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细粒赤铁矿精矿絮凝沉降行为和动力学

董鑫, 芦川, 张芹

董鑫, 芦川, 张芹. 细粒赤铁矿精矿絮凝沉降行为和动力学[J]. 矿产综合利用, 2024, 45(6): 124-130. DOI: 10.3969/j.issn.1000-6532.2024.06.019
引用本文: 董鑫, 芦川, 张芹. 细粒赤铁矿精矿絮凝沉降行为和动力学[J]. 矿产综合利用, 2024, 45(6): 124-130. DOI: 10.3969/j.issn.1000-6532.2024.06.019
DONG Xin, LU Chuan, ZHANG Qin. Flocculation and settlement behavior and dynamic of fine-grained hematite concentrate[J]. Multipurpose Utilization of Mineral Resources, 2024, 45(6): 124-130. DOI: 10.3969/j.issn.1000-6532.2024.06.019
Citation: DONG Xin, LU Chuan, ZHANG Qin. Flocculation and settlement behavior and dynamic of fine-grained hematite concentrate[J]. Multipurpose Utilization of Mineral Resources, 2024, 45(6): 124-130. DOI: 10.3969/j.issn.1000-6532.2024.06.019

细粒赤铁矿精矿絮凝沉降行为和动力学

基金项目: 国家自然科学基金面上项目(50974099)
详细信息
    作者简介:

    董鑫(2000-),男,硕士研究生,从事细粒赤铁矿精矿絮凝沉降研究

    通讯作者:

    张芹(1966-),女,教授,硕士研究生导师,从事矿物加工理论与工艺研究。

  • 中图分类号: TD951

Flocculation and Settlement Behavior and Dynamic of Fine-grained Hematite Concentrate

  • 摘要:

    这是一篇矿物加工工程领域的论文。本文以细粒赤铁矿精矿为研究对象,探究了絮凝剂分子量、絮凝剂配制浓度、絮凝剂用量和矿浆温度对细粒赤铁矿精矿絮凝沉降效果的影响,同时通过动力学分析探究符合细粒赤铁矿精矿的絮凝沉降动力学模型。结果表明:分子量为900万、配制浓度为0.05%和用量为60 g/t的APAM有好的沉降效果和较低的成本,添加APAM会明显改善沉降效果,高APAM分子量,会得到快的沉降速度和低的最终底流浓度;低APAM配制浓度,会得到快的沉降速度和高的最终底流浓度;APAM用量的增加使沉降速度变快、絮团粒径增大,但二者增加速度均逐渐变缓,最终底流浓度下降;随着矿浆温度的升高,沉降速度增加,但增速不太明显,最终底流浓度升高;APAM对细粒赤铁矿精矿絮凝沉降的动力学分析,双曲线动力学模型方程取得了较好的拟合度,APAM对细粒赤铁矿精矿絮凝沉降动力学可以优先使用双曲线动力学模型1/(1-$ \dfrac{{c}}{{c}_{{0}}} $)=$ \dfrac{{k}}{{t}}+b $来描述。

    Abstract:

    This is an article in the field of mining processing engineering. The article takes fine-grained hematite concentrate as the research object, explores the effects of the molecular weight of flocculant, the concentration of flocculant, the dosage of flocculant, and slurry temperature on the flocculation and settlement effect of fine-grained hematite concentrate. At the same time, through dynamic analysis, the flocculation and settlement dynamic model that conforms to fine-grained hematite concentrate is explored. The results show that APAM with a molecular weight of 9 million, a preparation concentration of 0.05%, and a dosage of 60 g/t has a good settling effect and lower cost. The addition of APAM significantly improves the settling effect, while high APAM molecular weight results in fast settling speed and low final underflow concentration. Low APAM concentration results in fast settling speed and high final underflow concentration. Increasing the APAM dosage results in a faster settling speed and an increase in floc particle size, but both rates of increase gradually slow down, ultimately resulting in a decrease in underflow concentration. As the temperature of the slurry increases, the settling speed increases, but the growth rate is not very significant, and the concentration of underflow increases. The dynamic analysis of flocculation and settlement of fine-grained hematite concentrate by APAM shows that the hyperbolic dynamic model equation has achieved good fitting. APAM can prioritize using the hyperbolic dynamic model 1/(1-$ \dfrac{{c}}{{{c}}_{{0}}} $)=$ \dfrac{{k}}{{t}}+b $ to describe the flocculation and settlement dynamic of fine-grained hematite concentrate.

