Control Volume and Reynolds Transport Theorem 10-11-2013 Final

57:020 Fluids Mechanics Fall2013 Control Volume and Reynolds Transport Theorem 10. 11. 2013 Hyunse Yoon, Ph.D. Assistant Research Scientist IIHR-Hydroscience & Engineering University of Iowa 1 57:020 Fluids Mechanics Fall2013 Reynolds Transport Theorem (RTT) • An analytical tool to shift from describing the laws governing fluid motion using the system concept to using the control volume concept 2 57:020 Fluids Mechanics Fall2013 System vs. Control Volume • System: A collection of matter of fixed identity – Always the same atoms or fluid particles – A specific, identifiable quantity of matter • Control Volume (CV): A volume in space through which fluid may flow – A geometric entity – Independent of mass 3 57:020 Fluids Mechanics Fall2013 Examples of CV Fixed CV CV fixed at a nozzle Moving CV CV moving with ship Deforming CV CV deforming within cylinder 4 57:020 Fluids Mechanics Fall2013 Laws of Mechanics 1. Conservation of mass: 2. Conservation of linear momentum: 3. Conservation of angular momentum: 4. Conservation of Energy: 𝑑𝑑𝑑𝑑 =0 𝑑𝑑𝑑𝑑 𝐹𝐹 = 𝑚𝑚𝑎𝑎 = 𝑚𝑚 𝑀𝑀 = 𝑑𝑑𝐻𝐻 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑄𝑄̇ − 𝑊𝑊̇ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑉𝑉 𝑑𝑑 = 𝑚𝑚𝑉𝑉 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 • The laws apply to either solid or fluid systems • Ideal for solid mechanics, where we follow the same system • For fluids, the laws need to be rewritten to apply to a specific region in the neighborhood of our product (i.e., CV) 5 57:020 Fluids Mechanics Fall2013 Extensive vs. Intensive