Heat Transfer

You can find information on thermal conductivity at MatWeb and other materials sites.

Hello,

Heat required for raising the temperature of the steel billets :

Q=m(15 tons)*Cp(steel billets)*Delta T(1250-45)

Volume of Coal gas required

Q/6270

However, you would also need to consider the heat lost from the Reheating furnace. That would be a function of the 1250 deg C, the insulation of the furnace (k,thermal conductivity, thickness)and the surface area (A)over which this happens. Apart from that, other heat losses at locations where billets are moved in and out of the furnace.

Thanks

Thanks for your comment. I am considering no loss as I want compare the heat required for Stainless Steel and Mild Steel. Incidentally, can you suggest the average heat losses in the reheating furnace?

Rao

All steels have the same conductivity and Sp.Ht. The limiting parameter is the heat transfer between the hot gas and the metal through the interface.

I'd suggest that you look at the numbers closely before making incorrect blanket statements such as yours.

Heat loss should be taken into consideration since it is usually (for purposes as stated) in the range 8 - 15% but don't forget to look at your system carefully. I would go a little deeper from what stoic80 and Lyn have pointed.

In some heating furnace working with LPG or NG the fuel efficiency is about 25%, but the calorific value of these gases is much larger than the coal gas reported by you. The thermal and fuel efficiencies depend on the furnace design. For a better evaluation it would need more information such as:

1 - What is the composition and density of gas?
2 - What are the internal dimensions of the furnace?
3 - Are there injection of air/oxygen? If so, how much and at what temperature?
4 - What is the average thickness of the furnace wall?
5 - Heating is done by continuous process? If in batches, how many blanks are heated at a time?

Yes I agree with you entirely, let us take the furnace or fuel efficiency as 25%. Since I am interested in the comparsion of THERMAL CONDUCTIVITY of both the steel, so as to ascertain how much more or less fuel is required for heating the ingots of similar size, as we have 8000 Nm3/hr of producer Gas of 5225 KJ/Nm3 calorific value. The efficiency of furnace of similar sizes varies to great extent from place to place, therefore, I would not be able to generalise the efficiency.

Govind

According to the MatLab:

1 - Stainless steel AISI 302 has:
thermal conductivity "K" = 18.5 kcal / hm ° C
density "d" = 7860 kg/m3
specific heat "C" = 0.119 kcal / kg º C
thermal diffusivity "a" = 0.0198 m2 / h [a = K / (dc)]

2 - Carbon steel AISI 1010 has:
thermal conductivity "K" = 44.9 kcal / hm ° C
density "d" = 7870 kg/m3
specific heat "C" = 0.197 kcal / kg º C
thermal diffusivity "a" = 0.0291 m2 / h [a = K / (dc)]

Putting these data into a program that I developed that simulates a metallurgical furnace and considering the same emissivity (0.75) and the same furnace, we would have:

Specific consumption of gas (Nm3 of gas per weight of heated material):

Case 1 (Inox): 0.5273 Nm3/kg

Case 2 (Steel): 0.8697 Nm3/kg

You mean to say that given the size of the both the steel as same and are heated in the same environment , The carbon Steel "consumes" more heat than Stainless Steel?

Yes, assuming the same blank temperature. Struck me as odd too, but it was the result found. I'll check the calculations looking for some mistake. I'll contact you soon.

Could have something to do with the specific heat of steel #2 being almost double that of steel #1...

Actually the ratio is 1.653 and the ratio of the gas estimates is 1.649

So I think the 2% discrepancy, which could be due to the difference in the thermal conductivity and or heating time and resultant losses, doesn't discredit the correlation.

Sorry for the delay, I was in a work travel.

The used spreadsheet is for the design of heating furnaces for mechanical processing of metals. In this case, some considerations are made in order to improve, in the first place, the productivity, thus the main mechanism of heat transfer to the load is the irradiation (95%), which is more efficient for this purpose, and the convection is only 5% ;

The thermal conductivity in a transient heating will allow to evaluate the "speed of temperature input" into the blank, however, for a "good" forming, in practice it is desirable that the temperature difference between surface and center of the blank keep within 5% (Biot number <0.1), this will prevent burning the surface or getting cold the center, limiting the global coefficient of thermal exchange and influencing fuel consumption.

Thus, for a transient regime:

(1) T - Ta = (To - Ta) * exp (- Bi * Fo)

Where:

T = final temperature of the blank (1240 ° C)

Ta= average temperature of the furnace (~ 1370 º C)

To = initial temperature of the blank (45 º C)

Bi = Biot number (dimensionless)

Fo = Fourier number (dimensionless)

Bi= HL/K

Fo= Kt/rcL²

H = average coefficient of heat transfer (in this case involves radiation and convection) (kcal/hm2 C)

L (m) = (V) Volume (0,015625 m³) / (As) Exposed Surface Area (0,40625 m²)

K = Thermal conductivity (kcal / hm ° C)

t = time (h)

r = density (kg/m3)

c = specific heat (kcal / kg º C)

Thus, for the analysis of the two materials, the Biot number was kept at 0.1 resulting in:

(Stainless steel) H = 48.1 (C kcal/hm2)

(Carbon steel) H = 116.74 (C kcal/hm2)

Substituting these values ​​in (1), time to reach 1240 ° C will be:

(Stainless steel) t =1.73 h

(Carbon steel) t =1.2 h

The amount of heat transmitted in the time interval t, considering 96 blanks (an arbitrary number from my study)  is:

Q (kcal) = rV c (Ta - To)*[1 - exp ( - H*t /crL)] ( Q = m c DT * transient)

Q (Carbon steel ) = 2,787,315 kcal --> 11805 kg = 236 kcal / kg

According to current information, for carbon steel: 8000 Nm3 / h * 1248 kcal/Nm3 = (9,984,000 kcal / h) / (15000 kg / h) = 665.6 kcal / kg, which would imply in an efficiency of 236/665, 6 = 35%

Q (Stainless steel) = 1,675,852 kcal --> 11790 kg = 142 kcal / kg

Considering the same efficiency (*): 142 / 0.35 = 406 kcal / kg

For 15000 kg / h: 15000 * 406 = 6,090,000 kcal / h

Consumption would be: 6090000 / 1248 = 4880 Nm3 / h for stainless steel. (61% of carbon steel)

These numbers are consistent with the observation made by GordieGii, i.e., in this case is more important the difference within the specific heats and densities than within the thermal conductivities.

(*) Recall that to verify this in the practice it can be necessary to change some operating variables of the furnace [to reduce the average coefficient of heat transfer (H) in 41.2% (48.1/116.74)] , that is, if you leave the furnace operating in the same way as with carbon steel, gas consumption for stainless steel may be even higher or heating may cause damage to the material.

Thanks a lot for your information! its so much in detail that I would be taking quite a bit of time to assimilate the logic. I have just strayed into Chemical Engineering a year back. Let me get my fundamentals clear before I ask few questions on this, if I dont understand it. Once again thanks a lot.

Govind

Sir,

Would you help to cast an spread sheet for the above calculation? it would be helpful for me to evaluate other steels in different environment. Its an earnest request.