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Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Lecture 1 Introduction to Flow machines and Advanced Fluid Dynamics Dr. Mohammed Rabie Assistant Professor – Mansoura Universit...

Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Lecture 1 Introduction to Flow machines and Advanced Fluid Dynamics Dr. Mohammed Rabie Assistant Professor – Mansoura University  Activity Score Quizzes & Attend.. Grading system Section Project Midterm Final Course Outlines Introduction Control Volume Analysis Mass conservation Linear momentum conservation Angular momentum conservation Energy conservation Differential analysis Mass conservation momentum conservation Energy conservation Boundary layer Potential flow Introduction to flow machines What is Fluid? A material that continuously deforms (moves or changes its shape), when subject to a given stress. What is Fluid Mechanics? The study of different effects of forces on the behavior of fluids at rest (fluid statics) or in motion (fluid dynamics). Importance of Fluid Mechanics? Classification of Fluid Flow  Advanced Fluid Dynamics Reynolds Transport Theorem  Reynolds Transport Theorem CS B → Any system property 𝑑𝐵 → For Example: 𝑑𝑚 If B is mass, B = m,  =  If B is momentum, B = mv,  = v And so on….  System ➔ we can apply laws of mechanics CV (Control volume) ➔ Specific region of interest (to study) RTT ➔ Is used to convert system analysis to control volume analysis  ➔ Density ➔ Volume V ➔ Velocity A ➔ Area  Flow rate Volume flow rate (m3/s) Mass flow rate (kg/s) When average velocity is used: Q =AV 12 kg/s Initial mass in the tank = m1 = 20 kg Mass flow rate inlet: 𝐦ሶ 𝐢 = 𝟏𝟐 𝐤𝐠/𝐬 Mass flow rate outlet: 𝐦ሶ 𝐨 = 𝟏𝟎 𝐤𝐠/𝐬 After 5 sec (t = 5 s) 20 kg 10 kg/s Final mass in the tank = m2 = ?? kg 10 kg/s Initial mass in the tank = m1 = 20 kg Mass flow rate inlet: 𝐦ሶ 𝐢 = 𝟏𝟎 𝐤𝐠/𝐬 Mass flow rate outlet: 𝐦ሶ 𝐨 = 𝟏𝟐 𝐤𝐠/𝐬 After 5 sec (t = 5 s) 20 kg 12 kg/s Final mass in the tank = m2 = ?? kg 12 kg/s Initial mass in the tank = m1 = 20 kg Mass flow rate inlet: 𝐦ሶ 𝐢 = 𝟏𝟐 𝐤𝐠/𝐬 Mass flow rate outlet: 𝐦ሶ 𝐨 = 𝟏𝟐 𝐤𝐠/𝐬 After 5 sec (t = 5 s) 20 kg 12 kg/s Final mass in the tank = m2 = ?? kg 12 kg/s Initial mass in the tank = m1 = 20 kg Mass flow rate inlet: 𝐦ሶ 𝐢 = 𝟏𝟐 𝐤𝐠/𝐬 Mass flow rate outlet: 𝐦ሶ 𝐨 = 𝟏𝟎 𝐤𝐠/𝐬 After 5 sec (t = 5 s) 20 kg 10 kg/s Final mass in the tank = m2 = ?? kg 𝑑𝑚 = 𝑚ሶ 𝑖 − 𝑚ሶ 𝑜 𝑑𝑡 𝑚2 − 𝑚1 = 𝑚ሶ 𝑖 − 𝑚ሶ 𝑜 Δ𝑡 𝑚2 − 20 = 12 − 10 𝐦𝟐 = 𝟑𝟎 𝐤𝐠 5 10 kg/s Initial mass in the tank = m1 = 20 kg Mass flow rate inlet: 𝐦ሶ 𝐢 = 𝟏𝟎 𝐤𝐠/𝐬 Mass flow rate outlet: 𝐦ሶ 𝐨 = 𝟏𝟐 𝐤𝐠/𝐬 After 5 sec (t = 5 s) 20 kg 12 kg/s Final mass in the tank = m2 = ?? kg 𝑑𝑚 = 𝑚ሶ 𝑖 − 𝑚ሶ 𝑜 𝑑𝑡 𝑚2 − 𝑚1 = 𝑚ሶ 𝑖 − 𝑚ሶ 𝑜 Δ𝑡 𝑚2 − 20 = 10 − 12 𝐦𝟐 = 𝟏𝟎 𝐤𝐠 5 12 kg/s Initial mass in the tank = m1 = 20 kg Mass flow rate inlet: 𝐦ሶ 𝐢 = 𝟏𝟐 𝐤𝐠/𝐬 Mass flow rate outlet: 𝐦ሶ 𝐨 = 𝟏𝟐 𝐤𝐠/𝐬 After 5 sec (t = 5 s) 20 kg 12 kg/s Final mass in the tank = m2 = ?? kg 𝑑𝑚 = 𝑚ሶ 𝑖 − 𝑚ሶ 𝑜 𝑑𝑡 𝑚2 − 𝑚1 = 𝑚ሶ 𝑖 − 𝑚ሶ 𝑜 Δ𝑡 𝑚2 − 20 = 12 − 12 𝐦𝟐 = 𝟐𝟎 𝐤𝐠 5 1. Conservation of Mass B is mass, ⸫ B = m,  =  ➔ Mass conservation ➔ If the control volume has a definite number of inlets and outlets: ➔ ➔ If the flow within the control volume is steady: ➔0 ➔ If the control volume has a definite number of inlets and outlets:  ⅆ𝑚 + න 𝜌 𝑣. 𝑛 𝑑𝐴 = 0 ⅆ𝑡 𝑐𝑠 ⅆ𝑣 Incompressible Fluid, deformable control volume ➔ 𝜌 ⅆ𝑡 ⅆ𝜌 ෍ 𝑚ሶ 𝑜𝑢𝑡 − ෍ 𝑚ሶ 𝑖𝑛 Compressible Fluid, Fixed control volume ➔ 𝑣 ⅆ𝑡 For definite number of ⅆ𝜌𝑣 inlets and outlets Compressible Fluid, deformable control volume ➔ ⅆ𝑡 ⅆ𝑚 Incompressible Fluid, Fixed control volume ➔ =0 ⅆ𝑡 Example 1:  Solution 𝒅𝝆 𝐂 𝝆 𝐂 = − 𝒅𝒕 𝝆 𝒗 𝐥𝐧 =− 𝒕 𝝆𝟎 𝒗 Example 2: Solution Example 3: Solution  Solution ⅆ𝑚 + න 𝜌 𝑣. 