Heat Transfer (01): Introduction to heat transfer, conduction, convection, and radiation
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- Опубліковано 11 лют 2025
- 0:00:15 - Introduction to heat transfer
0:04:30 - Overview of conduction heat transfer
0:16:00 - Overview of convection heat transfer
0:24:51 - Overview of radiation heat transfer
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/meo...
This lecture series was recorded live at Cal Poly Pomona during Spring 2020. The textbook is "Fundamentals of Heat and Mass Transfer", Bergman et al., 8th Edition.
This man had got me through fluid mechanics, now he’s carrying me through heat transfer. I owe this man a whole degree
This man has helped me too
FACTS, only if he taught intermediate strength
same :)
REALLY enjoyed fluid mechanics with him
Maybe we should start a Patreon for Prof. Biddle. :P
@@CPPMechEngTutorials facts
I literally owe this man my life. I spent an entire class time doing this and he explained it to me in 10 minutes. Man is built different
You dumbass
Professor Biddle is an absolute godsend. It's a crime I'm paying my university thousands when really, I'm learning from him.
Amen??
haha true
I am just having a laugh at myself, because I considered Cal Poly Pomona when I was transferring, but opted not to apply because I had good shot at getting into a "better school". The funny bit is now that I'm actually in one, I'm watching lectures from Cal Poly Pomona and UC Boulder because all my teachers are ass
hahhaha@@NeatoBandido
pay him now
I passed my thermodynamics class because of Professor David Miller lectures. I got shocked when i saw how many students fail the Heat Transfer class at my college. But with your amazing lectures i'm sure i'm in good hands. Thanks in advance!
If you liked Prof. Miller, you are gonna love Prof. Biddle.
Did you pass?
@@CPPMechEngTutorials Yes :)
Which university?
Wow.
See timing
Amazing class. Just started for 5 mins but his love and way of telling this subject made me to watch full. This is really amazing.
Halfway through February of spring semester and I've already absorbed more from this video than my lectures 3 times a week in person.
I'm a Brasilian student, and i dont understend so much english, but in this video, with this teacher, heat transference is very much easy! Congratulation for your job.
Helping people in every hemisphere is what we do. :D
I decided to pursue energy engineering and am now watching His videos to prepare for the elective on Heat transfer that I inevitably will have to take. I am one of those people who don't have the confidence to do anything without having a strong theoretical side as a backup. I didn't know what Cal Poly Pomona was before I learned Numerical Methods through Dr. Shokoufeh Mirzaei's tutorials. To undergraduates in this university, you are very fortunate to have such great teachers who are absolutely talented at delivering their lectures in an easily understood manner.
wow this professor is so good, im learning thermodynamics from him now.
way he describes and expresses the abstarct topics just amazed me. Thank you a lot sir. God bless you.
:D
Man, I can't say how grateful I am to Professor Biddle for his exceptional teaching of this course. I failed the Heat and Mass Transfer course in my sophomore year, but then, in my third year, I got an A- in this course after finding this playlist, all thanks to this man.
I watched lots of videos before came here, and this man omg he is literally godsend!! I only watched this lecture but i am pretty sure i will understand others before midterm!! Thank you professor may Allah bless you :)))
Thanks!
@@CPPMechEngTutorials
Boss.
What textbooks did the professor use please?
@@onekobo7937 Check the show notes.
Biddle is the best. He helped me understand Transport Phenomena concepts.
He's pretty good. :)
Good evening
Right in time for my heat transfer class. Thank you 😭😭😭
The videos will be slowly rolled out over the semester. Unfortunately, at this time we only have 3/4 of the course recorded. We're hoping the university will allow us to complete the lecture series in the next couple months.
this guy got me through fluid mechanics last semester. just started my heat transfer class and i'll be watching every single one of these videos 🫡
You won't be sorry.
Most articulated course on HT available on YT, will recommend as best introductory course for HT.
Aura of prof is very calming and helps in learning.
I'm an ASU student and our heat transfer professor does NOT teach. I'm so grateful to have Dr. Biddle's lectures to learn from. :-)
:D
First time seeing his lecture, ive never seen someone teaches a science topic like this. He is a real Guru❤
The thankful words would not be enough for what you are doing Sir, I graduated 12 years ago and your outstanding lectures make it easy to me to remember scientific information. thanks a lot dear.
