Flow Machines & Advanced Fluid Dynamics Lectures 1-4 PDF

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 Tutorial 1 Introduction to Flow machines and Advanced Fluid Dynamics Dr. Mohammed Rabie Assistant Professor – Mansoura University  ⅆ𝑚 + න  𝑣. [ Q𝐴 = 0 ⅆ a 𝑐𝑠 ⅆ𝑣 Incompressible Fluid, deformable control volume ➔  ⅆ a ⅆ  ෍ 𝑚ሶ \ b a − ෍ 𝑚ሶ 𝑖𝑛 Compressible Fluid, Fixed control volume ➔ 𝑣 ⅆ a For definite number of ⅆ 𝑣 inlets and outlets Compressible Fluid, deformable control volume ➔ ⅆ a ⅆ𝑚 Incompressible Fluid, Fixed control volume ➔ =0 ⅆ a  Solution Q𝑚 For Steady Flow ➔ =0 0 = 𝑚ሶ 𝑖 − 𝑚ሶ \ 𝑚ሶ 𝑖 = 𝑚ሶ \ Q a Water ➔ density is constant 𝑄𝑖 = 𝑄 \ b Q1 + Q w = Q 2 𝜋 𝜋 12 0.08 + (0.15)(0.3016) = V2 0.082 2 4 4 𝐕 Ð = Ð𝟏 𝐦/𝐬 If h = 1m – 30 cm = 1- 0.3 = 0.7m → find t Δℎ 0.7 Δ a = = = 45.75 𝑠 R𝑐 Qℎ 0.0153 Q a 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 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 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 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 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 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 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 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 Tutorial 4 RTT Energy conservation Dr. Mohammed Rabie Assistant Professor – Mansoura University Energy Thanks