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تطبيقات المعادلات التفاضلية
- 1. الورشة عنوان تطبيقات المعادالت التفاضلية المحاضر : م . د العلي سلمان رزاق محمد كربالء جامعة - الصرفة للعلوم التربية كلية / الرياضيات قسم An example is the application of partial differential equations in fluid mechanics الموا ميكانيكا في الجزئية التفاضلية المعادالت تطبيقات ذلك على مثال ئع Website: www.contstudies.uokerbala.edu.iq
- 2. 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
- 3. 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
- 4. 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
- 5. 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)
- 6. 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)
- 7. 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
- 8. 𝜕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)
- 9. 𝜕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 + ⋯
- 10. 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
- 11. 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
- 13. Pressure rise Heat transfer coefficient
- 15. For Listening