𝑛 𝑑𝐴 = 0 ⅆ𝑡 𝑐𝑠 ⅆ𝑚 ⅆ𝑚 + - ⅆ𝑡 water ⅆ𝑡 air = 0 (constant air mass) ⅆ𝑚 = = ⅆ𝑡 water Thanks Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Lecture 2 RTT Momentum conservation (Linear) Dr. Mohammed Rabie Assistant Professor – Mansoura University  Advanced Fluid Dynamics Reynolds Transport Theorem  Reynolds Transport Theorem CS B → Any system property Q𝐵 → For Example: Q𝑚 If B is mass, B = m,  =  If B is momentum, B = mv,  = v And so on….  System ➔ we can apply laws of mechanics CV (Control volume) ➔ Specific region of interest (to study) RTT ➔ Is used to convert system analysis to control volume analysis  ➔ Density ➔ Volume V ➔ Velocity A ➔ Area 1. Conservation of Mass B is mass, ⸫ B = m,  =  ➔ Mass conservation ➔ If the control volume has a definite number of inlets and outlets: ➔ ➔ If the flow within the control volume is steady: ➔0 ➔ If the control volume has a definite number of inlets and outlets:  1. Linear Momentum B is Linear momentum, ⸫ B = mV,  = V The entire equation is a vector relation; thus, it must have three components, x, y, and z components  1. Linear Momentum equation ⅆ ෍ 𝐹𝑥 = (න b𝑥 𝜌 ⅆ𝜈) + න b𝑥 𝜌(𝑉 ⋅ 𝑛) ⅆ𝐴 ⅆ𝑡 𝑐𝑣 6𝑆 ⅆ ෍ 𝐹 f = (න b f 𝜌 ⅆ𝜈) + න b f 𝜌(𝑉 ⋅ 𝑛) ⅆ𝐴 ⅆ𝑡 𝑐𝑣 6𝑆 ⅆ ෍ 𝐹 g = (න b g 𝜌 ⅆ𝜈) + න b g 𝜌(𝑉 ⋅ 𝑛) ⅆ𝐴 ⅆ𝑡 𝑐𝑣 6𝑆 If the cross section is one-dimensional, v and  are uniform over the area, and the control volume has one dimensional inlets and outlets: It also must be applied for the specific direction  Reaction forces Reaction applied to the CV Gauge pressure Pressure force Positive on CV Pressure force= PA Downward direction Weight (vertical) Pressure force= mg Frictional force Example You can get the y component alone… Example Example Thanks Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Lecture 3 RTT Momentum conservation (Angular) Dr. Mohammed Rabie Assistant Professor – Mansoura University  Fluid Mechanics II Reynolds Transport Theorem Angular Momentum & Energy Prof. A.R. Dohina Dr. M. Rabie  b. Angular Momentum B is Angular momentum, ⸫ B = m(r X V),  = (r X V)  Reaction forces Reaction applied to the CV Gauge pressure Pressure force Positive on CV Pressure force= PA PA RP Downward direction Weight (vertical) Weight force= mg Mg RW Frictional force  Clockwise rotation ➔ Negative Torque Counterclockwise rotation ➔ Positive Torque Example 3. Energy equation One dimensional energy flux term The steady-flow Energy Equation Power as a function of head: 𝐏 = 𝛄  𝐇 = 𝛒 𝐠  𝐇 = 𝐦ሶ 𝐠 𝐇 𝐏 ).𝐦 ) = 𝛄  𝐇 ).𝐦 ) = 𝛒 𝐠  𝐇 ).𝐦 ) = 𝐦ሶ 𝐠 𝐇 ).𝐦 ) 𝐏𝐓.𝐫𝐛𝐢 '𝐞 = 𝛄  𝐇𝐓.𝐫𝐛𝐢 '𝐞 = 𝛒 𝐠  𝐇𝐓.𝐫𝐛𝐢 '𝐞 = 𝐦ሶ 𝐠 𝐇𝐓.𝐫𝐛𝐢 '𝐞 𝐏𝐅𝐫𝐢 𝐭𝐢 ( ' = 𝛄  𝐇𝐅𝐫𝐢 𝐭𝐢 ( ' = 𝛒 𝐠  𝐇𝐅𝐫𝐢 𝐭𝐢 ( ' = 𝐦ሶ 𝐠 𝐇𝐅𝐫𝐢 𝐭𝐢 ( ' Thanks Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Lecture 4 RTT Energy conservation Dr. Mohammed Rabie Assistant Professor – Mansoura University  Fluid Mechanics II Reynolds Transport Theorem Angular Momentum & Energy Prof. A.R. Dohina Dr. M. Rabie 4. Energy equation One dimensional energy flux term The steady-flow Energy Equation Power as a function of head: 𝐏 = 𝛄  𝐇 = 𝛒 𝐠  𝐇 = 𝐦ሶ 𝐠 𝐇 𝐏 ).𝐦 ) = 𝛄  𝐇 ).𝐦 ) = 𝛒 𝐠  𝐇 ).𝐦 ) = 𝐦ሶ 𝐠 𝐇 ).𝐦 ) 𝐏𝐓.𝐫𝐛𝐢 '𝐞 = 𝛄  𝐇𝐓.𝐫𝐛𝐢 '𝐞 = 𝛒 𝐠  𝐇𝐓.𝐫𝐛𝐢 '𝐞 = 𝐦ሶ 𝐠 𝐇𝐓.𝐫𝐛𝐢 '𝐞 𝐏𝐅𝐫𝐢 𝐭𝐢 ( ' = 𝛄  𝐇𝐅𝐫𝐢 𝐭𝐢 ( ' = 𝛒 𝐠  𝐇𝐅𝐫𝐢 𝐭𝐢 ( ' = 𝐦ሶ 𝐠 𝐇𝐅𝐫𝐢 𝐭𝐢 ( ' Thanks Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Lecture 5 Differential relations for a fluid particle Dr. Mohammed Rabie Assistant Professor – Mansoura University  Chapter 4 Differential relations for a fluid particle Conservation of mass: Unsteady, compressible, and three-dimensional velocity field Write the conservation of mass for: 1- Unsteady, Compressible, 1-D 2- Steady, Compressible two dimensional 3- Incompressible, 2-D, steady Momentum equation in three directions Newtonian & Incompressible flow