  • 我国铁矿资源丰富,储存量多,但我国铁矿石存在品位低、嵌布粒度细和矿物组成复杂的特点,简称“贫”“细”“杂”,且这类铁矿石占比大。为了更加有效地进行分选,必须让其得到充分单体解离,从而导致细粒矿物的增多[1]。但随着细粒矿物的增多,会对后续的浓缩过滤产生不良的影响,例如沉降时间长、过滤速率低、穿漏现象严重、滤饼含水率高等问题[2]。因此为解决这一问题,通常加入絮凝剂,如聚丙烯酰胺、明矾、聚合氯化铝、聚合氯化铁等,通过絮凝剂的吸附架桥、电性中和、网捕卷扫等作用来改善沉降效果[3]

    研究发现,细粒颗粒絮凝形成絮团需要满足两个条件,一是颗粒之间必须产生碰撞,二是必须有一定比率的颗粒碰撞使颗粒粘结在一起。影响碰撞的因素有三种作用:布朗运动、悬浮颗粒的不等速沉降和悬浮液体内的速度梯度[4]。胡宗岗等[5]以赤铁矿精矿为对象,采用正交实验,考查不同因素对细粒赤铁矿精矿絮凝沉降的影响,结果表明影响因素对平均沉降速度影响程度大小顺序为:絮凝剂单耗>絮凝剂溶液浓度>入料浓度>絮凝剂分子量。周丹等[6]通过分析颗粒污泥沉降过程的特点,采用颗粒物质干涉沉降规律对颗粒污泥沉降过程进行描述,结合Allen阻力公式,建立了颗粒污泥沉降动力学模型。陈洪松等[7]用吸管法研究了不同浓度NaCl对细颗粒泥沙静水絮凝沉降的影响,结果表明细颗粒泥沙相对浓度随时间的变化符合双曲线动力学模式。关许为等[8]在比较一、二级动力学模式后认为:二者都可用于絮凝速率系数和平均沉降速度问题的研究,但二级动力学模式用于数据拟合的相关性更好。唐海香等[9]以煤泥水为对象,采用正交实验,探究了动力学因素对煤泥水的絮凝沉降效果影响很大。

    大量资料探究了泥沙、污泥等细颗粒动力学,但有关细粒赤铁矿精矿相关研究资料较少。本研究将以细粒赤铁矿精矿为研究对象,探索不同条件—絮凝剂分子量、絮凝剂浓度、絮凝剂用量和矿浆温度对赤铁矿精矿絮凝沉降效果的影响,并对赤铁矿精矿絮凝沉降进行动力学分析。

    实验所用原料取自于宝武钢铁武汉工业港铁精矿,密度为4 150 kg/m3,试样化学多元素分析见表1

    表  1  试样多元素分析/%
    Table  1.  Chemical multi-element analysis of samples
    TFeSiO2Al2O3MnPSZnCaO
    64.508.4101.5690.1500.0420.0210.0040.012
    下载: 导出CSV 
    | 显示表格

    将原理用XMCQ-Φ290 mmⅹ90 mm瓷衬球磨机磨至-150 μm备用,用湿筛的方法对铁精矿进行粒度分析,结果见表2

    表  2  试样粒度分析
    Table  2.  Particle size analysis of samples
    粒级/μm产率/%累积产率/%
    +1500.000.00
    -150+7411.0011.00
    -74+4515.6026.60
    -45+384.2530.85
    -3869.15100.00
    下载: 导出CSV 
    | 显示表格

    表1可知实验矿样全铁品位为64.32%,由表2可知实验矿样-38 μm粒级69.15%,细粒矿物含量非常高。

    实验所用絮凝剂为有机高分子阴离子型聚丙烯酰胺(APAM),选取三种不同分子量的聚丙烯酰胺,分别为500万、900万和1 300万,分析级。

    实验验过程中主要使用的仪器与设备有纯水仪、250 mL沉降瓶、数字搅拌器、200 mL量筒、精密电子称、烧杯、秒表、分析天平、药匙等。

    用精密电子秤称取12.5 g矿样置于烧杯中,然后用量程为200 mL的量筒量取180 mL去离子水倒入烧杯中,再将数字搅拌器放入矿浆中,以360 r/min的转速搅拌10 min,使得矿浆充分搅拌均匀。搅拌完成后迅速将矿浆倒入沉降瓶中,后加入一定量的絮凝剂溶液再定容至250 mL,匀速上下翻转十次混合均匀,静置于水平实验台后开始计时,记录底层悬浊液高度及矿浆的沉降高度,并计算出悬浊液的浓度,计算公式见式(1):

    $$ c= \frac{{m}_{0} \cdot {c}_{0}}{{m}_{0}-\left({v}_{0}-v\right){{\rho }}_{\mathrm{水}}} \times 100{\text{%}}$$ (1)

    式中,c为记录点底流悬浊液浓度,g/L;m0为初始矿浆质量,g;c0为初始矿浆浓度,g/L;v0为初始矿浆体积,mL;v为记录点底流悬浊液体积,mL;ρ水为水的密度,g/cm3

    在絮凝剂配制浓度为0.05%,用量为60 g/t,温度为室温(20 ℃)的条件下,探究不同分子量大小的絮凝剂对铁精矿沉降效果的影响,结果见图1

    图  1  絮凝剂分子量大小对底流悬浊液浓度的影响
    Figure  1.  Effect of molecular weight of flocculants on the concentration of underflow suspension

