Stress tensor in a Newtonian fluid

Lectures of Transport Phenomena course at Olin College
The Newtonian stress tensor and connection to fluid kinematics.

[@manueljenkin95]

Thank you so much professor. I’m an EE student, studying turbulence modelling just out of curiosity, and your video has helped me bridge some of that gap.

[@amiraadel4941]

Thanks for the clear explanation. Could you please explain what will change in the momentum equation in case we have a compressible fluid?

[@xXBaller4LifeXx]

what an amazing quick explanation. thank you :)

[@MrAmgadHasan]

@17:30 Why is it 2*du/dx and 2*dv/dy and not only du/dx and dv/dy?

[@arts5852]

Good video, very informative👍. Can you explain how does stress tensor in compressible fluids differ from the stress tensor in incompressible fluids. And what is it a volume viscosity ?

[@brianstorey7830]

In compressible flow, there is an additional term in the stress tensor that is proportional to the divergence of the velocity field. I neglect that here for simplicity. You can find the basic form in any fluid mechanics text - or even wikipedia. This term would be referred to as the volume viscosity - as it is irreversible friction ocurring during volume changes. There is no new parameter as the coefficient in front of the divergence term is 2/3 mu where mu is the fluids viscosity

[@JamesVestal-dz5qm]

Was there supposed to be a strain rotation relationship in viscous heat transfer?

[@gah6864]

Excellent lecture - thanks. If a slip condition was allowed does that just mean that 𝝏u/𝝏x will have a specific value (and hence 𝝏v/𝝏y = - 𝝏u/𝝏x) ? Or does anything else change?

[@brianstorey7830]

If there is a constant slip, the du/dx is still zero. There are cases in what is called electrokinetics where there can be an effective (not true) slip at the surface that can vary in a way that du/dx is not zero. In that case, just as you mean that means that du/dx = -dv/dy - but not much else can be said in a general way!

[@JamesVestal-dz5qm]

If momentum has 9 equations and mass has 1 does that mean velocity does 90 percent of the work?

[@renjithjames8779]

what is the physical significance or meaning of 'normal viscous stress' contributed by the diagonal elements of strain tensor multiplied with dynamic viscosity? Also, whether this constitutive relation is based on experiments or derived from any theories? pls clarify. thanks in advance.

[@brianstorey7830]

Take a bowl of syrup. Hold a spoon horizontal and push the round back end into the syrup a little bit. Lift up. The syrup will come with the spoon and be pulled out of the bowl. That is the "normal viscous stress". The constitutive relationship for a Newtonian fluid can be derived for a gas. The derivation is long and detailed and requires you to know quite a bit of statistical physics and mathematics. The full theory of constitutive laws for gases is the Chapman Enskog theory . This theory does yield the actual law and the coefficient of viscosity, based on molecular parameters of the gas. The theory is complex, but the final results are pretty simple and yield accurate predictions. en.wikipedia.org/wiki/Chapman%E2%80%93Enskog_theory For a liquid like water, the Newtonian law is really based on experiments. There is no simple theory for liquids. There is some theoretical underpinning, but it is largely phenomenological and the coefficient of viscosity is ultimately measured. www.amazon.com/Non-Equilibrium-Thermodynamics-Dover-Books-Physics/dp/0486647412

[@randallpatton2941]

At 17:30, the off diagonal terms show du/dx+dv/dy. Shouldn't it be du/dy+dv/dx as at 20:00?

[@brianstorey7830]

Yes. you are correct. I just added an annotation here in the video pointing out the error. Thanks!

[@MAth4379]

good ...can you explane mor about non newtonian fluid ... bingham model ??

[@fatimaiqbal2600]

Establish a relation between stress tensor components and the deformation field by considering the newtonian fluid????

[@andresyesidmorenovilla7888]

Hi, a question, what do you mean by "generalization" It isn't clear to me when can I use the first constitutive relation that the shear stresses are just proporctional to the velocity's spatial derivative, and when can I use the whole stress tensor.Could you maybe explain me plis?

