and the form of the velocity profile. The boundary conditions are:

From B.C (1),

(1) 𝑦 = 0, 𝑢 = 0 (2) 𝑦 = ℎ, 𝑢 = 0 (3) 𝑦 = ℎ⁄2 , 𝑢 = 𝑢𝑚𝑚𝑚 𝑢(0) = 0 = 𝑢𝑚𝑚𝑚 𝐶

From B.C (2),

𝐶=0

From B.C (3),

𝑢(ℎ) = 0 = 𝑢𝑚𝑚𝑚 (𝐴ℎ2 + 𝐵ℎ)

Thus

ℎ2 ℎ ℎ 𝑢 � � = 𝑢𝑚𝑚𝑚 = 𝑢𝑚𝑚𝑚 �𝐴 + 𝐵 � 2 2 4 𝐴=−

4 ℎ2

𝐵 = −𝐴ℎ =

4 ℎ

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We have:

𝑢 = 𝑢𝑚𝑚𝑚 (𝐴𝑦 2 + 𝐵𝐵 + 𝐶) = 𝑢𝑚𝑚𝑚 �−4 ℎ

𝑦2 𝑦 𝑦 𝑦 2 + 4 � = 4𝑢𝑚𝑚𝑚 � − � � � 2 ℎ ℎ ℎ ℎ ℎ

ℎ

𝑦 𝑦 2 𝑦2 𝑦3 𝑄 = � 𝑢𝑢𝑢𝑢 = � 4𝑢𝑚𝑚𝑚 � − � � � 𝑏𝑏𝑏 = 4𝑢𝑚𝑚𝑚 𝑏 � − 2 � ℎ ℎ 2ℎ 3ℎ 0 0 0 ℎ ℎ 2 𝑄 = 4𝑢𝑚𝑚𝑚 𝑏 � − � = 𝑢𝑚𝑚𝑚 𝑏ℎ 2 3 3

Since

𝑄 2 = 𝑢 ℎ 𝑏 3 𝑚𝑚𝑚

�⃗ 𝐴 = 𝑉 �⃗ 𝑏ℎ 𝑄=𝑉

So we have for the average velocity:

2 𝑄 �⃗ ℎ = 𝑢𝑚𝑚𝑚 ℎ =𝑉 3 𝑏 �⃗ 𝑉

𝑢𝑚𝑚𝑚

=

2 3

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8.11 A hydraulic jack supports a load of 9000 𝑘𝑘. The following data are given: 8.4 Diameter of piston

Radial clearance between piston and cylinder

100 𝑚𝑚

0.05 𝑚𝑚

120 𝑚𝑚

Length of piston

Estimate the rate of leakage of hydraulic fluid past the piston, assuming the fluid is SAE 30 oil at 30 ℃. Assumptions: The flow as steady, fully developed laminar flow between stationary parallel plates.

Solution: Use the expression for flow between parallel plates: 𝑄 𝑎3 ∆𝑝 = 𝑙 12𝜇𝜇

where 𝑙 = 𝜋𝜋.

From Fig. A.2 at 𝑇 = 30 ℃, 𝜇 = 3.0 × 10−1

𝑁∙𝑠 . 𝑚2

∆𝑝 = 𝑝1 − 𝑝𝑎𝑎𝑎

𝑝1 =

𝑝1 =

𝑊 𝑚𝑚 4𝑚𝑚 = = 𝐴 𝐴 𝜋𝐷 2

4 × 9000 𝑘𝑘 × 9.81 𝜋 × (0.1 𝑚)2

𝑚 𝑠 2 = 11.2 𝑀𝑀𝑀

𝜋 −5 3 6 𝑁 𝑚3 𝜋𝜋𝑎3 ∆𝑝 12 × (0.1 𝑚) × (5 × 10 𝑚) × 11.2 × 10 𝑚2 = = 1.01 × 10−6 𝑄= 𝑁∙𝑠 12𝜇𝜇 𝑠 3.0 × 10−1 × 0.12 𝑚 𝑚2