Property Governing Differential Equations (GDE’s): 𝑑𝑑 𝑚𝑚, 𝑚𝑚𝑉𝑉, 𝐸𝐸 = 0, 𝐹𝐹, 𝑄𝑄̇ − 𝑊𝑊̇ 𝑑𝑑𝑑𝑑 𝐵𝐵 • 𝐵𝐵 = The amount of 𝑚𝑚, 𝑚𝑚𝑉𝑉, or 𝐸𝐸 contained in the total mass of a system or a CV; Extensive property – Dependent on mass • 𝛽𝛽 (or 𝑏𝑏) = The amount of 𝐵𝐵 per unit mass; Intensive property – Independent on mass 𝑑𝑑𝑑𝑑 for nonuniform 𝐵𝐵) 𝛽𝛽 or 𝑏𝑏 = 𝐵𝐵/𝑚𝑚 (= 𝑑𝑑𝑑𝑑 𝐵𝐵 = 𝛽𝛽 ⋅ 𝑚𝑚 = � 𝛽𝛽 𝜌𝜌𝜌𝜌𝜌𝜌 � for nonuniform 𝛽𝛽 𝑉𝑉 =𝑑𝑑𝑑𝑑 𝐸𝐸 𝑒𝑒 6 57:020 Fluids Mechanics Fall2013 Fixed CV At time 𝑡𝑡: SYS = CV 𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡 = 𝐵𝐵𝐶𝐶𝐶𝐶 (𝑡𝑡) At time 𝑡𝑡 + 𝛿𝛿𝛿𝛿: SYS = (CV – I) + II 𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡 + 𝛿𝛿𝛿𝛿 = 𝐵𝐵𝐶𝐶𝐶𝐶 𝑡𝑡 + 𝛿𝛿𝛿𝛿 − 𝐵𝐵𝐼𝐼 𝑡𝑡 + 𝛿𝛿𝛿𝛿 + 𝐵𝐵𝐼𝐼𝐼𝐼 𝑡𝑡 + 𝛿𝛿𝛿𝛿 7 57:020 Fluids Mechanics Fall2013 Time Rate of Change of 𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝛿𝛿𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡 + 𝛿𝛿𝛿𝛿 − 𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡) = 𝛿𝛿𝛿𝛿 𝛿𝛿𝛿𝛿 𝐵𝐵𝐶𝐶𝐶𝐶 𝑡𝑡 + 𝛿𝛿𝛿𝛿 − 𝐵𝐵𝐼𝐼 𝑡𝑡 + 𝛿𝛿𝛿𝛿 + 𝐵𝐵𝐼𝐼𝐼𝐼 𝑡𝑡 + 𝛿𝛿𝛿𝛿 = 𝛿𝛿𝛿𝛿 ∴ − 𝐵𝐵𝐶𝐶𝐶𝐶 𝑡𝑡 𝛿𝛿𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝐵𝐵𝐶𝐶𝐶𝐶 𝑡𝑡 + 𝛿𝛿𝛿𝛿 − 𝐵𝐵𝐶𝐶𝐶𝐶 𝑡𝑡 𝐵𝐵𝐼𝐼𝐼𝐼 (𝑡𝑡 + 𝛿𝛿𝛿𝛿) 𝐵𝐵𝐼𝐼 𝑡𝑡 + 𝛿𝛿𝛿𝛿 = + − 𝛿𝛿𝛿𝛿 𝛿𝛿𝛿𝛿 𝛿𝛿𝛿𝛿 𝛿𝛿𝛿𝛿 1) Change of 𝐵𝐵 within CV over 𝛿𝛿𝛿𝛿 2) Amount of 𝐵𝐵 flowing out through CS over 𝛿𝛿𝛿𝛿 Now, take limit of 𝛿𝛿𝛿𝛿 → 0 to Eq. (1) term by term 3) Amountt of 𝐵𝐵 flowing in through CS over 𝛿𝛿𝛿𝛿 Eq. (1) 8 57:020 Fluids Mechanics Fall2013 LHS of Eq. (1) 𝛿𝛿𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡 + 𝛿𝛿𝛿𝛿 − 𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡) lim = lim = 𝛿𝛿𝛿𝛿→0 𝛿𝛿𝛿𝛿 𝛿𝛿𝛿𝛿→0 𝛿𝛿𝛿𝛿 𝑑𝑑𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑𝑑𝑑 Time rate of change of 𝐵𝐵 within the system 𝐷𝐷𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 or, = ; material derivative 𝐷𝐷𝐷𝐷 9 57:020 Fluids Mechanics Fall2013 First term of RHS of Eq.