    图1可知,沉降300 s时,不添加APAM的底流浓度为1 041.7 g/L, APAM分子量为500万的底流浓度为1 136.4 g/L,APAM分子量为900万的底流浓度为1 116.7 g/L,APAM分子量为1 300万的底流浓度为1 041.7 g/L,其曲线更靠左,到达 625 g/L所用时间更短。随着APAM分子量的增加,沉降速度变快,最终底流浓度降低,阻滞空间会更大,表明这会对后续过滤脱水产生不良的影响。这可能是因为APAM的长分子链中含有一定数量的极性基团,能够吸附矿浆中的矿物颗粒,使颗粒间链接架桥或通过电荷中和凝聚形成大的絮团。而高分子量的APAM的链会更长,在与赤铁矿精矿作用时产生的絮团会更大,得到的絮团结构会更加松散[10-11]。综合考虑,选择分子量为900万的APAM。

    在絮凝剂分子量为900万,用量为60 g/t,温度为室温(20 ℃)的条件下,探究不同絮凝剂配制浓度对铁精矿沉降效果的影响,结果见图2

    图  2  絮凝剂配制浓度对底流悬浊液浓度的影响
    Figure  2.  Effect of concentration of flocculants solution on the concentration of underflow suspension

    图2可知,沉降300 s时,APAM配制浓度为0.05%时底流浓度为1 116.7 g/L,APAM配制浓度为0.10%时底流浓度为1 157.4 g/L,APAM配制浓度为0.15%时底流浓度为1 190.5 g/L,APAM配制浓度为0.20%时底流浓度为1 225.5 g/L。随着APAM配制浓度的增加,沉降速度明显下降,最终底流浓度则升高,阻滞空间小,表明这对后续的过滤脱水作业更有利。高分子絮凝剂溶液在一般的操作浓度下黏度比较高,其均匀分布在整个悬浮液中需要一定的时间,絮凝沉降作用的发生有一个滞后过程。当絮凝剂用量一定时,达到最大固液分离的效率取决于絮凝剂溶与悬浮液的混合均匀程度。APAM为高分子絮凝剂,其溶液黏度很高,在翻转过程中在矿浆中均匀分散需要一定时间,低配制浓度的APAM溶液更加容易均匀分散在矿浆中,提高了赤铁矿颗粒与APAM 之间的作用效率,而高配制浓度的APAM溶液由于没有分散更均匀,部分赤铁矿颗粒未与APAM作用,沉降后得到的絮团孔隙率小[12]。综合考虑,选用0.05%配制浓度的APAM。

    在絮凝剂分子量为900万,配制浓度为0.05%,温度为室温(20 ℃),探究不同絮凝剂用量对铁精矿沉降效果的影响,结果见图3

    图  3  絮凝剂用量对底流悬浊液浓度的影响
    Figure  3.  Effect of dosage of flocculants on the concentration of underflow suspension

    图3可知,沉降300 s时,当APAM用量分别为20、40、60、80 g/t时,其底流浓度分别为1 190.5、1 136.4、1 116.7、1 106.9 g/L。APAM用量的增加,赤铁矿颗粒与APAM形成的絮团会变大,从而使得沉降速度变快,絮团之间会更加松散[13],最终底流浓度则降低,阻滞空间变大,表明这将不利于后续过滤脱水作业。絮凝剂用量越高经济成本越高,综合经济成本和实验效果考虑,选用APAM用量为60 g/t。

    因为铁精矿浓缩过滤在浓缩机中进行,矿浆温度容易因环境温度的改变而改变,所以铁精矿浓缩脱水实际生产过程中应考虑温度的影响。在絮凝剂分子量为900万,配制浓度为0.05%,用量为60 g/t的条件下,探究不同矿浆温度对铁精矿沉降效果的影响,结果见图4

    图  4  温度对底流悬浊液浓度的影响
    Figure  4.  Effect of pulp temperature on the concentration of underflow suspension

    图4可知,沉降300 s时,矿浆温度为10 ℃的底流浓度为1 106.2 g/L,矿浆温度提升到20 ℃时底流浓度为1 116.7 g/L,矿浆温度提升到30 ℃时底流浓度为1 136.4 g/L ,矿浆温度提升到40℃时底流浓度为1 190.5 g/L,矿浆温度提升到50 ℃时底流浓度为1 275.5 g/L。矿浆温度的升高降低了矿浆的粘性并加剧了赤铁矿颗粒的布朗运动,促进了赤铁矿颗粒与APAM的碰撞几率,有利于APAM与赤铁矿颗粒形成絮团,从而使沉降速度变快,但增速不太明显,最终底流浓度升高,阻滞空间变小,表明这对后续过滤脱水作业更有利。温度对沉降速度没有明显的影响,但对最终底流浓度影响较大。

