Calculating heat loss through soaker oven lid

[meyeal]

Hi all,

I'm currently exploring energy improvements on our soaking pits, which heat steel before rolling. The pits are refractory brick lined with a refractory concrete cast lid. This lid is currently my focus, and I am trying to justify changing this refractory to a composite roof consisting of a layer of refractory and then a layer of insulation.

I have a thermal conductivity value, k, of 1.75 W/mK for the refractory, and I know my internal (1300) and external air (5) temperatures. I also know the outside surface temperature is 200 degrees C.

Using the equations for heat transfer through a plane wall,

At the boundary of the internal surface: q = h1(T1 - Ts1)

(where h1 = heat transfer coefficient inside the soaker oven, T1 = internal temp., Ts1 = temp. of inside surface of lid.)

For the refractory lid: q = [k(Ts1-Ts2)/L]

(where k = thermal conductivity, Ts2 = temp. of exterior surface of lid, L = thickness of lid)

At the boundary of the external surface: q = h2(Ts2 - T2)

(where h2 = heat transfer coefficient outside the soaker oven, T2 = external temp.)

As you can see I know all of the inputs apart from the values of Ts1, h1 and h2. Could somebody please advise what approximate values to use for h1 and h2? I've seen 10-100 recommended for air but obviously this factor of 10 variation is going to affect my justification when I need to recommend action steps to improve the heat loss.

Any advise is much appreciated.

 

[corus]

On the hot face you'll have radiation to 1300C as the dominant factor. On the outer surface you'll also have radiation plus natural convection to an ambient temperature. All of these conditions are equivalent to a temperature dependent h1 or h2, so no constant value can be used.
Given that you already have known materials and a known outer surface temperature, why don't you estimate the values of h1 amd h2 until your surface temperature matches the measreud value, and then apply the same conditions with your new materials. You could also simply assume that the hot face is at the ambent temperture of 1300C, and just calculate h2. These mathods would give you an estimate of the outer temperature. A more precise method, allowing for radiation and natural convection, can be obtained using iterative finite difference or finite element methods.

Hoping to say Tata

 

[ione]

Assuming the lid is a flat plate, the order of magnitude for the heat transfer coefficient (natural convection) for the outside surface is about 10 W/(m^2*°C). Probably it is even less depending on the lid dimensions. Further to the inside surface I would assume the temperature is 1300 °C.

Anyway as already pointed out by Corus do not neglect radiation in your calculation.

 

[meyeal]

Corus, Ione,

Thanks for the input. I'd previously approximated the inner surface to be 1300C, which yielded a conducted heat transfer of 9.6kW/m^2. Incidentally, given the dimensions of the lid (approx 15m^2), this heat loss translates to a cost (assuming 2p per kWh of energy from gas) which is greater than our actual gas bill. Hence the confusion and my hoping to do some calcs using h1 or h2.

I've calculated radiation at the outer surface to be much lower, of the order 0.08kW/m^2.

I'm thinking that my delta T for the soaker lid needs to be much lower for the heat transfer rate to make sense.

Using a h2 value of 10W/(m^2/degC), I calculated the convected + radiated heat loss at the outer surface to be 2.03kW/m^2. This is much more realistic than the first calculation but still yields an energy cost of approximately half our gas bill, which seems much too high. Solving for the inner surface temp now gives me 223 degrees C, which seems rather low, and a heat transfer coefficient of 1.88 for h1.

I would appreciate your opinion on these figures if you could spare me 5 minutes?

Thanks.

 

[ione]

I am a bit puzzled about the radiation heat transfer from the outside surface (it should have approx. the same order of magnitude of convection heat transfer)

The specific (per square metre) radiated heat is

Q = epsilon*sigma*(Ts2^4 – Tamb^4) = 2.25 kW/m^2

Epsilon = 0.9 (refractory brick emissivity to be checked)
Sigma = 5.67 * 10^(-8) W/m^2/K^4 (Stefan-Boltzman constant)
Ts2 = 473.15 K (200 °C)
Tamb = 278.15 K (5 °C)

So adding this value to the convective heat transfer, you should be capable to match your gas bill.