Glad we could help.
Ah discovered the newer version of this course just in time. I watched Dr. Biddle's fluid mechanics course and it really helped with in-depth understanding and intuition. Thanks for the empowering lectures!
Our pleasure.
This is actually an entire class worth of lectures posted online for free, amazing
Prof. Biddle certainly is amazing. :D
Watched every single one of your fluid videos and managed to get an A. I am here to run it back. Thank you for this!
Glad it helped! Congrats on the A.
I say it with all sincerity, you explain it very well. Thanks a lot Professor.
What a great prof . i am not your physical student but i am your student on You Tube .. may god bless you for making subjects easier than its
thank you so much
ua-cam.com/video/K6yQJhLXm9s/v-deo.html
pl visit and comment
This man is brilliant, you dont always need someone from MIT to list out the basics for ALL to comprehend. You can tell he enjoys his job and has dedicated his life for it!
40+ years!
Oh my God this man is great. From this distant place I am I wish you were my lecturer... You're simply amazing
Thanks!
I can't thank him enough, he teaches things so fluently💗
voy a aplicar a un programa de doctorado y estaba buscando clases de transferencia de calor, para refrescar mi memoria jajaja, creo que nunca tuve una clase tan buena como esta... I was looking for heat transfer classes and I don't remember thar I had taken a good enough heat transfer classes such this, I really enjoy it.
God bless the Students that listen to this Professor due to the fact of keeping his feet down “ toes Un stepped on “ this guy will provide future students with the natural Gifts of the Holy Spirit because he don’t fear really because he really knows some science yo!! 👍👍😉👌
this is great i had heat transfer in spring of 2020 when covid hit this is a good refresher
I watch you from middle east
All thanks to you
We're glad you are enjoying the video.
I love this lecturer. literally.
Big thanks professor for that course, that really amazing, you have changed my world 🌎
:D
@@CPPMechEngTutorials where is professor J.Biddle? Is all good with him?
I have already failed Heat Transfer with a 50 once. Starting off just as bad. However this guy makes things better. Hopefully I will pass now
Good luck!
I also failed this course an year ago but this sem through this playlist got an A- on it. This is the best course for HMT I am telling you, all the best man.
@@adityavikramsingh321 didn't work out. The first test was weighed too much and the final was brutal, top score 72.
@@nightwings8984 ahh, sorry for you, bud. Hopefully, things will sort out. Keep going, it's a beautiful life we are blessed with.
Excellent demo and examples on the complex topic.
this man got has got me through heat transfer , thanks😄😄
I teach heat transfer and I used to watch this vids every year
I also started to teach this semesiter so i had to watch it .
Very special kind of handwriting. I think it should have a special name like "Times New Roman" :)
Thanks a lot sir
Finally i got a video to understand the topic for exams
Hooray!
Great lectures Mr. Biddel.
Please share also lectures about Classical Mechanics (Strength of Materials, Lattice etc.). Best Regards
We have a Strength of Materials lecture series. www.cpp.edu/meonline/strength-of-materials.shtml
He has simplified my life. Well explained.
Is he a professor or a god? Thank you. You r the best
Thank you for the knowledge Professor. God bless you.
It's really amazing class by professor Biddle , i may Give my best on My B.tech exam on 8th August . I will go through other classes for more information
Thank you 🤗
Thanks to you now I feel ready for exam:)
Am already in love with these lectures. I will love to get the lecture materials, i.e the book, he refers to.
Book is Fundamentals of Heat and Mass Transfer, 8th Edition
By which author😢
Astonishing , absolutely amazing.
Thanks ! Wonderful Heat Transfer lectures !
Bless up Prof Biddle 🙏 🙌 ✨️
Thank you for uploading first video
More videos will be slowly released over the semester.
Nice teacher
this is some sign from god, i was only looking for chem 1 heat transfer and ended up watching the whole thing
amazing lecture, loved it.
Hi, are you following the Fundamentals of Heat and Mass Transfer book by Frank P. Incorpera? And which edition? If not, which textbook are you following? Thank you ☺️
Thanks a lot for these amazing lectures. May I ask you which textbook are you using for this course? Thanks in advance
Fundamentals of Heat and Mass Transfer, 8th Edition
His lectures are awesome!