If you are interested in the instructor solution manual and / or PDF ebook Order it Now! Contact email: markrainsun(@)gmail(.)com Please send your request via e-mail.. The leakage flow is It is NOT free 𝑄 = 1.01 × 10−3

Check the Reynolds number

𝑅𝑅 =

𝐿 𝑠

�⃗ 𝑎 𝑉 �⃗ 𝑎 𝜌𝑉 = 𝜇 𝜐

𝜐 = 2.8 × 10−4

𝑚2 𝑠

𝑚3 1.01 × 10−6 𝑄 𝑄 𝑚 𝑠 �⃗ = = = = 0.0643 𝑉 −5 𝐴 𝑙𝑙 𝜋 × 0.1 𝑚 × 5 × 10 𝑚 𝑠 𝑅𝑅 =

𝑚 × 5 × 10−5 𝑚 𝑠 = 0.011 𝑚2 2.8 × 10−4 𝑠

0.0643

The flow is definitely laminar.

Check whether we can neglect the motion of the piston. It moves down at speed 𝑈 and displaces liquid at rate 𝑄 where: 𝑄=

Since

𝜋𝐷 2 𝑈 4

𝑚3 4 × 1.01 × 10−6 4𝑄 𝑠 = 1.29 × 10−4 𝑚 = 𝑈= 2 2 𝑠 𝜋𝐷 𝜋 × (0.1 𝑚) −4 𝑚 𝑈 1.29 × 10 𝑠 = 𝑚 = 0.002 �𝑉⃗ 0.0643 𝑠

So the motion of piston can be neglected.

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8.5 A horizontal laminar flow occurs between two infinite parallel plates that are 0.3 m apart. The velocity at the centerline is 2.7 m/s. Determine the flow 𝑚 rate through a section 0.9 m wide, the the velocity at theatmidpoint between plates 2.7 shearing . Calculate (a) the rate through cross velocity gradient the surface of thethe plate, theiswall stress, and flow the pressure drop afor a 𝑠 30 length. The fluid viscosity is 1.44 N s/m2. section 0.9 𝑚 wide, (b) the velocity gradient at the surface of the plate, (c) the wall shearing stress if the

8.13 When a horizontal laminar flow occurs between two parallel plates of infinite extent 0.3 𝑚 apart, fluid has viscosity 1.44 𝑃𝑃 ∙ 𝑠, (d) the pressure drop in each 30 𝑚 along the flow.

Assumptions Flow is steady, fully established, and incompressible.

Solution: Use the expressions for the velocity profile for laminar flow between parallel plate. For this laminar flow we have the velocity profile in terms of position and pressure gradient as: 𝑢=

The velocity gradient is

𝑎2 𝜕𝜕 𝑦 2 𝑦 � � �� � − � �� 𝑎 𝑎 2𝜇 𝜕𝜕

2𝑦 1 𝑑𝑑 𝑎2 𝜕𝜕 = � � �� 2 � − � �� 𝑎 𝑎 𝑑𝑑 2𝜇 𝜕𝜕

In this particular case:

𝑎 = 0.3 𝑚

For the velocity at the midpoint we have:

𝑦 = 0.15 𝑚

(0.3 𝑚)2 𝜕𝜕 𝑎2 𝜕𝜕 𝑦 2 𝑦 0.15 𝑚 2 0.15 𝑚 𝑚 � � �� � − � �� = � � �� � −� �� = 2.7 𝑉𝑐 = 𝑎 𝑎 𝜕𝜕 0.3 𝑚 0.3 𝑚 𝑠 2𝜇 𝜕𝜕 2𝜇

Thus the pressure gradient is

1 𝜕𝜕 � �= 2𝜇 𝜕𝜕

𝑚 1 𝑠 = −120 2 𝑚∙𝑠 0.15 𝑚 0.15 𝑚 (0.3 𝑚)2 × �� � −� �� 0.3 𝑚 0.3 𝑚 2.7

(a) The average velocity is then:

