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تطبيقات المعادلات التفاضلية
- 1. الرح الرحمن هللا بسم يم ( ) صدق هللا العلي العظيم ( البقرة / 32 )
- 2. Applications of Differential Equations المعادالت تطبيقات التفاضلية An example is the application of partial differential equations in fluid mechanics ف الجزئية التفاضلية المعادالت تطبيقات ذلك على مثال ي الموائع ميكانيكا BY Mohammed R. Salman
- 3. Solids Gasses Resist a shear stress Science of flow (rheological) Hydraulics Hydrodynamics Experimental studies theoretical studies Fluid Mechanics Introduction liquids Matter
- 4. Peristalsis is defined as a wave of relaxation contraction imparted to the walls of a flexible conduit effecting the pumping of enclosed material. Peristaltic phenomenon Practical applications Physiology Bio- mechanical Examples
- 6. Types of Fluids Newtonian Fluids Non-Newtonian Fluids Water Blood Silicon ketchup molasses n du = + B dy Alcohol Gasoline
- 7. The Flexible Channels Co-axial pipes Curved pipes Straight pipes Tapered pipes Symmetric motion Anti-symmetric motion uniform Non-uniform 0 ≤ ∅ ≤ 𝜋
- 8. Elementary concepts and basic definitions Fluid mechanics Is that branch of applied mathematics which deal with the behavior of fluid at rest and in motion. Statics Kinematic Dynamic Relation between the stress and the rateof strain Classification of fluid