Yes.
we are the best teatcher of heat transfer
I am indian and I clearly understand this lecture ❤
Wonderful explanation from Grandpa
That's Professor Grandpa.
Thank you Sir..🙏
He is a genius ❤❤
THANK YOU PROFESSOR
I'm on a general review right now, before starting my final years project.
Thank you so much doctor for your tutorials.. make for us tutorials for material science..
We are discussing the possibility of making a lecture series on engineering materials. But it wouldn't be ready for a while.
What is the name of the book, professor mentioned during lectures?
Will you be deleting the previously recorded course from the channel or will they both exist?
Same question 😕😕???
@@yousifameen3607 Both the F2015 and S2020 lecture series will be public for at least a couple years. Eventually the older lecture series will become "unlisted" but still viewable if you saved the links.
I'm watching the earlier version of this course right now HAHAHA, and I'm gonna catch up the latest one :)
The content should be very similar, but this version is higher quality.
Awesome teacher!
:D
I lost my engineering degree , I could not Crack this subject
And there was not one decent professor who could take the class like him
Never mind , I have moved on
Anyways that was a long time back
Three decades
Untuk bentuk bola
Saya hanya fokus pada perbandingan jari jari selubung bola terhadap ketebalan bola. Jika ditemukan R/ dR = 1, maka rumus tersebut haru dikaji ulang. Dengan persamaan umumnya adalah :
Q = - k. A. (dT)/ (dR ), Hukum fourrier
Q = k. 4π R2. (∆ T)/ ( R2 - R1 ). ………. 1
Q = k. 4π R1. R2. (∆ T)/ ( R2 - R1 ). ………. 2
Q = k. 4π R1 . R2.(∆ T)/ ( R2 - R1 ) = k. 4π R2.(∆ T)/ ( R2 - R1 ). Pada rumus ini menekankan pada perbandingan jari jari selubung bola terhadap ketebalan bola. Sehingga persamaannya adalah :
k. 4π R1 . R2.(∆ T)/ ( R2 - R1 ) = k. 4π R2. (∆ T)/ ( R2 - R1 ), Sehingga bagian yang tereliminasi adalah k .4. π.
R1 . R2. (∆ T)/ ( R2 - R1 ) = R2.(∆ T)/ ( R2 - R1 ).
( R1 . R2.)0.5 (∆ T)/ ( R2 - R1 ) = ( R2 )0.5.(∆ T)/ ( R2 - R1 ).
( R1 . R2.)0.5 . (∆ T)/ ( R2 - R1 ) = R.(∆ T)/ ( R2 - R1 ). Sehingga didapatkan perbandingan jari jari selubung bola terhadap ketebalan bola.
( R1 . R2.)0.5 / ( R2 - R1 ) = R/ ( R2 - R1 ). Sehingga perbandingan jari jari selubung bola terhadap ketebalan bola
( R1 . R2.)0.5 : ( R2 - R1 ), adalah perbandingan jari jari selubung bola terhadap ketebalan bola. …….… 3
Dimana,
dR : Ketebalan sesaat. (R2 - R1 ) atau ( outer -inner )
dT : Temperature sesaat
K : Konduktifitas thermal materi
Q : Nilai kalor
R : Panjang jari jari selubung bola
∆T : Hasil integral dari dT
∆R : Hasil ( R2 - R1 ) = ( outer - inner ) = tebal
4πR2 : Luas kulit bola
R/ dR : Sebanding dengan R / ( R2 - R1 ) Perbandingan panjang jari jari selubung bola terhadap ketebalan bola.
Keberatan saya adalah :
1. R2 : R2 .- R1
R2 : R2 .- R1 ~ R2. R1 : R2 .- R1,
R2 = R2. R1
Komentar saya adalah bahwa R2 ≠ R1 x R2. R yang dimaksud adalah memiliki nilai yang sama, yaitu R x R.
Uraian penalaran :
R2 : R2 - . R1
R2 : R2 .- R1, untuk R2 = R x R.
R2 = R x R
R. R = R1 x R2 …???
R . R = R1 x R2, oleh karena R bernilai sama, maka R1 = R2. Penulisan ini akan bermasalah dengan persamaan :
Q = k. 4π R2.L (∆ T)/ ( R2 - R1 ), dimana R2 = R1.