(1) 𝑑𝑑 𝐵𝐵𝐶𝐶𝐶𝐶 𝑡𝑡 + 𝛿𝛿𝛿𝛿 − 𝐵𝐵𝐶𝐶𝐶𝐶 (𝑡𝑡) 𝑑𝑑𝐵𝐵𝐶𝐶𝐶𝐶 = = � 𝛽𝛽𝛽𝛽𝛽𝛽𝛽𝛽 𝛿𝛿𝛿𝛿→0 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝐶𝐶𝑉𝑉 𝛿𝛿𝛿𝛿 lim Time rate of change of 𝐵𝐵 withich CV 10 57:020 Fluids Mechanics Fall2013 2nd term of RHS of Eq.(1) and 𝛿𝛿𝑚𝑚𝑜𝑜𝑜𝑜𝑜𝑜 = 𝜌𝜌𝜌𝜌𝑉𝑉 𝛿𝛿𝑉𝑉 = 𝛿𝛿𝛿𝛿 ⋅ 𝛿𝛿ℓ𝑛𝑛 = 𝛿𝛿𝛿𝛿 ⋅ 𝛿𝛿𝛿 � cos 𝜃𝜃 = 𝛿𝛿𝛿𝛿 ⋅ (𝑉𝑉𝑉𝑉𝑉𝑉 cos 𝜃𝜃) =𝑉𝑉𝑉𝑉𝑉𝑉 Thus, the amount of 𝐵𝐵 flowing out of CV through 𝛿𝛿𝛿𝛿 over a short time 𝛿𝛿𝛿𝛿: ∴ 𝛿𝛿𝐵𝐵𝑜𝑜𝑜𝑜𝑜𝑜 = 𝛽𝛽𝛽𝛽𝑚𝑚𝑜𝑜𝑜𝑜𝑜𝑜 = 𝛽𝛽𝛽𝛽𝛽𝛽 cos 𝜃𝜃 𝛿𝛿𝛿𝛿𝛿𝛿𝛿𝛿 11 57:020 Fluids Mechanics Fall2013 12 2nd term of RHS of Eq.(1) – Contd. By integrating 𝛿𝛿𝐵𝐵𝑜𝑜𝑜𝑜𝑜𝑜 over the entire outflow portion of CS, Thus, 𝐵𝐵𝐼𝐼𝐼𝐼 𝑡𝑡 + 𝛿𝛿𝛿𝛿 = � 𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢 𝑑𝑑𝐵𝐵𝑜𝑜𝑜𝑜𝑜𝑜 = � 𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢 𝛽𝛽𝛽𝛽𝛽𝛽 cos 𝜃𝜃 𝛿𝛿𝛿𝛿𝛿𝛿𝛿𝛿 𝐵𝐵𝐼𝐼𝐼𝐼 (𝑡𝑡 + 𝛿𝛿𝛿𝛿) 1 = lim 𝛿𝛿𝛿𝛿 � 𝛽𝛽𝛽𝛽𝛽𝛽 cos 𝜃𝜃 𝑑𝑑𝑑𝑑 𝛿𝛿𝛿𝛿→0 𝛿𝛿𝛿𝛿→0 𝛿𝛿𝛿𝛿 𝛿𝛿𝛿𝛿 𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢 lim =� 𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢 �, Note that 𝑉𝑉 cos 𝜃𝜃 = 𝑉𝑉 ⋅ 𝒏𝒏 ∴ 𝐵𝐵̇out = � 𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢 𝛽𝛽𝛽𝛽 𝑉𝑉 cos 𝜃𝜃 𝑑𝑑𝑑𝑑 =𝑉𝑉𝑛𝑛 ≡ 𝐵𝐵̇𝑜𝑜𝑜𝑜𝑜𝑜 i.e., Out flux of 𝐵𝐵 through CS �𝑑𝑑𝑑𝑑 𝛽𝛽𝛽𝛽𝑉𝑉 ⋅ 𝒏𝒏 𝑎𝑎 ⋅ 𝑏𝑏 = 𝑎𝑎 𝑏𝑏 cos 𝜃𝜃 57:020 Fluids Mechanics Fall2013 3rd term of RHS of Eq.(1) and 𝛿𝛿𝑉𝑉 = 𝛿𝛿𝛿𝛿 ⋅ 𝛿𝛿ℓ𝑛𝑛 = 𝛿𝛿𝛿𝛿 ⋅ 𝛿𝛿𝑚𝑚𝑖𝑖𝑖𝑖 = 𝜌𝜌𝜌𝜌𝑉𝑉 𝛿𝛿𝛿 � =𝑉𝑉𝑉𝑉𝑉𝑉 − cos 𝜃𝜃 <0 = 𝛿𝛿𝛿𝛿 ⋅ (−𝑉𝑉𝑉𝑉𝑉𝑉 cos 𝜃𝜃) Thus, the amount of 𝐵𝐵 flowing out of CV through 𝛿𝛿𝛿𝛿 over a short time 𝛿𝛿𝛿𝛿: ∴ 𝛿𝛿𝐵𝐵𝑖𝑖𝑖𝑖 = 𝛽𝛽𝛽𝛽𝑚𝑚𝑖𝑖𝑖𝑖 = −𝛽𝛽𝛽𝛽𝛽𝛽 cos 𝜃𝜃 𝛿𝛿𝛿𝛿𝛿𝛿𝛿𝛿 13 57:020 Fluids Mechanics Fall2013 3rd term of RHS of Eq.