    以不同絮凝剂用量对沉降的影响为例,绘制沉降曲线,见图5。由图5可以看出赤铁矿精矿絮凝沉降在60 s前已经基本沉降完全,后面为沉降压缩点后的压缩层的压缩,故取沉降前30 s和60 s来研究细粒赤铁矿精矿絮凝沉降动力学。

    图  5  不同絮凝剂用量沉降曲线
    Figure  5.  Settling curves for different dosage of flocculants

    图1图2图3图4各沉降时间与底流悬浊液浓度的关系曲线,采用一级动力学方程模型、二级动力学方程模型和双曲线动力学方程模型进行拟合,结果见表3表4

    表  3  沉降时间30 s动力学模型拟合参数
    Table  3.  Settlement time of 30 s of the parameters for dynamic model
    条件 一级动力学模型方程lnc=-kt+lnc0 二级动力学模型方程$ \dfrac{{1}}{{c}}={k}{t}+\dfrac{{1}}{{{c}}_{{0}}} $ 双曲线动力学模型方程1/(1-$ \dfrac{{c}}{{{c}}_{{0}}} $)=$ \dfrac{{k}}{{t}} $+b
    动力学方程 R2 动力学方程 R2 动力学方程 R2
    0 lnc=0.004 1 t+3.899 2 0.944 6 $ \dfrac{1}{c} $=-0.000 1 t+0.020 2 0.952 6 1/(1-$ \dfrac{{c}}{{{c}}_{{0}}} $)=$ \dfrac{-706.860\; {0}}{{t}} $+19.143 6 0.996 0
    絮凝剂
    分子
    量/万
    500 lnc=0.086 2 t+3.566 0 0.929 3 $ \dfrac{1}{{c}} $=-0.000 7 t+0.020 8 0.980 6 1/(1-$ \dfrac{{c}}{{{c}}_{{0}}} $)=$ \dfrac{{-144.402}\;{0}}{{t}} $+7.134 2 0.906 5
    900 lnc=0.112 7 t+3.672 4 0.937 1 $ \dfrac{1}{{c}} $=-0.000 7 t+0.019 0 0.894 1 1/(1-$ \dfrac{{c}}{{{c}}_{{0}}} $)=$ \dfrac{-69.573\;7}{{t}} $+3.484 1 0.908 7
    1 300 lnc=0.101 5 t+4.147 2 0.849 5 $ \dfrac{1}{{c}} $=-0.000 6 t+0.015 5 0.761 0 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-12.656}{{t}} $+0.564 4 0.938 8
    絮凝剂
    浓度/%
    0.05 lnc=0.112 7 t+3.762 4 0.937 1 $ \dfrac{1}{{c}} $=-0.000 7 t+0.019 0 0.894 1 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-69.573\;7}{{t}} $+3.484 1 0.908 7
    0.1 lnc=0.048 0 t+3.720 8 0.933 3 $ \dfrac{1}{{c}} $=-0.000 3 t+0.021 3 0.983 5 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-144.171\;5}{{t}} $+6.217 9 0.961 7
    0.15 lnc=0.043 2 t+3.755 9 0.951 2 $ \dfrac{1}{{c}} $=-0.000 5 t+0.021 1 0.985 9 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-144.402\;0}{{t}} $+6.155 6 0.960 5
    0.2 lnc=0.034 2 t+3.800 5 0.957 2 $ \dfrac{1}{{c}} $=-0.000 4 t+0.020 9 0.990 2 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-141.786\;5}{{t}} $+5.778 1 0.959 0
    絮凝剂
    用量/
    (g/t)
    20 lnc=0.010 3 t+3.897 3 0.970 6 $ \dfrac{1}{{c}} $=-0.000 2 t+0.020 1 0.984 6 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-126.464\;9}{{t}} $+1.218 8 0.999 2
    40 lnc=0.087 5 t+3.542 9 0.905 3 $ \dfrac{1}{{c}} $=-0.000 7 t+0.020 7 0.990 7 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-69.375\;0}{{t}} $+3.002 2 0.966 2
    60 lnc=0.112 7 t+3.762 4 0.937 1 $ \dfrac{1}{{c}} $=-0.000 7 t+0.019 0 0.894 1 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-69.573\;7}{{t}} $+3.484 1 0.908 7
    80 lnc=0.112 6 t+3.934 7 0.916 1 $ \dfrac{1}{{c}} $=-0.000 6 t+0.016 5 0.830 9 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-15.643\;0}{{t}} $+0.685 1 0.965 0
    温度/℃ 10 lnc=0.108 3 t+3.586 7 0.940 1 $ \dfrac{1}{{c}} $=-0.000 7 t+0.019 0 0.935 4 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-69.647\;0}{{t}} $+3.364 1 0.928 7
    20 lnc=0.112 7 t+3.762 4 0.937 1 $ \dfrac{1}{{c}} $=-0.000 7 t+0.019 0 0.894 1 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-69.573\;7}{{t}} $+3.484 1 0.908 7
    30 lnc=0.113 6 t+3.668 1 0.936 4 $ \dfrac{1}{{c}} $=-0.000 7 t+0.018 9 0.890 6 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-69.506\;0}{{t}} $+3.493 1 0.905 1
    40 lnc=0.114 3 t+3.703 3 0.935 4 $ \dfrac{1}{{c}} $=-0.000 7 t+0.018 4 0.882 4 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-44.401\;0}{{t}} $+2.163 9 0.923 7
