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 = �⃗ 𝜌𝜋𝑉 𝜇 �⃗ 𝑄 = 𝐴𝑉 �⃗ 𝑚̇ = 𝜌𝐴𝑉 𝐴= Then Also 𝑅𝑟 = 𝜋𝜋 2 4 �⃗ 𝜌𝜋 𝑄 4𝜌𝜋 𝑄 4𝜌𝑄 4𝑄 𝜌𝜋𝑉 = = = = 2 𝜇 𝐴 𝜇 𝜋𝜋 𝜇𝜋𝜋 𝜐𝜋𝜋 𝜇 𝑅𝑟 = �⃗ 𝐴 4𝜋 𝑚̇ 4𝑚̇ 𝜋 𝜌𝑉 = = 2 𝜇 𝜋𝜋 𝜋𝜇𝜋 𝜇 𝐴 We have the following equation from above: 𝑄= 𝜐𝜋𝜋𝑅𝑟 4 For the same flow rate in section with different channel diameter, 𝜋1 𝑅𝑟1 = 𝜋2 𝑅𝑟2 𝑅𝑟2 = 𝜋1 10 𝑚𝑚 𝑅𝑟1 = × 1800 = 3000 6 𝑚𝑚 𝜋2 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 8.2 It is NOT free (Difficulty 2) 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 𝜕𝑠𝑚𝑟 and 100 ℉ ? If the pressure is raised to 60 𝜕𝑠𝑚𝑟, 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 = 𝑉𝜌𝜋 𝑉𝜋 = 𝜈 𝜇 𝑚 = 𝜌̇ 𝐴 𝑉 𝜌= We have 𝜕 𝑅𝑇 𝜕 = 30 𝜕𝑠𝑚𝑟 = 4320 𝑅 = 1716 𝑙𝑢𝑙 ∙ 𝑙𝜕 𝑠𝑙𝑢𝜌 ∙ °𝑅 𝑇 = 100 ℉ = 560 °𝑅 Thus the density is 𝜕 𝜌= = 𝑅𝑇 𝑙𝑢𝑙 𝑙𝜕 2 𝑙𝑢𝑙 𝑠𝑙𝑢𝜌 𝑙𝑢𝑙 ∙ 𝑠2 𝑙𝜕 2 = 0.0045 = 0.0045 𝑙𝑢𝑙 ∙ 𝑙𝜕 𝑙𝜕 3 𝑙𝜕 4 1716 × 560 °𝑅 𝑠𝑙𝑢𝜌 ∙ °𝑅 4320 For the maximum laminar flow we have the Reynolds number at the critical value: For this situation 𝑅𝑟𝑐𝑐𝑖𝑎 = 2300 𝜋 = 4 𝑚𝑚 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.. 𝑙𝑢𝑙 ∙ 𝑠 𝜇 = 3.94 × 10−7 It is NOT free 2 𝑙𝜕 The maximum velocity is then: The cross section area is: −7 𝑙𝑢𝑙 ∙ 𝑠 × 2300 𝜇𝑅𝑟 3.94 × 10 𝑙𝜕 𝑙𝜕 2 𝑉= = = 0.605 2 𝑙𝑢𝑙 ∙ 𝑠 4 𝜌𝜋 𝑠 � 𝑙𝜕� × 0.0045 12 𝑙𝜕 4 𝐴= The maximum flow rate: 𝑚̇ = 𝜌𝑉𝐴 = 0.0045 𝜋𝜋 2 4 = 𝜋×� 2 4 𝑙𝜕� 12 = 0.0873 𝑙𝜕 2 4 𝑙𝜕 𝑠𝑙𝑢𝜌 𝑠𝑙𝑢𝜌 × 0.605 × 0.0873 𝑙𝜕 2 = 2.38 × 10−4 3 𝑠 𝑠 𝑙𝜕 If the pressure is raised up to 60 𝜕𝑠𝑚 = 8640 𝑙𝑙𝑙 , 𝑙𝑎 2 the density of the air will become: 𝑙𝑢𝑙 𝑠𝑙𝑢𝜌 𝑙𝑢𝑙 ∙ 𝑠2 𝑙𝜕 2 𝜌= = 0.009 = 0.009 𝑙𝑢𝑙 ∙ 𝑙𝜕 𝑙𝜕 3 𝑙𝜕 4 1716 × 560 °𝑅 𝑠𝑙𝑢𝜌 ∙ °𝑅 8640 The maximum velocity in this case is: −7 𝑙𝑢𝑙 ∙ 𝑠 × 2300 𝜇𝑅𝑟 3.94 × 10 𝑙𝜕 𝑙𝜕 2 𝑉= = = 0.302 2 𝑙𝑢𝑙 ∙ 𝑠 4 𝜌𝜋 𝑠 � 𝑙𝜕� × 0.009 12 𝑙𝜕 4 And the maximum flow rate: 𝑚̇ = 𝜌𝑉𝐴 = 0.009 𝑠𝑙𝑢𝜌 𝑙𝜕 𝑠𝑙𝑢𝜌 × 0.302 × 0.0873 𝑙𝜕 2 = 2.38 × 10−4 3 𝑙𝜕 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 = 𝜌𝑉𝐴 = 𝜇𝑅𝑟𝐴 𝜇𝑅𝑟 𝜌𝐴 = 𝜋 𝜌𝜋 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 𝑙𝑢𝑙 ∙ 𝑠 𝑙𝜕 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 𝑙𝑢𝑙 ∙ 𝑠 𝑠𝑙𝑢𝜌 4.49 × 10 𝑠𝑙𝑢𝜌 𝑙𝜕 2 × = 2.71 × 10−4 𝑙𝑢𝑙 ∙ 𝑠 𝑠 𝑠 3.94 × 10−7 𝑙𝜕 2 The reason for the increase is that the viscosity is a function of temperature. 