Furnace wall temperature 4

Take the look at the attached paper. Hope it could give you some inputs

The worst case would be steady state with an inner temperature fixed at 1300C. If they've used that method then your observed temperatures would be much lower as they neglect the transient change of temperature through the layers of material. For insulating material it may take a considerable amount of time to approach steady state, and the maximum shell temperature.

ex-corus (semi-detached)

They have stated in their emails that the calculation is based on steady state, but they have not provided details of the calcuation.

The temperature we are observing is higher than they predict. I expected this as the new liner has a higher conductivity (lower resistance) than the previous liner.

The bit I am struggling with is how to calculate the shell temperature. Do you have to assume a heat flow figure?

You have to equal the heat transferred by conduction throughout the multilayer from the inner to the outer surface (assume Tinner = 1300 °C and Touter is unknown) to the heat dissipated by convection and radiation from the outer surface. The paper I have attached before seemed quite clear.

Please forgive me if I'm not understanding something simple here? I have had a good look at the attached paper, thanks for that, but does it not require an outside wall temperature for the equation for the heat dissipated from the outside surface?

ProcessJim,

I didn’t mean to offend you but if you’ve taken my previous post as an offence please accept my apologies.


Not only the heat dissipated from the outside surface does require the outside wall temperature, but also the heat transferred by conduction.
The outside wall temperature is your unknown in the equation that equals the heat transferred by conduction through the multilayer wall and the heat dissipated to ambient (convection + radiative heat transfer).
You have to proceed by iteration until the two amounts of heat (Qcond and Qdissipated) converge, as the outside surface temperature reaches an equilibrium when the two amounts of heat are equal.

This is the approach I would follow if I had to face this problem, anyway you’ll find in this forum people more qualified than me and prone to help you with valuable inputs.

This is just a typical heat transfer application and calculation of multilayer wall found in text books but to get the results right you must get the conditions right first. For example, does the supplier assume some wind velocity of the 20 °C ambient air? This affects greatly the heat exchanging coefficient at the outer surface and shell's temperature.

One other possibility I can think of is that most suppliers assume some fouling factors on the inside surfaces which adds heat resistances but foulings might not immediately happen to new furnaces.

The liner (refractory layer, I guess) heat resistance that you suspected could also be the reason. The actual properties of the material might be different from what's given to the engineer to calculate. I would suspect more about the moisture content being still high inside the liner. The initial baking and curing procedure cannot normally drive all the water out resulting a lower heat resistance value. If this is the case you should see the shell temperature coming down slowly after using the furnace for a while.

For your particular application I would recommend a CD from refractory supplier Harbison-Walker and it contains a software to calculate outer surface temperatures with both data from their standard products and mannual inputs. I just don't remember if it contains applications for more layers than just insulation/refractory plus outer metal liner/jacket. You should be able to request this CD by contacting H-W.

Another factor is that the lining will expand and tend to crush the insulation board and so produce a lower thickness than as-built. Of course you could have some penetration of the molten material into the bricks to produce hot spots.

To calculate the wall temperature, calculate the resistance of the layers by summing x/k. The flux across the layers is the temperature difference between the shell and 1300C divided by this resistance. This is equivalent to the heat loss to the ambient which is h.(Tshell-20). As an estimate, take a value of about 10 W/m^2 C for h and solve for Tshell.

ex-corus (semi-detached)

Is the brick on the outside and steel on the inside?

TTFN

FAQ731-376

Assuming steel is on the inside, I get a surface temp of about 360°C for a natural convective of 2.5W/m^2 K, which also assumes that all surfaces of the furnace behave the same.



TTFN

FAQ731-376

Corus,

Have you deliberately omitted radiative heat losses or have you accounted for this component assuming 10 W/m^2 C for h? I think radiation should not be omitted in this case.

My bad, too.

If you add radiated loss, the surface temp drops to 116°C. Note, however, the temperature could be higher, depending on the emissivity of the brick. Ratio of radiated to convected is about 2.5:1

TTFN

FAQ731-376

Oh, the ratio was with low emissivity. With standard brick emissivity, the ratio os about 3.4:1. For comparison, the incremental heat transfer coefficient worked out to be about 11 W/m^2-K.


TTFN

FAQ731-376

ione, As I said, 10 W/m^2 K is an estimate, ie. for a rough hand calculation. It would include radiation as well as natural convection. Of course if the steel temperature is of the order of 300C then the value would be higher. You could produce a spreadsheet of values for natural convection plus radiation for various surface temperatures to improve this rough calculation.

ex-corus (semi-detached)

IRStuff,
I'm not sure where you get your material properties from to calculate your temperatures, but for sure the conductivity of steel is about 50 W/m K, decreasing with higher temperatures. The steel shell would be on the outside, radiating to ambient, with the brick seeing the 1300C.

ex-corus (semi-detached)

Thanks, that's why I had asked the question earlier. However, steel's emissivity, unless it's painted or coated is substantially poorer than brick's. So, with an emissivity of 0.38 and a thermal conductivity of 20 W/m-K, the surface temperature goes to 184°C.

But, the point was to get the OP to do the calculation himself, wasn't it?

TTFN

FAQ731-376

The value which appears in Irstuff’s pdf seems more likely to be that typical of iron.

I agree with corus that steel shell is probably on the outside. Anyway note that if it is a stainless steel shell conductivity is lower than that indicated by corus.

Ione, no offence taken, I am always just wary that I am wasting people's time by asking for solutions to problems that are simple.

I thing I understand the iterative approach now. Thank you all for your contributions.

ProcessJim,

I think no one is wasting time here: we are not forced to give answers and hints.

Back to your problem, both corus and IRstuff suggested an approach aligned to what I have suggested. What matters, in the end, is that you have earnt some useful answers to solve your problem.