You'll teach for generations, my friend.. you tube and science should congrat you for that!!
@MichelvanBiezen3 ай бұрын
Thanks!
@mahad12034 ай бұрын
Thank you sir, I was very confused about what area and length should be taken for its elemental form but you have cleared it.
@MichelvanBiezen4 ай бұрын
Glad it was helpful.
@ABHISHEKSHARMA-mt2cf3 жыл бұрын
Concept booster video sir🔥🔥🔥... Thanks a lot... From India
@MichelvanBiezen3 жыл бұрын
Most welcome!
@tasnimultanim466710 ай бұрын
great sir👍 Thank you from Bangladesh.
@MichelvanBiezen10 ай бұрын
Glad you found our videos. Welcome to the channel! 😊
@L.NEVER.LOSES.5 ай бұрын
beautiful
@MichelvanBiezen5 ай бұрын
Thank you! 😊
@AJ-et3vf5 ай бұрын
Great video. Thank you!
@MichelvanBiezen5 ай бұрын
Glad you liked it!
@1morrel3 ай бұрын
How can dQ/dt remain constant if there is heat transfer through the pipe. Wouldn't the temperature inside the pipe change with time therefore making dQ/dt variable and not constant?
@MichelvanBiezen3 ай бұрын
The assumption we must make is that there is a source that continues to push the fluid through the piple that contains heat keeping the temorature in the pipe constant
@eirikschjlset6898 Жыл бұрын
Tanku botie man, always can rely on you 😍🥰
@MichelvanBiezen Жыл бұрын
Glad you found our videos. 🙂
@AboAli-gu2iv3 жыл бұрын
Thank you very much from Egypt ❤️
@MichelvanBiezen3 жыл бұрын
Welcome to the channel!
@JamesVestal-dz5qm7 ай бұрын
Sayandeep is right that I have OCD attacks when I poop in the middle of the day. I am still vaping nicotine and suffering from OCD, but I am speeding up in chemical engineering. I did the transport and reactors homework a few times to learn from my mistakes, and I re read my own notes to understand errors in my own thinking. I labeled him Dmitri the illustrious, the chosen gucci boy for all the females to save his life. I knew that Shana banana was one of my accusers, but she didn't have the street smarts to land it in the school.
@filomenamendes53492 жыл бұрын
Fantastic! A life saver. Thank you
@MichelvanBiezen2 жыл бұрын
Glad you found our videos.
@PACO_MAX2 жыл бұрын
Hi! When you say that dQ/dt is constant, you mean that heat flux doesn't change with time? or that the heat flux is uniform in all the surface? thanks!
@MichelvanBiezen2 жыл бұрын
dQ/dt is the amount of heat that flows through the walls per unit time and it is the same for any segment along the path, and it remains constant as long as the temperature doesn't change.
@PACO_MAX2 жыл бұрын
@@MichelvanBiezen Thank you for your effort and being so fast answering. I have an exercise of a cylinder inside the ground, ground temperature changes with depth and cylinder heat flow from inside to the ground decreases with time. I have to model it on Fluent, I thought to use the equation in the video to know the initial cylinder temperature, and suppose it uniform in all the cylinder. With that initial point, insert the q(t) function given where the q decreases with time, the T(z) function of the ground, and give the time steps and characteristic time. Would it be a good approach for a conduction only problem? Or should I use another transient approach? I suppose that the heat transfer happens only in 1-D and it's axisymmetric. Sorry about the long question, but it's a more difficult exercise than the ones given as example by the professor.
@user-xq2wi6uh3m6 ай бұрын
I'm little bit confused the formula are teacher gave us is a little bit different
@MichelvanBiezen6 ай бұрын
what formula is in your text book?
@prettymkhonto27863 жыл бұрын
thank you sir
@MichelvanBiezen3 жыл бұрын
Most welcome
@Az-y1xb7 ай бұрын
1:17 will the area be calculated using the radius of the inner wall of the pipe or the radius of the outer wall of the pipe? Been confused about this for a long time
@MichelvanBiezen7 ай бұрын
The area is calculated using the variable r which will be integrated from the inner surface to the outer surface, which become the limits of integration.
@ashishpingle32443 жыл бұрын
DQDTEEEEEEEEEE :) although nice explanation sir.
