Online lesson for EME 303 at Penn State Hazleton. This lesson follows the derivation of the Reynolds Transport Theorem. We will apply this theorem in several subsequent lessons. License: CC-BY-SA 4.0 (creativecommons.org/licenses/...)
Hi there, I really like the way you relate and explain everything. Very clear. From now on, I will study my Fluid Mechanic class just watching your videos.
Thank you! Really great explanation, I have everything clear know. Hope to see new videos! - Universidad Técnica Federico Santa Maria, Valparaiso, Chile.-
Thank you so much. It's really helpful. Soon I will be uploading a little more visualization based explanation of RTT based on what I learnt through your video. Thanks again.
What about that application of RTT in the nozzle and cylindrical pipe? Didn't get what that means. Could you be more specific like the other explanations you gave?
great exposition, clear, simple, within the range of konwledge of an averaged physic student: i see, you´ve got six videos in a row about CFD, are you going to add new videos ?? thanks
Than you so so so much. My native language is not English, so I find it trouble to understand what my prof said cause he spoke at a high speed. Now I understand!!!
Thank you very much for the great video! I still have same issues understanding the lagrangian idea. If we would talk about a solid material or one particle, I would totally understand it. But how is it possible for fluids, that the particles, that are considered at time t in the system volume are the same particles at time t +dt and are not mixed up with other particles. So basically the question is, how is it possible to follow a package of mass as it could could comprise of other particles a little time later?
One year later but I hope I can help. That's because of continuum hypothesis. The volume we have is big enough to keep particles inside it and small enough to measure the same quantities in every point. Besides of that, there's no mass crossing a system (or a volume of fluid) , so it can be deformed and its shape can be changed but it will still have the same particles. And because it can be deformed, in Fluid Mechanics it's much more useful to work with control volumes, which is completely defined by you in the sense of geometry, size and so on.
Hello professor. Im a bit confused of the illustration of the control volume which is the red part. At time t and t+dt is cv is still the same? Thank you. And if yes, can we just cancel out the those cv in the further equations? Thank you sir! :)
The idea is that when we allow dt to become very very small (as the limit approaches zero in calculus language), the two parts of the sketch become the same region in space. That doesn't mean we can cancel them out though.
6:42 "And you can probably see that we're going to get into calculus here". Bruh I can't even see a "d" in a word without thinking I'm about to be calcced
Tim Nguyen Not totally sure what you mean, but Bcv(t+dt) does not equal Bcv(t). We say the system and the control volume are equivalent only at the beginning instant (t) because they occupy the same space. The second equation is where we account for the changes over the interval dt.
I'm still confused about the derivative parts. Why do we use material derivative for the B_sys? What I've learnt is this: For B_sys, it only got derived by time, not with position, so it can't be written as material derivative right? Why don't we use partial derivative notation here?
Hi Kevin, you have a valid point. The material derivative is not applicable to B_sys. The operator D/Dt can be applied only to field variables (like velocity, temperature, density etc) and not to variables associated with identifiable entities of mass. Hence, it is appropriate to use the operator d/dt for B_sys where B_sys = integral of (rho*b*dV) over the material volume. Also, please note that the volume integral corresponding to B_sys at 3:20 is an integral over a material volume whereas the volume integral corresponding to B_cv = integral of (rho*b*dV) is an integral over a volume which is a geometric entity (not associated with matter).
Hi Joe, excellent explanation. Actually understood what was happening for once! Just a question, 11:18 was this supposed to be "Flow rate of B(out) of the CV, and not system? Thanks!
No it does not. That term represents the amount of b inside the CV, and so velocity doesn't enter into it. Remember though that in the case of momentum, b represents the velocity.
It really just means that you're fixed to something that's independent of the flow. So perhaps the control volume is "all the water in a bath tub." The size of that control volume will increase and decrease along with the water level but the physical thing it represents remains fixed. In Into Thermo we would describe the difference most basically as: a system follows the same mass while a control volume allows mass to enter and leave to accommodate representation of some convenient device.
@@EME303HN ok. By now I have never encountered any case where the size of control volume changes. In fact I have seen some well known people in fluid mechanics area saying that the volume of a control volume region doesn't change ( that is why I asked about size and not shape). Thank you😊.
+Sonnie Chan It's due to the dot product in the (V dot n_hat). On an inlet, the surface normal (n_hat) points outward from the surface, which is opposite the direction of the velocity. So there's 180° between the velocity and surface normal vectors, which produces a -1 when the dot product is computed.
