fig.3.5.which体现。这些曲线是由经验公式表示。我们将使用下

fig.3.5.which体现。这些曲线是由经验公式表示。我们将使用下列符号，指量表示为每100g熟料痣：K =总氧化钾；N =总氧化钠；S =总SO3；KS，NS，CS、SS为K2O，Na2O，CaO和SO3为硫酸盐。
（1）计算的比率为/（钾、氮）和钾的比例。
（2）计算不锈钢；
（一）如果R≤0.8，SS = S
（b）if0.8＜R≤2，SS = [ 10。0.25（r-0.8）]的
（c）如果R＞2，SS = 0.7
（3）计算KS和NS：
（d）如果R≤0.8和K＜3.67，KS = 1.12rk和NS = 0.56rn
（e）如果R＞0.8和K＜3.67，KS = 0.9k和NS = 0.45n
（f）如果K≥3.67，按照（d）或（e）但相乘结果（K + N）/（1.12k + 0.56n）。这保证了（生理盐水+ KS）≤SS
（4）计算CS = SS KS NS。
如果水溶性钾和N2O的内容是众所周知的，它只能从他们计算，KS和NS和R，然后实施阶段2和4。从KS，NS和CS，单个硫酸盐相量可估计为第3.5.6.in给出的例子说明，计算结果表明那硫酸相占所有Na2O K2O so3,0.07 %，0.04草0.28%。这些可能发生钙langbeinite aphthitalite（0.6%）和（0.1%）。
4.4.4估计的主要阶段
从钙含量，扣除游离石灰和任何曹作为硫酸盐或碳酸钙。在没有更详细的数据，假设的不溶残渣67% SiO2和33% al2o3.and扣除相应。这给al2o3,5.6 sio2,21.5 %；%；曹，65.3%。
假设四个主要阶段有成分表1.2给出，即阿利特在注1，表表明MgO和Fe2O3含量的熟料为根据。
设置四个线性联立方程，矩阵形式表示为内= b，其中aij是氧化组分I相J的重量分数，许继在熟料相J的重量百分比，二是修正后的重量百分比组成我在熟料中氧化；
0.7185x1 + 0.635x2 + 0.566x3 + 0.475x4 = b1 = 65.3（4.7）
0.2515x1 + 0.315x2 + 0.037x3 + 0.036x4 = b2 = 21.5（4.8）

結果 (中文) 3:

Fig.3.5.which. These curves are expressed by empirical formulas. We will use the following symbols, which said as per 100g clinker nevus: k = total potassium oxide; n = total Na2O; s = total SO3; KS, NS, CS, SS for K2O, Na2O, Cao and SO3 to sulfate.
(1) the ratio of the calculated rate is / (k, n) and the ratio of potassium.

(2) calculation of stainless steel; (a) if R = 0.8, SS = S
(b) if0.8 < R < 2, SS = 10. 0.25 (r-0.8)]

(c) if r > 2. SS = 0.7
(3) calculated KS and NS:
(d) if R is less than or equal to 0.8 and K < 3.67, KS = 1.12rk and ns = 0.56rn
(E) if r > 0.8 and K < 3.67, KS = 0.9k and ns = 0.45n
(f) if K is greater than or equal to 3.67, according to (d) or (E) but results multiplication (k n) / (1.12k 0.56n). This ensures that (NS KS) or SS
(4) CS = SS KS NS calculation. If the contents of water soluble potassium and N2O are well known, it can only be calculated from them, KS and NS and R, and then the implementation stages 2 and 4. From KS, NS, and CS, a single sulfate phase volume can be estimated as an example given by 3.5.6.in, and the results show that the sulfuric acid accounts for all K2O so3,0.07 Na2O%, 0.04 0.28%.These are likely to occur in aphthitalite langbeinite (0.6%) and (0.1%).
4.4.4 was estimated at the main stage of

from the calcium content, after deducting free lime and any Cao as a sulfate or calcium carbonate. In the absence of more detailed data, assuming the insoluble residue 67% SiO2 and 33% al2o3.and are deducted. This gives sio2,21.5 al2o3,5.6%;%; Cao, 65.3%.
hypothesis there are four main stages in composition are given in Table 1.2, alite in note 1, shows that the MgO and Fe2O3 content of clinker on. Four linear simultaneous equations are set up, and the matrix form is expressed as an inner = B, where AIJ is the weight fraction of the I phase J,XJ in clinker phases J weight percentage, and the second is modified components in percentage by weight: I in the clinker oxidation;
0.7185x1 0.635x2 0.566x3 0.475x4 B1 = = 65.3 (4.7)
0.2515x1 0.315x2 0.037x3 0.036x4 = B2 = 21.5 (4.8)