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 8.5 It is NOT free Problem 8.3 (Difficulty: 2) 8.38.5 An incompressible fluid flows between two infinite stationary parallel plates. The velocity profile is given by 𝑢 = 𝑢𝑚𝑚𝑠 (𝐴𝑦 2 + 𝐵𝑦 + 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 �⃗ �𝑢𝑚𝑚𝑠 . 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 = 𝑢𝑚𝑚𝑠 𝑢(0) = 0 = 𝑢𝑚𝑚𝑠 𝐶 From B.C (2), 𝐶=0 From B.C (3), 𝑢(ℎ) = 0 = 𝑢𝑚𝑚𝑠 (𝐴ℎ2 + 𝐵ℎ) Thus ℎ2 ℎ ℎ 𝑢 � � = 𝑢𝑚𝑚𝑠 = 𝑢𝑚𝑚𝑠 �𝐴 + 𝐵 � 2 2 4 𝐴=− 4 ℎ2 𝐵 = −𝐴ℎ = 4 ℎ 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.. Then It is NOT free We have: 𝑢 = 𝑢𝑚𝑚𝑠 (𝐴𝑦 2 + 𝐵𝑦 + 𝐶) = 𝑢𝑚𝑚𝑠 �−4 ℎ 𝑦 𝑦 𝑦 2 𝑦2 + 4 � = 4𝑢𝑚𝑚𝑠 � − � � � 2 ℎ ℎ ℎ ℎ ℎ ℎ 𝑦 𝑦 2 𝑦2 𝑦3 𝑄 = � 𝑢𝑢𝑢𝑦 = � 4𝑢𝑚𝑚𝑠 � − � � � 𝑢𝑢𝑦 = 4𝑢𝑚𝑚𝑠 𝑢 � − 2 � ℎ ℎ 2ℎ 3ℎ 0 0 0 2 ℎ ℎ 𝑄 = 4𝑢𝑚𝑚𝑠 𝑢 � − � = 𝑢𝑚𝑚𝑠 𝑢ℎ 3 2 3 Since 𝑄 2 = 𝑢 ℎ 𝑢 3 𝑚𝑚𝑠 �⃗ 𝐴 = 𝑉 �⃗ 𝑢ℎ 𝑄=𝑉 So we have for the average velocity: 2 𝑄 �⃗ ℎ = 𝑢𝑚𝑚𝑠 ℎ =𝑉 3 𝑢 �⃗ 𝑉 𝑢𝑚𝑚𝑠 = 2 3 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 8.11 It is NOT free Problem 8.4 (Difficulty: 3) 8.11 A hydraulic jack supports a load of 9000 𝑘𝜌. 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 𝑟3 ∆𝜕 = 𝑙 12𝜇𝐿 where 𝑙 = 𝜋𝜋. From Fig. A.2 at 𝑇 = 30 ℃, 𝜇 = 3.0 × 10−1 𝑁∙𝑠 . 𝑚2 ∆𝜕 = 𝜕1 − 𝜕𝑚𝑎𝑚 𝜕1 = 𝜕1 = 𝑊 𝑚𝜌 4𝑚𝜌 = = 𝐴 𝜋𝜋 2 𝐴 4 × 9000 𝑘𝜌 × 9.81 𝜋 × (0.1 𝑚)2 𝑚 𝑠 2 = 11.2 𝑀𝑃𝑟 𝜋 3 6 𝑁 −5 𝑚3 𝜋𝜋𝑟3 ∆𝜕 12 × (0.1 𝑚) × (5 × 10 𝑚) × 11.2 × 10 𝑚2 = = 1.01 × 10−6 𝑄= 𝑁∙𝑠 12𝜇𝐿 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 �⃗ 𝑟 𝑉 �⃗ 𝑟 𝜌𝑉 = 𝜇 𝜐 𝜐 = 2.8 × 10−4 𝑚2 𝑠 𝑚3 1.01 × 10−6 𝑄 𝑄 𝑚 𝑠 �⃗ = = = = 0.0643 𝑉 −5 𝐴 𝑙𝑟 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 𝜋𝜋 2 𝑈 4 𝑚3 4 × 1.01 × 10−6 4𝑄 𝑠 = 1.29 × 10−4 𝑚 = 𝑈= 2 2 𝑠 𝜋𝜋 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. 