I don't know what really to say, these lectures are gems, this professor is wonderful and I am really very happy to see something like that for free. Thank you very much!
Finding these lectures through a global pandemic trying to understand a teacher that is as confused as us on ZOOM is literally a life saver. I cannot thank you enough for this Sir. You are what got me through this weird time
In question 3.52 you said that heat generated by the heater will eventually get out by convection in the air. My question is that why heat is not being conducted by the tube and then convected in the air? There should be conduction in the tube first before going outside in the air. Waiting for the reply thanks.
For Part B of the first example problem (around 9:00) he has the electrical resistance for convection as (1/h) which gives units of (m^2k/w), but with the convection resistance, you get (1/((W/mk)*(m^2)) which equals (1/(Wm/K)), which equals (K/wm). You should have the same units, and i believe that K/Wm should be the units due to it being per unit length. Did I do something wrong w/ my unit analysis?
the convective heat transfer coeff., h, typically has units of [ W / (m^2*K) ] but can be described without K as well. He mustve made a mistake - i havent gone to look. The units you are describing are for thermal conductivity, k, which is associated with conductive heat transfer [ W / (m*K) ]
It was a pleasure taking Professor Biddle during my undergrad. All these "Techie" new age PHD professors from UCI are nowhere near the competence of this man! cough cough
In the case where both ends are being heated with different temperatures also you are given surrounding temperature, assuming the fin has a finite length and a defined cross-sectional area. How do you find the heat loss from the bar?
The first problem in this video i don't understand quite fully. I assumed the resistance for conduction per unit length of would need [ln(r2/r1)]/(2(pi)k) not 1/(kA)
I know you asked a while ago but I figure someone else might ask. It is because he is doing q prime which is per unit length so you divide by the length on all the parts.
Fins are made of metals with very high k like aluminium or copper and their y and z lengths are very very small compared to their x length. This means the drop in temperature in y and z directions is very very small compared to the x direction.
We consider the heat flow in the y and the z direction. The heat flow in the vertical direction ultimately is lost due to convection which is considered.
I don't know what really to say, these lectures are gems, this professor is wonderful and I am really very happy to see something like that for free. Thank you very much!
No problem.
SHUT UP
Finding these lectures through a global pandemic trying to understand a teacher that is as confused as us on ZOOM is literally a life saver. I cannot thank you enough for this Sir. You are what got me through this weird time
These videos are 10x better than anything I get from my lectures. Thank you!
Our pleasure.
27:40 how did he changed d²T/dx² to d²ϴ/dx²
while ϴ = T - T∞
Guess that it's because T∞ is a constant, so when you derive it it goes to zero.
dϴ/dx = d(T(x) - T∞)dx = dT(x)/dx
Pedro Franco
True!
For extended surfaces, move to 14:32. Thank you.
This is the type of professor you just want to work hard for.
Nobody is talking about how the camera man deserves a nomination for 'Best lens' award...
In question 3.52 you said that heat generated by the heater will eventually get out by convection in the air. My question is that why heat is not being conducted by the tube and then convected in the air? There should be conduction in the tube first before going outside in the air. Waiting for the reply thanks.
wow.....awesome
never seen a teacher like him
We like him. :)
For Part B of the first example problem (around 9:00) he has the electrical resistance for convection as (1/h) which gives units of (m^2k/w), but with the convection resistance, you get (1/((W/mk)*(m^2)) which equals (1/(Wm/K)), which equals (K/wm). You should have the same units, and i believe that K/Wm should be the units due to it being per unit length. Did I do something wrong w/ my unit analysis?
the convective heat transfer coeff., h, typically has units of [ W / (m^2*K) ] but can be described without K as well. He mustve made a mistake - i havent gone to look. The units you are describing are for thermal conductivity, k, which is associated with conductive heat transfer [ W / (m*K) ]
I am so grateful for these videos. Please come teach at ASU !!!
I agree, heat transfer profs here at ASU arent great!
It was a pleasure taking Professor Biddle during my undergrad. All these "Techie" new age PHD professors from UCI are nowhere near the competence of this man! cough cough
these videos be saving my GPA ...thanks sooo much
In the case where both ends are being heated with different temperatures also you are given surrounding temperature, assuming the fin has a finite length and a defined cross-sectional area. How do you find the heat loss from the bar?
Mistake on 8:40 - should be Q0=(Th-Tinf)/(1/(2hPiR2)+Rb)
Yeah, I think so.
I think area cancels out, that's why he didn't put an A
@@AminaXXX3 No area doesnt cancel out. Length does cancel out. Hence, He should have put PiR2 next to h term.
RB should also be R'B.
Wonderful Heat Transfer lectures ! ! !
God bless you sir! A true life saver!
If there are 4 fins of different heights (t) then in calculation what should we consider for (t) ?
I prefer watching this videos than go to pass class in the university
How we can use thermal resistance concept when there is a heat source present inside the tube like you did in the very first example??
where can I get a complete and detailed version of the derivation of the energy balance? hard to get it....
Oh my god I have to derive the differential equations in the exam, and you give them like that. I hate my life
Great lecture as always
and lol...I have this same shirt Dr. Biddle is wearing :)
amazing, incredible lecture
The first problem in this video i don't understand quite fully. I assumed the resistance for conduction per unit length of would need [ln(r2/r1)]/(2(pi)k) not 1/(kA)
I got Th equals to 8.47 .. did you get the same value of Th that I got?
im guessing its using the conduction for Cylinder so yes use the ln function. and rmb to eliminate any L since its find heat per unit length
Where is the basic lecture from Dr.Biddle sir about extenden surface?
thank you for all of them
God bless this Sir
We're glad it helped.
I have a question. Why is the surface area is 2*pi* r_2 times L divided by L (as per the professor's script) .
I know you asked a while ago but I figure someone else might ask. It is because he is doing q prime which is per unit length so you divide by the length on all the parts.
what is the name of the professor ???
How did u get RB value?
Rb=Ln(R2/R1)/(2PiK) - conductive cylinder resistance
He gave the solution of Rb in the last lecture.
sir why it is one dimensional heat conduction in fin material while it is transferring in y and z direction too?
Fins are made of metals with very high k like aluminium or copper and their y and z lengths are very very small compared to their x length. This means the drop in temperature in y and z directions is very very small compared to the x direction.
Mohamed Gamal but still fin is loosing heat from these coordinate directions. But what are the basis that allow to neglect heat from the directions...
@@iamzed2009 1 dimensional heat transfer only when Biot number is less than 1/6.
We consider the heat flow in the y and the z direction. The heat flow in the vertical direction ultimately is lost due to convection which is considered.
TARİKAT KUR MURİDİ OLAK BİDİL REİS.
Tables are the same in 11th. ed. Fundementals of heat and mass transfer 11th.
@14:38 fins
problem 3.52 here is 3.41 in sixth edition.
قال النبي محمد ( زكاة العلم نشره )
palpatine???????????? sus