Fluid Mechanics Lesson 04A: The Material Derivative
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- เผยแพร่เมื่อ 2 ส.ค. 2022
- Fluid Mechanics Lesson Series - Lesson 04A: The Material Derivative
In this 12.5-minute video, Professor Cimbala compares the Lagrangian description to the Eulerian description and shows how to change from one to the other using the material derivative. He concentrates on the material acceleration, but the material derivative applies to any flow variable.
This video incorporates material from Section 4-1 of the Fluid Mechanics textbook by Cengel and Cimbala.
An Excel file listing of all the videos in this series can be found at
www.me.psu.edu/cimbala/Cengel... .
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Dr. John M. Cimbala is Professor of Mechanical Engineering at Penn State. He is an educator, textbook author, Christian author, husband, father, and grandfather. He also created and maintains a website for helping people grow in their faith called Christian Faith Grower at www.christianfaithgrower.com/ His TH-cam channel is at / @johncimbala
Thank you. Wish I had paid more attention in calculus. I think trying to understand fluid mechanics without strong fundamentals of calculus makes it way more difficult than it needs to be.
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This video helps me a lot, with very clear explanation and examples. Thank you!
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Great explanation 👏👏👏👏 thank you so much🌹🌹🌹🌹🌹
Hi sir thank a lot. I have a question. How can we take the material derivative of velocity in cylindrical coordinates? Can you make an example of it. I couldnt find any on the internet
The left-hand side of the Navier-Stokes equation is essentially the material derivative of velocity, which is acceleration of a fluid particle following the fluid particle. So... look at the equation sheet in cylindrical coordinates and the left-hand side is the material derivative in cylindrical coordinates!
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Nice video but I dont understand what would be the answer if in the example problem V=3xy*i-3y*j? So the first component of the vector field also depends on y for example.
That is why you take PARTIAL derivatives instead of total derivatives. delV/dely = 3x*i - 3j in your example. And delV/delx = 3y*i since there is no x in the second term of your example. When all is said and done, collect all the i terms and all the j terms separately. Hope this helps.
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I'm not sure what u,v and w are? At 7:00 what happens to u,v, and w in the operator, also i'm not sure of this combined operator and why it's not just del dotted into v. It seems wrong to me.
u, v, and w are the velocity components of V, the velocity vector. Del dot V would become a scalar equal to delu/delx +delv/dely + delw/delz, whereas V dot Del is still an operator that can be applied to any scalar, vector, or even tensor.
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