Recent content by SebastianRM

That is correct Sir, I think. By doing ##\delta u = \ell du/dy = \ell V/(h+2l) ## we can arrive to the same form. Thank you!

I was trying to derive the determinant itself, I know that is the correct form. I was using the definition of a cross product to do the curl; however, the curl is not really a cross product of vectors in an orthonormal basis.

I have a vector in cylindrical Coordinates: $$\vec{V} = \left < 0 ,V_{\theta},0 \right> $$ where ##V_\theta = V(r,t)##. The Del operator in ##\{r,\theta,z\}$ is: $\vec{\nabla} = \left< \frac{\partial}{\partial r}, \frac{1}{r}\frac{\partial}{\partial \theta}, \frac{\partial}{\partial z}...

I see, that makes sense. Would you say du/dy is the usual slope V/h ? While ##\partial u / \partial y ## accounts for the velocity V affected by slip and the "new" height, such that ## (V - 2\delta u)/(h + 2b) ## ?

The diagram says *No-slip* but that is a typo (the teacher said so).

That is for no-slip condition, this is Slip condition. The velocity at the top is not V, because there is slip. Same at the bottom, the fluid is not attached to the plate, as there is slip.

The problem states: Two parallel plates separated by distance h, the plate at the top moves with velocity V, while the one at the bottom remains stationary. My initial approach was: I considered, ##du/dy = V/h## and for the shear stress ##\tau = \mu \frac{\partial u}{\partial y}## For...

We looked at linear deformations in the x,y and z direction, as infinitesimal displacement, then by rearranging terms we got $$\frac{1}{V} \frac{dV}{dt} = \nabla \cdot \vec{v}$$ We talked about infinitesimal displacements and we rearranged them (by treating them like ##\Delta x_i##), so we ended...

in class we derived the following relationship: $$\frac{1}{V}\frac{dV}{dt}= \nabla \cdot \vec{v}$$ This was derived though the analysis of linear deformation for a fluid-volume, where: $$dV = dV_x +dV_y + dV_z$$ I understood the derived relation as: 1/V * (derivative wrt time) = div (velocity)...

As it turns out the static effect = 1/2 (Term sum). Which fixes all issues. The author did not say this explicitly so I misinterpreted him.

I am reading a book of Fundamental Energy Systems. The author describes the rate of change in head for a turbomachine as: $$ \frac{1}{2}[(V_1^2-V_2^2)+(U_1^2-U_2^2)+(V_{R2}^2-V_{R1}^2)] = H =U_1V_{u1} - U_2V_{u2} $$ and the static effect as: $$SE =(U_1^2-U_2^2)+(V_{R2}^2-V_{R1}^2) $$ However...

I was looking at Kirchoffs Laws: "A solid, liquid or dense gas produces a continuous spectrum". I would expect objects to produce an emission spectrum since we would be observing the photons that come from spontaneous emission of electrons in excited states. This photons are specific to the...

I am unsure as to how to apply binding energy since we did not see this in class. I did take his advice into account. But got the same answer in the end.

What is the author meaning by that, I literally just read in section 5.1, that the depending on the energy loss of the electrons that make up the substance, this energy will be released as photons. Since this is particular for each atom, it explained how we can tell which element is wich. So if...

Yeah it was supposed to be MeV. My mistake there. I did check several websites including Wikipedia, it states that the mass is less than 0.12 eV/c^2. So since it is not a concrete number and the mass is so small it seems mass less I arrived to the conclusion that perhaps I should not include it...