Q = k. 4π R2.L (∆ T)/ ( R2 - R2 ).
Q = k. 4π R2.L (∆ T)/ ( 0 ).
Q = + ∞ … ???.
Alternatifnya adalah :
R2 : R2 - . R1 ,
R2 : R2 .- R1, untuk R2 = R x R.
(R . R}0.5 = ( R2 . R1 )0.5
R = ( R2 . R 1 )0.5 … ???.
Apakah jari jari selubung bola ( R ) adalah ( R2 . R1 )0.5 … ???
Jika demikian halnya. Apa yang bisa dijelaskan dari ( R2 .R1)0.5 secara visualisasi … ???.
Komentar saya adalah kesimpulannya R2 ≠ R1. R2
2. Saya akan paksakan bahwa R2 = R1. R2. untuk membuktikan keberatan saya. Selanjutnya dengan melakukan substitusi ke persamaan awalnya, saat R1 = 0. ( Solid ).
Q = - k. A(dT)/ (dR ). Hukum fourrier
Q = k. 4π R2.L (∆T)/ (∆R)
Q = k. 4π R2.L (∆T)/ ( R2 - R1 )
R2.adalah jari jari outer dan R1 adalah jari jari inner
R2 = R1.R2
Q = k. 4π R2.L (∆T)/ ( R2 - R1 ).
Q = k. 4π R2.L (∆T)/ ( R2 - R1 ). saya fokus pada persamaan yang hurufnya bold.
Bagaimana jika R1 bernilai 0 ?. Saya substitusikan rumus dasarnya.
Q = k. 4π R2.L (∆T)/ ( R2 - R1 ).
Q = k. 4π R1. R2.L (∆T)/ ( R2 - R1 ).
Q = k. 4π 0. R2.L (∆T)/ ( R2 - 0 ).
Q = k. 4π 0. .L (∆T)/ ( R2 ).
Q = k. 4π 0. .L (∆T)/ ( R2 ). Pada persamaan ini muncul 2 jawaban.
Dalam hal ini akan muncul 2 alternatif jawaban.
Q = k. 4π 0. L (∆T)/ ( R2 ), 0/R2 = 0,
Sehingga,
Q = 0 ( alternatif 1 ).
Komentar saya adalah apakah saat bola dalam keadaan solid, maka Q = 0 ???. hal ini sangat tidakl masuk akal.
Atau altenatif lainnya adalah
Q = k. 4π R2.L (∆T)/ ( R2 - R1 ), R2 = R1. R2
Q = k. 4π R1. R2.L (∆T)/ ( R2 - R1 ), 0 = R1
Q = k. 4π 0. R2.L (∆T)/ ( R2 - 0 ), R2 - 0 = R2
Q = k. 4π 0. L (∆T)/ ( R2 ).
Q = k. 4π 0. .L (∆T)/ ( R2 ), 0/R2 = 0 pindah ruas kiri
Q / 0 = k. 4π .L (∆T)
+∞ = k. 4π .L (∆T ), k. 4π.L adalah nilai konstan ( C ), sehingga
+∞ = C ((∆T ), sehingga,
+∞ = ( ∆T ), ∆T = T1 - T2
+∞ = T1 - T2, , T1 = +∞ dan T2 diketahui, sehingga
+∞ = ( +∞ - T2 ), satuan sisi kiri adalah watt dan satuan sisi kanan temperature sehingga tidak boleh dikurangi.
+∞ = T1
+∞ = (∆T ), ∆T = ( +∞ - T2 ) ( alternatif 2 )
+∞ = T1, ∆T = ( +∞ - T2 )
+∞ = T1,
Komentar saya adalah apakah saat bola dalam keadaan solid, maka (∆ T ) = (+∞ ) atau (T1 ) = (+∞ ) ???, hal ini sangat tidakl masuk akal.
continue....
17:10, can someone explain this graph? I am not familiar with Fluid Mechanics so I don't know what the 'u' variable stands for.
What I got from the lecture:-
The horizontal line is a hard surface (say, concrete slab), the sloid line going up from it is a wall and we fill in the space between this wall and the vertical axis with a fluid (say, water). Now, what is u and u-infinity?
(I am watching these videos for pure curiosity, and am not enrolled in any class).