Navier-Stokes Equations Couette flow between a fixed and moving plates: Assumptions for Couette flow: Two dimensional Steady Incompressible Viscous flow Very wide and long parallel plates One can neglect gravity effect One plate is fixed and the other moves with velocity V: Flow due to pressure gradient between two fixed plates: Flow rate, mean velocity, and maximum velocity: 𝑄 = න 𝑢 ⅆ𝐴 𝐴 𝑄 𝑢𝑚𝑒𝑎𝑛 = 𝐴 ⅆ𝑢 To get maximum velocity location :Umax @ =0 ⅆ𝑦 (get y for max velocity) Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Lecture 6 Revision Dr. Mohammed Rabie Assistant Professor – Mansoura University  Solution 𝑑𝑚 For Steady Flow ➔ =0 0 = 𝑚ሶ 𝑖 − 𝑚ሶ 𝑜 𝑚ሶ 𝑖 = 𝑚ሶ 𝑜 𝑑𝑡 Water ➔ density is constant 𝑄𝑖 = 𝑄𝑜 Example V2 = ……… Thanks Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Tutorial 1 Introduction to Flow machines and Advanced Fluid Dynamics Dr. Mohammed Rabie Assistant Professor – Mansoura University  ⅆ𝑚 + න 𝜌 𝑣. 𝑛 𝑑𝐴 = 0 ⅆ𝑡 𝑐𝑠 ⅆ𝑣 Incompressible Fluid, deformable control volume ➔ 𝜌 ⅆ𝑡 ⅆ𝜌 ෍ 𝑚ሶ 𝑜𝑢𝑡 − ෍ 𝑚ሶ 𝑖𝑛 Compressible Fluid, Fixed control volume ➔ 𝑣 ⅆ𝑡 For definite number of ⅆ𝜌𝑣 inlets and outlets Compressible Fluid, deformable control volume ➔ ⅆ𝑡 ⅆ𝑚 Incompressible Fluid, Fixed control volume ➔ =0 ⅆ𝑡  Solution 𝑑𝑚 For Steady Flow ➔ =0 0 = 𝑚ሶ 𝑖 − 𝑚ሶ 𝑜 𝑚ሶ 𝑖 = 𝑚ሶ 𝑜 𝑑𝑡 Water ➔ density is constant 𝑄𝑖 = 𝑄𝑜 b Q1 + Q w = Q 2 𝜋 𝜋 12 0.08 + (0.15)(0.3016) = V2 0.082 2 4 4 𝐕𝟐 = 𝟐𝟏 𝐦/𝐬 If h = 1m – 30 cm = 1- 0.3 = 0.7m → find t Δℎ 0.7 Δ𝑡 = = = 45.75 𝑠𝑒𝑐 𝑑ℎ 0.0153 𝑑𝑡 Thanks Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Tutorial 2 RTT Momentum conservation (Linear) Dr. Mohammed Rabie Assistant Professor – Mansoura University  Advanced Fluid Dynamics Reynolds Transport Theorem  1. Linear Momentum B is Linear momentum, ⸫ B = mV,  = V The entire equation is a vector relation; thus, it must have three components, x, y, and z components  1. Linear Momentum equation ⅆ ෍ 𝐹𝑥 = (න 𝑢𝑥 𝜌 ⅆ𝜈) + න 𝑢𝑥 𝜌(𝑉 ⋅ 𝑛) ⅆ𝐴 ⅆ𝑡 𝑐𝑣 𝐶𝑆 ⅆ ෍ 𝐹𝑦 = (න 𝑢𝑦 𝜌 ⅆ𝜈) + න 𝑢𝑦 𝜌(𝑉 ⋅ 𝑛) ⅆ𝐴 ⅆ𝑡 𝑐𝑣 𝐶𝑆 ⅆ ෍ 𝐹𝑧 = (න 𝑢𝑧 𝜌 ⅆ𝜈) + න 𝑢𝑧 𝜌(𝑉 ⋅ 𝑛) ⅆ𝐴 ⅆ𝑡 𝑐𝑣 𝐶𝑆 If the cross section is one-dimensional, v and  are uniform over the area, and the control volume has one dimensional inlets and outlets: It also must be applied for the specific direction  Reaction forces Reaction applied to the CV Gauge pressure Pressure force Positive on CV Pressure force= PA Downward direction Weight (vertical) Pressure force= mg Frictional force Example You can get the y component alone… Example Example Thanks Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Tutorial 3 RTT Momentum conservation (Angular) Dr. Mohammed Rabie Assistant Professor – Mansoura University  b. Angular Momentum B is Angular momentum, ⸫ B = m(r X V),  = (r X V)  Reaction forces Reaction applied to the CV Gauge pressure Pressure force Positive on CV Pressure force= PA PA RP Downward direction Weight (vertical) Weight force= mg Mg RW Frictional force  Clockwise rotation ➔ Negative Torque Counterclockwise rotation ➔ Positive Torque Power as a function of head: 𝐏 = 𝛄  𝐇 = 𝛒 𝐠  𝐇 = 𝐦ሶ 𝐠 𝐇 𝐏 ).𝐦 ) = 𝛄  𝐇 ).𝐦 ) = 𝛒 𝐠  𝐇 ).𝐦 ) = 𝐦ሶ 𝐠 𝐇 ).𝐦 ) 𝐏𝐓.𝐫𝐛𝐢 '𝐞 = 𝛄  𝐇𝐓.𝐫𝐛𝐢 '𝐞 = 𝛒 𝐠  𝐇𝐓.𝐫𝐛𝐢 '𝐞 = 𝐦ሶ 𝐠 𝐇𝐓.𝐫𝐛𝐢 '𝐞 𝐏𝐅𝐫𝐢 𝐭𝐢 ( ' = 𝛄  𝐇𝐅𝐫𝐢 𝐭𝐢 ( ' = 𝛒 𝐠  𝐇𝐅𝐫𝐢 𝐭𝐢 ( ' = 𝐦ሶ 𝐠 𝐇𝐅𝐫𝐢 𝐭𝐢 ( ' Thanks Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Tutorial 4 RTT Energy conservation Dr. Mohammed Rabie Assistant Professor – Mansoura University Energy Thanks Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Tutorial 5 Differential relations for a fluid particle Dr. Mohammed Rabie Assistant Professor – Mansoura University  Chapter 4 Differential relations for a fluid particle Conservation of mass: Unsteady, compressible, and three-dimensional velocity field Momentum equation in