This is very insightful thank you! Especially liking the speeding up during the writing of equations. Also, may I ask where the mass equation you've used initially comes from?
Thanks Brian. Is it true that in fully-developed pipe flow enthalpy is conserved (ignoring any elevation). If so can we replace h=cpT? I notice that in heat transfer for incompressible flows, they do that in textbooks. But here I don't think h=cpT would apply as I think the temperature would have to rise due to irreversible effects. Isn't this a double standard?
Excellent set of notes. Thanks for that. I wish the section on pipe flow continued. What's your take on substituting for enthalpy -> h=cpT. I see it in many engineering textbooks but if pressure is dropping how could this be valid?
The notes are a work in progress and kind of drop off in the later chapters. One day... I didnt introduce enthalpy here as I was not trying to get too deep into thermodynamics. If you want a real thorough explanation of the different forms of the energy equation, see the class book by Bird Stewart and Lightfoot. It is more complete than anything I could ever hope to do.
Why we need to integrate the velocity along the domain y to find the volume flow rate. Because according to my understanding, the volumn flow rate is the velocity X cross-sectional area. so it should be integral velocity by dA. Does my understanding wrong?
What is the difference between this link (ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-mtoHgKdXLJk.html) and this video? Why the volume flow rate calculation method different?
+Jay Tee Yes, I am using the term "steady state" to imply that things don't change with time. Of course, this is an assumption that things are not changing in time. It can certainly be true experimentally with these flows that you can have inlet conditions which are constant in time but the flow itself is fluctuating. In this analysis, we seek a solution consistent with our assumptions. For this flow, if we set up a quick experiment in the lab and keep the Reynolds number less than ~100 and keep the channel height to be small compared to the length, it is the actually the flow we would observe. We could also observe the parabolic flow at higher Reynolds number (~1000) if we were a little more careful. At some point, however, the flow loses stability no matter how careful we are!
+Brian Storey Thank you for the reply. It's just the way you stated it, at steady state, no change in time. which is not true because the constant thing is another variable like mass over the change in time.
+Hdt Pdr Yes, of course the pressure gradient is needed to drive the flow and dP/dx must stay around. Hopefully the whole video is clear about the assumption that the flow doesn't change down the channel, but the pressure does drop continuously and dP/dx is not zero. I choose to talk about d/dx terms being zero as that seems perhaps more intuitive as to what you are assuming (that all sections of the pipe are the same) than saying that lets assume u(y) only. Of course they end up being equivalent. Hopefully the explanation is clear since with the videos it is not easy to go back a redo one little part to make something clearer!