1 −120 1 𝜕𝜕 2 𝑚 ∙ 𝑠 × (0.3 𝑚)2 = 1.8 𝑚 𝑉� = − � �𝑎 = − 12𝜇 𝜕𝜕 𝑠 6 The volumetric flow rate for width 𝑑 = 0.9 𝑚 is: 𝑄 = 𝑉�𝐴 = 1.8

𝑚 𝑚3 × 0.3 𝑚 × 0.9 𝑚 = 0.486 𝑠 𝑠

If you are interested in the instructor solution manual and / or PDF ebook Order it Now! Contact email: markrainsun(@)gmail(.)com Please send your request via e-mail.. (b) The velocity gradient at the surface of the plate 𝑦 = 0 or 𝑦 = 0.3 𝑚. It is NOT free At 𝑦 = 0: 2𝑦 1 1 1 1 𝑑𝑑 𝑎2 𝜕𝜕 = � � �� 2 � − � �� = (0.3 𝑚)2 × �−120 � �− � �� = 36 𝑎 𝑎 𝑚∙𝑠 0.3 𝑚 𝑠 𝑑𝑑 2𝜇 𝜕𝜕

At 𝑦 = 0.3 𝑚

2𝑦 1 1 2 × 0.3𝑚 1 1 𝑑𝑑 𝑎2 𝜕𝜕 = � � �� 2 � − � �� = (0.3 𝑚)2 × �−120 � �� � − � �� = −36 (0.3 𝑚)2 𝑎 𝑎 𝑚∙𝑠 0.3𝑚 𝑠 𝑑𝑑 2𝜇 𝜕𝜕

(c) For the shear stress of the wall we have: 𝑑𝑑 1 𝜏𝑤 = 𝜇 = 1.44 𝑃𝑃 ∙ 𝑠 × 36 = 51.8 𝑃𝑃 𝑑𝑑 𝑠 (d) As the viscosity we have:

𝜇 = 1.44 𝑃𝑃 ∙ 𝑠

Thus

𝜕𝜕 1 𝑃𝑃 � � = �−120 � × 2 × 1.44 𝑃𝑃 ∙ 𝑠 = −346 𝜕𝜕 𝑚∙𝑠 𝑚

For the length we have is:

The pressure drop is:

∆𝑥 = 30 𝑚

𝜕𝜕 𝑃𝑃 ∇𝑝 = � � ∆𝑥 = −346 × 30 𝑚 = −10380 𝑃𝑃 = −10.38 𝑘𝑘𝑘 𝜕𝜕 𝑚

If you are interested in the instructor solution manual and / or PDF ebook Order it Now! [Difficulty: 2] Problem Problem 8.11 8.6 Contact email: markrainsun(@)gmail(.)com Please send your request via e-mail.. fully developed and laminar flow of oil occurs It is 8.6 NOTA free between parallel plates. The pressure gradient creating the y 2h flow is 1:25 kPa/m of length and the channel half-width is x 1.5 mm. Calculate the magnitude and direction of the wall shear stress at the upper plate surface. Find the volume flow rate through the channel. The viscosity is 0.50 N s/m2. Laminar flow between flat plates Given: Find:

Shear stress on upper plate; Volume flow rate per width

Solution: du τyx = μ⋅ dy

Then

τyx =

At the upper surface

y=h

The volume flow rate is

⌠ h 2 ⌠ ⌠ h ⋅ b dp ⎮ Q = ⎮ u dA = ⎮ u ⋅ b dy = − ⋅ ⋅⎮ ⌡ 2⋅ μ dx ⎮ ⌡ −h ⌡

−h 2

2

⋅

u(y) = −

dp dx

⋅⎛−

2⋅ y ⎞

⎜ 2 ⎝ h ⎠

= −y ⋅

h

2

Basic equation

⋅

⎡

dp

2⋅ μ dx

⋅ ⎢1 −

⎣

b

=−

(from Eq. 8.7)

