Why is a resistor for convection not included in the thermal circuit solution?

[theBEAST]

Homework Statement

Here is a picture of the question with the solution:

When I attempted this I added a resistor for the convection of the fluid inside. But the solution does not do this, does anyone know why?

 

[maajdl]

Model these two situations:

1: without foam
2: with foam

Where then have you used the convection coefficient?

 

[Chestermiller]

Homework Statement

Here is a picture of the question with the solution:

When I attempted this I added a resistor for the convection of the fluid inside. But the solution does not do this, does anyone know why?

The problem statement says the wall is at 200 C, so, for whatever reason, they don't expect you to consider the resistance inside the surface.

Chet

 

[maajdl]

The last question must be some kind of hint, to compensate for the opaque and tricky statement of the problem.

My understanding is that the heat rate (W) does not change, as it is determined by the chemical reaction.
Therefore, no heat transfer coefficient is needed, neither convective in air, nor conductive in the foam.
It is only the outer diameter that matters.
It must be such that the heat flow (W/m²) is reduced from hcv(200-25) to hcv(40-25).
Therefore the surface must increase by a factor 11.66 .
Therefore, the diameter must be sqrt(11.66) = 3.41 larger, which gives the thickness of the foam = 2.41/2 = 1.2 .

Question:
Would it be possible to replace the foam by a shell around the tank, with no convective heat transfer, but mainly radiation?

 

[Chestermiller]

The last question must be some kind of hint, to compensate for the opaque and tricky statement of the problem.

My understanding is that the heat rate (W) does not change, as it is determined by the chemical reaction.
Therefore, no heat transfer coefficient is needed, neither convective in air, nor conductive in the foam.
It is only the outer diameter that matters.
It must be such that the heat flow (W/m²) is reduced from hcv(200-25) to hcv(40-25).
Therefore the surface must increase by a factor 11.66 .
Therefore, the diameter must be sqrt(11.66) = 3.41 larger, which gives the thickness of the foam = 2.41/2 = 1.2 .

Question:
Would it be possible to replace the foam by a shell around the tank, with no convective heat transfer, but mainly radiation?

I think you're reading more into this problem than there is. Yes, in real life you would have to design it so that the same amount of heat is removed (as you indicated). But this is just a contrived problem where they state that the surface temperature is somehow kept at 200C. They are just trying to give the OP practice on a simple (admittedly unrealistic) problem. This is attested to by the fact that the solution they provide is consistent with this assumption.

Chet

 

[maajdl]

They did not say that the wall is maintained at a fixed temperature!
And that's clearly a very badly conceived exercise, imho!

 

[Chestermiller]

They did not say that the wall is maintained at a fixed temperature!

They kind of did. "The wall... is at 200C...". What they meant was that it is maintained at that temperature.

And that's clearly a very badly conceived exercise, imho!

I totally agree.

Chet

 

[maajdl]

I understand your point of view.
However, note that, if the heat rate is assumed to be constant, it would also be necessary to provide data about the wall temperature, since the heat flux (W/m²) is hcv.(Twall-Tamb) .
Therefore, any student with a perfect understanding of the topic would have to chose between two interpretations of why the wall temperature was provided and what is the purpose of the exercise.

 

[Chestermiller]

I understand your point of view.
However, note that, if the heat rate is assumed to be constant, it would also be necessary to provide data about the wall temperature, since the heat flux (W/m²) is hcv.(Twall-Tamb) .
Therefore, any student with a perfect understanding of the topic would have to chose between two interpretations of why the wall temperature was provided and what is the purpose of the exercise.

I can see that this problem really annoys you. It annoys me too.

Chet