three directions Newtonian & Incompressible flow Navier-Stokes Equations Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Lecture 7 Differential relations for a fluid particle Cylindrical Coordinates Dr. Mohammed Rabie Assistant Professor – Mansoura University Differential relations for a fluid particle Cylindrical Coordinates Flow rate, mean velocity, and maximum velocity: 𝑄 = න 𝑢 ⅆ𝐴 𝐴 𝑄 𝑢𝑚𝑒𝑎𝑛 = 𝐴 ⅆ𝑢 To get maximum velocity location :Umax @ =0 ⅆ𝑦 (get y for max velocity) Cylindrical Polar coordinates Velocity in r direction → 𝜐𝑟 Velocity in  direction → 𝜐𝜃 Velocity in z direction → 𝜐𝑧  Navier–Stokes equations for cylindrical polar coordinates Continuity: General Incompressible Momentum r − Direction:  Navier–Stokes equations for cylindrical polar coordinates Momentum  − Direction: 𝜕𝜐𝜃 𝜕𝜐𝜃 𝜐𝜃 𝜕𝜐𝜃 𝜐𝜃 𝜐𝑟 𝜕𝜐𝜃 𝜌 + 𝜐𝑟 + + + 𝜐𝑧 𝜕𝑡 𝜕𝑟 𝑟 𝜕𝜃 𝑟 𝜕𝑧 1 𝜕𝑝 𝜕 1 𝜕(𝑟𝜐𝜃 ) 1 𝜕 2 𝜐𝜃 2 𝜕𝜐𝑟 𝜕 2 𝜐𝜃 =− + 𝜌𝑔𝜃 + 𝜇 + 2 2 + 2 + 𝑟 𝜕𝜃 𝜕𝑟 𝑟 𝜕𝑟 𝑟 𝜃 𝑟 𝜕𝜃 𝜕𝑧 2 Momentum z − Direction: Steady Axisymmetric flow: 𝝏∎ Steady → =𝟎 𝝏𝒕 𝝏∎ Axisymmetric → =𝟎 𝝏𝜽 Types of Axisymmetric flow: Axisymmetric Rectilinear Axisymmetric Torsional Axisymmetric Radial 𝜐𝑟 = 𝜐𝜃 = 0 𝜐𝑟 = 𝜐𝑧 = 0 𝜐𝜃 = 𝜐𝑧 = 0 𝜐𝑧 ≠ 0 𝜐𝜃 ≠ 0 𝜐𝑟 ≠ 0  1- Axisymmetric Rectilinear 𝝊𝒓 = 𝝊𝜽 = 𝟎 𝝊𝒛 ≠ 𝟎 Continuity: 0 0 Get that: 𝝏𝝊𝒛 =𝟎 𝝏𝒛 Momentum Eq. in Z direction: 0 𝝊𝒓 = 𝟎 𝝊𝜽 = 𝟎 𝝏𝝊𝒛 𝝏𝝊𝒛 𝝏𝝊𝒛 =𝟎 =𝟎 =𝟎 Steady 𝝏𝒛 𝝏𝜽 𝝏𝒛 Axisymmetric  𝝏𝑷 𝟏 𝝏 𝝏𝝊𝒛 − + 𝝆𝒈𝒛 + 𝝁 𝒓 =𝟎 𝝏𝒛 𝒓 𝝏𝒓 𝝏𝒓 𝟏 𝝏 𝝏𝝊𝒛 𝝏𝑷 𝝁 𝒓 = − 𝝆𝒈𝒛 𝒓 𝝏𝒓 𝝏𝒓 𝝏𝒛 𝝏 𝝏𝝊𝒛 𝟏 𝝏𝑷 𝒓 = − 𝝆𝒈𝒛 𝒓 Integrate 𝝏𝒓 𝝏𝒓 𝝁 𝝏𝒛 𝝏𝝊𝒛 𝟏 𝝏𝑷 𝒓 = − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 Eqn. 1→ /r 𝝏𝒓 𝟐𝝁 𝝏𝒛 𝝏𝝊𝒛 𝟏 𝝏𝑷 𝑪𝟏 = − 𝝆𝒈𝒛 𝒓 + Integrate 𝝏𝒓 𝟐𝝁 𝝏𝒛 𝒓 𝟏 𝝏𝑷 𝝊𝒛 = − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 Eqn. 2 𝟒𝝁 𝝏𝒛  𝝏𝝊𝒛 𝟏 𝝏𝑷 𝒓 = − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 Eqn. 1→ /r 𝝏𝒓 𝟐𝝁 𝝏𝒛 𝟏 𝝏𝑷 𝝊𝒛 = − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 𝝏𝒛 Pressure Gravity cause flow cause flow  b- Fully developed flow in annulus Fluid flow vertically downward around a pipe of radius a due to gravity 𝟏 𝝏𝑷 𝝊𝒛 = − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 𝝏𝒛 As flow due to gravity only, so neglect 𝝏𝑷 pressure gradient =𝟎 𝝏𝒛 𝟏 𝝊𝒛 = 𝟎 − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 𝝆𝒈𝒛 𝟐 𝝊𝒛 = − 𝒓 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁  𝝆𝒈𝒛 𝟐 𝝊𝒛 = − 𝒓 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 Apply boundary conditions to get C1 and C2 𝝏𝝊𝒛 @ r = b, no shear stress (maximum velocity) , = 0, Apply in the previous eqn. 1 𝝏𝒓 𝝏𝝊𝒛 𝟏 𝒓 = −𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 𝝏𝒓 𝟐𝝁 𝟏 𝝆𝒈𝒛 𝒃𝟐 𝒃 𝟎 = −𝝆𝒈𝒛 𝒃𝟐 + 𝑪𝟏 → get C1 = 𝟐𝝁 𝟐𝝁 At r = a, 𝝊𝒛 = 𝟎 𝝆𝒈𝒛 𝟐 𝝊𝒛 = − 𝒓 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 𝝆𝒈𝒛 𝟐 𝝆𝒈𝒛 𝒃𝟐 𝟎=− 𝒂 + 𝒍𝒏 𝒂 + 𝑪𝟐 𝟒𝝁 𝟐𝝁 𝝆𝒈𝒛 𝒂𝟐 𝝆𝒈𝒛 𝒃𝟐 𝑪𝟐 = − 𝒍𝒏 𝒂 𝟒𝝁 𝟐𝝁  𝝆𝒈𝒛 𝟐 𝝊𝒛 = − 𝒓 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 𝝆𝒈𝒛 𝒃𝟐 𝝆𝒈𝒛 𝒂𝟐 𝝆𝒈𝒛 𝒃𝟐 C1 = 𝑪𝟐 = − 𝒍𝒏 𝒂 𝟐𝝁 𝟒𝝁 𝟐𝝁 𝝆𝒈𝒛 𝟐 𝝆𝒈𝒛 𝒃𝟐 𝝆𝒈𝒛 𝒂𝟐 𝝆𝒈𝒛 𝒃𝟐 𝝊𝒛 = − 𝒓 + 𝒍𝒏 𝒓 + − 𝒍𝒏 𝒂 𝟒𝝁 𝟐𝝁 𝟒𝝁 𝟐𝝁 𝒈𝒛 = 𝒈 𝝆𝒈 𝟐 𝝆𝒈𝒃𝟐 𝝆𝒈𝒂𝟐 𝝆𝒈𝒃𝟐 𝝊𝒛 = − 𝒓 + 𝒍𝒏 𝒓 + − 𝒍𝒏 𝒂 𝟒𝝁 𝟐𝝁 𝟒𝝁 𝟐𝝁 Report: Find the flow rate, average velocity, maximum velocity location, maximum velocity value 2- Axisymmetric Torsional flow 𝑢𝑟 = 0 𝑢𝑧 = 0 𝑢𝜃 = 𝑓 𝑟 𝜕 =0 𝜕𝜃 No gravity in 𝜃 direction No pressure gradient  Flow between long concentric cylinders 𝜕𝑢𝜃 𝜕𝑢𝜃 𝑢𝜃 𝜕𝑢𝜃 𝑢𝜃 𝑢𝑟 𝜕𝑢𝜃 1 𝜕𝑝 𝜕 1 𝜕(𝑟𝑢𝜃 ) 1 𝜕 2 𝑢𝜃 2 𝜕𝑢𝑟 𝜕 2 𝑢𝜃 𝜌 + 𝑢𝑟 + + + 𝑢𝑧 =− + 𝜌𝑔𝜃 + 𝜇 + 2 + + 𝜕𝑡 𝜕𝑟 𝑟 𝜕𝜃 𝑟 𝜕𝑧 𝑟 𝜕𝜃 𝜕𝑟 𝑟 𝜕𝑟 𝑟 𝜕𝜃 2 𝑟 2 𝜕𝜃 𝜕𝑧 2 0 𝑢𝑟 =0 𝑢 =0 𝑁𝑜 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑔𝜃 = 0 𝜕 𝑢𝑟 =0 = 0 𝑢𝑟 =0 𝑧 𝜕𝑢𝜃 𝜕 =0 Steady 𝜕𝜃 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝜕𝜃 𝜕𝑧 𝜕 1 𝜕(𝑟𝑢𝜃 ) = 0 𝜕𝑟 𝑟 𝜕𝑟 1 𝜕(𝑟𝑢𝜃 ) = C1 𝑟 𝜕𝑟 𝜕(𝑟𝑢𝜃 ) = C1r 𝜕𝑟  Flow between long concentric cylinders 𝜕(𝑟𝑢𝜃 ) = C1r 𝜕𝑟 𝑟2 𝑟𝑢𝜃 = C1 + C2 2 𝑐2 𝑢𝜃 = C1 