(1) – Contd. By integrating 𝛿𝛿𝐵𝐵𝑜𝑜𝑜𝑜𝑜𝑜 over the entire outflow portion of CS, Thus, 𝐵𝐵𝐼𝐼 𝑡𝑡 + 𝛿𝛿𝛿𝛿 = � 𝐶𝐶𝑆𝑆𝑖𝑖𝑖𝑖 𝑑𝑑𝐵𝐵𝑖𝑖𝑖𝑖 = � 𝐶𝐶𝑆𝑆𝑖𝑖𝑖𝑖 −𝛽𝛽𝛽𝛽𝛽𝛽 cos 𝜃𝜃 𝛿𝛿𝛿𝛿𝛿𝛿𝛿𝛿 𝐵𝐵𝐼𝐼 (𝑡𝑡 + 𝛿𝛿𝛿𝛿) 1 = lim 𝛿𝛿𝛿𝛿 � −𝛽𝛽𝛽𝛽𝛽𝛽 cos 𝜃𝜃 𝑑𝑑𝑑𝑑 𝛿𝛿𝛿𝛿→0 𝛿𝛿𝛿𝛿→0 𝛿𝛿𝛿𝛿 𝛿𝛿𝛿𝛿 𝐶𝐶𝑆𝑆𝑖𝑖𝑖𝑖 lim = −� 𝐶𝐶𝑆𝑆𝑖𝑖𝑖𝑖 �, Note that 𝑉𝑉 cos 𝜃𝜃 = 𝑉𝑉 ⋅ 𝒏𝒏 ∴ 𝐵𝐵̇in = − � 𝐶𝐶𝑆𝑆𝑖𝑖𝑖𝑖 𝛽𝛽𝛽𝛽 𝑉𝑉 cos 𝜃𝜃 𝑑𝑑𝑑𝑑 =−𝑉𝑉𝑛𝑛 �𝑑𝑑𝑑𝑑 𝛽𝛽𝛽𝛽𝑉𝑉 ⋅ 𝒏𝒏 ≡ 𝐵𝐵̇𝑖𝑖𝑖𝑖 i.e., influx of 𝐵𝐵 through CS 14 57:020 Fluids Mechanics Fall2013 RTT for Fixed CV Now the relationship between the time rate of change of 𝐵𝐵 for the system and that for the CV is given by, 𝐷𝐷𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑 �𝑑𝑑𝑑𝑑 − − � 𝛽𝛽𝛽𝛽𝑉𝑉 ⋅ 𝒏𝒏 �𝑑𝑑𝑑𝑑 = � 𝛽𝛽𝛽𝛽𝛽𝛽𝛽𝛽 + � 𝛽𝛽𝛽𝛽𝑉𝑉 ⋅ 𝒏𝒏 𝐷𝐷𝐷𝐷 𝑑𝑑𝑑𝑑 𝐶𝐶𝑉𝑉 𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢 𝐶𝐶𝑆𝑆𝑖𝑖𝑖𝑖 𝐵𝐵̇𝑜𝑜𝑜𝑜𝑜𝑜 𝐵𝐵̇𝑖𝑖𝑖𝑖 With the fact that 𝐶𝐶𝐶𝐶 = 𝐶𝐶𝑆𝑆𝑜𝑜𝑜𝑜𝑜𝑜 + 𝐶𝐶𝑆𝑆𝑖𝑖𝑖𝑖 , 𝐷𝐷𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑 �𝑑𝑑𝑑𝑑 = � 𝛽𝛽𝛽𝛽𝛽𝛽𝛽𝛽 + � 𝛽𝛽𝛽𝛽𝑉𝑉 ⋅ 𝒏𝒏 𝐷𝐷𝐷𝐷 𝑑𝑑𝑑𝑑 𝐶𝐶𝑉𝑉 𝐶𝐶𝑆𝑆 Time rate of change of 𝐵𝐵 within a system = Time rate of change of 𝐵𝐵 within CV + Net flux of 𝐵𝐵 through CS = 𝐵𝐵̇𝑜𝑜𝑜𝑜𝑜𝑜 − 𝐵𝐵̇𝑖𝑖𝑖𝑖 15 57:020 Fluids Mechanics Fall2013 Example 1 16 57:020 Fluids Mechanics Fall2013 Example 1 - Contd. 𝐵𝐵̇𝑜𝑜𝑜𝑜𝑜𝑜 = � 𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢 �𝑑𝑑𝑑𝑑 𝜌𝜌𝜌𝜌𝑽𝑽 ⋅ 𝒏𝒏 (4.16) � = 𝑉𝑉 cos 𝜃𝜃 , With 𝛽𝛽= 1 and 𝑽𝑽 ⋅ 𝒏𝒏 𝐵𝐵̇𝑜𝑜𝑜𝑜𝑜𝑜 = � 𝜌𝜌𝜌𝜌 cos 𝜃𝜃 𝑑𝑑𝑑𝑑 = 𝜌𝜌𝜌𝜌 cos 𝜃𝜃 � 𝑑𝑑𝑑𝑑 𝐶𝐶𝐷𝐷 where = 𝜌𝜌𝜌𝜌 cos 𝜃𝜃 𝐴𝐴𝐶𝐶𝐶𝐶 𝐴𝐴𝐶𝐶𝐶𝐶 = ℓ × (2 𝑚𝑚) = 0.5 𝑚𝑚 cos 𝜃𝜃 2 𝑚𝑚 = 𝐶𝐶𝐷𝐷 =𝐴𝐴𝐶𝐶𝐶𝐶 1 𝑚𝑚2 cos 𝜃𝜃 17 57:020 Fluids Mechanics