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 8.13 It is NOT free Problem 8.5 (Difficulty 2) 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 𝑃𝑟 ∙ 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 𝑟2 𝜕𝜕 𝑦 2 𝑦 � � �� � − � �� 𝑟 𝑟 2𝜇 𝜕𝑑 2𝑦 1 𝑢𝑢 𝑟2 𝜕𝜕 = � � �� 2 � − � �� 𝑟 𝑟 𝑢𝑦 2𝜇 𝜕𝑑 In this particular case: 𝑟 = 0.3 𝑚 For the velocity at the midpoint we have: 𝑦 = 0.15 𝑚 (0.3 𝑚)2 𝜕𝜕 𝑦 2 𝑦 0.15 𝑚 2 0.15 𝑚 𝑚 𝑟2 𝜕𝜕 � � �� � − � �� = � � �� � −� �� = 2.7 𝑉𝑐 = 𝑟 𝑟 𝜕𝑑 0.3 𝑚 0.3 𝑚 𝑠 2𝜇 2𝜇 𝜕𝑑 Thus the pressure gradient is 1 𝜕𝜕 � �= 2𝜇 𝜕𝑑 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� = − � �𝑟 = − 𝑠 12𝜇 𝜕𝑑 6 The volumetric flow rate for width 𝑢 = 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: 1 2𝑦 1 1 1 𝑢𝑢 𝑟2 𝜕𝜕 = � � �� 2 � − � �� = (0.3 𝑚)2 × �−120 � �− � �� = 36 𝑚∙𝑠 𝑟 𝑟 0.3 𝑚 𝑠 𝑢𝑦 2𝜇 𝜕𝑑 At 𝑦 = 0.3 𝑚 1 2𝑦 1 2 × 0.3𝑚 1 1 𝑢𝑢 𝑟2 𝜕𝜕 = � � �� 2 � − � �� = (0.3 𝑚)2 × �−120 � �� � − � �� = −36 (0.3 𝑚)2 𝑚∙𝑠 𝑟 𝑟 0.3𝑚 𝑠 𝑢𝑦 2𝜇 𝜕𝑑 (c) For the shear stress of the wall we have: 𝑢𝑢 1 𝜏𝑤 = 𝜇 = 1.44 𝑃𝑟 ∙ 𝑠 × 36 = 51.8 𝑃𝑟 𝑢𝑦 𝑠 (d) As the viscosity we have: 𝜇 = 1.44 𝑃𝑟 ∙ 𝑠 Thus 𝜕𝜕 1 𝑃𝑟 � � = �−120 � × 2 × 1.44 𝑃𝑟 ∙ 𝑠 = −346 𝜕𝑑 𝑚∙𝑠 𝑚 For the length we have is: The pressure drop is: ∆𝑑 = 30 𝑚 𝜕𝜕 𝑃𝑟 ∇𝜕 = � � ∆𝑑 = −346 × 30 𝑚 = −10380 𝑃𝑟 = −10.38 𝑘𝑃𝑟 𝜕𝑑 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 ⎞ h 2 Basic equation ⋅ ⎡ dp 2⋅ μ dx ⋅ ⎢1 − ⎣ 2 ⎛ y ⎞ ⎥⎤ ⎜h ⎝ ⎠⎦ (from Eq. 8.7) dp ⎜ 2 = −y⋅ dx ⎝ h ⎠ 1⋅ m 3 N τyx = −1.5⋅ mm × × 1.25 × 10 ⋅ 2 1000⋅ mm m ⋅m h −h 3 ⎡ ⎢1 − ⎣ 2 ⎛ y ⎞ ⎥⎤ dy ⎜h ⎝ ⎠⎦ 3 Q= − 2 1⋅ m ⎞ m 3 N = − × ⎛⎜ 1.5⋅ mm × × × 1.25 × 10 ⋅ 1000⋅ mm ⎠ 0.5⋅ N⋅ s b 3 ⎝ 2 m ⋅m Q 2 τyx = −1.88Pa 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 𝜏𝑦𝑠 = 𝜇 Then and the torque is Solving for the viscosity 𝜏𝑦𝑠 = 𝜇 𝑈 𝜇𝜇𝑟𝑖 = ∆𝑟 ∆𝑟 𝑇 = 𝑟𝑖 �2𝜋𝑟𝑖 𝐿𝜏𝑦𝑠 � = 2𝜋𝑟𝑖2 𝐿𝜏𝑦𝑠 𝜇= 𝜇= 𝑢𝑢 𝑢𝑦 ∆𝑟𝑇 2𝜋𝜇𝑟𝑖3 𝐿 2𝜋𝜇𝜇𝑟𝑖3 