I continued watching the video after my comment, and feel like I grasped the concept well enough.
The 'u' variable wasn't necessary after the 17:10 graph. There is fluid (liquid or gas) on a solid surface with a surface temperature as well as fluid temperature. We are trying to find out at what rate will heat transfer between the fluid and the solid surface depending upon their temperature difference, the inherent nature of the fluid, and geometry & size of the solid surface.
Wwooww, He is such a Good Lecurer
is this ideal lecture for gate/ese/isro preparation ?
Perfect timing for my PE exam this summer
what does PE stand for?
@@oximas-oe9vf professional engineer
It's common for chemical
well explained thank you
Am enjoying this class than my school class
Let others know about your experience. :D
best video. Thank you!!!
Hey, thanks for helping me with heat transfer! I just want to ask if you have videos for mass transfer...
No. Mass transfer is less necessary in mechanical engineering compared to chemical engineering.
I am truly thankful for the lectures of the Professor and the content was so helpful for me. I would appreciate it if you could tell me what was the textbook the professor used as a reference in this course?
It's in the video notes.
@@CPPMechEngTutorials I am deeply grateful for your assistance, and I'm sorry about my carelessness.
@@rasasaibani668 No problem. We get asked this question all the time. :)
Fundamental of Heat and Mass Transfer by Theodore L. Bergman and Frank and David P , 8TH EDITION DETAILED SOLUTION
SOLUTION: Fundamental of Heat and Mass Transfer || 8TH Edition || step by step solution - Studypool
Fundamental of Heat and Mass Transfer by Theodore L. Bergman and Frank and David P , 8TH EDITION DETAILED SOLUTION
SOLUTION: Fundamental of Heat and Mass Transfer || 8TH Edition || step by step solution - Studypool
When Will other video lectures regarding subjects like mechanical engineering design, engineering drawing, refrigeration and air conditioning, dynamics and mathematical techniques be uploaded?
In half a year or so?
Btw thanks for these lectures,you people are helping alot.
It takes a lot of time and resources to make these videos. We create them when time permits.
Why can't I have professors like him
Excellent !
Nice video sir
What are the recommended text books for this module...please
Is a mirror a black surface, body, or perfect emitter?
Thank you sir..
Does anyone have "Fundamentals of Heat and Mass Transfer", Bergman et al., 8th Edition, PDF version for free, please? 🙂
❤❤❤❤❤❤❤ thank you so much
I wish this guy was our teacher
For ball shape
I only focus on the ratio of the radius of the ball casing to the thickness of the ball. If R/ dR = 1 is found, then the formula must be reviewed. The general equation is:
Q = - k. A. (dT)/ (dR ), Fourrier law
Q = k. 4π R2. (∆ T)/ ( R2 - R1 ). ………. 1
Q = k. 4π R1. R2. (∆ T)/ ( R2 - R1 ). ………. 2
Q = k. 4π R1 . R2.(∆ T)/ ( R2 - R1 ) = k. 4π R2.(∆ T)/ ( R2 - R1 ). This formula emphasizes the ratio of the radius of the ball casing to the thickness of the ball. So the equation is:
k. 4π R1 . R2.(∆ T)/ ( R2 - R1 ) = k. 4π R2. (∆ T)/ ( R2 - R1 ), So the eliminated part is k .4. π.
R1 . R2. (∆ T)/ ( R2 - R1 ) = R2.(∆ T)/ ( R2 - R1 ).
( R1 . R2.)0.5 (∆ T)/ ( R2 - R1 ) = ( R2 )0.5.(∆ T)/ ( R2 - R1 ).
( R1 . R2.)0.5 . (∆ T)/ ( R2 - R1 ) = R.(∆ T)/ ( R2 - R1 ). So we get the ratio of the radius of the ball casing to the thickness of the ball.
( R1 . R2.)0.5 / ( R2 - R1 ) = R/ ( R2 - R1 ). So the ratio of the radius of the ball casing to the thickness of the ball. ( R1 . R2.)0.5 : ( R2 - R1 ), ), is the ratio of the radius of the ball casing to the thickness of the ball. ………. 3
Such,
dR : Instantaneous thickness. (R2 - R1) or (outer -inner)
dT : Instantaneous temperature
K : Thermal conductivity of the material
Q : Calorific value
R : Length of the radius of the ball casing
∆T : Integral result of dT
∆R : Result (R2 - R1) = (outer - inner) = thickness
4πR2 : Area of the sphere's shell
R/ dR : Proportional to R / (R2 - R1) The ratio of the length of the radius of the ball casing to the thickness of the ball.