r + 𝑟 Apply boundary conditions to get constants At r=ri 𝑢𝜃 = i ri At r=ro 𝑢𝜃 =  Flow between long concentric cylinders  ρVD Reynolds Number → Re = μ ρ → Fluid density, kg/m3 D → Pipe diameter, m ( D = 2R) V → Average flow velocity, m/s (From Hagen – Poiseuille flow) 1 𝜕P 2 UAverage = − R 8μ 𝜕Z 𝜕P ΔP = → Pressure gradient, Pa/m 𝜕Z L μ → dynamic viscosity, N. s/m2 = Pa.s = kg/(m. s),…. Re < 2000, Laminar flow Re > 4000, Turbulent flow 2000 < Re < 4000, Transition Example: Solution: 𝟏 𝝏𝑷 𝝊𝒛 = − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 𝝏𝒛 𝝏𝑷 − 𝝆𝒈𝒛 = 𝟎 𝝏𝒛 𝝊𝒛 = 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 At r = a, 𝝊𝒛 = U 𝑼 = 𝑪𝟏 𝒍𝒏 𝒂 + 𝑪𝟐 →1 At r = b, 𝝊𝒛 = 0 𝟎 = 𝑪𝟏 𝒍𝒏 𝒃 + 𝑪𝟐 →2 𝑼 Eqn 1 – Eqn 2 → 𝑼 = 𝑪𝟏 𝒍𝒏 𝒂/𝒃 𝑪𝟏 = 𝒍𝒏 𝒂/𝒃 Put C1 in Eqn 2 → 𝑼 𝑼 𝟎= 𝒍𝒏 𝒃 + 𝑪𝟐 𝑪𝟐 = − 𝒍𝒏 𝒃 𝒍𝒏 𝒂/𝒃 𝒍𝒏 𝒂/𝒃  𝝊𝒛 = 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝑼 𝑼 𝝊𝒛 = 𝒍𝒏 𝒓 − 𝒍𝒏 𝒃 𝒍𝒏 𝒂/𝒃 𝒍𝒏 𝒂/𝒃 𝑼 𝝊𝒛 = 𝒍𝒏 𝒓 − 𝒍𝒏 𝒃 𝒍𝒏 𝒂/𝒃 𝑼𝒍𝒏 𝒓/𝒃 𝝊𝒛 = 𝒍𝒏 𝒂/𝒃 Given:  = 870 kg/m3,  = 0.104 kg/(m.s) Solution Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Lecture 8 Boundary Layer Dr. Mohammed Rabie Assistant Professor – Mansoura University  Boundary Layer A boundary layer is the thin layer of fluid close to a solid surface where the flow is influenced by the surface's friction. Within this region, the velocity of the fluid changes from zero at the surface (due to the no-slip condition) to the free-stream velocity away from the surface. U = U͚ → free stream velocity u→ velocity in the boundary layer = f(x,y)  ρUx Ux Local Reynolds number → Rex = = μ υ ρ → fluid density. (kg/m3) U → Flow stream velocity, (m/s) x → some definite section, (m) μ → Dynamic viscosity, (kg/m.s), (Pa s), … υ → Kinematic viscosity, (m2/s) For flow over flat plate: Rex < 5 x 105 → Laminar flow Rex > 5 x 105 → Turbulent flow 𝛿 5 = → Laminar flow 𝑥 𝑅𝑒𝑥 Exact solution: 𝛿 0.16 = 1/7 → Turbulent flow 𝑥 𝑅𝑒𝑥  𝐲 d = d(x) u = u(x ,y) h= 𝛅 First step, assume velocity profile (Polynomial): 𝐮 = 𝐚𝟎 + 𝐚𝟏 𝐲 𝟏 + 𝐚𝟐 𝐲 𝟐 +𝐚𝟑 𝐲 𝟑 +𝐚𝟒 𝐲 𝟒 + ⋯ Or in dimensionless form: 𝐮 = 𝐚𝟎 + 𝐚𝟏 𝛈𝟏 + 𝐚𝟐 𝛈𝟐 +𝐚𝟑 𝛈𝟑 +𝐚𝟒 𝛈𝟒 + ⋯ 𝐔 To get constants (ao, a1, a2,…) apply the boundary conditions: 1- @ y = 0, u = 0 → @ y/d = 0/d, u/U = 0/U → @ h = 0, u/U = 0 2- @ y = d, u = U → @ y/d = d /d, u/U = U /U → @ h = 1, u/U = 1 𝛛𝐮 𝐔𝛛𝐮/𝐔 𝛛 𝐮/𝐔 3- @ y = d, = 0 → @ y/d = d /d, = 0 → @ h = 1, =𝟎 𝛛𝐲 𝛅𝛛𝐲/𝛅 𝛛𝛈 And so on for more boundary conditions…. Example: Assume first order for velocity distribution, 𝐮 𝐮 = 𝐚𝟎 + 𝐚𝟏 𝒚 = 𝐚𝟎 + 𝐚𝟏 𝛈𝟏 𝐔 @ y = 0, u = 0 → ao = 0 @ h = 0, u/U = 0 → 𝟎 = 𝐚𝟎 + 𝐚𝟏 𝟎 → ao = 0 @ y = d, u = U → 𝐔 = 𝟎 + 𝐚𝟏 d → 𝐚𝟏 = U/ d @ h = 1, u/U = 1 → 𝟏 = 𝟎 + 𝐚𝟏 (𝟏) → 𝐚𝟏 =1 𝐮 𝐮 = 𝟎 + (U/ d) 𝒚 𝐔 = 𝐚𝟎 + 𝐚𝟏 𝛈𝟏 = 𝟎 + 𝜼 𝑼 𝐮 𝐮= 𝒚 = 𝜼 𝜹 𝐔  Definitions 1- Displacement thickness 𝛅∗ 𝛅 𝐮 𝛅⋆ =න 𝟏− 𝐝𝐲 𝟎 𝐔 𝟏 𝛅⋆ 𝐮 =න 𝟏− 𝐝𝛈 𝛅 𝟎 𝐔 𝛅∗ → Displacement thickness 𝛅 → Boundary Layer thickness 𝐮 → Velocity profile, f(y) or f(𝛈) 𝐔  Definitions 2- Momentum thickness  𝛅 𝐮 𝐮 𝛉=න 𝟏− 𝐝𝐲 𝟎 𝐔 𝐔 𝟏 𝛉 𝐮 𝐮 =න 𝟏− 𝐝𝛈 𝛅 𝟎 𝐔 𝐔 𝛉→ Momentum thickness 𝛅 → Boundary Layer thickness 𝐮 → Velocity profile, f(y) or f(𝛈) 𝐔  Definitions 3- Wall shear stress 𝛕𝐰  𝐝𝐮 𝛕𝐰 = 𝛍 At the wall, y = 0 𝐝𝐲 𝐮 𝐮 𝐔𝐝( ) 𝐔 𝐔) 𝐝( 𝛕𝐰 = 𝛍 𝐔 At the wall, 𝛈 = 0 𝐲 = 𝛍 𝛅 𝐝(𝛈) 𝛅𝐝( ) 𝛅 𝛕𝐰 → Wall shear stress 𝛍→ Dynamic viscosity 𝐮 → Velocity profile, f(y) or f(𝛈) 𝐔  Momentum integral equation 𝐝𝛉 𝛕𝐰 = 𝐝𝐱 𝛒 𝐔 𝟐 It will be used to get d as a function of x We can get  as a function of d, And also, can get 𝛕𝐰 as a function of constant parameters, Then we could integrate the equation relative to d and x @x=0→d=0 @x=x→d=d 𝛿 𝑥 න ∎∎∎ 𝑑𝛿 = න ∎∎∎ 𝑑𝑥 0 0 Friction Coefficient Cf 𝛕𝐰 𝐂𝐟 = 𝟎. 𝟓𝛒 𝐔 𝟐 Drag Force D L D = න 𝛕𝐰 dA 0 L dA = Bdx Drag Coefficient CD B 𝐃 x 𝐂𝐃 = 𝟎. 