The raw material conditions and their supply are known, and the alkalinity of the sinter is determined to be 1.2. It is required to calculate the fuel, flux, mixture, returning, and the amount of water added to the mixture, the annual supply of the sinter and its composition.
S smokeless coal component content of SO 2 in said low, no conversion.
calculation steps:
1) The amount of fuel (Q k), 10% of iron-containing material (q) Total:
Q k =Q A ×q+Q B ×q+Q C ×q=ΣQ i ×q i (1)
Q k = (150 + 100 + 80 + 14) × 0.1 = 34,400 tons / year 2) flux annual limestone effective: CaO' N = CaO N - RSiO 2 N = 46.3 - 1.2 × 1.7 = 44.26
Concentrate required: CaO' 4 =RSiO 2 A-CaO A =1.2×12.66-1.12=14.072
CaO' B = RSiO 2 B-CaO B = 1.2 × 12.8-0.45 = 14.91
CaO' C = RSiO 2 c-CaOc = 1.2 × 4.1 - 2.3 = 2.62
Required for blast furnace dust: CaO' D =RSiO 2 D-CaO D =1.2×13.3-10.63=5.33
CaO' K = RSiO 2 K-CaO K = 1.2 × 7.8 - 2.11 = 7.25 for anthracite
The amount of limestone needed:
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4) Return amount Q o =Q mixed ×q o =704.2309×25%=176.0577 tons/year 5) Water consumption of mixture (excluding evaporation loss during transportation)
The water content of the mixture itself:
Q mixed qH 2 O=704.2309×0.08=56.3385
QH 2 O=56.3385-(16.6667+13.6363+7.9121+0.7368+3.3888+2.1957)
=11.8021 million tons/year or 117.802 million meters 2 years/year 6) Sintering amount Set the FeO A of the sintered ore to 18% and the desulfurization rate of sintering is 90%.
Total residual material 烧结Q i (1-Ig-0.9S) during sintering
Considering that the raw material becomes sinter, the increase or decrease of oxygen due to the change in the amount of FeO is 1/gFeO, which is calculated as follows:
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Calculate the denominator term: 9+FeO A =9+0.18=9.18
Molecular item:
Concentrate A
Q A [9(1-Ig-0.9S A )+FeO A ]=150[9(1-1.04%-0.9×0.086%)+25.7%]=1374.0726
Concentrate B
Q B [9(1-Ig-0.9S B )+FeO B ]=100[9(1-0.60%-0.9×0.04%)+25.23%]=919.536
Concentrate C
Q c [9(1-Ig-0.9S c )+FeO c ]=80[9(1-3%-0.9×0.37%)+25.77%]=716.6184
Blast Furnace D
Q D [9(1-Ig-0.9S D )+FeO D ]=14[9(1-19.18%-0.9×0.3%)+10.2%]=102.543
Limestone N
Q N [9(1-Ig-0.9S N )+FeO N ]=93.4347[9(1-42.18%-0.9×0)+0%]=486.2155
Anthracite K
Q K [9(1-Ig-1.9S K )+FeO K ]=34.4[9(1-82.84%-0.9×0.7%)+1.56%]=51.7135
Concentrate A brought in Fe=Q A ×Fe A %=150×60.84%=91.26
Concentrate B brought in Fe=Q B ×Fe B %=100×61.94%=61.94
Concentrate C brought in Fe=Q c ×Fe c %=80×61.82%=49.456
Fe=Q D ×Fe D %=14×38.1%=5.334
Fe=Q N ×Fe N %=93.4347×0=0
Fe = Q K × Fe K %=34.4×2.165%=0.7448
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