My objections are:
R2 : R2 .- R1
R2 : R2 .- R1 ~ R2. R1 : R2 .- R1,
R2 = R2. R1
My comment is that R2 ≠ R1 x R2. The R in question has the same value, namely R x R.
Description of reasoning:
1. R2 : R2 - . R1
R2 : R2 .- R1, for R2 = R x R.
R2 = R x R
R. R = R1 x R2 , …???
R . R = R1 x R2, because R has the same value, then R1 = R2.
This writing will have problems with the equation:
Q = k. 4π R2.L (∆ T)/ ( R2 - R1 ), where R2 = R1.
Q = k. 4π R2.L (∆ T)/ ( R2 - R2 ).
Q = k. 4π R2.L (∆ T)/ ( 0 ).
Q = + ∞ … ???.
The alternatives are:
R2 : R2 - . R1 ,
R2 : R2 .- R1, untuk R2 = R x R.
(R . R}0.5 = ( R2 . R1 )0.5
R = ( R2 . R1 )0.5 … ???.
Is the radius of the spherical casing (R) (R2. R1)0.5...???
If that's the case. What can be explained from ( R2 .R1)0.5 in visualization ... ???
My comment is that the conclusion is R2 ≠ R1. R2
2. I will insist that R2 = R1. R2. to prove my objection. Next, by substituting into the initial equation, when R1 = 0. (Solid).
Q = - k. A(dT)/ (dR ). Fourrier law
Q = k. 4π R2.L (∆T)/ (∆R)
Q = k. 4π R2.L (∆T)/ ( R2 - R1 )
R2. Is outer radius dan R1 is inner radius
R2 = R1.R2
Q = k. 4π R2.L (∆T)/ ( R2 - R1 ). I focus on the equations in bold.
What if R1 is 0? I substituted the basic formula.
Q = k. 4π R2.L (∆T)/ ( R2 - R1 ).
Q = k. 4π R1. R2.L (∆T)/ ( R2 - R1 ).
Q = k. 4π 0. R2.L (∆T)/ ( R2 - 0 ).
Q = k. 4π 0. .L (∆T)/ ( R2 ).. I focus on the equations in bold
What if R1 is 0? I substituted the basic formula.
In this equation, 2 answers appear.
In this case, 2 alternative answers will appear.
Q = k. 4π 0. L (∆T)/ ( R2 ), 0/R2 = 0,
Then,
Q = 0 ( alternatif 1 ).
My comment is whether when the ball is in a solid state, then Q = 0???. this is very unreasonable.
Or second opinions is :
Q = k. 4π R2.L (∆T)/ ( R2 - R1 ), R2 = R1. R2
Q = k. 4π R1. R2.L (∆T)/ ( R2 - R1 ), 0 = R1
Q = k. 4π 0. R2.L (∆T)/ ( R2 - 0 ), R2 - 0 = R2
Q = k. 4π 0. L (∆T)/ ( R2 ).
Q = k. 4π 0. .L (∆T)/ ( R2 ), 0/R2 = 0, turn left side
Q / 0 = k. 4π .L (∆T)
+∞ = k. 4π .L (∆T ), k. 4π.L is constant( C ), and then
+∞ = C ((∆T ), and then
+∞ = ( ∆T ), ∆T = T1 - T2
+∞ = T1 - T2, , T1 = +∞ dan T2 known, and then
+∞ = ( +∞ - T2 ), left side is watt and right side is temperature. so it cannot be reduced
+∞ = T1
+∞ = (∆T ), ∆T = ( +∞ - T2 ) (second opinions )
+∞ = T1, ∆T = ( +∞ - T2 )
+∞ = T1,
My comment is whether when the ball is in a solid state, then (∆ T ) = (+∞ ) or (T1 ) = (+∞ ) ???, this doesn't really make sense.
Well the Professor of course Gid Bkess Always 💕💕 love you dude no HoMO!!!!!