    50 lnc=0.113 0 t+3.849 7 0.944 4 $ \dfrac{1}{{c}} $=-0.000 7 t+0.017 2 0.846 3 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-24.109\;6}{{t}} $+1.134 1 0.938 6
    下载: 导出CSV 
    | 显示表格
    表  4  沉降时间60 s动力学模型拟合参数
    Table  4.  Settlement time of 60 s of the parameters for dynamic model
    条件 一级动力学模型方程lnc=-kt+lnc0 二级动力学模型方程$ \dfrac{{1}}{{c}}{=kt+}\dfrac{{1}}{{{c}}_{{0}}} $ 双曲线动力学模型方程1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{k}{t} $+b
    动力学方程 R2 动力学方程 R2 动力学方程 R2
    0 lnc=0.005 4 t+3.882 1 0.966 3 $ \dfrac{1}{c} $=-0.000 1 t+0.020 4 0.979 2 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-667.834 \;9}{t} $+13.900 1 0.987 7
    絮凝剂
    分子
    量/万
    500 lnc=0.056 8 t+3.969 4 0.884 5 $ \dfrac{1}{c} $=-0.000 3 t+0.016 2 0.770 5 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-61.288\; 0}{t} $+2.018 1 0.918 4
    900 lnc=0.054 8 t+4.417 1 0.732 0 $ \dfrac{1}{c} $=-0.000 3 t+0.013 8 0.607 6 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-59.988\; 6}{t} $+2.193 6 0.858 4
    1 300 lnc=0.043 7 t+4.881 6 0.604 4 $ \dfrac{1}{c} $=-0.000 2 t+0.010 8 0.478 1 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-10.993 \;1}{t} $+0.340 5 0.886 3
    絮凝剂
    浓度/%
    0.05 lnc=0.054 8 t+4.417 1 0.732 0 $ \dfrac{1}{c} $=-0.000 3 t+0.013 8 0.607 6 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-59.988 \;6}{t} $+2.193 6 0.858 4
    0.1 lnc=0.057 5 t+3.642 4 0.944 3 $ \dfrac{1}{c} $=-0.000 4 t+0.018 8 0.897 8 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-59.988\; 6}{t} $+2.193 6 0.858 4
    0.15 lnc=0.053 7 t+3.627 4 0.968 1 $ \dfrac{1}{c} $=-0.000 4 t+0.018 9 0.921 4 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-129.025\; 3}{t} $+4.181 9 0.932 3
    0.2 lnc=0.044 9 t+3.660 1 0.976 6 $ \dfrac{1}{c} $=-0.000 3 t+0.019 4 0.959 8 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-127.943\; 5}{t} $+3.915 6 0.935 5
    絮凝剂
    用量/(g/t)
    20 lnc=0.003 10 t+3.611 9 0.823 1 $ \dfrac{1}{c} $=-0.000 3 t+0.021 6 0.956 9 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-130.129\; 4}{t} $+1.715 9 0.997 3
    40 lnc=0.059 1 t+3.930 4 0.876 6 $ \dfrac{1}{c} $=-0.000 3 t+0.016 5 0.780 7 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-61.922\; 0}{t} $+2.000 2 0.934 3
    60 lnc=0.054 8 t+4.417 1 0.732 0 $ \dfrac{1}{c} $=-0.000 3 t+0.013 8 0.607 6 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-59.988\; 6}{t} $+2.193 6 0.858 4
    80 lnc=0.050 1 t+4.730 8 0.663 2 $ \dfrac{1}{c} $=-0.000 3 t+0.011 6 0.533 5 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-13.749\; 5}{t} $+0.430 3 0.9224
    温度/℃ 10 lnc=0.0555 t+4.296 7 0.775 8 $ \dfrac{1}{c} $=-0.000 3 t+0.014 0 0.668 4 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-22.273\; 0}{t} $+0.630 4 0.970 6
    20 lnc=0.054 8 t+4.417 1 0.732 0 $ \dfrac{1}{c} $=-0.000 3 t+0.013 8 0.607 6 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-59.988\; 6}{t} $+2.193 6 0.858 4
    30 lnc=0.055 5 t+4.416 3 0.733 8 $ \dfrac{1}{c} $=-0.000 3 t+0.013 8 0.608 2 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-60.001\; 0}{t} $+2.195 5 0.858 5
    40 lnc=0.054 7 t+4.469 0 0.720 2 $ \dfrac{1}{c} $=-0.000 3 t+0.013 4 0.598 0 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-38.699\; 8}{t} $+1.361 9 0.885 9
    50 lnc=0.053 2 t+4.614 4 0.716 8 $ \dfrac{1}{c} $=-0.000 3 t+0.012 2 0.554 7 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-21.068\; 9}{t} $+0.713 9 0.892 6
    下载: 导出CSV 
    | 显示表格

