heat exchange of a tank

[mir]

Hi.
They are struggling with a heat exchange. these are the data:

testerna = 45 °C
= 30 °C

type insulation: polystyrene
polystyrene lambda = 0.02 w/(m k)
thickness insulation=0.16 m

area = 226.19 sq m (total dispersion area covered by insulation)

then calculate the heat that is dispersed (in stationary regime clearly):

q = lambda * area * (outside - t inside) / insulation thickness = 424.1 w

ie 365.29 kcal/h.

Does the procedure seem right?

Thank you.
♪

 

[AMinati]

in reality the thermal resistance is not only the one relative to the layer of polystyrene, but also those of the laminary (and/or turbulent) layers of the internal and external surfaces with the internal and external environment of more difficult evaluation. besides that (very low?) relative to the tank itself, presumably metallic. If what you are interested is only an indicative value of maximum I would say that as first approximation and in favor of safety can go well, otherwise not. Moreover, if the tank is small diameter, you should also take into account the radial distibution and not flat of the heat flow.

aminati

 

[mir]

Thank you. That was exactly what I thought. we say that I have neglected the thermal resistance of the sheet (they are tenths of centimeter) and that the tank is on 6 meters in diameter.

But does the polystyrene lamb look right?

 

[mir]

I think of something else: the tank rests with the bottom floor on a plate of cls ... on this side there is no isolation but only a plate of 7 mm ... with an area of 28 sq m... according to you is negligible? How can I calculate the dispersion on the ground?

Thank you.
♪

 

[mir]

then calculate the heat that is dispersed (in stationary regime clearly):

q = lambda * area * (outside - t inside) / insulation thickness = 424.1 w

This is the thermal power that is lost!

 

[AMinati]

in net I found this data, in kcal / (h × m × °c):
expanded polystyrene 0.026 ÷ 0.028
polyurethane foam 0.0020

Other:
extruded polystyrene panel, type polyfoam c-350, with extrusion skin, expanded with
gas according to law, co2, thermal conductivity λd 0.033 w/mk

the radial distribution can be overlooked given the diameter, while the effects at the ends definitely not: the dispersion to the lower floor and that to the outside environment at the top I think are at least equal to the lateral one (unless the tank is not high several dozen meters).

there are several regulations regarding tanks outside the ground 650 bees and en 14015 for example and which also treat isolated tanks.
If the tank is taller than a few feet I would go slowly to say that the thicknesses are only a few millimeters. . .

aminati

 

[AMinati]

follows:
the norm en 14015 for example for diameters from 4 to 10 meters minimum thickness of 5 mm for carbon steel and 3 for stainless steel

aminati

 

[mir]

I'm a little puzzled ... I tried to calculate it as a cylinder and came to this result:

rcrit = k / h

place k polystyrene = 0.032564 [W / (m × K)]ed h in aria = 5 [W / (m2 × K)]I get a rcrit= 6.52 mm...

I have a reservoir of 6 meters therefore re >>>> r> rcrit ... this, if I have not understood badly, does it mean that the more I add isolation the lower thermal dispersions?

 

[SOLID]

I have a reservoir of 6 meters therefore re >>>> r> rcrit ... this, if I have not understood badly, does it mean that the more I add isolation the lower thermal dispersions?

Sure. .
in the case of cylindrical geometries the increase of the insulating thickness increases the conductive resistance, but also reduces the external convective resistance (in practice increases the convective exchange area).
the critical radius of insulation is defined just as the radius that maximizes the heat dissipated power.
so if the radius of the insulator is greater than the critical one you have no problem if you want to isolate. (and it is almost always so, except for example in electric conductors where the electrical insulator acts as a heat sink).

 

[mir]

He thinks that when I had studied I had not associated his limit to electrical conductors.... that means that on large dimensions the rcritic should be studied only when the insulating (conductive resistance) isolates little or when the external environment determines a very low convective convection

 

[SOLID]

large size the rcritic should be studied only when the insulating (conductive resistance) isolates little or when the external environment determines a very low convective conduct

However if you think that the lowest value of h is about 5[W/m^2*K) e mediamente la K di un isolante è di 0.05[W/m*K]therefore the critical radius for practical applications is always very small, so the addition of insulating can do nothing but always reduce the heat dissipated.