@eivindbjrndalen7588 Жыл бұрын
THnakyou Bow tie man mashalla
@MichelvanBiezen Жыл бұрын
You are welcome. 🙂
@sammalik63552 жыл бұрын
Does metal melt in variable state? If yes Then how it will be possible at constant temperature
@MichelvanBiezen2 жыл бұрын
Not as long as the temperature remains below the melting point of the metal. Variable simply means that the heat transfer per unit distance changes as you go from the inside to the outside of the cylinder.
@yatiqueperwejshorts36383 жыл бұрын
Aws explanation sir
@MichelvanBiezen3 жыл бұрын
Thanks and welcome
@mostafaabdelrauof31858 ай бұрын
hello sir can you help me in a case which an insulated pipe and fluid inside it and heat transfer with the surrounding air , And I have calculated the heat loss rate already but can't calculate the final temp. !!?
@mostafaabdelrauof31858 ай бұрын
I have seen many videos on you tube talking about thermal equilibrium , but for a peace of alu in water using Q=m.C.delta T , but it's not my case.
@MichelvanBiezen8 ай бұрын
Based on your description, we cannot form a picture of what is happening or what you are asking. Any liquid or gas that flows through a pipe will lose energy through the walls of the pipe. Therefor the temperature of the fluid will drop in reference to the starting point at presumably a specific temperature. The rate of heat loss depends on the difference in temperature between the inside and the outside, and the thickness and type of material the walls are made of. Thus the rate of heat loss deminished with greater distance from the source point.
@mostafaabdelrauof31858 ай бұрын
@MichelvanBiezen I've used ansys and meshing method to get the final temperature. If it was 90 degrees it would be 78, but I can't imagine ..did he use integration or what !! .. by your equations, I got the q per meter as energy loss How then I translate this to temp and calculate temperature degrees that will be dropped ! Like the aluminum peace in jar of water example Thank you, and I hope you can understand me
@pietervb7166 Жыл бұрын
Why does dQ/dt has to be a constant?
@MichelvanBiezen Жыл бұрын
Once steady state is reached, the amount of heat flowing through any section of the pipe is the same as the heat flowing through any other section of the pipe. (It is the same as water flowing through a hose).
@l052_jeeveshpathak84 жыл бұрын
awesome job
@luqman45533 жыл бұрын
Thank You for the video. could you help to calculate heat transfer for double wall cup. Thank You.
@MichelvanBiezen3 жыл бұрын
That is solved the same way as a double pane window (examples are in the heat transfer playlist).
@luqman45533 жыл бұрын
@@MichelvanBiezen Alright. Thank you for the reply. Best regards
@awoldn2 жыл бұрын
Could Please show how estimate dQ/dt if k is a variable value like for real insulation?
@MichelvanBiezen2 жыл бұрын
It depends on how k varies. I would suspect that k would be a function of r and thus it would be multiplied with the 2 pi r L term in the numerator.
@awoldn2 жыл бұрын
@@MichelvanBiezen if I will apply k(T) = aT^4+bT^3+cT^2+dT+e. I could estimate average value as integral that function by dT and divide by (T2-T1) but if I will put that function into the thermal resistance it could be locks not the same I assume.
@MichelvanBiezen2 жыл бұрын
I don't think you need the first 2 terms T^4 and T^3.
@awoldn2 жыл бұрын
@@MichelvanBiezen maybe you are right but when it used the R^2 value much close to 1.
@pikulicluka2 жыл бұрын
Now that's what I call an intuitive, clear and accessible explanation! Thank you for making these terrific videos! I've got one question though: what if the temperature outside (T_b in this case) is higher than the temperature inside the hollow cylinder (T_a)? I reckon the derivation is similar, except there's no minus sign in front of the thermal conductivity k? Also; why did the L-term disappear in the infinitesimal shell dQ/dt formula and get replaced by a dr-term? The dr-term is not a substitute for the length L, it's in a wholly different direction.
@MichelvanBiezen2 жыл бұрын
If it was warmer outside than inside the pipe, the heat flow would simply be in the opposite direction and we would solve the problem in exactly the same way.
