MAP_STEEL_ALUM estimates the amount of aluminium that remains in solid solution in ferrite in a low-alloy steel arc weld deposit.
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Product form: | Source code |
The atomic weight of aluminium is 26.98 and that of oxygen is 16. At the beginning of reaction we have w(Al) of aluminium and w(O) of oxygen. A quantity x of aluminium reacts with the oxygen. For the reaction:-
2Al + 3O = Al2O3We therefore have [w(Al) - x] of aluminium left unreacted, and [w(O) - {{16*3}/{26.98*2}}x] of oxygen left unreacted. The equilbrium constant is therefore:-
K = [1/{(w(Al) - x)2(w(O) - 0.89x)3}]In the subroutine, w(O) = OXYWT, w(Al)=ALTOT and K0.5 = CONST. The value of CONST is derived by matching with data from [2-4].
1. H.K.D.H. Bhadeshia, Unpublished research, (1990).
2. G. Thewlis, Joining and Materials, 2, (1989), 25-31.
3. G. Thewlis, Joining and Materials, 2, (1989), 125-129.
4. F.C. Liao and S. Liu, American Welding Journal, 71, (1992), 94s-104s.
IFAIL set to 1 indicates that the calculations are outside the validated range of experimental data.
Predictiblity of soluble aluminium is ±50 parts per million by weight.
None.
To calculate the soluble aluminium concentration when the total aluminium concentration is 0.02 by percentage weight, and the total oxygen concentration is 0.03 by percentage weight.
DOUBLE PRECISION ALTOT, ALSOL, OXYWT, CONST INTEGER MAXI, IU, IFAIL READ (5,*) ALTOT, OXYWT, MAXI, CONST IFAIL=0 CALL MAP_STEEL_ALUM(OXYWT, ALTOT, ALSOL, MAXI, IU, IFAIL, CONST) WRITE(6,1) ALTOT, ALSOL, OXYWT, IU 1 FORMAT('Total aluminium =',F8.5,' by weight percent' & 'Dissolved aluminium =',F8.5,' by weight percent' & 'Total oxygen =',F8.5,' by weight percent' & I4,' iterations were performed') STOP END
0.02 0.03 20 0.616D-04
Total aluminium = 0.0200 by weight percent Dissolved aluminium = 0.00495 by weight percent Total oxygen = 0.03000 by weight percent 6 iterations were performed
None.
aluminium, solid solution, ferrite
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