 

[meyeal]

Ione,

The radiation heat transfer equation that I used was simply Q = epsilon*sigma*(Ts2^4) = 0.08 kW/m^2, the 'radiation heat flux from a real surface'. I thought the secondary Tsurface term was only required when there was another surface to absorb the energy, not simply the air radiating it away?

I would actually expect the heat loss through the soaker lid to be a smaller percentage of the gas bill. This is due to the very nature of the operation, i.e. putting immense quantities of heat into ingots and blooms to heat them to a suitable rolling temperature. I'd expect the majority of the heat energy to be spent here, and the rest to be lost to the surroundings through both refractory surfaces and also gaps in seals etc. My project is to minimise these losses wherever possible.

Thanks

 

[ione]

Hey meyeal,

You have to use absolute temperature scale and NOT Celsius scale.

0,9*5,67*10^(-8)*(473,15^4) = 2557,531 W/m^2 = 2,557 kW/m^2

 

[corus]

For a surface temperature of 200 C then the overall heat transfer coefficient to air is about h=20 W/m^2 C, with an emissivity of 0.9, the surface facing up, and an ambient of 20C. Given the hot face is at 1300C then calculate the total thermal resistance through the layers, and calculate the cold face temperature as it is now using this h value If it doesn't come out at about 200C then there's something wrong somewhere.

Hoping to say Tata

 

[IRstuff]

How do you know that your only source of heat loss is the lid? There should be the specific heat of the steel, as well any possible losses through the other 5 sides of the pit, and any losses associated with the heaters themselves.

Even if the lid loss were zero, you have to HEAT something, so the gas cost cannot ever be zero.

While your environment is not a forced convection problem, the fact that your temperature is 5°C means that your HVAC is hard at work, unless you're in the Arctic. This could mean that your convective coefficient is higher than the usual natural convection problem. Moreover, radiated loss is often, at best, a SWAG, given that you don't know for sure what the spectral emissivities really are, and you don't know what you're actually emitting into.

TTFN

FAQ731-376

 

[zekeman]

I,m puzzled over the almost 5:1 ratio of conducted heat through the lid and the radiated heat out of the lid to atmosphere.

I would think that the conducted value would be more accurate, since the inner wall temperature estimate looks OK to me and the total h to atmosphere is also a good estimate, but could be slightly too high if the emmisivity is much smaller than 0.9, making the conductive value 9.6KW/m^2 you got even less credible.

When I see such a disparity of results, of what I think are reasonable assumptions , I would start all over

I would check everything again-- temperature measurements,the gas usage again, check the K for your concrete and verify the thickness of the lid watching your units.Also, if possible get an inner surface temperature measurement.

 

[meyeal]

Dear all,

Many thanks for your various pieces of advice. I started the calcs again, and doubel checked my parameters. I've changed my material k value to be 1.2, from 1.75. When I actually measured the refractory depth myself, it was 50% thicker than quoted by the guys who install it. I was also advised that the emissivity of the refractory is 0.8.

Using these new figures, at the upper surface of the refractory the radiation heat loss calculated as 1938W/m^2. The convected heat loss came out as 1800W/m^2. This 3.74kW/m^2 figure compares favourably with the calculations of heat transfer through the refractory material; assuming that the hot face is at 1200 degrees, the total heat transfer in that regard calculates as 4kW/m^2.

Bearing in mind the accuracy of my parameters, I'm happy that I am on the right track. Now when the datasheet arrives for the suggested insulative material, I can calculate energy savings with a composite layer.

I shall post them on here when I have them, if anybody is interested.

Thanks again.

 

[ione]

meyeal,

Eventually glad to see the radiated heat loss are now comparable to the convection heat loss

By the way in your Stefan-Boltzman equation you should keep into account the temperature of the surrounding ambient (this very slightly changes your result, anyway it should be considered).

Q = epsilon*sigma*(Ts2^4 – Tamb^4)

Temperature expressed in Kelvin!