𝟓𝛒 𝐀 𝐔 𝟐 A=BL dx  Example Assume first order velocity distribution 𝐮 = 𝐚𝟎 + 𝐚𝟏 𝛈 𝐔 @ h = 0, u/U = 0 → 𝟎 = 𝐚𝟎 + 𝐚𝟏 (𝟎) → 𝐚𝟎 = 0 @ h = 1, u/U = 1 → 𝟏 = 𝟎 + 𝐚𝟏 𝟏 → 𝐚𝟏 = 𝟏 So: 𝐚𝟎 = 𝟎 𝐚𝟏 = 𝟏 𝐮 =𝜼 𝐔 1- Displacement thickness 𝛅∗ 𝟏 𝛅⋆ 𝐮 =න 𝟏− 𝐝𝛈 𝛅 𝟎 𝐔 𝟏 𝛅⋆ = න 𝟏 − 𝜼 𝐝𝛈 𝛅 𝟎 1 𝛅⋆ 𝛈𝟐 = 𝜼− 𝛅 𝟐 0 𝛅⋆ 𝟏 =𝟏− 𝛅 𝟐 𝛅⋆ 𝟏 = 𝛅 𝟐 𝛅 𝛅⋆ = 𝟐 2- Momentum thickness  𝟏 𝛉 𝐮 𝐮 =න 𝟏− 𝐝𝛈 𝛅 𝟎 𝐔 𝐔 𝟏 𝛉 = න 𝜼 𝟏 − 𝜼 𝐝𝛈 𝛅 𝟎 𝟏 𝛉 = න 𝜼 − 𝛈𝟐 𝐝𝛈 𝛅 𝟎 1 𝛉 𝛈 𝟐 𝛈𝟑 𝛉 𝟏 𝟏 𝟏 = − = − = 𝛅 𝟐 𝟑 𝛅 𝟐 𝟑 𝟔 0 𝛉 𝟏 𝟏 = 𝛉= 𝛅 𝛅 𝟔 𝟔 3- Wall shear stress 𝛕𝐰  𝐮 𝐔 𝐔) 𝐝( 𝛕𝐰 = 𝛍 𝛅 𝐝(𝛈) 𝛈=𝟎 𝐮 𝐝(𝒖/𝑼) 𝐝(𝒖/𝑼) =𝜼 =𝟏 =𝟏 𝐔 𝐝(𝛈) 𝐝(𝛈) 𝛈 = 𝟎 𝐔 𝛕𝐰 = 𝛍 𝛅 4- Momentum integral equation 𝐝𝛉 𝛕𝐰 = 𝐝𝐱 𝛒 𝐔 𝟐 𝟏 𝐔 𝛉= 𝛅 𝛕𝐰 = 𝛍 𝟔 𝛅 𝟏 𝐔 𝐝 𝟔 𝛅 𝛍 = 𝛅𝟐 𝐝𝐱 𝛒𝐔  𝟏 𝐔 𝐝 𝟔 𝛅 𝛍 = 𝛅𝟐 𝐝𝐱 𝛒𝐔 𝟏𝐝 𝛅 𝛍 = 𝟔 𝐝𝐱 𝛒𝐔𝛅 𝟔𝛍 𝛅𝐝 𝛅 = 𝐝𝐱 𝛒𝐔 𝛅 𝒙 𝟔𝛍 න 𝛅𝐝 𝛅 = න 𝐝𝐱 𝟎 𝟎 𝛒𝐔 𝛅 𝟔𝛍 𝒙 න 𝛅𝐝 𝛅 = න 𝐝𝐱 𝟎 𝛒𝐔 𝟎 d x 𝜹𝟐 𝟔𝛍 = 𝒙 𝟐 𝛒𝐔 0 0  𝜹𝟐 𝟔𝛍 = 𝒙 𝟐 𝛒𝐔 𝜹𝟐 𝟔𝛍 𝟐 = 𝒙 × 𝟐 𝟐 𝛒𝐔 𝒙 𝜹𝟐 𝟏𝟐𝛍 𝟏𝟐 𝟐 = = 𝒙 𝛒𝐔𝒙 𝑹𝒆𝒙 𝜹 𝟏𝟐 = 𝒙 𝑹𝒆𝒙 𝜹 𝟑. 𝟒𝟔𝟒 = 𝒙 𝑹𝒆𝒙 𝟑. 𝟒𝟔𝟒𝒙 𝜹= 𝑹𝒆𝒙 The calculated wall shear stress was: 𝐔 𝛕𝐰 = 𝛍 𝛅 We can get it as a function of local position x: 𝐔 𝟑. 𝟒𝟔𝟒𝒙 𝛕𝐰 = 𝛍 𝐔 𝑹𝒆𝒙 𝜹≅ 𝟑. 𝟒𝟔𝟒𝒙 𝛕𝐰 = 𝛍 𝑹𝒆𝒙 𝟑. 𝟒𝟔𝟒𝒙 𝑹𝒆𝒙 5- Friction Coefficient Cf 𝛕𝐰 𝐂𝐟 = 𝛍𝐔 𝑹𝒆𝒙 𝟎. 𝟓𝟕𝟕 𝛍 𝑹𝒆𝒙 𝟎. 𝟓𝟕𝟕 𝑹𝒆𝒙 𝟎. 𝟓𝛒 𝐔𝟐 𝐂𝐟 = 𝐂𝐟 = 𝐂𝐟 = 𝟎. 𝟓𝛒 𝐔 𝟐 𝟑. 𝟒𝟔𝟒𝒙 𝛒𝐔𝒙 𝑹𝒆𝒙 𝟎. 𝟓𝟕𝟕 𝐂𝐟 = 𝑹𝒆𝒙 𝟎. 𝟓𝟕𝟕 𝐂𝐟 = 𝑹𝒆𝒙 6- Drag Force D L D = න 𝛕𝐰 dA dA = Bdx 0 𝟎. 𝟓𝟕𝟕 𝛕𝐰 𝛕𝐰 𝒂 𝐂𝐟 = = 𝟐 = Where a = 0.577 𝑹𝒆𝒙 𝟎. 𝟓𝛒 𝐔 𝟐 𝟎. 𝟓𝛒 𝐔 𝑹𝒆𝒙 𝒂 𝛕𝐰 = 𝟎. 𝟓𝛒 𝐔𝟐 = 𝟎. 𝟓𝛒 𝐔𝟐 𝒂 𝑹𝒆𝒙 −𝟎.𝟓 𝑹𝒆𝒙 𝟎.𝟓 𝟐 𝝁 𝛕𝐰 = 𝟎. 𝟓𝛒 𝐔 𝒂 𝛒𝐔𝒙 L 𝟎.𝟓 L 𝟎.𝟓 𝟐 𝝁 𝟐 𝝁 D = න 𝟎. 𝟓𝛒 𝐔 𝒂 Bdx = න 𝟎. 𝟓𝛒 𝐔 𝒂 B𝒙−0.5 dx 0 𝛒𝐔𝒙 0 𝛒𝐔 𝟎.𝟓 L 𝝁 D = 𝟎. 𝟓𝒂B𝛒 𝐔𝟐 න 𝒙−0.5 dx 𝛒𝐔 0 𝟎.𝟓 L 𝝁 𝒙0.5 D = 𝟎. 𝟓𝒂B𝛒 𝐔𝟐 𝛒𝐔 𝟎. 𝟓 0  𝟎.𝟓 𝝁 𝟎.𝟓 0.5 𝑳0.5 𝝁 𝑳0.5 D = 𝒂B𝛒 𝐔𝟐 𝑳 × D = 𝟎. 𝟓𝒂B𝛒 𝐔𝟐 𝛒𝐔 𝑳0.5 𝛒𝐔 𝟎. 𝟓 𝟎.𝟓 𝝁 D = 𝒂B𝑳𝛒 𝐔𝟐 𝛒𝐔𝑳 𝟎.𝟓 𝝁 D = 𝒂B𝑳𝛒 𝐔𝟐 𝛒𝐔𝑳 𝝁 A = BL, a=0.577, 𝑹𝒆𝑳 = 𝛒𝐔𝑳 𝒂𝐴𝛒 𝐔𝟐 𝟐𝒂𝐴 𝟎. 𝟓𝛒 𝐔𝟐 D= D= 𝑹𝒆𝑳 𝑹𝒆𝑳 7- Drag Coefficient CD: 𝟐𝒂𝐴 𝟎. 𝟓𝛒 𝐔𝟐 D 𝟐𝒂 D= = = 𝑪𝑫 𝑹𝒆𝑳 𝟎. 𝟓𝛒𝐔𝟐 𝐴 𝑹𝒆𝑳 D 𝟐𝒂 𝟐(𝟎. 𝟓𝟕𝟕) 𝟏. 𝟏𝟓𝟒 𝟏. 𝟏𝟓𝟒 𝑪𝑫 = = = = 𝑪𝑫 = 𝟎. 𝟓𝛒𝐔𝟐 𝐴 𝑹𝒆𝑳 𝑹𝒆𝑳 𝑹𝒆𝑳 𝑹𝒆𝑳 Dimensionless profile shape factor H: 𝛅⋆ 𝛅⋆ /𝛅 𝟏/𝟐 𝑯= = = 𝛉 𝛉/𝛅 𝟏/𝟔 𝑯=𝟑 Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Lecture 8 Boundary Layer Dr. Mohammed Rabie Assistant Professor – Mansoura University  Boundary Layer A boundary layer is the thin layer of fluid close to a solid surface where the flow is influenced by the surface's friction. Within this region, the velocity of the fluid changes from zero at the surface (due to the no-slip condition) to the free-stream velocity away from the surface. U = U͚ → free stream velocity u→ velocity in the boundary layer = f(x,y)  ρUx Ux Local Reynolds number → Rex = = μ υ ρ → fluid density. (kg/m3) U → Flow stream velocity, (m/s) x → some definite section, (m) μ → Dynamic viscosity, (kg/m.s), (Pa s), … υ → Kinematic viscosity, (m2/s) For flow over flat plate: Rex < 5 x 105 → Laminar flow Rex > 5 x 105 → Turbulent flow 𝛿 5 = → Laminar flow 𝑥 𝑅𝑒𝑥 Exact solution: 𝛿 0.16 = 1/7 → Turbulent flow 𝑥 𝑅𝑒𝑥  𝐲 d = d(x) u = u(x ,y) h= 𝛅 First step, assume velocity profile (Polynomial): 𝐮 = 𝐚𝟎 + 𝐚𝟏 𝐲 𝟏 + 𝐚𝟐 𝐲 𝟐 +𝐚𝟑 𝐲 𝟑 +𝐚𝟒 𝐲 𝟒 + ⋯ Or in dimensionless form: 𝐮 = 𝐚𝟎 + 𝐚𝟏 𝛈𝟏 + 𝐚𝟐 𝛈𝟐 +𝐚𝟑 𝛈𝟑 +𝐚𝟒 𝛈𝟒 + ⋯ 𝐔 To get constants (ao, a1, a2,…) apply the boundary conditions: 1- @ y = 0, u = 0 → @ y/d = 0/d, u/U = 0/U → @ h = 0, u/U = 0 2- @ y = d, u = U → @ y/d = d /d, u/U = U /U → @ h = 1, u/U = 1 𝛛𝐮 𝐔𝛛𝐮/𝐔 𝛛 𝐮/𝐔 3- @ y = d, = 0 → @ y/d = d /d, = 0 → @ h = 1, =𝟎 𝛛𝐲 𝛅𝛛𝐲/𝛅 𝛛𝛈 And so on for more boundary conditions…. Example: Assume first order for velocity distribution, 𝐮 𝐮 = 𝐚𝟎 + 𝐚𝟏 𝒚 = 𝐚𝟎 + 𝐚𝟏 𝛈𝟏 𝐔 @ y = 0, u = 0 → ao = 0 @ h = 0, u/U = 0 → 𝟎 = 𝐚𝟎 + 𝐚𝟏 𝟎 → ao = 0 @ y = d, u = U → 𝐔 = 𝟎 + 𝐚𝟏 d → 𝐚𝟏 = U/ d @ h = 1, u/U = 1 → 𝟏 = 𝟎 + 𝐚𝟏 (𝟏) → 𝐚𝟏 =1 𝐮 𝐮 = 𝟎 + (U/ d) 𝒚 𝐔 = 𝐚𝟎 + 𝐚𝟏 𝛈𝟏 = 𝟎 + 𝜼 𝑼 𝐮 𝐮= 𝒚 = 𝜼 𝜹 𝐔  Definitions 1- Displacement thickness 𝛅∗ 𝛅 𝐮 𝛅⋆ =න 𝟏− 𝐝𝐲 𝟎 𝐔 𝟏 𝛅⋆ 𝐮 =න 𝟏− 𝐝𝛈 𝛅 𝟎 𝐔 𝛅∗ → Displacement thickness 𝛅 → Boundary Layer