- 10. Effect of convective conditions in a radiative peristaltic flow of pseudoplastic nanofluid through a porous medium in a tapered an inclined asymmetric channel للمائع الشعاعي الجريان على الحدودية الشروط تأثير النانوي خ الكاذب الل متماثلة غير مائلة مستدقة قناة في مسامي وسط d d T0 b 2 b 1 H 2 H 1 T1 X Y 𝑌 = 𝐻2 𝑋, 𝑡 = 𝑑 + 𝑚𝑋 + 𝑏2 𝑆𝑖𝑛 2𝜋 𝜆 𝑋 − 𝑐𝑡 𝑌 = 𝐻1 𝑋, 𝑡 = −𝑑 − 𝑚𝑋 − 𝑏1 𝑆𝑖𝑛 2𝜋 𝜆 𝑋 − 𝑐𝑡 + 𝜙
- 11. Calculation of Lorentz force 𝑉 × 𝐵 = 𝑖 𝑗 𝑘 𝑢 𝑣 0 0 𝐵0 0 = 𝐵0𝑢𝑘 𝐽 = 𝜎 𝑉 × 𝐵 = 𝜎𝐵0𝑢 𝑘 𝐽 × 𝐵 = 𝑖 𝑗 𝑘 0 0 𝜎𝐵0𝑢 0 𝐵0 0 = −𝜎𝐵0 2 𝑢 𝑖 . Both walls move towards outward or inward simultaneously and further 𝑏1,𝑏2,𝑑 and 𝜙 satisfy the following condition at the inlet of divergent channel 𝑏1 2 + 𝑏2 2 + 2𝑏1𝑏2 𝐶𝑜𝑠(𝜙) ≤ 𝑑1 + 𝑑2 2
- 12. Constitutive Equations 𝑺 + 𝜆1 𝐷𝑺 𝐷𝑡 + 1 2 𝜆1 − 𝜇1 𝐴1𝑺 + 𝑺𝐴1 = 𝜇𝐴1 𝐴1 = 𝛻𝑉 + 𝛻𝑉 𝑇 𝑑𝑺 𝑑𝑡 = 𝜕𝑺 𝜕𝑡 + 𝑽. 𝛻𝑺 𝐷𝑺 𝐷𝑡 = 𝑑𝑺 𝑑𝑡 − 𝛻𝑉 𝑺 − 𝑺 𝛻𝑉 𝑇 𝐷𝑺 𝐷𝑡 = 𝜕𝑺 𝜕𝑡 + 𝑽. 𝛻𝑺 − 𝛻𝑉 𝑺 − 𝑺 𝛻𝑉 𝑇 𝑺 = 𝑆𝑋𝑋 𝑆𝑋𝑌 𝑆𝑌𝑋 𝑆𝑌𝑌 = 𝑎11 𝑎12 𝑎21 𝑎22 , 𝐴1 = 2 𝜕𝑈 𝜕𝑋 𝜕𝑈 𝜕𝑌 + 𝜕𝑉 𝜕𝑋 𝜕𝑉 𝜕𝑋 + 𝜕𝑈 𝜕𝑌 2 𝜕𝑉 𝜕𝑌
