You can solve these problems in many different ways. I showcase one method, but as long as you get to the same answer, the path you take doesn't necessarily matter much.
Professor I do not understand something, at 3:11 it is said that _diffuser_ increases pressure while reducing velocity_ what's the logic behind that ? I mean how pressure is increasing as velocity decreasing?
I am not sure how to explain the "logic" behind this without going into proofs. But to increase pressure, the velocity needs to drop in a diffuser. For a nozzle, to increase the velocity, the pressure drops. In other words, when we reduce the cone at the right end, the velocity goes us, but the pressure starts to drop. Please see your textbook for proofs :)
you initially said delta h =Cp(T2-T1) but changed it in the diffuser question to Cp(T1-T2). How do you know which way to put the temperatures? Does it matter?
@@Jon-xu7jlAhh, I see, it should be h2-h1 corresponding to T2-T1. They correspond to each other. At 5:30, when we did the energy balance, we got h1-h2, which gives us the corresponding T1-T2.
Please see this video (ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-rKbjRG4Y-HM.html) on how to figure out if a substance is superheated, or a compressed liquid. Especially around 5:50 min mark, I go into how to figure it out.
This process is called unity conversions. You can see a video of it here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-shXni5PxQ0w.html 1 j/kg = m^2/s^2 If you know that, you can basically convert it whenever you want. Please make sure to understand unity conversions, it'll make your life a lot easier :)
I think when you increase pressure velocity increases, and the reverse happens when you decrease the pressure. This is because big area means small pressure, and small area means big pressure
Sir at the @ 10:59 you have considered mass flow rate and since also there only one inlet and one outlet and can I solve the problem without considering mass flow rate 😊
@@QuestionSolutions my question I believe is general, but for the sake of the video, it's on 5:40. I thought we can only use specific heat cp, if the pressure is constant, but it changes in diffusers and nozzles.
So this is just calculating enthalpy with the assumption that it's an ideal gas. For enthalpy, you use cp, and generally, you will find that the difference in pressure is minimal that it makes little difference. You can see a different problem solved here: pressbooks.bccampus.ca/thermo1/chapter/5-5-application-of-the-mass-and-energy-conservation-equations-in-steady-flow-devices/ check out example number 3 if you need further clarifications. @@itsyogurt9247
I don't have a channel for strength, I also don't know any to recommend. Did you try googling it to see what comes up? There are so many really good, educational channels that teach pretty much anything you want :)
@@QuestionSolutions I know but they don't show as you do. I like this channel because thanks to you I can imagine what is going on in the problem :D Thank you tho