thickness 𝐮 → Velocity profile, f(y) or f(𝛈) 𝐔  Definitions 2- Momentum thickness  𝛅 𝐮 𝐮 𝛉=න 𝟏− 𝐝𝐲 𝟎 𝐔 𝐔 𝟏 𝛉 𝐮 𝐮 =න 𝟏− 𝐝𝛈 𝛅 𝟎 𝐔 𝐔 𝛉→ Momentum thickness 𝛅 → Boundary Layer thickness 𝐮 → Velocity profile, f(y) or f(𝛈) 𝐔  Definitions 3- Wall shear stress 𝛕𝐰  𝐝𝐮 𝛕𝐰 = 𝛍 At the wall, y = 0 𝐝𝐲 𝐮 𝐮 𝐔𝐝( ) 𝐔 𝐔) 𝐝( 𝛕𝐰 = 𝛍 𝐔 At the wall, 𝛈 = 0 𝐲 = 𝛍 𝛅 𝐝(𝛈) 𝛅𝐝( ) 𝛅 𝛕𝐰 → Wall shear stress 𝛍→ Dynamic viscosity 𝐮 → Velocity profile, f(y) or f(𝛈) 𝐔  Momentum integral equation 𝐝𝛉 𝛕𝐰 = 𝐝𝐱 𝛒 𝐔 𝟐 It will be used to get d as a function of x We can get  as a function of d, And also, can get 𝛕𝐰 as a function of constant parameters, Then we could integrate the equation relative to d and x @x=0→d=0 @x=x→d=d 𝛿 𝑥 න ∎∎∎ 𝑑𝛿 = න ∎∎∎ 𝑑𝑥 0 0 Friction Coefficient Cf 𝛕𝐰 𝐂𝐟 = 𝟎. 𝟓𝛒 𝐔 𝟐 Drag Force D L D = න 𝛕𝐰 dA 0 L dA = Bdx Drag Coefficient CD B 𝐃 x 𝐂𝐃 = 𝟎. 𝟓𝛒 𝐀 𝐔 𝟐 A=BL dx  Example Assume first order velocity distribution 𝐮 = 𝐚𝟎 + 𝐚𝟏 𝛈 𝐔 @ h = 0, u/U = 0 → 𝟎 = 𝐚𝟎 + 𝐚𝟏 (𝟎) → 𝐚𝟎 = 0 @ h = 1, u/U = 1 → 𝟏 = 𝟎 + 𝐚𝟏 𝟏 → 𝐚𝟏 = 𝟏 So: 𝐚𝟎 = 𝟎 𝐚𝟏 = 𝟏 𝐮 =𝜼 𝐔 1- Displacement thickness 𝛅∗ 𝟏 𝛅⋆ 𝐮 =න 𝟏− 𝐝𝛈 𝛅 𝟎 𝐔 𝟏 𝛅⋆ = න 𝟏 − 𝜼 𝐝𝛈 𝛅 𝟎 1 𝛅⋆ 𝛈𝟐 = 𝜼− 𝛅 𝟐 0 𝛅⋆ 𝟏 =𝟏− 𝛅 𝟐 𝛅⋆ 𝟏 = 𝛅 𝟐 𝛅 𝛅⋆ = 𝟐 2- Momentum thickness  𝟏 𝛉 𝐮 𝐮 =න 𝟏− 𝐝𝛈 𝛅 𝟎 𝐔 𝐔 𝟏 𝛉 = න 𝜼 𝟏 − 𝜼 𝐝𝛈 𝛅 𝟎 𝟏 𝛉 = න 𝜼 − 𝛈𝟐 𝐝𝛈 𝛅 𝟎 1 𝛉 𝛈 𝟐 𝛈𝟑 𝛉 𝟏 𝟏 𝟏 = − = − = 𝛅 𝟐 𝟑 𝛅 𝟐 𝟑 𝟔 0 𝛉 𝟏 𝟏 = 𝛉= 𝛅 𝛅 𝟔 𝟔 3- Wall shear stress 𝛕𝐰  𝐮 𝐔 𝐔) 𝐝( 𝛕𝐰 = 𝛍 𝛅 𝐝(𝛈) 𝛈=𝟎 𝐮 𝐝(𝒖/𝑼) 𝐝(𝒖/𝑼) =𝜼 =𝟏 =𝟏 𝐔 𝐝(𝛈) 𝐝(𝛈) 𝛈 = 𝟎 𝐔 𝛕𝐰 = 𝛍 𝛅 4- Momentum integral equation 𝐝𝛉 𝛕𝐰 = 𝐝𝐱 𝛒 𝐔 𝟐 𝟏 𝐔 𝛉= 𝛅 𝛕𝐰 = 𝛍 𝟔 𝛅 𝟏 𝐔 𝐝 𝟔 𝛅 𝛍 = 𝛅𝟐 𝐝𝐱 𝛒𝐔  𝟏 𝐔 𝐝 𝟔 𝛅 𝛍 = 𝛅𝟐 𝐝𝐱 𝛒𝐔 𝟏𝐝 𝛅 𝛍 = 𝟔 𝐝𝐱 𝛒𝐔𝛅 𝟔𝛍 𝛅𝐝 𝛅 = 𝐝𝐱 𝛒𝐔 𝛅 𝒙 𝟔𝛍 න 𝛅𝐝 𝛅 = න 𝐝𝐱 𝟎 𝟎 𝛒𝐔 𝛅 𝟔𝛍 𝒙 න 𝛅𝐝 𝛅 = න 𝐝𝐱 𝟎 𝛒𝐔 𝟎 d x 𝜹𝟐 𝟔𝛍 = 𝒙 𝟐 𝛒𝐔 0 0  𝜹𝟐 𝟔𝛍 = 𝒙 𝟐 𝛒𝐔 𝜹𝟐 𝟔𝛍 𝟐 = 𝒙 × 𝟐 𝟐 𝛒𝐔 𝒙 𝜹𝟐 𝟏𝟐𝛍 𝟏𝟐 𝟐 = = 𝒙 𝛒𝐔𝒙 𝑹𝒆𝒙 𝜹 𝟏𝟐 = 𝒙 𝑹𝒆𝒙 𝜹 𝟑. 𝟒𝟔𝟒 = 𝒙 𝑹𝒆𝒙 𝟑. 𝟒𝟔𝟒𝒙 𝜹= 𝑹𝒆𝒙 The calculated wall shear stress was: 𝐔 𝛕𝐰 = 𝛍 𝛅 We can get it as a function of local position x: 𝐔 𝟑. 𝟒𝟔𝟒𝒙 𝛕𝐰 = 𝛍 𝐔 𝑹𝒆𝒙 𝜹≅ 𝟑. 𝟒𝟔𝟒𝒙 𝛕𝐰 = 𝛍 𝑹𝒆𝒙 𝟑. 𝟒𝟔𝟒𝒙 𝑹𝒆𝒙 5- Friction Coefficient Cf 𝛕𝐰 𝐂𝐟 = 𝛍𝐔 𝑹𝒆𝒙 𝟎. 𝟓𝟕𝟕 𝛍 𝑹𝒆𝒙 𝟎. 𝟓𝟕𝟕 𝑹𝒆𝒙 𝟎. 𝟓𝛒 𝐔𝟐 𝐂𝐟 = 𝐂𝐟 = 𝐂𝐟 = 𝟎. 𝟓𝛒 𝐔 𝟐 𝟑. 𝟒𝟔𝟒𝒙 𝛒𝐔𝒙 𝑹𝒆𝒙 𝟎. 𝟓𝟕𝟕 𝐂𝐟 = 𝑹𝒆𝒙 𝟎. 𝟓𝟕𝟕 𝐂𝐟 = 𝑹𝒆𝒙 6- Drag Force D L D = න 𝛕𝐰 dA dA = Bdx 0 𝟎. 𝟓𝟕𝟕 𝛕𝐰 𝛕𝐰 𝒂 𝐂𝐟 = = 𝟐 = Where a = 0.577 𝑹𝒆𝒙 𝟎. 𝟓𝛒 𝐔 𝟐 𝟎. 𝟓𝛒 𝐔 𝑹𝒆𝒙 𝒂 𝛕𝐰 = 𝟎. 𝟓𝛒 𝐔𝟐 = 𝟎. 𝟓𝛒 𝐔𝟐 𝒂 𝑹𝒆𝒙 −𝟎.𝟓 𝑹𝒆𝒙 𝟎.𝟓 𝟐 𝝁 𝛕𝐰 = 𝟎. 𝟓𝛒 𝐔 𝒂 𝛒𝐔𝒙 L 𝟎.𝟓 L 𝟎.𝟓 𝟐 𝝁 𝟐 𝝁 D = න 𝟎. 𝟓𝛒 𝐔 𝒂 Bdx = න 𝟎. 𝟓𝛒 𝐔 𝒂 B𝒙−0.5 dx 0 𝛒𝐔𝒙 0 𝛒𝐔 𝟎.𝟓 L 𝝁 D = 𝟎. 𝟓𝒂B𝛒 𝐔𝟐 න 𝒙−0.5 dx 𝛒𝐔 0 𝟎.𝟓 L 𝝁 𝒙0.5 D = 𝟎. 𝟓𝒂B𝛒 𝐔𝟐 𝛒𝐔 𝟎. 𝟓 0  𝟎.𝟓 𝝁 𝟎.𝟓 0.5 𝑳0.5 𝝁 𝑳0.5 D = 𝒂B𝛒 𝐔𝟐 𝑳 × D = 𝟎. 𝟓𝒂B𝛒 𝐔𝟐 𝛒𝐔 𝑳0.5 𝛒𝐔 𝟎. 𝟓 𝟎.𝟓 𝝁 D = 𝒂B𝑳𝛒 𝐔𝟐 𝛒𝐔𝑳 𝟎.𝟓 𝝁 D = 𝒂B𝑳𝛒 𝐔𝟐 𝛒𝐔𝑳 𝝁 A = BL, a=0.577, 𝑹𝒆𝑳 = 𝛒𝐔𝑳 𝒂𝐴𝛒 𝐔𝟐 𝟐𝒂𝐴 𝟎. 𝟓𝛒 𝐔𝟐 D= D= 𝑹𝒆𝑳 𝑹𝒆𝑳 7- Drag Coefficient CD: 𝟐𝒂𝐴 𝟎. 𝟓𝛒 𝐔𝟐 D 𝟐𝒂 D= = = 𝑪𝑫 𝑹𝒆𝑳 𝟎. 𝟓𝛒𝐔𝟐 𝐴 𝑹𝒆𝑳 D 𝟐𝒂 𝟐(𝟎. 𝟓𝟕𝟕) 𝟏. 𝟏𝟓𝟒 𝟏. 𝟏𝟓𝟒 𝑪𝑫 = = = = 𝑪𝑫 = 𝟎. 𝟓𝛒𝐔𝟐 𝐴 𝑹𝒆𝑳 𝑹𝒆𝑳 𝑹𝒆𝑳 𝑹𝒆𝑳 Dimensionless profile shape factor H: 𝛅⋆ 𝛅⋆ /𝛅 𝟏/𝟐 𝑯= = = 𝛉 𝛉/𝛅 𝟏/𝟔 𝑯=𝟑 Aeronautical & Aerospace Engineering Flow Machines & Advanced Fluid Dynamics MEC351 Lecture 7 Differential relations for a fluid particle Cylindrical Coordinates Dr. Mohammed Rabie Assistant Professor – Mansoura University Differential relations for a fluid particle Cylindrical Coordinates Flow rate, mean velocity, and maximum velocity: 𝑄 = න 𝑢 ⅆ𝐴 𝐴 𝑄 𝑢𝑚𝑒𝑎𝑛 = 𝐴 ⅆ𝑢 To get maximum velocity location :Umax @ =0 ⅆ𝑦 (get y for max velocity) Cylindrical Polar coordinates Velocity in r direction → 𝜐𝑟 Velocity in  direction → 𝜐𝜃 Velocity in z direction → 𝜐𝑧  Navier–Stokes equations for cylindrical polar coordinates Continuity: General Incompressible Momentum r − Direction:  Navier–Stokes equations for cylindrical polar coordinates Momentum  − Direction: 𝜕𝜐𝜃 𝜕𝜐𝜃 𝜐𝜃 𝜕𝜐𝜃 𝜐𝜃 𝜐𝑟 𝜕𝜐𝜃 𝜌 + 𝜐𝑟 + + + 𝜐𝑧 𝜕𝑡 𝜕𝑟 𝑟 𝜕𝜃 𝑟 𝜕𝑧 1 𝜕𝑝 𝜕 1 𝜕(𝑟𝜐𝜃 ) 1 𝜕 2 𝜐𝜃 2 𝜕𝜐𝑟 𝜕 2 𝜐𝜃 =− + 𝜌𝑔𝜃 + 𝜇 + 2 2 + 2 + 𝑟 𝜕𝜃 𝜕𝑟 𝑟 𝜕𝑟 𝑟 𝜃 𝑟 𝜕𝜃 𝜕𝑧 2 Momentum z − Direction: Steady Axisymmetric flow: 𝝏∎ Steady → =𝟎 𝝏𝒕 𝝏∎ Axisymmetric → =𝟎 𝝏𝜽 Types of Axisymmetric flow: Axisymmetric Rectilinear Axisymmetric Torsional Axisymmetric Radial 𝜐𝑟 = 𝜐𝜃 = 0 𝜐𝑟 = 𝜐𝑧 = 0 𝜐𝜃 = 𝜐𝑧 = 0 𝜐𝑧 ≠ 0 𝜐𝜃 ≠ 0 𝜐𝑟 ≠ 0  1- Axisymmetric Rectilinear 𝝊𝒓 = 𝝊𝜽 = 𝟎 𝝊𝒛 ≠ 𝟎 Continuity: 0 0 Get that: 𝝏𝝊𝒛 =𝟎 𝝏𝒛 Momentum Eq. in Z direction: 0 𝝊𝒓 = 𝟎 𝝊𝜽 = 𝟎 𝝏𝝊𝒛 𝝏𝝊𝒛 𝝏𝝊𝒛 =𝟎 =𝟎 =𝟎 Steady 𝝏𝒛 𝝏𝜽 𝝏𝒛 Axisymmetric  𝝏𝑷 𝟏 𝝏 𝝏𝝊𝒛 − + 𝝆𝒈𝒛 + 𝝁 𝒓 =𝟎 𝝏𝒛 𝒓 𝝏𝒓 𝝏𝒓 𝟏 𝝏 𝝏𝝊𝒛 𝝏𝑷 𝝁 𝒓 = − 𝝆𝒈𝒛 𝒓 𝝏𝒓 𝝏𝒓 𝝏𝒛 𝝏 𝝏𝝊𝒛 𝟏 𝝏𝑷 𝒓 = − 𝝆𝒈𝒛 𝒓 Integrate 𝝏𝒓 𝝏𝒓 𝝁 𝝏𝒛 𝝏𝝊𝒛 𝟏 𝝏𝑷 𝒓 = − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 Eqn. 1→ /r 𝝏𝒓 𝟐𝝁 𝝏𝒛 𝝏𝝊𝒛 𝟏 𝝏𝑷 𝑪𝟏 = − 𝝆𝒈𝒛 𝒓 + Integrate 𝝏𝒓 𝟐𝝁 𝝏𝒛 𝒓 𝟏 𝝏𝑷 𝝊𝒛 = − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 Eqn. 2 𝟒𝝁 𝝏𝒛  𝝏𝝊𝒛 𝟏 𝝏𝑷 𝒓 = − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 Eqn. 1→ /r 𝝏𝒓 𝟐𝝁 𝝏𝒛 𝟏 𝝏𝑷 𝝊𝒛 = − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 𝝏𝒛 Pressure Gravity cause flow cause flow  b- Fully developed flow in annulus Fluid flow vertically downward around a pipe of radius a due to gravity 𝟏 𝝏𝑷 𝝊𝒛 = − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 𝝏𝒛 As flow due to gravity only, so neglect 𝝏𝑷 pressure gradient =𝟎 𝝏𝒛 𝟏 𝝊𝒛 = 𝟎 − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 𝝆𝒈𝒛 𝟐 𝝊𝒛 = − 𝒓 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁  𝝆𝒈𝒛 𝟐 𝝊𝒛 = − 𝒓 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 Apply boundary conditions to get C1 and C2 𝝏𝝊𝒛 @ r = b, no shear stress (maximum velocity) , = 0, Apply in the previous eqn. 1 𝝏𝒓 𝝏𝝊𝒛 𝟏 𝒓 = −𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 𝝏𝒓 𝟐𝝁 𝟏 𝝆𝒈𝒛 𝒃𝟐 𝒃 𝟎 = −𝝆𝒈𝒛 𝒃𝟐 + 𝑪𝟏 → get C1 = 𝟐𝝁 𝟐𝝁 At r = a, 𝝊𝒛 = 𝟎 𝝆𝒈𝒛 𝟐 𝝊𝒛 = − 𝒓 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 𝝆𝒈𝒛 𝟐 𝝆𝒈𝒛 𝒃𝟐 𝟎=− 𝒂 + 𝒍𝒏 𝒂 + 𝑪𝟐 𝟒𝝁 𝟐𝝁 𝝆𝒈𝒛 𝒂𝟐 𝝆𝒈𝒛 𝒃𝟐 𝑪𝟐 = − 𝒍𝒏 𝒂 𝟒𝝁 𝟐𝝁  𝝆𝒈𝒛 𝟐 𝝊𝒛 = − 𝒓 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 𝝆𝒈𝒛 𝒃𝟐 𝝆𝒈𝒛 𝒂𝟐 𝝆𝒈𝒛 𝒃𝟐 C1 = 𝑪𝟐 = − 𝒍𝒏 𝒂 𝟐𝝁 𝟒𝝁 𝟐𝝁 𝝆𝒈𝒛 𝟐 𝝆𝒈𝒛 𝒃𝟐 𝝆𝒈𝒛 𝒂𝟐 𝝆𝒈𝒛 𝒃𝟐 𝝊𝒛 = − 𝒓 + 𝒍𝒏 𝒓 + − 𝒍𝒏 𝒂 𝟒𝝁 𝟐𝝁 𝟒𝝁 𝟐𝝁 𝒈𝒛 = 𝒈 𝝆𝒈 𝟐 𝝆𝒈𝒃𝟐 𝝆𝒈𝒂𝟐 𝝆𝒈𝒃𝟐 𝝊𝒛 = − 𝒓 + 𝒍𝒏 𝒓 + − 𝒍𝒏 𝒂 𝟒𝝁 𝟐𝝁 𝟒𝝁 𝟐𝝁 Report: Find the flow rate, average velocity, maximum velocity location, maximum velocity value 2- Axisymmetric Torsional flow 𝑢𝑟 = 0 𝑢𝑧 = 0 𝑢𝜃 = 𝑓 𝑟 𝜕 =0 𝜕𝜃 No gravity in 𝜃 direction No pressure gradient  Flow between long concentric cylinders 𝜕𝑢𝜃 𝜕𝑢𝜃 𝑢𝜃 𝜕𝑢𝜃 𝑢𝜃 𝑢𝑟 𝜕𝑢𝜃 1 𝜕𝑝 𝜕 1 𝜕(𝑟𝑢𝜃 ) 1 𝜕 2 𝑢𝜃 2 𝜕𝑢𝑟 𝜕 2 𝑢𝜃 𝜌 + 𝑢𝑟 + + + 𝑢𝑧 =− + 𝜌𝑔𝜃 + 𝜇 + 2 + + 𝜕𝑡 𝜕𝑟 𝑟 𝜕𝜃 𝑟 𝜕𝑧 𝑟 𝜕𝜃 𝜕𝑟 𝑟 𝜕𝑟 𝑟 𝜕𝜃 2 𝑟 2 𝜕𝜃 𝜕𝑧 2 0 𝑢𝑟 =0 𝑢 =0 𝑁𝑜 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑔𝜃 = 0 𝜕 𝑢𝑟 =0 = 0 𝑢𝑟 =0 𝑧 𝜕𝑢𝜃 𝜕 =0 Steady 𝜕𝜃 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝜕𝜃 𝜕𝑧 𝜕 1 𝜕(𝑟𝑢𝜃 ) = 0 𝜕𝑟 𝑟 𝜕𝑟 1 𝜕(𝑟𝑢𝜃 ) = C1 𝑟 𝜕𝑟 𝜕(𝑟𝑢𝜃 ) = C1r 𝜕𝑟  Flow between long concentric cylinders 𝜕(𝑟𝑢𝜃 ) = C1r 𝜕𝑟 𝑟2 𝑟𝑢𝜃 = C1 + C2 2 𝑐2 𝑢𝜃 = C1 r + 𝑟 Apply boundary conditions to get constants At r=ri 𝑢𝜃 = i ri At r=ro 𝑢𝜃 =  Flow between long concentric cylinders  ρVD Reynolds Number → Re = μ ρ → Fluid density, kg/m3 D → Pipe diameter, m ( D = 2R) V → Average flow velocity, m/s (From Hagen – Poiseuille flow) 1 𝜕P 2 UAverage = − R 8μ 𝜕Z 𝜕P ΔP = → Pressure gradient, Pa/m 𝜕Z L μ → dynamic viscosity, N. s/m2 = Pa.s = kg/(m. s),…. Re < 2000, Laminar flow Re > 4000, Turbulent flow 2000 < Re < 4000, Transition Example: Solution: 𝟏 𝝏𝑷 𝝊𝒛 = − 𝝆𝒈𝒛 𝒓𝟐 + 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝟒𝝁 𝝏𝒛 𝝏𝑷 − 𝝆𝒈𝒛 = 𝟎 𝝏𝒛 𝝊𝒛 = 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 At r = a, 𝝊𝒛 = U 𝑼 = 𝑪𝟏 𝒍𝒏 𝒂 + 𝑪𝟐 →1 At r = b, 𝝊𝒛 = 0 𝟎 = 𝑪𝟏 𝒍𝒏 𝒃 + 𝑪𝟐 →2 𝑼 Eqn 1 – Eqn 2 → 𝑼 = 𝑪𝟏 𝒍𝒏 𝒂/𝒃 𝑪𝟏 = 𝒍𝒏 𝒂/𝒃 Put C1 in Eqn 2 → 𝑼 𝑼 𝟎= 𝒍𝒏 𝒃 + 𝑪𝟐 𝑪𝟐 = − 𝒍𝒏 𝒃 𝒍𝒏 𝒂/𝒃 𝒍𝒏 𝒂/𝒃  𝝊𝒛 = 𝑪𝟏 𝒍𝒏 𝒓 + 𝑪𝟐 𝑼 𝑼 𝝊𝒛 = 𝒍𝒏 𝒓 − 𝒍𝒏 𝒃 𝒍𝒏 𝒂/𝒃 𝒍𝒏 𝒂/𝒃 𝑼 𝝊𝒛 = 𝒍𝒏 𝒓 − 𝒍𝒏 𝒃 𝒍𝒏 𝒂/𝒃 𝑼𝒍𝒏 𝒓/𝒃 𝝊𝒛 = 𝒍𝒏 𝒂/𝒃 Given:  = 870 kg/m3,  = 0.104 kg/(m.s) Solution