- 13. The Components of Stress Tensor 𝑆𝑋𝑋 + 𝜆1 𝜕 𝜕𝑡 + 𝑈 𝜕 𝜕𝑋 + 𝑉 𝜕 𝜕𝑌 𝑆𝑋𝑋 − 2𝑆𝑋𝑋 𝜕𝑈 𝜕𝑋 − 2𝑆𝑋𝑌 𝜕𝑈 𝜕𝑌 + 1 2 𝜆1 − 𝜇1 4𝑆𝑋𝑋 𝜕𝑈 𝜕𝑋 + 2𝑆𝑋𝑌 𝜕𝑈 𝜕𝑌 + 𝜕𝑉 𝜕𝑋 = 2𝜇 𝜕𝑈 𝜕𝑋 ……(1) 𝑆𝑋𝑌 + 𝜆1 𝜕 𝜕𝑡 + 𝑈 𝜕 𝜕𝑋 + 𝑉 𝜕 𝜕𝑌 𝑆𝑋𝑌 − 𝑆𝑋𝑋 𝜕𝑉 𝜕𝑋 − 𝑆𝑌𝑌 𝜕𝑈 𝜕𝑌 + 1 2 𝜆1 − 𝜇1 𝑆𝑋𝑋 + 𝑆𝑌𝑌 𝜕𝑈 𝜕𝑌 + 𝜕𝑉 𝜕𝑋 = 𝜇 𝜕𝑈 𝜕𝑌 + 𝜕𝑉 𝜕𝑋 …….(2) 𝑆𝑌𝑌 + 𝜆1 𝜕 𝜕𝑡 + 𝑈 𝜕 𝜕𝑋 + 𝑉 𝜕 𝜕𝑌 𝑆𝑌𝑌 − 2𝑆𝑌𝑋 𝜕𝑉 𝜕𝑋 − 2𝑆𝑌𝑌 𝜕𝑉 𝜕𝑌 + 1 2 𝜆1 − 𝜇1 2𝑆𝑋𝑌 𝜕𝑈 𝜕𝑌 + 𝜕𝑉 𝜕𝑋 + 4𝑆𝑌𝑌 𝜕𝑉 𝜕𝑌 = 2𝜇 𝜕𝑉 𝜕𝑌 ……(3)
- 14. Governing Equations 𝜕𝑈 𝜕𝑋 + 𝜕𝑉 𝜕𝑌 = 0 …(4) 𝜌𝑓 𝜕𝑈 𝜕𝑡 + 𝑈 𝜕𝑈 𝜕𝑋 + 𝑉 𝜕𝑈 𝜕𝑌 = − 𝜕𝑃 𝜕𝑋 + 𝜕 𝜕𝑋 𝑆𝑋𝑋 + 𝜕 𝜕𝑌 𝑆𝑋𝑌 − 𝝈𝐵𝟎 2 𝑈 − 𝜇 𝐾0 𝑈 +𝜌𝑓 𝑔 𝑠𝑖𝑛𝛼 𝜌𝑓 𝜕𝑉 𝜕𝑡 + 𝑈 𝜕𝑉 𝜕𝑋 + 𝑉 𝜕𝑉 𝜕𝑌 = − 𝜕𝑃 𝜕𝑌 + 𝜕 𝜕𝑋 𝑆𝑌𝑋 + 𝜕 𝜕𝑌 𝑆𝑌𝑌 − 𝜇 𝐾0 𝑉 −𝜌𝑓 𝑔 𝑐𝑜𝑠𝛼 …(6) (𝜌𝑐)𝑓 𝜕𝑇 𝜕𝑡 + 𝑈 𝜕𝑇 𝜕𝑋 + 𝑉 𝜕𝑇 𝜕𝑌 = 𝜅 𝜕2𝑇 𝜕𝑋2 + 𝜕2𝑇 𝜕𝑌2 + (𝜌𝑐)𝑝 𝐷𝑇 𝑇𝑚 𝜕𝑇 𝜕𝑋 2 + 𝜕𝑇 𝜕𝑌 2 …(7) …(5)
- 15. Fixed frame (𝑿,𝒀,𝒕) Wave frame 𝒙,𝒚 𝑥 = 𝑋 − 𝑐𝑡 ,𝑦 = 𝑌,𝑢 𝑥,𝑦 = 𝑈 𝑋,𝑌,𝑡 − 𝑐 ,𝑣 𝑥,𝑦 = 𝑉 𝑋,𝑌,𝑡 , 𝑝 𝑥,𝑦 = 𝑃 𝑋,𝑌,𝑡 The coordinates and velocities in the two frames are related by Dimensionless Analysis 𝑥 = 𝑥 𝜆 , 𝑦 = 𝑦 𝑑 , 𝑡 = 𝑐𝑡 𝜆 , 𝑢 = 𝑢 𝑐 , 𝑣 = 𝑣 𝑐 , 𝛿 = 𝑑 𝜆 , ℎ1 = 𝐻1 𝑑 , ℎ2 = 𝐻2 𝑑 , 𝜃 = 𝑇 − 𝑇° 𝑇1 − 𝑇° , 𝑝 = 𝑑2𝑝 𝜆𝜇𝑐 , 𝑆𝑖𝑗 = 𝑑 𝑐𝜇 𝑆𝑖𝑗 , 𝜆1 ∗ = 𝜆1𝑐 𝑑 , 𝜇1 ∗ = 𝜇1𝑐 𝑑 , 𝐾 = 𝐾0 𝑑2 , 𝑅𝑒 = 𝜌𝑓𝑐𝑑 𝜇 , 𝑎 = 𝑏1 𝑑 , 𝑏 = 𝑏2 𝑑 , m = mλ d , Pr = μcf κ , Nt = τDT(T1 − T0) Tm𝓋 , M = σ μ dB0, u = 𝜕ψ 𝜕y , v = −δ 𝜕ψ 𝜕x
- 16. 𝜕𝑝 𝜕𝑥 = 𝜕 𝜕𝑦 𝑆𝑥𝑦 − 𝑀2 + 1 𝐾 𝜕𝜓 𝜕𝑦 + 1 𝜕𝑝 𝜕𝑦 = 0 𝜕2 𝜃 𝜕𝑦2 + 𝑃𝑟𝑁𝑡 𝜕𝜃 𝜕𝑦 2 = 0 long-wavelength approximation i.e.(𝜹 ≪ 𝟏) and the low Reynolds number i.e. (𝑹𝒆 → 𝟎). (8) (9) (10) 𝑆𝑥𝑥 = 𝜆1 + 𝜇1 𝑆𝑥𝑦 𝜕2𝜓 𝜕𝑦2 (11) 𝑆𝑥𝑦 + 1 2 𝜆1 − 𝜇1 𝑆𝑥𝑥 + 𝑆𝑦𝑦 𝜕2𝜓 𝜕𝑦2 − 𝜆1𝑆𝑦𝑦 𝜕2𝜓 𝜕𝑦2 = 𝜕2𝜓 𝜕𝑦2 (12) 𝑆𝑦𝑦 = − 𝜆1 − 𝜇1 𝑆𝑥𝑦 𝜕2 𝜓 𝜕𝑦2 (13)