    表3可知在赤铁矿精矿絮凝沉降前30 s过程中,双曲线动力学模型方程的拟合相关系数R2比二级动力学模型方程大,较于一级动力学模型方程相差不大,而且各沉降过程的R2基本都大于0.9;由表4可知在赤铁矿精矿絮凝沉降前60 s过程中,双曲线动力学模型方程的拟合相关系数R2较于一级动力学模型方程和二级动力学模型方程,其拟合相关系数R2都大于0.85且远大于其他两种模型。在两个时间段进行拟合时,双曲线动力学模型方程相关性比其他两种动力学模型方程更好,即双曲线动力学模型方程取得了比一级动力学模型方程和二级动力学模型方程更好的拟合度,因此APAM对细粒赤铁矿精矿絮凝沉降可以优先使用双曲线动力学模型来描述。

    沉降速度是衡量赤铁矿絮凝沉降的重要参数,沉降速度的计量表准有三种:中值、时间加权平均和重量加权平均,本文采用时间加权平均沉降速度v来代表赤铁矿的沉降速度,v可表示为:

    $$ \mathit{v=h/t} $$ (2)

    式中h为沉降高度,mm,t为该高度所对应的时间,s,本文统一取澄清层从250 mL处到沉降筒20 mL处来计算平均沉降速度。进一步应用Stokes公式可以求得与沉降速度v所对应的平均粒径dd可表示为:

    $$ d= \frac{\mu v}{2kg(\rho -{\rho }_{0})} $$ (3)

    式中,h为取样点的深度,mm;μ为水介质黏度,Pa·s;k为形状系数;g为重力加速度,m/s;ρ为铁精矿颗粒比重,kg/m3ρ0为水介质比重,kg/m3[14]

    以不同絮凝用量为例,探究不同絮凝剂用量的沉降速度和絮团粒径的变化,见图6

    图  6  沉降速度与絮团粒径的变化趋势
    Figure  6.  Variation trend of settlement velocity and particle size of floc

    图6可知,絮凝剂用量从20 g/t增加到80 g/t,v从3.04 mm/s增加到12.47 mm/s,增加了9.43 mm/s,d从0.344 mm增加到1. 413 mm,增加了1.069 mm。随着絮凝剂用量的增加,沉降速度和絮团粒径增加,但增加速度逐渐变缓。

    (1)絮凝剂的添加对赤铁矿精矿的沉降效果有很明显的改善,综合实验结果和经济价值考虑,APAM的使用选择分子量为900万、配置浓度为0.05%和用量为60 g/t。在此APAM使用条件和室温20 ℃条件下,会得到快的沉降速度和低的最终底流浓度,极大地改善了沉降效果,有利于后续过滤脱水作业。

    (2)通过APAM对细粒赤铁矿精矿絮凝沉降的动力学分析,APAM对细粒赤铁矿精矿絮凝沉降动力学可以优先使用双曲线动力学模型1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{k}{t}+b $来描述。

    (3)絮凝剂用量较少时,沉降速度慢、絮团粒径小,絮凝剂用量的增加能够促进絮团的增长,加快沉降速度,改善沉降性能。

  • 图  1   絮凝剂分子量大小对底流悬浊液浓度的影响

    Figure  1.   Effect of molecular weight of flocculants on the concentration of underflow suspension

    图  2   絮凝剂配制浓度对底流悬浊液浓度的影响

    Figure  2.   Effect of concentration of flocculants solution on the concentration of underflow suspension

    图  3   絮凝剂用量对底流悬浊液浓度的影响

    Figure  3.   Effect of dosage of flocculants on the concentration of underflow suspension

    图  4   温度对底流悬浊液浓度的影响

    Figure  4.   Effect of pulp temperature on the concentration of underflow suspension

    图  5   不同絮凝剂用量沉降曲线

    Figure  5.   Settling curves for different dosage of flocculants

    图  6   沉降速度与絮团粒径的变化趋势

    Figure  6.   Variation trend of settlement velocity and particle size of floc

    表  1   试样多元素分析/%

    Table  1   Chemical multi-element analysis of samples

    TFeSiO2Al2O3MnPSZnCaO
    64.508.4101.5690.1500.0420.0210.0040.012
    下载: 导出CSV

    表  2   试样粒度分析

    Table  2   Particle size analysis of samples

    粒级/μm产率/%累积产率/%
    +1500.000.00
    -150+7411.0011.00
    -74+4515.6026.60
    -45+384.2530.85
    -3869.15100.00
    下载: 导出CSV

