Heat Transfer: Thermal Radiation Network Examples (16 of 26)

UPDATED SERIES AVAILABLE WITH NEW CONTENT:
• Heat Transfer (2020) -...
You can find the syllabus for the course here: www.cpp.edu/~me...
Want to see more mechanical engineering instructional videos? Visit the Cal Poly Pomona Mechanical Engineering Department's video library, ME Online: www.cpp.edu/meo...

[@buckeyefan635]

I wish you were my heat transfer professor. Thank you sir.

[@CPPMechEngTutorials]

You're welcome.

[@piunikhaghnazarian1524]

WOW! my professor attempted to teach this in just 2 slides with just 3 equations of the nodes . Dr. Biddle bless you because you just shined a whole new light on this! Thank you!

[@Shubhamkumar-cq5wt]

Thank you sir! This lecture has cleared all my doubts.👌

[@thuyavansathiamurthi3592]

This is just straight god tier teaching. Holy fuck, i'm gonna get an "A" on my exam because of this. Amazing

[@arandomguymarra8612]

Come work for UF. We are in dire need for a better heat transfer professor!!! You are GREAT! Thanks!

[@aiart655]

Helped me a lot! Thanks!

[@jipengelhart531]

Bob Ross of heat transfer, thank you!

[@CPPMechEngTutorials]

Even happy little trees emit thermal radiation.

[@debjeetdutta9433]

Thanks .. I found the lecture very useful to solve numericalsss

[@ab-xo4yc]

Excellent explanation

[@santhoshkumar.s.1435]

Thank u sir,very helpfull videos and very informative

[@SureshKesraniOfficial]

very well explained

[@modeleola8862]

Thank you very much sir

[@SquishyBrained]

SO GOOD!!! Thank you so much.

[@user-tg3wb7vt6p]

Thank you so much, it was really helpful.

[@vineetlakhera1183]

Thank you sir.

[@mrzombiesbo]

So all of this theory about space and surface resistances assumes opaque, grey surfaces, correct?

[@shashwatchopra6154]

Too good

[@shiekhmajhrul1891]

Thank u so much sir 🙏

[@abelmedina7879]

At 22:00 to know which direction your q is going are we saying that if it is negative that it goes away from Eb3?

[@hadimalekzadeh4567]

Thank you for this useful teaching

[@CPPMechEngTutorials]

:)

[@jeremymanus7229]

How does R23=R32? A2 does not equal A3.

[@lucaswernerkuipers8120]

By reciprocity... AiFij = AjFji, as resistance is defined: Rij = (AiFij)^-1 ... Raising the reciprocity expression to -1 does not change it... So if A2F23 = A3F32 -> 1/(A2F23) = 1/(A3F32).

[@entropyz5242]

That’s not the reason, although it is a true statement. It’s because of symmetry.

[@Amba97]

so R23=1/(9*0.8), R32=1/(36*F32). since F23 = 0.8 what surface 3 "sees" of surface 2 has to be 0.2. Therefore R23=1/(9*0.8) = 5/36, R32=1/(36*0.2)= 5/36. So they're equal. I do agree he wasn't very clear when he explained why they're equal.