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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

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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.

@@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.

what formula is in your text book?

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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.

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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.

Thanks and welcome

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.

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.

@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

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).

That is solved the same way as a double pane window (examples are in the heat transfer playlist).

@@MichelvanBiezen Alright. Thank you for the reply. Best regards

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.

@@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.

I don't think you need the first 2 terms T^4 and T^3.

@@MichelvanBiezen maybe you are right but when it used the R^2 value much close to 1.

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.

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).

@@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?

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.

@@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.

Yes. Essentially it works like an electric circuit, when you have zero radius then you have an "open circuit" which has infinite resistance.

Ohk thanks for explaining 🙂🙂

@@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?

@@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.

There would be no cylinder

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

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!)

@@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.

No we haven't covered that topic yet here.

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.

You are welcome. Glad the video helped.

So nice of you