    表  3   沉降时间30 s动力学模型拟合参数

    Table  3   Settlement time of 30 s of the parameters for dynamic model

    条件 一级动力学模型方程lnc=-kt+lnc0 二级动力学模型方程$ \dfrac{{1}}{{c}}={k}{t}+\dfrac{{1}}{{{c}}_{{0}}} $ 双曲线动力学模型方程1/(1-$ \dfrac{{c}}{{{c}}_{{0}}} $)=$ \dfrac{{k}}{{t}} $+b
    动力学方程 R2 动力学方程 R2 动力学方程 R2
    0 lnc=0.004 1 t+3.899 2 0.944 6 $ \dfrac{1}{c} $=-0.000 1 t+0.020 2 0.952 6 1/(1-$ \dfrac{{c}}{{{c}}_{{0}}} $)=$ \dfrac{-706.860\; {0}}{{t}} $+19.143 6 0.996 0
    絮凝剂
    分子
    量/万
    500 lnc=0.086 2 t+3.566 0 0.929 3 $ \dfrac{1}{{c}} $=-0.000 7 t+0.020 8 0.980 6 1/(1-$ \dfrac{{c}}{{{c}}_{{0}}} $)=$ \dfrac{{-144.402}\;{0}}{{t}} $+7.134 2 0.906 5
    900 lnc=0.112 7 t+3.672 4 0.937 1 $ \dfrac{1}{{c}} $=-0.000 7 t+0.019 0 0.894 1 1/(1-$ \dfrac{{c}}{{{c}}_{{0}}} $)=$ \dfrac{-69.573\;7}{{t}} $+3.484 1 0.908 7
    1 300 lnc=0.101 5 t+4.147 2 0.849 5 $ \dfrac{1}{{c}} $=-0.000 6 t+0.015 5 0.761 0 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-12.656}{{t}} $+0.564 4 0.938 8
    絮凝剂
    浓度/%
    0.05 lnc=0.112 7 t+3.762 4 0.937 1 $ \dfrac{1}{{c}} $=-0.000 7 t+0.019 0 0.894 1 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-69.573\;7}{{t}} $+3.484 1 0.908 7
    0.1 lnc=0.048 0 t+3.720 8 0.933 3 $ \dfrac{1}{{c}} $=-0.000 3 t+0.021 3 0.983 5 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-144.171\;5}{{t}} $+6.217 9 0.961 7
    0.15 lnc=0.043 2 t+3.755 9 0.951 2 $ \dfrac{1}{{c}} $=-0.000 5 t+0.021 1 0.985 9 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-144.402\;0}{{t}} $+6.155 6 0.960 5
    0.2 lnc=0.034 2 t+3.800 5 0.957 2 $ \dfrac{1}{{c}} $=-0.000 4 t+0.020 9 0.990 2 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-141.786\;5}{{t}} $+5.778 1 0.959 0
    絮凝剂
    用量/
    (g/t)
    20 lnc=0.010 3 t+3.897 3 0.970 6 $ \dfrac{1}{{c}} $=-0.000 2 t+0.020 1 0.984 6 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-126.464\;9}{{t}} $+1.218 8 0.999 2
    40 lnc=0.087 5 t+3.542 9 0.905 3 $ \dfrac{1}{{c}} $=-0.000 7 t+0.020 7 0.990 7 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-69.375\;0}{{t}} $+3.002 2 0.966 2
    60 lnc=0.112 7 t+3.762 4 0.937 1 $ \dfrac{1}{{c}} $=-0.000 7 t+0.019 0 0.894 1 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-69.573\;7}{{t}} $+3.484 1 0.908 7
    80 lnc=0.112 6 t+3.934 7 0.916 1 $ \dfrac{1}{{c}} $=-0.000 6 t+0.016 5 0.830 9 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-15.643\;0}{{t}} $+0.685 1 0.965 0
    温度/℃ 10 lnc=0.108 3 t+3.586 7 0.940 1 $ \dfrac{1}{{c}} $=-0.000 7 t+0.019 0 0.935 4 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-69.647\;0}{{t}} $+3.364 1 0.928 7
    20 lnc=0.112 7 t+3.762 4 0.937 1 $ \dfrac{1}{{c}} $=-0.000 7 t+0.019 0 0.894 1 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-69.573\;7}{{t}} $+3.484 1 0.908 7
    30 lnc=0.113 6 t+3.668 1 0.936 4 $ \dfrac{1}{{c}} $=-0.000 7 t+0.018 9 0.890 6 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-69.506\;0}{{t}} $+3.493 1 0.905 1
    40 lnc=0.114 3 t+3.703 3 0.935 4 $ \dfrac{1}{{c}} $=-0.000 7 t+0.018 4 0.882 4 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-44.401\;0}{{t}} $+2.163 9 0.923 7
    50 lnc=0.113 0 t+3.849 7 0.944 4 $ \dfrac{1}{{c}} $=-0.000 7 t+0.017 2 0.846 3 1/(1-$ \dfrac{{c}}{{{c}}_{0}} $)=$ \dfrac{-24.109\;6}{{t}} $+1.134 1 0.938 6
    下载: 导出CSV