[@brianstorey7830]

I am not certain I understand your question. Can you point to a time stamp where it is unclear and I will try to answer. Generally, the assumption of a Newtonian fluid is something that comes from experimental observation. Water and simple oils follow the law well. Gases generally do to. Polymeric liquids - not so much. The assumption of Euler's equation (no viscosity) is something that no fluid obeys (other than bizzarre states of superfluid Helium maybe). However, Euler equation is a useful approximation due to simplicity of the equations.

[@adityarathi1984]

Why are we adding the transpose?

[@brianstorey7830]

+Aditya Rathi Good question. The Newtonian constitutive "law" for a fluid like water is really an empirical equation. It is not something we can really prove. We *can* show that the constitutive laws cant have an arbitrary form. One restriction is that the tensor must be symmetric. This requirement comes from conservation of angular momentum (and not something I proved in this lecture, but is pretty easy to demonstrate). Other limitations on the form of the stress tensor come from requirements such as wanting the result to not depend upon our reference frame or if we need the fluid to be isotropic (same in all directions). In this lecture I just write down the constitutive law - it is not a proof.

[@MrMohamedhamid]

good ,thanks

[@JamesVestal-dz5qm]

Man I'm tryna think about if 9 stress tensor components is like 9 classes at a time?

[@rommelron9626]

Beautiful !

[@fantacake]

7:02 could you please explain a bit about how the (∇^2)v comes from? THX

[@brianstorey7830]

+Kriss Lee It is probably easiest to show this in component form. Look back around 4:45 where I work out the components on the Grad(v) + Grad(v)^T tensor. Now take the divergence of this tensor to get the vector that shows up in the momentum equation. To do this, take d/dx of the (1,1) element, d/dy of the (2,1) element, and d/dz of the (3,1) element then you would have the terms that show up in the x component of the momentum. You will see that you have some of the terms for the Laplacian of the x component of the velocity but there is some extra stuff. The extra stuff will look like d^2 v/dxdy and d^2 w/dxdz. These can be written as d/dx (dv/dy +_dw/dy). Using conservation of mass (du/dx = -dv/dy - dw/dz) you would get -d^u/dx^x. This would subtract off the term with the 2 in front of it. If you can't figure this out, I can write it out on paper and attach an image. It is always hard to explain in these small text boxes!

[@fantacake]

+Brian Storey Thank you for the reply. I finally figured it out with wikipedia upload.wikimedia.org/math/d/8/7/d87ab8ac795c8c13e659e902a1bc5a1a.png. Really appreciate all your videos!

[@fsteam1]

Thank you for this answer which is surprisely difficult to find in many books on fluid dynamics

[@redfc11]

Brian Storey is a Real G

[@samuelosei-somuah3072]

Wow

[@sharankadyamada4104]

at 7:07 how do we write the stress tensor to cartesian?

[@brianstorey7830]

I write out the components around 4:30-5:00. Is this what you are looking for?

[@sharankadyamada4104]

Yeah ,after adding (gradv +T(gradv)) the final answer is ( dui/dxj +duj/dxi)*ij. Which I am not able to arrive at

[@starryfolks]

In non newtonian fluid?

[@simhadivya237]

why did you write ∇.T as the ∇P? Why it can't be the ∇.P

[@brianstorey7830]

So if P is the scalar value of pressure at a point, then divergence of a scalar is an undefined operation. In vector calculus, we aren't allowed to take the divergence of a scalar function. In our tensor notation, the piece of the stress tensor with pressure in it is taken be P*I where I would be the identity matrix (i.e. P*I is a 3x3 matrix with the scalar value of P at that point along the diagonal of the matrix). When I take the divergence of P*I and work out all the terms, I would find this is equivalent to gradient of P. The gradient of a scalar gives me a vector just like the divergence of a tensor gives me a vector. The simplification is just that the pressure only acts normal to the face, thus it is a special tensor where all the off-diagonal components are zero. Since the three values of P on the diagonal are the same, that also makes things simpler. Hope this helps.

[@ملاك-ظ9ج6ك]

سلام عليكم