Fall2013 Example 1 - Contd. Thus, with 𝜌𝜌 = 1,000 kg/m3 for water and 𝑉𝑉 = 3 m/s, 𝐵𝐵̇𝑜𝑜𝑜𝑜𝑜𝑜 = 1,000 kg m3 m 1 3 cos 𝜃𝜃 m2 = 3,000 kg/s s cos 𝜃𝜃 With 𝛽𝛽 = 1/𝜌𝜌, 𝐵𝐵̇𝑜𝑜𝑜𝑜𝑜𝑜 = � 𝑉𝑉 cos 𝜃𝜃 𝑑𝑑𝑑𝑑 = 𝑉𝑉 cos 𝜃𝜃 � 𝑑𝑑𝑑𝑑 = 𝑉𝑉 cos 𝜃𝜃 𝐴𝐴𝐶𝐶𝐶𝐶 𝐶𝐶𝐷𝐷 𝐶𝐶𝐷𝐷 =𝐴𝐴𝐶𝐶𝐶𝐶 =1/ cos 𝜃𝜃 m 1 = 3 cos 𝜃𝜃 m2 = 3 m3 ⁄s (𝑖𝑖. 𝑒𝑒. , volume flow rate) cos 𝜃𝜃 s Note: These results are the same for all 𝜃𝜃 values 18 57:020 Fluids Mechanics Fall2013 19 Special Case: 𝑉𝑉= constant over discrete CS’s 𝐵𝐵̇𝑖𝑖𝑖𝑖 = � 𝐶𝐶𝑆𝑆𝑖𝑖𝑖𝑖 𝐵𝐵̇𝑜𝑜𝑜𝑜𝑜𝑜 = � � 𝑑𝑑𝑑𝑑 = � 𝛽𝛽𝑖𝑖 𝜌𝜌𝑖𝑖 𝑉𝑉𝑖𝑖 𝐴𝐴𝑖𝑖 𝛽𝛽𝛽𝛽 𝑉𝑉� ⋅ 𝒏𝒏 𝐶𝐶𝑆𝑆𝑜𝑜𝑜𝑜𝑜𝑜 � 𝑑𝑑𝑑𝑑 = � 𝛽𝛽𝑗𝑗 𝜌𝜌𝑗𝑗 𝑉𝑉𝑗𝑗 𝐴𝐴𝑗𝑗 𝛽𝛽𝛽𝛽 𝑉𝑉� ⋅ 𝒏𝒏 𝐷𝐷𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑 ∴ = � 𝛽𝛽𝛽𝛽𝛽𝛽𝛽𝛽 + � 𝛽𝛽𝑗𝑗 𝜌𝜌𝑗𝑗 𝑉𝑉𝑗𝑗 𝐴𝐴𝑗𝑗 𝐷𝐷𝐷𝐷 𝑑𝑑𝑑𝑑 𝐶𝐶𝑉𝑉 𝑗𝑗 𝑖𝑖 constant 𝑖𝑖𝑖𝑖 𝑚𝑚̇𝑗𝑗 𝑗𝑗 constant 𝑜𝑜𝑜𝑜𝑜𝑜 − � 𝛽𝛽𝑖𝑖 𝜌𝜌𝑖𝑖 𝑉𝑉𝑖𝑖 𝐴𝐴𝑖𝑖 𝑖𝑖 𝑚𝑚̇𝑖𝑖 𝑖𝑖𝑖𝑖 𝑜𝑜𝑜𝑜𝑜𝑜 57:020 Fluids Mechanics Fall2013 Example 2 Given: • Water flow (𝜌𝜌 = constant) • 𝐷𝐷1 = 10 cm; 𝐷𝐷2 = 15 cm • 𝑉𝑉1 = 10 cm/s • Steady flow Find: 𝑉𝑉2 = ? Mass conservation: • 𝐷𝐷𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 /𝐷𝐷𝐷𝐷 = 0 • 𝛽𝛽 = 1 • 𝜌𝜌1 = 𝜌𝜌2 = 𝜌𝜌 𝜌𝜌1 𝐴𝐴1 𝜌𝜌 ∴ 𝑉𝑉2 = 𝑉𝑉1 = 𝜌𝜌2 𝐴𝐴2 𝜌𝜌 Steady flow 0= 𝑑𝑑 � 𝜌𝜌𝜌𝜌𝜌𝜌 + 𝜌𝜌2 𝑉𝑉2 𝐴𝐴2 − 𝜌𝜌1 𝑉𝑉1 𝐴𝐴1 𝑑𝑑𝑑𝑑 𝐶𝐶𝑉𝑉 or, 𝜌𝜌1 𝑉𝑉1 𝐴𝐴1 = 𝜌𝜌2 𝑉𝑉2 𝐴𝐴2 𝐷𝐷1 𝐷𝐷2 2 𝑉𝑉1 = 1 10 cm 15 cm 2 10 cm = 4.4 cm/s s 20 57:020 Fluids Mechanics Fall2013 Example 3 Given: • 𝐷𝐷1 = 5 cm; 𝐷𝐷2 = 7 cm • 𝑉𝑉1 = 3 m/s • 𝑄𝑄3 = 𝑉𝑉3 𝐴𝐴3 = 0.01 m3/s • ℎ = constant (i.e., steady flow) • 𝜌𝜌1 = 𝜌𝜌2 = 𝜌𝜌3 = 𝜌𝜌water Find: 𝑉𝑉2 = ? = 0; steady flow 𝑑𝑑 0 = � 𝜌𝜌𝜌𝜌𝜌𝜌 + 𝜌𝜌2 𝑉𝑉2 𝐴𝐴2 − 𝜌𝜌1 𝑉𝑉1 𝐴𝐴1 − (𝜌𝜌3 𝑉𝑉3 𝐴𝐴3 ) 𝑑𝑑𝑑𝑑 𝐶𝐶𝑉𝑉 or, 𝑉𝑉2 𝐴𝐴2 = 𝑉𝑉1 𝐴𝐴1 + 𝑉𝑉� 3 𝐴𝐴3 = 𝑄𝑄3 𝑉𝑉1 𝐴𝐴1 + 𝑄𝑄3 3 𝜋𝜋 0.05 2 /4 + (0.01) ∴ 𝑉𝑉2 = = = 4.13 m/s 