3. Apakah jari jari selubung bola dibanding ketebalan kulit bola pernah bernilai 1? Jika ditemukan rasio jari jari selubung bola terhadap ketebalan kulit bola memiliki nilai 1 , maka rumus yang digunakan perlu untuk dikaji ulang. Alasannya adalah jari jari selubung bola bergerak tidak linier terhadap ketebalan kulit bola. Sedangkan jari jari selubung bola bergerak berdasarkan fungsi hiperbola tertentu( tidak menyatakan garis linier. Yang bermakna R2 / ( R2 - R1 ) ~ R2 R1 ( R2 - R1 ).
Sehingga,
R / ( R2 - R1 ) = ( R2 R1 )0.5 / ( R2 - R1 ) adalah perbandingan jari jari selubung bola terhadap ketebalan kulit ( R2 - R1 ).
(R1 . R2 )0.5 / R2 -. R1 tidak pernah bernilai 1. Ditulis dalam persamaan adalah (R1 . R2)0.5 / R2 -. R1 ≠ 1.
Mari kita buktikan bahwa (R1 . R2)0.5 / R2 -. R1 = 1, jika bisa ditemukan , maka rumus sebelumnya harus dikaji ulang. Sebagai catatan bahwa saya akan paksakan bahwa R2 = R2. R1 dan R = ( R2 . R1)0.5. Sehingga perbandingan jari jari selubung bola terhadap ketebalan kulit bola ditulis dalam persamaan :
( R1 . R2 )0.5 / R2 -. R1 = 1 .……. Persamaan 1
Q ` = k. 4π R2.L (∆ T)/ ( R2 - R1 ).
Q = k. 4π R1.R2.L (∆ T)/ tebal.
Fokus pada R2 / ( R2 - R1 ).
R2 / ( R2 - R1 )
R2 = R2. R1
( R2 )0.5 = ( R2. R1 )0.5
R = ( R2. R1 )0.5
Yang dicari adalah jari jari selubung bola terhadap ketebalan kulit bola.
R / ( R2 - R1 ) = 1, R = ( R2. R1 )0.5
( R2. R1 )0.5 / ( R2 - R1 ) = 1, sehingga
( R2. R1 )0.5 = 1.( R2 - R1 )
( R2. R1 )0.5 = ( R2 - R1 ), persamaan di kuadratkan sehingga
( R2. R1 ) = ( R2 - R1 )2, diuraikan sehingga
( R2. R1 ) = ( R22 - 2 R2 . R1 + R12 ), pindah ruas
( R22 - 2 R2 . R1 + R12 ) = ( R2. R1 ), persamaan ( R2. R1 ) pindah ruas, sehingga
( R22 - 2 R2 . R1 + R12 ) / ( R2. R1 ) = 1, dilakukan pembagian, sehingga
R2 / R1 - 2 + R1 / R2 = 1,
R2 / R1 + R1 / R2 = 1 + 2
R2 / R1 + R1 / R2 = 3, persamaan dikalikan dengan R1 . R2, sehingga
R22 + R12 = 3 R1 . R2
R22 - 3. R1 . R2 + R12 = 0
(R2 - 0.381969.R1 ) (R2 - 2.618031.R1 ) = 0
R2 - 0.381969.R1 = 0, harus memenuhi R2 > R1
R2 = 0.381969.R1
R2 : R1 = 0.381969 : 1, ternyata R2 < R1. Persamaan ini tidak bisa digunakan.
Dan berikutnya adalah persamaan yang ke dua
R2 - 2.618031.R1 = 0
R2 = 2.618031.R1
R2 : R1 = 2.618031 : 1, persamaan yang digunakan adalah
(R2 - 2.618031.R1 ), memenuhi R2 > R1.
Jadi untuk sembarang R2 ( outer ) akan selalu memenuhi persamaan
( R1 . R2 )0.5 / R2 -. R1 = 1, saat R2 : R1 = 2.618031 : 1
R2 : R1 = 2.618031 : 1, adalah rasio outer terhadap inner
Kembali pada persamaan 3
( R1 . R2.)0.5 : ( R2 - R1 ), adalah perbandingan jari jari selubung bola terhadap ketebalan bola.
R2 = 2.618031
R1 = 1.
( 1. .2.618031 )0.5 / ( 2.618031 -. 1) = 1.
