INFORME TECNICAS.pdf
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Transcript of INFORME TECNICAS.pdf
₂
₂ ₂ ₃
₂
₂ ₃
₃
₂ ₂
₂
₂ ₃
₃
₃
₂ ₃
₂
₂
% 𝐻𝑢𝑚𝑒𝑑𝑎𝑑 =2,0169 − 1,7870
2,0169∗ 100
𝐼 =100 − 0,0001
(2,0169)2
∗ √(2,0169)2 + (1,7870)2
%𝐶𝑒𝑛𝑖𝑧𝑎 =1,6547
1,9569 ∗ (100 − 11,3987)
∗ 10000
𝐼 = 10000 ∗ 0.0001
1.9569 ∗ (100 − 11,3987)∗
√
(0.0001)2 ∗ (1 +1.65472
1.95692) + (66.2424)2 ∗
(1.6547
100 − 11.3987)
2
𝐼 = 7,1352 ∗ 10−3
SiO₂
SiO₂
SiO₂
%SiO₂ =1,3832
1,6547∗ 100
%SiO₂ =
%CaO =(𝑃𝑒𝑠𝑜 𝑃𝑟𝑒𝑐𝑖.∗ 𝐹) (
25050
) ∗ 100
𝑊 𝑖𝑛𝑖𝑐𝑖𝑎𝑙
𝐹 =𝑃. 𝐶𝑎𝑜
𝑃. 𝐶𝑎𝐶𝑂₃(250
50 )
%CaO =0.0139 ∗ (
59100 )(
25050
)
1,6547
%CaO = 2,352
₂ ₃
%𝑅₂𝑂₃ =𝑃𝑒𝑠𝑜 𝑀𝑢𝑒𝑠𝑡𝑟𝑎 𝐶𝑎𝑙𝑐𝑖𝑛𝑎𝑑𝑎 𝑥 𝐴𝑙𝑖𝑐𝑜𝑡𝑎
𝑃𝑒𝑠𝑜 𝑀𝑢𝑒𝑠𝑡𝑟𝑎 𝑖𝑛𝑖𝑐𝑖𝑎𝑙∗ 100
%𝑅₂𝑂₃ =(0,0433) (
25050
)
1,6547∗ 100
%𝑅2𝑂3 = 13,39
𝑌−0.0024
0.003
𝑚𝑔
𝐿 𝐴𝑙
50
5)
𝑚𝑔
𝐿 𝐴𝑙
1.6547𝑔
250𝑚𝐿→
1654.7𝑚𝑔
0,250𝐿= 6618,8
𝑚𝑔
𝐿
𝑚𝑔
𝐿
𝑚𝑔
𝐿
𝑚𝑔
𝐿
𝑚𝑔
𝐿 𝐴𝑙
𝑌+0.0015
0.00217
𝑚𝑔
𝐿 𝐹𝑒
50
10) (
50
1)
𝑚𝑔
𝐿 𝐹𝑒
𝑚𝑔
𝐿
𝑚𝑔
𝐿
𝑚𝑔
𝐿
𝑚𝑔
𝐿 𝐹𝑒
𝑅₂𝑂₃
%𝑅₂𝑂₃ = ₂ ₃ ₂ ₃
%𝐴𝑙₂𝑂₃ = % 𝐴𝑙(Al₂O₃
𝐴𝑙) (
1
2)
%𝐴𝑙₂𝑂₃ = 4.714 %(102
27) (
1
2)
%𝐴𝑙₂𝑂₃ = 8,9
%𝐹𝑒₂𝑂₃ = % 𝐹𝑒(Fe₂O₃
𝐹𝑒) (
1
2)
%𝐹𝑒₂𝑂₃ = 2,439 %(159,6
55,8) (
1
2)
%𝐹𝑒₂𝑂₃ = 3,4
%𝑅₂𝑂₃ = 8,9 + 3,4
%𝑅₂𝑂₃ = 12,3
𝑌+0,0014
0.4597
𝑚𝑔
𝐿 𝑀𝑔
50
10) (
50
1)
𝑚𝑔
𝐿 𝑀𝑔
𝑚𝑔
𝐿
𝑚𝑔
𝐿
𝑚𝑔
𝐿
𝑚𝑔
𝐿 𝑀𝑔
2,352
13,39