dp dx

1⋅ m 3 N τyx = −1.5⋅ mm × × 1.25 × 10 ⋅ 2 1000⋅ mm m ⋅m h

−h

Q

2 ⎛ y ⎞ ⎤⎥ ⎜ ⎝h⎠ ⎦

2 3

× ⎛⎜ 1.5⋅ mm ×

⎝

1⋅ m

3

⎡ ⎢1 − ⎣

2⎤ ⎛ y ⎞ ⎥ dy ⎜ ⎝h⎠ ⎦

τyx = −1.88Pa

3

Q= −

2

⎞ × 1.25 × 103⋅ N × m 2 0.5⋅ N⋅ s 1000⋅ mm ⎠ m ⋅m

Q b

2⋅ h ⋅ b dp ⋅ 3⋅ μ dx

= −5.63 × 10

2 −6 m

s

If you are interested in the instructor solution manual and / or PDF ebook Order it Now! Contact email: markrainsun(@)gmail(.)com Please send your request via e-mail.. Problem 8.16 It is NOT free Problem 8.7 (Difficulty: 2)

8.78.16 A sealed journal bearing is formed from concentric cylinders. The inner and outer radii are 25 and

26 mm, the journal length is 100 mm, and it turns at 2800 rpm. The gap is filled with oil in laminar motion. The velocity profile is linear across the gap. The torque needed to turn the journal is 0.2 𝑁 ∙ 𝑚. Calculate the viscosity of the oil. Will the why torque or decrease time?or Why? Explain theincrease torque will increase, with decrease, stay the same with time.

Assumptions: Linear velocity profile in the gap and the flow is laminar Solution: Since the gap is small the flow is that between parallel plates. Apply Newton’s law of viscosity.

Newton’s law of viscosity is 𝜏𝑦𝑦 = 𝜇

Then

and the torque is

Solving for the viscosity

𝜏𝑦𝑦 = 𝜇

𝑈 𝜇𝜇𝑟𝑖 = ∆𝑟 ∆𝑟

𝑇 = 𝑟𝑖 �2𝜋𝑟𝑖 𝐿𝜏𝑦𝑦 � =

2𝜋𝑟𝑖2 𝐿𝜏𝑦𝑦

𝜇=

𝜇=

𝑑𝑑 𝑑𝑑

∆𝑟𝑟 2𝜋𝜋𝑟𝑖3 𝐿

2𝜋𝜋𝜋𝑟𝑖3 𝐿 = ∆𝑟

𝑚𝑚𝑚 1 1 𝑟𝑟𝑟 𝑠 1 × 0.001 𝑚 × 0.2 𝑁 ∙ 𝑚 × × × × × 60 2800 𝑟𝑟𝑟 (0.025 𝑚)3 0.1 𝑚 2𝜋 𝑟𝑟𝑟 𝑚𝑚𝑚 2𝜋

If you are interested in the instructor solution manual and / or PDF ebook Order it Now! Contact email: markrainsun(@)gmail(.)com Please send your request via e-mail.. 𝑁∙𝑠 𝜇 = 0.0695 It is NOT free 2 𝑚

Because th bearing is sealed, so oil temperature will increase as energy is dissipated by friction. For liquids, 𝜇 decreases T increases. Thus torque will decreases, since it is propotional to 𝜇.

If you are interested in the instructor solution manual and / or PDF ebook Order it Now! [Difficulty: 3] Problem 8.25 Problem 8.8 Contact email: markrainsun(@)gmail(.)com Please send your request via e-mail.. immiscible fluids are of equal thickness are contained between infinite parallel plates It is 8.8 NOTTwo free separated by a distance 2h. The lower plate is stationary and the upper plate moves at constant speed of 20 ft/s. The dynamic viscosity of the upper fluid is three times that of the lower fluid. The flow is laminar and the pressure gradient in the direction of flow is zero.