- 17. 𝜕𝑝 𝜕𝑥 = 𝜕3𝜓 𝜕𝑦3 + 𝜉 𝜕 𝜕𝑦 𝜕2𝜓 𝜕𝑦2 3 + 𝜉2 𝜕 𝜕𝑦 𝜕2𝜓 𝜕𝑦2 4 − 𝑁2 𝜕𝜓 𝜕𝑦 + 1 𝜕4 𝜓 𝜕𝑦4 + 3𝜉 𝜕2 𝜓 𝜕𝑦2 2 𝜕4 𝜓 𝜕𝑦4 + 2 𝜕3 𝜓 𝜕𝑦3 2 𝜕2 𝜓 𝜕𝑦2 − 𝑁2 𝜕2 𝜓 𝜕𝑦2 = 0 After using binomial expansion for small pseudoplastic fluid parameter 𝜉 ,where 𝜉 = (𝜇1 2 − 𝜆1 2 ) and 𝑁2 = 𝑀2 + 1 𝐾 we get: (14) (15) Perturbed System and Perturbation Solutions: In order to solve the present problem, we expand the flow quantities in a power series of 𝜉 𝑎𝑛𝑑 𝑃𝑟 𝑃 = 𝑃0 +𝜉𝑃1 + 𝜉2𝑃2 + ⋯ 𝜓 = 𝜓0 + 𝜉𝜓1 + 𝜉2 𝜓2 + ⋯ 𝐹 = 𝐹0 +𝜉𝐹1 + 𝜉2 𝐹2 + ⋯ 𝜃 = 𝜃0 +𝑃𝑟𝜃1 + 𝑃𝑟2 𝜃2 + ⋯
- 18. Zeroth- Order System 𝜕4𝜓0 𝜕𝑦4 − 𝑁2 𝜕2𝜓0 𝜕𝑦2 = 0 𝜕𝑝0 𝜕𝑥 = 𝜕3𝜓0 𝜕𝑦3 − 𝑁2 𝜕𝜓0 𝜕𝑦 + 1 = 0 𝜕2𝜃0 𝜕𝑦2 = 0 With corresponding convective boundary conditions 𝜓0 = − 𝐹0 2 , 𝜕𝜓0 𝜕𝑦 = 0 , 𝜃0 = 0, 𝑎𝑡 𝑦 = ℎ1 𝜓0 = 𝐹0 2 , 𝜕𝜓0 𝜕𝑦 = 0 , 𝜃0 = 1 , 𝑎𝑡 𝑦 = ℎ2
- 19. First- Order System 𝜕4𝜓1 𝜕𝑦4 + 3 𝜕4𝜓0 𝜕𝑦4 𝜕2𝜓0 𝜕𝑦2 2 + 6 𝜕3𝜓0 𝜕𝑦3 2 𝜕2𝜓0 𝜕𝑦2 − 𝑁2 𝜕2𝜓1 𝜕𝑦2 = 0 𝜕𝑝1 𝜕𝑥 − 𝜕3𝜓1 𝜕𝑦3 − 3 𝜕3𝜓0 𝜕𝑦3 𝜕2𝜓0 𝜕𝑦2 2 + 𝑁2 𝜕𝜓1 𝜕𝑦 = 0 𝜕2𝜃1 𝜕𝑦2 + 𝑁𝑡 𝜕𝜃0 𝜕𝑦 2 = 0 𝜓1 = − 𝐹1 2 , 𝜕𝜓1 𝜕𝑦 = 0, 𝜃1 = 0 𝑎𝑡 𝑦 = ℎ1 𝜓1 = 𝐹1 2 , 𝜕𝜓1 𝜕𝑦 = 0, 𝜃1 = 0 𝑎𝑡 𝑦 = ℎ2 With corresponding convective boundary conditions
- 21. Pressure rise Heat transfer coefficient
- 23. For Listening