Heat Transfer (14): Transient heat conduction, approx. solution model (spatial effects) and examples

0:00:15 - Review of previous lecture
0:01:26 - Spatial effects for transient heat conduction
0:20:52 - Example problem: Long cylinder with transient heat conduction
Note: This Heat Transfer lecture series (recorded in Spring 2020) will eventually replace the earlier Heat Transfer series (recorded in Fall 2015) in a couple years. The old lecture series had some errors and was lower quality.
Want to see more mechanical engineering instructional videos? Visit the Cal Poly Pomona Mechanical Engineering Department's video library, ME Online: www.cpp.edu/meonline​​​​
This lecture series was recorded live at Cal Poly Pomona during Spring 2020. The textbook is "Introduction to Heat Transfer" by Bergman et al.

[@briansimons2703]

This was absolutely fantastic. I watch these videos to understand the worked examples my professor does in class. Professor John Biddle is the MAN

[@apollo3597]

in my second year as a student in india at BITS Pilani, and i cannot thank you enough for these resources. a very big thank you to sir for helping us here! You're the best Prof John 🙏

[@manuboker1]

Thanks ! Wonderful Heat Transfer lectures !

[@blu3ey358]

Thanks a lot for sharing this content, it helped me a lot! :)

[@CPPMechEngTutorials]

:)

[@TRONN44]

I believe there is a mistake at 40:40 when writing the volume of the cylinder. It should be (pi*(D^2)*L)/4.

[@ryanhalloran6055]

I agree

[@jackeywackey9488]

This video helped me alot with my understanding, thank you.

[@CPPMechEngTutorials]

Our pleasure.

[@nicklarosa3816]

For the part of the problem where we solve for Q, all my values were negative? Is this a mistake or are we just taking taking the magnitude

[@anonymous-sk2pr]

where can i download the formulas from???

[@afonsojacinto7347]

thank you a lot, great video!

[@mustafaenesyaldiz]

why we took cyl volume as a Pi*D*L instead of (pi*D^2*L)/4

[@peronkop]

What is the ambient temperature is different on the two sides of the wall?

[@hishamsalameh5750]

What a great effort !!!!
Are the equations used when Bi>0.1 in final exam data package?
Plus could you please provide me with the final exam data package?

[@maxneale6334]

I find it so funny how all the problems and calculations are using metric because it makes it so much easier, then at the end when giving real world examples its back to feet and inches. Must be such a pain for american students to get their head around metric when they start studying engineering.

[@alisiraydemir]

Question: Is there any mistake with the volume of the cylinder, or did the lecturer just take the volume of the rectangle where the fluid contacts the cylinder, pi * d * L * 1 (unit thickness)?

[@Akkarsollum]

Great lecture but I wish the values were plug in and how he got it. Then the table just makes it ambiguous for anyone externally, having a poster size or even 11x17 showing the tables for on whiteboard would be awesome.

[@isaachiew4906]

Hi there, how is C1 = 1.048 rather than 1.0050? Since Bi = 0.02 (using Table 5.1 formula), the corresponding C1 value is 1.0050 and zeta = 0.1995. Moreover, how was J0 obtained? I understand we have to refer to the Table of Bessel Functions of the First Kind in Appendix B but how to determine which value of J0 we should take? Truly appreciate if someone can clarify this to me 😊

[@byungshin8773]

He says to solve for zeta*r in the J0 function. That's how x is determined, which you can use to find the J0 values from the table.

[@throttlejunky2211]

I think there is a mistake in the example of part b. I got a surface temp of 977c. if I don't multiply zeta1 and solve, I get the answer shown of 836c

[@CPPMechEngTutorials]

At what time?

[@throttlejunky2211]

@@CPPMechEngTutorials part b for the part leading up to the answer at 37:00

[@CPPMechEngTutorials]

@@throttlejunky2211 I think Prof. Biddle is correct. (T-1200)/(300-1200) = 0.444*0.91 Solve for T = 836

[@throttlejunky2211]

@@CPPMechEngTutorials ok thanks for checking. I thought right side should be =.444*.91*.611 as J=0.91 and zeta=.611. my mistake

[@mechanicalkiller1061]

@@throttlejunky2211 yes i think you are right, we must multiply by zeta!! since r* = 1

[@mohamedhamde6611]

Thank you sir

[@CPPMechEngTutorials]

You're welcome!

[@ryanhalloran6055]

Isn't Q supposed to be just Joules @ 43:02 ?

[@vannavareevannasurachat1945]

I think it should be J/m due to the unit of Q_0 but I am quite not sure about now. It is because there is no anyone notices that except me.

[@abdelrahmanmahdy9354]

@@vannavareevannasurachat1945 yes you are right, it is J/m. just a typo mistake.

[@mechanicalkiller1061]

@@vannavareevannasurachat1945 you are right

[@willcolletti4357]

Hello does anyone know what textbook they are using

[@lilmes3588]

incroperas heat transfer

[@martinjimenez1615]

@@lilmes3588 The textbook is "Introduction to Heat Transfer" by Bergman et al.

[@ulquiorras422]

@04:00 Corridor Digital Cap

[@FavouredElm2]

i see u r a person of culture as well

[@hs_house]

Fundamental of Heat and Mass Transfer by Theodore L. Bergman and Frank and David P , 8TH EDITION DETAILED SOLUTION
www.studypool.com/services/10789415

[@AdmiralSym]

nice catch

[@Werevertumoto]

Damn...

[@CPPMechEngTutorials]

In a good or a bad way?

[@AC-ul4xz]

geek

[@hs_house]

Fundamental of Heat and Mass Transfer by Theodore L. Bergman and Frank and David P , 8TH EDITION DETAILED SOLUTION
www.studypool.com/services/10789415