If your cylinder has thin walls, then you can just do Q = 2xQ of outside. If it is a solid cylinder, then only outside matters (since it is area-related). If it has thick walls, then you can calculate the nusselt twice (one for the outside wall, and one for the inside) just like in the video but changing the characteristic length. The only situation in which the Nu correlations from the book cannot be used, is when you have a void small enough in which the boundary layer of one part of the wall interacts with another part. Then you can use FEA

@@EngineeringHack Thank you for the prompt and valuable reply. Just a couple of more clarifications if that’s OK: 1- My cylinder has relatively thick walls, so the Nusselt number corresponding to the inside part of the cylinder should be computed. In this case, the characteristic length would be the pipe diameter (instead of the length), right? In this case, the Gr number in the condition [ (D/L) >= (35/Gr^0.25)] would be computed based on what characteristic length. 2- From what I understood from the video, in case the (D/L) >= (35/Gr^0.25) is not met, I should account for the flat horizontal plate. In my case, the top and bottom covers of the cylinder are adiabatic surfaces (my fluid is actually enclosed in a vertical cylinder with perfectly insulated top and bottom plates). So I need not worry whether this condition is met or not, right? 3- My cylinder diameter is around 95 mm with a length of 5 m. So the issue related to having “a void small enough in which the boundary layer of one part of the wall interacts with another part” could be applicable? How do I know if it is indeed applicable, because I would like to avoid Finite Element Analysis ☹. Kindly clarify 🙏.

You have flow energy, and a mass flow rate. So you need to account for that. Recall that h = u +/- PdV

This video might help kzbin.info/www/bejne/o4DWdppolqiGkK8si=z5a182mqVt8a6kW0

Not sure. At what time stamp is it blurry? Haven't seen it

Because we are using the dt/dx for aluminum. It wouldn't make sense to pair the k for water with the gradient for the aluminum. Also note, the h calculated there is related to the heat transfer as we go from conduction to convection. After the water molecules are carrying the energy (i.e., away from the boundary layer), the h will be different and based on the k of water

Check out the more recent videos on thermodynamics. There is always a link to the property tables in the video description. It is the same tables that were used here

I wrote both. It is the same for both units as it is a difference (delta) in temperature

Yes. You are spot on As a matter of fact, the math only works out if the 4.86 is in the bottom, dividing. Thanks for flagging, I'll update. Final numbers are correct though, I flipped during the presentation but kept the values I got when solving on my own

@@EngineeringHack No stress, just wanted to make sure I didn't mess up somewhere. Really appreciate your videos! They help a lot.

kzbin.info/www/bejne/pnPRpJ-madyBopI Thanks, mate

Yes, it is dynamics by beer, Johnson, Cornwell and Self

Thx sm

Not sure if you are asking about the power or the temperature. The power is in W, but it is 65. The temperature is 64 but is in C. Please clarify and I'll look into it for you

Sure. It is explained in detail @08:00 Any specific question?

Not anymore, mate. What are you needing help with?

Thank you. I do my best. I appreciate you taking the time to comment.

I'm glad it helped. Thanks for taking the time to comment

It is in the table, it just won't be a steam table. You'll look for a table that is only dependent on temperature and has Pr and Vr among others. Note this is not pressure and volume but proxies that are only temperature dependent

Sorry mate. This is a live recording of the class. New videos are hopefully better. I can record these again later

Thanks. Yeah, I'm not a fan of Chegg. Often incorrect and students have to unlearn something wrong to then learn it correctly...

You are welcome. I'm glad it helped

Pressure and volume or power and voltage? :)

@@EngineeringHack pressure and volume

It's dynamics by beer, Johnson, Cornwell and Self

@@EngineeringHack may i ask what chapter is this please, im not familiar with the book

It's clear

Thanks for commenting 😊

Right on

Wow! Rare exception!

Perfect. That is your brains system 2 though :)

So nice of you

Thanks for commenting. Hope it is helpful

Yunus A. Cengel & Michael A. Boles

Both :) A teacher never stops being a student m.kzbin.info/www/bejne/gXqtkGChrMqDgpo

Thanks

No... The question clearly states: "Taking into account the variation of specific heats with temperature" So you need to use a table. HOWEVER, if that wasn't the case, and knowing air can be treated as an ideal gas in most situations, you could use the specific heat and solve it without a table.

Two possibilities: either you grab another table (another source) that does have the temperature range you need, or you extrapolate. There is a video in the channel which I show that case. Let me see if I can find it

Check the case of Freon here kzbin.info/www/bejne/fYGXZKZ-r9SeiK8si=UqnR-ErsCqKh7dIL You can take the 15K delta and the 30k delta and extrapolate for things higher than 30k. The farther from 30K, the worst your estimate

Yep. It's dynamics by beer, Johnson, Cornwell and Self

Glad it was helpful! Thanks for taking the time to comment

You're welcome! thanks for commenting :)

Thanks for your comment. Which interpolation? The enthalpy of 562.26?

Yes. Nevermind, actually. The values are fine. Just realized you were interpolating for h, not T

Glad you like them! 😊

Really? How so?

I checked and I don't think it is :) But if you care to expand on why you think that, I can look into you.

Because we are solving for k, not for Sh

Yep. This is from Cengel and Yunus

You are welcome. Thanks for the comment ☺️

You need the molar weight if you use the ideal gas constant (R) in its general form (i.e., per mol). You'd then use the molar weight to convert the R into the specific R for oxygen. Because in our case we grabbed the specific R directly from the table, we were able to skip this step. Hope this helps. Happy studying

Glad you enjoyed it! Thanks for commenting

Yep. Check the UNSW question playlist and there are questions just like that

So nice of you

In which state? Problem statement says it is 300kPa...

State 3, I meant

@@yigitcan824 State 3 is also 300kPa because of the quasi-equilibrium situation we find when we have these frictionless pistons. A more comprehensive explanation can be found here kzbin.info/www/bejne/n6C8i3enhbdkeNEsi=ai9-E5iYSM2kngog&t=63

I can surely check. What did you get fro your answers?

@@EngineeringHack Giving you my answer so you can compare it to the textbook answer isn't going to confirm the calculus error. Understand this method of error evaluation is indeterminate, as I could have made an algebraic error from that point on-wards. The most effective way to see the error is if I directly show you the mistake. Reviewing your working from the 6.00 minute mark forward: You started with r(t)=(250/(t+4)) - agreed. You correctly calculated the radial velocity: r ̇(t)=((-250)/(t+4)^2 ) - agreed. Then when you calculated the radial acceleration: r ̈ (t)=d2r/dt2=d/dt {(-250/(t+4)^2 )}, you forgot the (-2) from the (-2)(-250)(t+4)^(-3) That is: d/dt {(-250/(t+4)^2 ) = d/dt {(-250*(t+4)^(-2)) =(-2)*(-250)*(t+4)^(-3) = (+500)*(t+4)^(-3) = (+500)/(t+4)^3, not (+250)/(t+4)^3. Hope this helps.

@@animalfarm7467 same lol good vid nonetheless