    表  4   沉降时间60 s动力学模型拟合参数

    Table  4   Settlement time of 60 s of the parameters for dynamic model

    条件 一级动力学模型方程lnc=-kt+lnc0 二级动力学模型方程$ \dfrac{{1}}{{c}}{=kt+}\dfrac{{1}}{{{c}}_{{0}}} $ 双曲线动力学模型方程1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{k}{t} $+b
    动力学方程 R2 动力学方程 R2 动力学方程 R2
    0 lnc=0.005 4 t+3.882 1 0.966 3 $ \dfrac{1}{c} $=-0.000 1 t+0.020 4 0.979 2 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-667.834 \;9}{t} $+13.900 1 0.987 7
    絮凝剂
    分子
    量/万
    500 lnc=0.056 8 t+3.969 4 0.884 5 $ \dfrac{1}{c} $=-0.000 3 t+0.016 2 0.770 5 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-61.288\; 0}{t} $+2.018 1 0.918 4
    900 lnc=0.054 8 t+4.417 1 0.732 0 $ \dfrac{1}{c} $=-0.000 3 t+0.013 8 0.607 6 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-59.988\; 6}{t} $+2.193 6 0.858 4
    1 300 lnc=0.043 7 t+4.881 6 0.604 4 $ \dfrac{1}{c} $=-0.000 2 t+0.010 8 0.478 1 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-10.993 \;1}{t} $+0.340 5 0.886 3
    絮凝剂
    浓度/%
    0.05 lnc=0.054 8 t+4.417 1 0.732 0 $ \dfrac{1}{c} $=-0.000 3 t+0.013 8 0.607 6 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-59.988 \;6}{t} $+2.193 6 0.858 4
    0.1 lnc=0.057 5 t+3.642 4 0.944 3 $ \dfrac{1}{c} $=-0.000 4 t+0.018 8 0.897 8 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-59.988\; 6}{t} $+2.193 6 0.858 4
    0.15 lnc=0.053 7 t+3.627 4 0.968 1 $ \dfrac{1}{c} $=-0.000 4 t+0.018 9 0.921 4 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-129.025\; 3}{t} $+4.181 9 0.932 3
    0.2 lnc=0.044 9 t+3.660 1 0.976 6 $ \dfrac{1}{c} $=-0.000 3 t+0.019 4 0.959 8 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-127.943\; 5}{t} $+3.915 6 0.935 5
    絮凝剂
    用量/(g/t)
    20 lnc=0.003 10 t+3.611 9 0.823 1 $ \dfrac{1}{c} $=-0.000 3 t+0.021 6 0.956 9 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-130.129\; 4}{t} $+1.715 9 0.997 3
    40 lnc=0.059 1 t+3.930 4 0.876 6 $ \dfrac{1}{c} $=-0.000 3 t+0.016 5 0.780 7 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-61.922\; 0}{t} $+2.000 2 0.934 3
    60 lnc=0.054 8 t+4.417 1 0.732 0 $ \dfrac{1}{c} $=-0.000 3 t+0.013 8 0.607 6 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-59.988\; 6}{t} $+2.193 6 0.858 4
    80 lnc=0.050 1 t+4.730 8 0.663 2 $ \dfrac{1}{c} $=-0.000 3 t+0.011 6 0.533 5 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-13.749\; 5}{t} $+0.430 3 0.9224
    温度/℃ 10 lnc=0.0555 t+4.296 7 0.775 8 $ \dfrac{1}{c} $=-0.000 3 t+0.014 0 0.668 4 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-22.273\; 0}{t} $+0.630 4 0.970 6
    20 lnc=0.054 8 t+4.417 1 0.732 0 $ \dfrac{1}{c} $=-0.000 3 t+0.013 8 0.607 6 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-59.988\; 6}{t} $+2.193 6 0.858 4
    30 lnc=0.055 5 t+4.416 3 0.733 8 $ \dfrac{1}{c} $=-0.000 3 t+0.013 8 0.608 2 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-60.001\; 0}{t} $+2.195 5 0.858 5
    40 lnc=0.054 7 t+4.469 0 0.720 2 $ \dfrac{1}{c} $=-0.000 3 t+0.013 4 0.598 0 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-38.699\; 8}{t} $+1.361 9 0.885 9
    50 lnc=0.053 2 t+4.614 4 0.716 8 $ \dfrac{1}{c} $=-0.000 3 t+0.012 2 0.554 7 1/(1-$ \dfrac{c}{{c}_{0}} $)=$ \dfrac{-21.068\; 9}{t} $+0.713 9 0.892 6
    下载: 导出CSV
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  • 收稿日期:  2024-03-23
  • 刊出日期:  2024-12-24

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