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 𝑟𝑟𝑢 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 ⎛ ∂ dx dx ⎞ ⎡ ∂ dy ⎛ ∂ dy ⎞⎤ ∂ ⋅ 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 = 𝑢𝑚𝑚𝑠 at 𝑦 = ℎ, And 𝑢= 𝜌𝜌 sin 𝜃 𝑦2 𝜌 sin 𝜃 𝑦2 �ℎ𝑦 − � = �ℎ𝑦 − � 𝜇 𝜐 2 2 𝑢𝑚𝑚𝑠 = 𝜌 sin 𝜃 2 ℎ2 𝜌 ℎ2 sin 𝜃 �ℎ − � = 𝜐 2𝜐 2 𝑚 2 −3 𝑚2 𝜌 ℎ2 sin 𝜃 9.81 𝑠2 × (0.8 × 10 𝑚) × sin 30° −4 = = 1 × 10 𝜐= 𝑚 2𝑢𝑚𝑚𝑠 𝑠 2 × �15.7 × 10−3 � 𝑠 𝑢 The plot is shown as: 𝑢𝑚𝑚𝑠 = �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 the spacing between the plates is a = 0.1 in. (a) Assume the air is incompressible and determine the flow rate per unit depth for the case of zero pressure gradient and the shear stress distribution across the channel. (b) Determine the magnitude and direction of the pressure gradient that will give zero shear stress at y=0:25a, and determine the magnitude and direction of the shear stress at both surfaces. Flow between parallel plates Given: Find: Shear stress on lower plate; Plot shear stress; Flow rate for pressure gradient; Pressure gradient for zero shear; Plot Solution: u(y) = From Section 8-2 U⋅ y a + ⎡ y 2 ⎞ − 2 ⋅ μ dx ⎣⎝ a ⎠ a 2 ⋅ dp ⋅ ⎢⎛⎜ y⎤ ⎥ a⎦ 3 ft a u = U⋅ For dp/dx = 0 a ⌠ ⌠ U⋅ a y = ⎮ u ( y ) dy = w⋅ ⎮ U⋅ dy = ⎮ ⌡ 2 l a 0 ⌡ y Q a 1 Q = 2 × 5⋅ ft × s 0.1 12 ⋅ ft Q = 0.0208⋅ s ft 0 τ = μ⋅ For the shear stress du dy = μ⋅ U − 7 lbf ⋅ s μ = 3.79 × 10 when dp/dx = 0 a ⋅ ft (Table A.9) 2 The shear stress is constant - no need to plot! τ = 3.79 × 10 − 7 lbf ⋅ s ⋅ ft 2 × 5⋅ ft s × 12 0.1⋅ ft × ⎛ 1⋅ ft ⎞ ⎜ 12⋅ in ⎝ ⎠ 2 −6 τ = 1.58 × 10 Q will decrease if dp/dx > 0; it will increase if dp/dx < 0. τ = μ⋅ For non- zero dp/dx: du dy = μ⋅ U a + a⋅ τ( y = 0.25⋅ a) = μ⋅ At y = 0.25a, we get U a dp dx ⋅ ⎛⎜ + a⋅ y − ⎝a dp dx ⋅ ⎛⎜ 1⎞ 2⎠ 1 ⎝4 − 1⎞ 2⎠ = μ⋅ U a − a dp ⋅ 4 dx lbf Hence this stress is zero when dp dx = 4 ⋅ μ⋅ U a 2 − 7 lbf ⋅ s = 4 × 3.79 × 10 ⋅ ft 2 × 5⋅ ft s × 2 2 ⎛ 12 ⎞ = 