A sticky question of heat transfer 1

[roaling]

I have a configuration like this.
- ambient air at -36°C (still)
- 38 mm insulation (foamglass K=0.045 W/m°C)
- electric heat tracing
- 25 mm steel wall
- 100 mm nitrogen wall (tank 2nd jacket, K=0.026 W/m°C)
- 25 mm steel wall
- glycol at 20°C
The question is: how much is the heat introduced with the electric tracing to maintain the glycol at 20°C ?
Thank you in advance.

 

[25362]

Speaking of the surfaces exposed to air, if I understood the query, it would suffice to keep the internal surface of the foamglass at 20[sup]o[/sup]C to avoid any heat transfer out of the glycol.

Thus one needs to estimate U, the overall heat transfer coefficient (OHTC) composed by conduction across the foamglass plus the convection and radiation effects on the surroundings. For a delta T of 20+36=56[sup]o[/sup]C, the heat load Q, per unit area A, would be fixed by the equation:

Q/A=U*56​

Please correct me if I'm wrong.

 

[IRstuff]

I wouldn't have implied a single coefficient, since that depends on the relative form factor of the tank.

Additionally, the deltaT applies primarily to the top surface of the glycol, while the transfer equation to the sides of the tank is more likely to have a positive heat flow from the heater with a smaller negative deltaT, while the heater has a larger positive deltaT to the environment.

For a 1ft square top surface and assuming h=4 W/m^2*K, I swag about 20 W each for convection and radiation.

TTFN

 

[roaling]

Awaiting other suggestions, I got the following result: 12 W/sqm. What do you think ? Thank again.
A clarification: the steel wall in contact with the glycol is the enclosure of a rectangular tank of approx 120 cum capacity. The second steel wall is the outside enclosure of the tank jacket in which nitrogen is used as buffer fluid. All tank (jacket included, is installed inside another steel enclosure and not subject to sun radiation.

 

[IRstuff]

But what are the dimensions of the tank?

Is construction identical on the top?

Is the tank filled?

TTFN

 

[roaling]

Dear IRstuff,
Tank dimensions are approx 9.5 x 4 x 4 m (LxWxH).
Top of the tank is identical to the bottom; all wall surfaces are subject to a small internal pressure, approx 0.2 barg plus liquid head where applicable.
The tank jacket is under higher pressure, approx 0.8 barg.
The tank is normally filled with glycol.
Thanks for your cooperation.

 

[IRstuff]

Heater on all 6 walls?

TTFN

 

[IRstuff]

Why is the heater so far away from the glycol?

The heater needs to be at a minimum of 20ºC, so you've got the maximum deltaT across the foam glass and your heater is primarily heating the foam glass.

TTFN

 

[30odd6]

I calculated 12 Watts per m^2. This problem looks like a 1D problem and you want to know what would be the required makeup heat to keep your glycol at 20 degrees C. My assumptions were:

1. The interface between the glycol and the steel is at 20 degrees C
2. The interface between the air and the insulation is at -36 degrees C. Normally, I would calculate the heat transfer coefficient for natural convection. However, you can assume the outer wall is at -36 degrees C for the worst case scenario.
3. No contact resistance around the heater.
4. No radiation heat transfer

Design Optimization
Now all you have to do is determine the cost of the heater. Then determine the cost increase by manipulating the design. For instance, how much does it cost to increase the thickness of the insulation or the N2 wall? This added cost to the overall design might lower your monthly electrical bill in the long run.

 

[25362]

Assuming that:

1. the tank is inside a housing such as a hangar
2. all tank surfaces are insulated and heated in the
same manner
3. there is some measure of moisture or ice intensification
factor by deposition on the external foamglass at
-36[sup]o[/sup]C
4. disregarding the fact that for the same surface-to-air
temperature differences, walls would lose heat by
natural convection in a different rate: top
face=1.3*vertical walls=2*bottom face
5. radiation effects to the surrounding air are small but
not negligible

I'd say that your estimate of 12 W/m[sup]2[/sup] is on the low side. A ballpark estimate would bring the heat loss to 10 times as much.

 

[IRstuff]

Given that worst case scenario of outside of foam at -36ºC

(0.045 W/mK)/38mm = 1.184W/m^2K

with 56°C delta, that results in 66 W/m^2



TTFN

 

[30odd6]

IRSTUFF is right. The heater position below the insulation layer only takes advantage of the insulation. If the heater was closer to the glycol, then you could use the other insulation layers to help reduce the heat load.

 

[IRstuff]

Using h_foam=1.184 W/m^2ºC and h_air=4 W/m^2°C

I get heat through foam of 57 W/m^2 with the outer surface of the foam at -28°C. This results in convective loss of 31 W/m^2 and radiative loss of 25 W/m^2.

This is obviously very rough, using ideal emissivities, etc., and applies only to the vertical sides. This also assumes that the glycol is essentially in an isothermal region, since it's surrounded by heaters on all sides and therefore, there is no loss from the glycol itself.

I need to get a life... ;-)

TTFN

 

[CRG]

IRstuff, are you confident with that high of radiative loss (44%)?

Seems to me that the convective losses should be much greater than the radiative losses.

 

[IRstuff]

Yeah... I agree, but I ran the blackbody model through a blackbody program as well as a separate Mathcad model. They both agreed as far as sigma(Ts^4-Ta^4) was concerned.

But that's a very gross model, since neither the foam nor the ambient are perfect blackbodies, hence my caveat of "ideal emissivities, etc."

Obviously, there are lots of factors involved that are not modeled well by the Stefan-Boltzmann radiation model, particularly, subtleties such as the high atmospheric absorption between 5 and 8 micrometers.

As a zero-order model, it's nearly a wash anyway because if you assume no radiation you get 66 W/m^2, while the perfect blackbody model gives you 57 W/m^2.

From a practical design perspective, you'd have to design with some margin and then run the thermal control with one or more temperature monitors in the glycol itself.


TTFN

 

[25362]

To IRstuff, is it true that your estimated value of 25 W/m[sup]2[/sup] for the radiation loss would correspond to an emissivity of 0.90-0.95 ?

 

[IRstuff]

It would correspond to an emissivity of 1.0 and no intervening atmospheric absorption, e.g., a infinite sink with infinite bandwidth.

TTFN

 

[25362]

The maximum-heat-loss case, by assuming the external surface being at the same temperature of the surrounding still air, discarded radiation and convection losses altogether. In practice, there are factors that may justify planning for higher losses such as those that may occur at insulating block junctions or at vessel discontinuities, such as nozzles, openings, manholes, etc. Let alone weather changes and operational changes.

 

[roaling]

I wish to thank all friends partecipating to this thread.
On the basis of info had from you and others from my knowledge and library, I have reached the following conclusion.

Disregarding the terms relevant to steel walls and glycol side, only three are partecipating to the heat transfer:
- convection+radiation on still air side, coeff. 4 W/sqm°C
- conductivity of foamglass, coeff. 1.18 W/sqm°C
- conductivity+radiation+convection, coeff. 2.22 W/sqm°C

Based on above, the overall heat transfer coefficient becames 0.646 W/sqm°C and heat loss through 1 sqm surface 36.2 W/sqm.

The critical point of this evaluation is the coefficient through the nitrogen filled double wall. In the coefficient assumed (calculated according to an engineeering manual), 14% is by conductivity, 58% by radiation and remaining by convection. Somebody knows a specific procedure and/or formula to evaluate this?

Again, thank for your help and partecipation.
Roal