Given:

Laminar flow of two fluids between plates

Find:

Velocity at the interface

Solution: Using the analysis of Section 8.2, the sum of forces in the x direction is

⎡ ∂ dy ⎛ ∂ dy ⎞⎤ ⎛ ∂ dx dx ⎞ ∂ ⋅ b ⋅ dy = 0 ⎢τ + τ ⋅ − ⎜ τ − τ ⋅ ⎥ ⋅ b ⋅ dx + ⎜ p − p ⋅ − p + p ⋅ ∂x 2 ⎠ ⎣ ∂y 2 ⎝ ∂y 2 ⎠⎦ ⎝ ∂x 2 Simplifying

dτ dy

=

dp dx

2

=0

μ⋅

or

dy

y=0

u1 = 0

2

=0

u 1 = c1 ⋅ y + c2

Applying this to fluid 1 (lower fluid) and fluid 2 (upper fluid), integrating twice yields We need four BCs. Three are obvious

d u

y = h u1 = u2

y = 2⋅ h

u 2 = c3 ⋅ y + c4

u2 = U

The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same y=h

du1 du2 μ1⋅ = μ2⋅ dy dy

Using these four BCs

0 = c2

c1⋅ h + c2 = c3⋅ h + c4

Hence

c2 = 0

From the 2nd and 3rd equations

c1⋅ h − U = −c3⋅ h

Hence

μ1 c1⋅ h − U = −c3⋅ h = − ⋅ h ⋅ c1 μ2

and

Hence for fluid 1 (we do not need to complete the analysis for fluid 2)

20⋅ Evaluating this at y = h, where u 1 = u interface

u interface =

ft s

⎛1 + 1 ⎞ ⎜ 3⎠ ⎝

U = c3⋅ 2⋅ h + c4

μ1 ⋅ c1 = μ2 ⋅ c3 c1 =

U

⎛

μ1 ⎞

⎝

μ2 ⎠

h⋅ ⎜ 1 +

u1 =

U

⎛ h ⋅⎜1 + ⎝

μ1 ⎞ μ2 ⎠

u interface = 15⋅

ft s

⋅y

μ1⋅ c1 = μ2⋅ c3

If you are interested in the instructor solution manual and / or PDF ebook Order it Now! Contact email: markrainsun(@)gmail(.)com Please send your request via e-mail.. Problem 8.22 It is NOT free Problem 8.9

(Difficulty: 2)

8.9 An incompressible viscous liquid flows steadily down an incline due to gravity. The flow is 8.22 Consider steady, incompressible, and fully developed laminar flow of a viscous liquid down an laminar and the velocity profile, derived in Example 5.9, is shown below, where q is the slope of incline with no pressure gradient. The velocity profile was derived in Example 5.9. Plot the velocity the incline and h is the thickness of the film. The fluid kinematic viscosity is 1x104 m2/s, the profile. Calculate the kinematic viscosity of the liquid if the film thickness on a 30° slope is 0.8 𝑚𝑚 and slope is 30o, and the film thickness is 0.8mm. Determine the maximum velocity and the flow rate the maximum velocity is 15.7 𝑚𝑚⁄𝑠. per meter of width. Solution:

𝑢 = 𝑢𝑚𝑚𝑚 at 𝑦 = ℎ,

And

𝑢=

𝜌𝜌 sin 𝜃 𝑦2 𝑔 sin 𝜃 𝑦2 �ℎ𝑦 − � = �ℎ𝑦 − � 𝜇 𝜐 2 2

𝑢𝑚𝑚𝑚 =

𝑔 sin 𝜃 2 ℎ2 𝑔 ℎ2 sin 𝜃 �ℎ − � = 𝜐 2𝜐 2

𝑚 −3 2 𝑚2 𝑔 ℎ2 sin 𝜃 9.81 𝑠2 × (0.8 × 10 𝑚) × sin 30° −4 = = 1 × 10 𝜐= 𝑚 2𝑢𝑚𝑚𝑚 𝑠 2 × �15.7 × 10−3 � 𝑠 𝑢

The plot is shown as:

𝑢𝑚𝑚𝑚

=

�ℎ𝑦 −

ℎ2 2

𝑦2 � 2

𝑦 𝑦 2 =2 −� � ℎ ℎ

If you are interested in the instructor solution manual and / or PDF ebook Order it Now! Problem 8.10 [Difficulty: 3] Problem 8.32 Contact email: markrainsun(@)gmail(.)com Please send your request via e-mail.. It is8.10 NOT There freeis a fully developed laminar flow of air between parallel plates. The upper plate moves at 5 ft/s and ...