Hallo Prof, In deriving the shear stresses, there is a mathematical manipulation carried out, namely multiplying du/dt with dy/dy. My question is why not multiply with dx/dx. It might seem a dumb question but just need to understand the logic behind this manipulation. Thanks
Really great question! I'm guessing it is just because it worked, we found that the skin friction scaled with flow gradient, which in this case is in the y-direction. There might be a more rigorous math reason, or a better explanation---in either case I do not know it.
Hi professor, thank you very much for your video. If the flow is 3dimensional, shouldn’t we express viscous forces in terms of the viscous stress tensor (or its components txx,txy,txz,etc.)? According to Stokes’ law of viscosity, T=mu*du/dy only applies to 1D flow, right? Thanks in advance!
Hey Roger! Do you mean around 12:18? Yep, for the full-on 3D approach you could start it out that way and then break it down. My approach here is a bit less exact but I feel it connects to the physical source of the force more clearly, and we add the other terms back in later on.
F=ma---or more generally F = d (m*v)/dt---is a form of conservation of momentum, saying if you apply force to a system it's momentum (m*v) must change. Navier-Stokes is really just a complicated form of F=ma for a fluid system.
Thankyou for the video sir. By the way I have a question, why do we use partial derivative (del) in fluid mechanics instead of the normal derivative (d)? Thank you sir
Thanks!! The partial derivative (symbol delta) is used in fluid mechanics because often the field of interest is functions of multiple variables, like x,y,z,t. So, when differentiating a function of multiple variables, you need to do the partial derivative. The "del" or "nabla" is slightly different, is a more complex mathematical operation that includes all spatial derivatives, it is often just used for shorthand notation that I tend to avoid.
@@mikeshan417 Hi! No dumb questions at all no worries. In this case, I think I have it correct. With two fractions multiplying, you can rearrange the numerator and denominator, kind of like the commutative property. A simpler example would be 3/4 x 4/3, which equals 1, but it's also equal to 4/4 x 3/3, and 12/12, etc. In your suggestion in the reply, I think it would only work if the two fractions were across an equals sign, for example if we originally had (delta u/delta t) = (delta y/delta y) then you could re-arrange for (delta y/delta t) = (delta y/delta u)
One more question sir, why does the total force of viscosity in the x direction have three components instead of two (~14:50)? I don't quite understand where the d2u/dx2 term come from, I thought the only contributors to the x force were tau xy on the y face and tau xz on the z face... Thank you.
Really excellent question Mike and thank you for following the videos so closely! It's a bit counterintuitive, because we normally think of the viscosity as being just shear on the tangential faces, but technically it also produces a normal force, in this cause through the x-derivative. Revisit the train analogy---two trains side-by-side going different velocities and you jump from one to the other so you impart a change in momentum. This physically explains the tangential forces. However, now picture cars in a line, like a highway lane, each going a different velocity. The cars are have different velocities in the direction they are moving, so there is a du/dx. Now, if you were to jump from one car, forward or backward to another car, you would impart a change momentum (just like in the trains). This is how viscosity produces normal force. Does this help?
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Thank you Professor for your efforts. This was a brilliant explanation!
Glad it was helpful!
you are a lifesaver mate thanks for the effort
No problem 👍
thank you so much for your valuable effort. It helps a lot :)
I'm glad it helped!
Thanks Professor
No problem!
Hallo Prof,
In deriving the shear stresses, there is a mathematical manipulation carried out, namely multiplying du/dt with dy/dy. My question is why not multiply with dx/dx. It might seem a dumb question but just need to understand the logic behind this manipulation. Thanks
Really great question! I'm guessing it is just because it worked, we found that the skin friction scaled with flow gradient, which in this case is in the y-direction. There might be a more rigorous math reason, or a better explanation---in either case I do not know it.
Hi professor, thank you very much for your video. If the flow is 3dimensional, shouldn’t we express viscous forces in terms of the viscous stress tensor (or its components txx,txy,txz,etc.)? According to Stokes’ law of viscosity, T=mu*du/dy only applies to 1D flow, right? Thanks in advance!
Hey Roger! Do you mean around 12:18? Yep, for the full-on 3D approach you could start it out that way and then break it down. My approach here is a bit less exact but I feel it connects to the physical source of the force more clearly, and we add the other terms back in later on.
Thank you!
Why is the Navier Stokes equation said as a conservation of momentum equation although the external forces are not zero?
F=ma---or more generally F = d (m*v)/dt---is a form of conservation of momentum, saying if you apply force to a system it's momentum (m*v) must change. Navier-Stokes is really just a complicated form of F=ma for a fluid system.
Thankyou for the video sir. By the way I have a question, why do we use partial derivative (del) in fluid mechanics instead of the normal derivative (d)? Thank you sir
Thanks!! The partial derivative (symbol delta) is used in fluid mechanics because often the field of interest is functions of multiple variables, like x,y,z,t. So, when differentiating a function of multiple variables, you need to do the partial derivative. The "del" or "nabla" is slightly different, is a more complex mathematical operation that includes all spatial derivatives, it is often just used for shorthand notation that I tend to avoid.
@@prof.vanburen Okay, thank you for the explanation sir
Hi sir, one very stupid question...how did you get from (delta u/delta t)(delta y/delta y) to (delta y/delta t)(delta u/delta y) around ~11:45 ?
Shouldn't it be (delta y/delta t)(delta y/delta u) ? Thank you
@@mikeshan417 Hi! No dumb questions at all no worries. In this case, I think I have it correct. With two fractions multiplying, you can rearrange the numerator and denominator, kind of like the commutative property. A simpler example would be 3/4 x 4/3, which equals 1, but it's also equal to 4/4 x 3/3, and 12/12, etc.
In your suggestion in the reply, I think it would only work if the two fractions were across an equals sign, for example if we originally had (delta u/delta t) = (delta y/delta y) then you could re-arrange for (delta y/delta t) = (delta y/delta u)
@@prof.vanburen Thank you
One more question sir,
why does the total force of viscosity in the x direction have three components instead of two (~14:50)? I don't quite understand where the d2u/dx2 term come from, I thought the only contributors to the x force were tau xy on the y face and tau xz on the z face...
Thank you.
Really excellent question Mike and thank you for following the videos so closely! It's a bit counterintuitive, because we normally think of the viscosity as being just shear on the tangential faces, but technically it also produces a normal force, in this cause through the x-derivative.
Revisit the train analogy---two trains side-by-side going different velocities and you jump from one to the other so you impart a change in momentum. This physically explains the tangential forces. However, now picture cars in a line, like a highway lane, each going a different velocity. The cars are have different velocities in the direction they are moving, so there is a du/dx. Now, if you were to jump from one car, forward or backward to another car, you would impart a change momentum (just like in the trains). This is how viscosity produces normal force.
Does this help?
@@prof.vanburen It does thank you very much