0.109 ⋅ ft = 7.58 × 10− 4 psi ⎜ 0.1⋅ ft ft ft ⎝ ⎠ 0.1 y (in) 0.075 0.05 0.025 −4 − 1× 10 0 −4 1× 10 Shear Stress (lbf/ft3) −4 2× 10 −4 3× 10 ⋅ psi 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 8.31 It is NOT free Problem 8.11 (Difficulty: 3) 8.11 8.31 A continuous belt, passing upward through a chemical bath at speed 𝑈0 , picks up a liquid film of thickness ℎ, density 𝜌, and viscosity 𝜇. Gravity tends to make the liquid drain down, but the movement of the belt keeps the liquid from running off completely. Assume that the flow is fully developed and laminar with zero pressure gradient, and that the atmosphere produces no shear stress at the outer surface of the film. State clearly the boundary conditions to be satisfied by the velocity at 𝑦 = 0 and 𝑦 = ℎ. Obtain an expression for the velocity profile. Assumptions: (1) 𝐹𝑠𝑠 due to shear forces only (2) steady flow (3) Fully-developed flow Solution: Apply x component of momentum equation. 𝐹𝑠𝑠 + 𝐹𝐵𝑠 = TCoose CV 𝑢𝑑𝑢𝑦𝑢𝑑 as shown. Then Or 𝜕 �⃗ ∙ 𝑢𝐴⃗ � 𝑢𝜌𝑢∀ + � 𝑢𝜌𝑉 𝜕𝜕 𝐶𝐶 𝐶𝐶 𝜏𝑦𝑠 = 𝜇 𝐹𝑠𝑠 + 𝐹𝐵𝑠 = 𝐹1 − 𝐹2 + 𝐹𝐵𝑠 = �𝜏 + 𝑢𝑢 =𝜏 𝑢𝑦 𝑢𝜏 𝑢𝑦 𝑢𝜏 𝑢𝑦 � 𝑢𝑑𝑢𝑑 − �𝜏 − � 𝑢𝑑𝑢𝑑 − 𝜌𝜌𝑢𝑑𝑢𝑦𝑢𝑑 𝑢𝑦 2 𝑢𝑦 2 𝑢𝜏 = 𝜌𝜌 𝑢𝑦 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.. Integrating, It is NOT free 𝜏 = 𝜌𝜌𝑦 + 𝑐1 = 𝜇 𝑢𝑢 𝜌𝜌𝑦 𝑐1 = + 𝜇 𝑢𝑦 𝜇 Integrating again, 𝑢𝑢 𝑢𝑦 𝜌𝜌𝑦 2 𝑐1 + 𝑦 + 𝑐2 𝑢= 2𝜇 𝜇 To evaluate the constants 𝑐1 and 𝑐2 , apply the boundary conditions: At 𝑦 = 0, 𝑢 = 𝑈0 , so 𝑐2 = 𝑈0 At 𝑦 = ℎ, 𝜏 = 0, so 𝑢𝑢 =0 𝑢𝑦 𝑐1 = −𝜌𝜌ℎ Substituting, 𝑢= Note that at 𝑦 = ℎ, 𝑢= 𝑢= 𝜌𝜌𝑦 2 𝜌𝜌ℎ − 𝑦 + 𝑈0 𝜇 2𝜇 𝜌𝜌 𝑦 2 � − ℎ𝑦� + 𝑈0 𝜇 2 ℎ2 𝜌𝜌 �− � + 𝑈0 ≠ 0 𝜇 2 Thus the solution is determined only when 𝑈0 and ℎ are known. 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.. If you are interested in the instructor solution manual and / or PDF ebook It is NOT free Order it Now! Contact email: markrainsun(@)gmail(.)com Please send your request via e-mail.. This service is NOT free