𝐴𝐴2 𝜋𝜋 0.07 2 /4 21 57:020 Fluids Mechanics Fall2013 Moving CV 22 57:020 Fluids Mechanics Fall2013 RTT for Moving CV (i.e., relative velocity 𝑉𝑉𝑟𝑟 ) 𝐷𝐷𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑 �𝑑𝑑𝑑𝑑 = � 𝛽𝛽𝜌𝜌𝜌𝜌𝜌𝜌 + � 𝛽𝛽𝜌𝜌𝑉𝑉𝑟𝑟 ⋅ 𝒏𝒏 𝐷𝐷𝐷𝐷 𝑑𝑑𝑑𝑑 𝐶𝐶𝑉𝑉 𝐶𝐶𝑆𝑆 23 57:020 Fluids Mechanics Fall2013 24 RTT for Moving and Deforming CV 𝑑𝑑𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑 � 𝑑𝑑𝑑𝑑 = � 𝛽𝛽𝛽𝛽𝛽𝛽𝛽𝛽 + � 𝛽𝛽𝛽𝛽 𝑽𝑽𝒓𝒓 ⋅ 𝒏𝒏 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝐶𝐶𝑉𝑉 𝐶𝐶𝑆𝑆 ∗ Both CV and CS change their shape and location with time 𝑽𝑽𝑟𝑟 = 𝑽𝑽(𝒙𝒙, 𝑡𝑡) − 𝑽𝑽𝑆𝑆 (𝒙𝒙, 𝑡𝑡) • 𝑉𝑉𝑆𝑆 (𝑥𝑥, 𝑡𝑡): Velocity of CS • 𝑉𝑉(𝑥𝑥, 𝑡𝑡): Fluid velocity in the coordinate system in which the 𝑉𝑉𝑠𝑠 is observed • 𝑉𝑉𝑟𝑟 : Relative velocity of fluid seen by an observer riding on the CV *Ref) Fluid Mechanics by Frank M. White, McGraw Hill 57:020 Fluids Mechanics Fall2013 RTT Summary (1) General RTT (for moving and deforming CV): 𝑑𝑑𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑 �𝑑𝑑𝑑𝑑 = � 𝛽𝛽𝛽𝛽𝛽𝛽𝛽𝛽 + � 𝛽𝛽𝛽𝛽𝑉𝑉𝑟𝑟 ⋅ 𝒏𝒏 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝐶𝐶𝑉𝑉 𝐶𝐶𝑆𝑆 Special Cases: 1) Non-deforming (but moving) CV 𝑑𝑑𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝜕𝜕 �𝑑𝑑𝑑𝑑 =� 𝛽𝛽𝛽𝛽 𝑑𝑑𝑉𝑉 + � 𝛽𝛽𝛽𝛽𝑉𝑉𝑟𝑟 ⋅ 𝒏𝒏 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝐶𝐶𝑉𝑉 𝐶𝐶𝑆𝑆 2) Fixed CV 𝑑𝑑𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝜕𝜕 �𝑑𝑑𝑑𝑑 =� 𝛽𝛽𝛽𝛽 𝑑𝑑𝑉𝑉 + � 𝛽𝛽𝛽𝛽𝑉𝑉 ⋅ 𝒏𝒏 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝐶𝐶𝑉𝑉 𝐶𝐶𝑆𝑆 3) Steady flow: 𝜕𝜕 =0 𝜕𝜕𝜕𝜕 4) Flux terms for uniform flow across discrete CS’s (steady or unsteady) �𝑑𝑑𝑑𝑑 = � 𝛽𝛽𝑚𝑚̇ � 𝛽𝛽𝛽𝛽𝑉𝑉 ⋅ 𝒏𝒏 𝐶𝐶𝑆𝑆 𝑜𝑜𝑜𝑜𝑜𝑜 − � 𝛽𝛽𝑚𝑚̇ 