( 2.618031 )0.5 / ( 2.618031 -. 1) = 1
Adalah suatu keadaan dimana panjang jari jari selubung bola terhadap ketebalan bola memilki panjang yang sama. Hal ini tidak pernah terjadi. Alasannya adalah jari jari selubung bola bergerak tidak linier terhadap jari jari bola. Sedangkan luas kulit bola bergerak berdasarkan fungsi hiperbola tertentu( tidak menyatakan garis linier).
Namun demikian untuk kasus perpindahan panas konduktor aliran steady state bentuk silinder dan bentuk bola. Ada syarat lain yang harus diikutii. Yaitu nilai jari jari selubung terhadap ketebalan bola harus selalu bernilai lebih dari 1. Mengapa harus selalu bernilai lebih dari 1 ? hal ini disebabkan adanya perubahan luas penampang yang tidak bergerak linier terhadap jari jari silinder maupun bentuk bola.
Kesimpulan :
Dengan menemukan persamaan jari jari selubung silinder terhadap ketebalan silinder = 1. maka saya merasa keberatan dan sebagai konsekuensinya adalah rumus yang sudah ada harus diuji di laboratorium
Dengan menemukan persamaan jari jari kulit bola terhadap ketebalan bola = 1. maka saya merasa keberatan dan sebagai konsekuensinya adalah rumus yang sudah ada harus diuji di laboratorium.
SUDUT PANDANG FILSAFAT TENTANG TEORI KEBENARAN
Berdasarkan teori kebenaran dalam rumus tersebut adalah merupakan teori kebenaran consensus. Yang disepakati bersama sama. Belum mencakup kebenaran secara universal dalam kondisi ketebalan. Hal ini memungkinkan adanya proposisi yang bertentangan dengan obyek dan hasil saat dilakukan konfrontir.
memahami Jenis teori kebenaran :
1. Teori Koherensi (The Consistence/Coherence Theory of Truth)
2. Korespondensi (The Corespondency Theory of Truth)
3. Pragmatisme (The Pragmatic Theory of Truth)
4. Teori Performatif
5. Teori consensus. Pendekatan teori kebenaran pada kasus ini masih bersifat kesepakatan para ilmuwan
Sumber-Sumber Pengetahuan :
1. Rasionalisme
2. Empirisme ( pada bagian ini saya belum melakukan pengujian fisik pada suhu material )
3. Kritisisme
4. Intuisisme
Secara umum kebenaran bersifat :
1. Rigid
2. Universal
3. Nonkontradiktif
4. Bisa dibuktikan
5. Radic
Quote philosophy :
It is with logic that one proves, it is with intuition that one invents ( Henri Poincare ).
My quote philosophy :
Matematika sanggup memprediksi kesalahan, namun belum sepenuhnya membuktikan kebenaran mutlak.
Saya tidak punya niat apapun selain memperbaiki pendapat akademik. Yang terlanjur menyebar luas sampai ke pelosok dunia ( khususnya pada materi perpindahan panas ). Dan kita bekerja sama untuk saling memperbaiki dan menguji kebenarannya. Alangkah indahnya dunia ini jika manusia saling belajar dan menjadi bijaksana untuk segala permasalahan yang dihadapi bersama di dunia ini.
My favourite music :
The lonely Shepherd by Andre rieu feat Gheorghe Zamfir
Marriage d’amor -paul de senneville II Jacob’s piano
The gael - last of the mohicans-Royal scots dragoon
Vincencius Sufijan Hadi
Could we get a link to the problem questions?
Which edition of the textbook is it? Does anyone else like going through these lectures to help with future classes?
I guess it is :Fundamentals of heat and mass Transfer, F. P. Incropera etc.
Amazing lecture by Prof Biddle .
Can I get lectures by Prof David Miller on IC engines ? I would be really grateful.
We recently uploaded a series of videos showing him running labs for his IC engine course.
from which book are they studying?(i just want the name of the book to buy it and watch this series for my basic)
I can't find your comment anymore but it said that the previous lectures were going to be delisted because of lower quality and errors. How significant are these errors? I've been studying that series pretty religiously and just noticed this version. Thanks!
The errors are primarily in the editing of the videos. Some content was accidentally omitted and the audio quality sometimes wasn't great.
If it werent for Prof. Biddle I would never be able to graduate.