@MichelvanBiezen2 жыл бұрын
The first equation is the general equation. The second equation is the same equation with the specifics of this example inserted. The "L" in the first equation is the "path lenght" and not the length of the cylinder. The path length is the thickness of the insulation. (Using L for the 2 different measurements is indeed confusing).
@pikulicluka2 жыл бұрын
@@MichelvanBiezen Thank you for answering so quickly, I really appreciate the tremendous dedication and effort you put into your teaching! I'm struggling to come to grips with how the opposite situation - i.e., that of the outside temperature T_b being higher than the inside temperature T_a - would play out the same. The (T_a - T_b) term that is present in your final expression for dQ/dt would then be negative, meaning that your heat flux would travel from low temperature to high temperature. Surely, that can't be right?
@MichelvanBiezen2 жыл бұрын
Heat always travels from the high temperature to the low temperature. I think you are stuck on that negative sign. Just ignore that. Just realize that heat will flow from hight to low temperature and find the magnitude of the heat transfer.
@foxlies0106 Жыл бұрын
@@pikulicluka Hi I think it's ok: if Ta-Tb is negative, such that it is warmer outside the cylinder than inside, then yes dQ/dt would be negative in the equation and that's ok. The reason is that dQ/dt implicitly has a vector direction imputed to it: in this case that direction is radially outward, along r vector. so if dQ/dt is negative, heat is flowing radially inward, in the negative-r direction. So that makes sense. It'd be like heat from an industrial process flowing radially into a cylindrical cooling pipe, that is full of fast moving chilled water; instead of the original Ta-Tb>0, like a radiator pipe heating a room of cool air.
@utkarshiitbhu42042 жыл бұрын
Sir on making smaller radius 0 Will it result in infinite thermal resistance?
@MichelvanBiezen2 жыл бұрын
Yes. Essentially it works like an electric circuit, when you have zero radius then you have an "open circuit" which has infinite resistance.
@utkarshiitbhu42042 жыл бұрын
Ohk thanks for explaining 🙂🙂
@foxlies0106 Жыл бұрын
@@MichelvanBiezen Could you say that if the larger radius was shrinking too, so that ln(b/a) stayed constant, then dQ/dt would stay the same? In other words, if a shrinks and b does proportionately, area for conduction = 2*pi*r*L, where r~a*(1+m)/2, is shrinking, where m=b/a, but the conduction path length a*(m-1), is shrinking at the same rate. So dQ/dt stays the same?
@carultch Жыл бұрын
@@foxlies0106 I'm not sure how you would set that up in practice, but if you keep the b to a ratio constant, and all other factors the same, you would keep the resistance of the cylindrical shell the same.
@gopalyelugu17944 жыл бұрын
what's about if a=0
@chinnudanturi97304 жыл бұрын
There would be no cylinder
@carultch Жыл бұрын
If a were to equal zero, then that means the source of heat is distributed within it, like a wire or a nuclear fuel rod. As opposed to localized to an inner surface, like it is with cladding, piping, or insulation. There is a solution to the heat conduction equation for heat being generated inside a solid cylinder, and the relationship between the rate of heat generation and the distribution of temperature. T(r) = Ts + S/(4*k) * (R^2 - r^2) S = volumetric heat generation rate R = radius r = radial position Ts = surface temperature
@foxlies0106 Жыл бұрын
the ratio b/a is crucial. If a approaches zero but b does so too, at the same time and proportionally, then the quantity ln(b/a) is constant, and dQ/dt will stay constant, as the cylinder shrinks.. (at least until a and b actually reach zero!)
@carultch Жыл бұрын
@@foxlies0106 The problem with a actually equaling zero, is that it would mean infinite heat per unit volume, generated in a way that is concentrated along the cylinder's axis. In reality, any real source (or sink) of heat is going to have a distribution in some form or another, whereby the heat per unit volume is finite.
@guenheejunheelee49834 жыл бұрын
Sir, Do you have any plans to cover PTC (Positive Temperature Coefficient) heater for your heat transfer-related lectures?
@MichelvanBiezen4 жыл бұрын
No we haven't covered that topic yet here.
@guenheejunheelee49834 жыл бұрын
Sir, PTC heater is a major concern in EVs as you know. Looking forward to your PTC heater lecture in the near future. I always appreciate.