𝑖𝑖𝑖𝑖 25 57:020 Fluids Mechanics Fall2013 RTT Summary (2) For fixed CV’s: Parameter (𝐵𝐵) 𝛽𝛽 = 𝐵𝐵/𝑚𝑚 Mass (𝑚𝑚) 1 0= 𝑉𝑉 � 𝐹𝐹 = 𝑒𝑒 𝑄𝑄̇ − 𝑊𝑊̇ = Momentum (𝑚𝑚𝑉𝑉) Energy (𝐸𝐸) RTT 𝑑𝑑 �𝑑𝑑𝐴𝐴 � 𝜌𝜌𝜌𝜌𝑉𝑉 + � 𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏 𝑑𝑑𝑡𝑡 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 𝑑𝑑 �𝑑𝑑𝐴𝐴 � 𝑉𝑉𝜌𝜌𝜌𝜌𝑉𝑉 + � 𝑉𝑉𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏 𝑑𝑑𝑡𝑡 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 𝑑𝑑 �𝑑𝑑𝐴𝐴 � 𝑒𝑒𝑒𝑒𝑒𝑒𝑉𝑉 + � 𝑒𝑒𝑒𝑒𝑉𝑉 ⋅ 𝒏𝒏 𝑑𝑑𝑡𝑡 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 Remark Continuity eq. (Ch. 5.1) Linear momentum eq. (Ch. 5.2) Energy eq. (Ch. 5.3) 26 57:020 Fluids Mechanics Fall2013 Continuity Equation (Ch. 5.1) RTT with 𝐵𝐵 = mass and 𝛽𝛽 = 1, or 𝐷𝐷𝑀𝑀𝑠𝑠𝑠𝑠𝑠𝑠 0= 𝐷𝐷𝐷𝐷 mass conservatoin 𝑑𝑑 �𝑑𝑑𝐴𝐴 = � 𝜌𝜌𝜌𝜌𝑉𝑉 + � 𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏 𝑑𝑑𝑡𝑡 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 �𝑑𝑑𝐴𝐴 � 𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏 𝐶𝐶𝐶𝐶 Net rate of outflow of mass across CS = − 𝑑𝑑 � 𝜌𝜌𝜌𝜌𝑉𝑉 𝑑𝑑𝑡𝑡 𝐶𝐶𝐶𝐶 Rate of decrease of mass within CV Note: Incompressible fluid (𝜌𝜌 = constant) 𝑑𝑑 �𝑑𝑑𝐴𝐴 = − � 𝑑𝑑𝑉𝑉 � 𝑉𝑉 ⋅ 𝒏𝒏 𝑑𝑑𝑡𝑡 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 (Conservation of volume) 27 57:020 Fluids Mechanics Fall2013 Simplifications 1. Steady flow �𝑑𝑑𝐴𝐴 = 0 � 𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏 𝐶𝐶𝐶𝐶 2. If 𝑉𝑉 = constant over discrete CS’s (i.e., one-dimensional flow) �𝑑𝑑𝑑𝑑 = � 𝜌𝜌𝜌𝜌𝜌𝜌 − � 𝜌𝜌𝑉𝑉𝐴𝐴 � 𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏 𝐶𝐶𝑆𝑆 𝑜𝑜𝑢𝑢𝑢𝑢 3. Steady one-dimensional flow in a conduit or For 𝜌𝜌 = constant 𝜌𝜌𝜌𝜌𝜌𝜌 𝑜𝑜𝑜𝑜𝑜𝑜 − 𝜌𝜌𝜌𝜌𝜌𝜌 𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 =0 𝜌𝜌2 𝑉𝑉2 𝐴𝐴2 − 𝜌𝜌1 𝑉𝑉1 𝐴𝐴1 = 0 𝑉𝑉1 𝐴𝐴1 = 𝑉𝑉2 𝐴𝐴2 (or 𝑄𝑄1 = 𝑄𝑄2 ) 28 57:020 Fluids Mechanics Fall2013 Some useful definitions Mass flux 𝑚𝑚̇ = � 𝜌𝜌𝑉𝑉 ⋅ 𝑑𝑑𝑑𝑑 𝐴𝐴 Volume flux �𝑑𝑑𝑑𝑑) (Note: 𝑉𝑉 ⋅ 𝑑𝑑𝑑𝑑 = 𝑉𝑉 ⋅ 𝒏𝒏 𝑄𝑄 = � 𝑉𝑉 ⋅ 𝑑𝑑𝐴𝐴 𝐴𝐴 Average velocity 𝐴𝐴̅ = Average density 𝜌𝜌̅ = 𝑄𝑄 1 = � 𝑉𝑉 ⋅ 𝑑𝑑𝐴𝐴 𝐴𝐴 𝐴𝐴 𝐴𝐴 1 � 𝜌𝜌𝜌𝜌𝜌𝜌 𝐴𝐴 𝐴𝐴 Note: 𝑚𝑚̇ ≠ 𝜌𝜌̅ 𝑄𝑄 unless 𝜌𝜌 = constant 29 57:020 Fluids Mechanics Fall2013 Example 4 Estimate the time required to fill with water a cone-shaped container 5 ft hight and 5 ft across at the top if the filling rate is 20 gal/min. Apply the conservation of mass (𝛽𝛽 = 1) 𝑑𝑑 �𝑑𝑑𝐴𝐴 0 = � 𝜌𝜌𝜌𝜌𝑉𝑉 + � 𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏 𝑑𝑑𝑡𝑡 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 For incompressible fluid (i.e., 𝜌𝜌 = constant) and one inlet, 0= 𝑑𝑑 � 𝑑𝑑𝑉𝑉 − 𝑉𝑉𝑉𝑉 𝑑𝑑𝑡𝑡 𝐶𝐶𝐶𝐶 =𝑉𝑉 𝑖𝑖𝑖𝑖 =𝑄𝑄𝑖𝑖𝑖𝑖 30 57:020 Fluids Mechanics Fall2013 Example 4 – Contd. Volume of the cone at time t, 𝜋𝜋𝐷𝐷2 𝑉𝑉 𝑡𝑡 = ℎ 𝑡𝑡 12 Flow rate at the inlet, gal 𝑄𝑄 = 20 min The continuity eq. becomes 𝜋𝜋𝐷𝐷2 𝑑𝑑𝑑 ⋅ = 𝑄𝑄 12 𝑑𝑑𝑑𝑑 in3 in3 231 � 1,728 3 = 2.674 ft 3 /min gal ft or 𝑑𝑑𝑑 12𝑄𝑄 = 𝑑𝑑𝑑𝑑 𝜋𝜋𝐷𝐷2 31 57:020 Fluids Mechanics Fall2013 Example 4 – Contd. Solve for ℎ(𝑡𝑡), 𝑡𝑡 12𝑄𝑄 12𝑄𝑄 ⋅ 𝑡𝑡 ℎ 𝑡𝑡 = � 𝑑𝑑𝑑𝑑 = 2 𝜋𝜋𝐷𝐷 𝜋𝜋𝐷𝐷2 0 Thus, the time for ℎ = 5 ft is 𝜋𝜋𝐷𝐷2 ℎ 𝜋𝜋 5 ft 2 (5 ft) 𝑡𝑡 = = = 12.2 min 12𝑄𝑄 (12)(2.674 ft 3 /min) 32

Control Volume and Reynolds Transport Theorem 10-11-2013 Final

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Control Volume and Reynolds Transport Theorem 10-11-2013 Final

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