Numerical Simulation of Titanium Slag Flow Fluid-bed Chlorination in Spout Fluidized Bed
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摘要:
基于双流体模型和Gidaspow方程建立了预测钛渣喷动床沸腾氯化的数学模型。首先，采用经验计算公式得到了钛渣最小喷吹速度和最大喷吹高度，并对最小吹动速度进行了修正；其次，最小喷吹速度和相关边界条件为基础，运用Fluent软件模拟了钛渣喷动床沸腾氯化床内气固两相流的流体动力学特征；最后，分析了不同喷吹速度下气固两相流的流动特征和分区特征。结果表明：Gisher && Mathur公式更适合于钛渣喷动床沸腾氯化的最小喷动速度的预测；修正后的最小喷吹速度能够准确的模拟出钛渣喷动床沸腾氯化气固两相流的流体动力学特征；数值模拟得到的床内流动特征、分区的云图和矢量图、固体颗粒轨迹图和最大喷吹高度与相关研究结果一致。
Abstract:Based on Eulerian two-fluid model and Gidaspow model, a mathematic model for predicting the fluidized chlorination of titanium slag spouting fluidized bed was established. The minimum spouting speed and maximum spouting height of titanium slag spouted bed were obtained by empirical formula, meanwhile the minimum spouting speed was corrected. Then, Based on the minimum spouting speed and the related boundary conditions, hydrodynamic characteristics of gas-solid two-phase flow in titanium slag spouting fluidized bed was simulated by fluent software. At last, the flow characteristics and zoning characteristics of titanium slag spouting fluidized bed at different injection rates were analyzed. The results show that the Gisher && Mathur formula is more suitable for the prediction of the minimum spouting speed in titanium slag spouting fluidized bed. The modified minimum spouting speed can accurately simulate the hydrodynamic characteristics of titanium slag spouted bed. The numerical simulation of the hydrodynamic characteristics, the partition cloud and vector, the solid particle trajectory and the maximum spouting height were consistent with the relevant research results.
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Key words:
- titanium slag /
- spouted bed /
- fluid-bed chlorination /
- numerical simulation
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表 1 模拟计算的边界条件
Table 1. Boundary conditions for simulation calculation
边界条件 值 钛渣密度/(kg·m-3) 4 800 钛渣直径/m 1×10-3 钛渣初始体积分数 0.32 氯气的密度/(g·L-1) 3.17 初始静床高/m 0.24 网格数 19 200 曳力模型 Gidaspow 湍流模型 RNGk-ε 求解算法 Simple 内摩擦角 30° 恢复系数 0.95 时间步长 10-5 
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表 2 最小喷吹速度
Table 2. Minimum spouting velocity
研究者 经验计算公式 计算结果/(m·s-1) Gisher && Mathur[16] ${\frac{{{U_{ms}}}}{{\sqrt {2gH} }} = \left( {\frac{{{d_p}}}{{{D_c}}}} \right){{\left( {\frac{{{D_i}}}{{{D_c}}}} \right)}^{1/3}}{{\left( {\frac{{{\rho _p} - {\rho _f}}}{{{\rho _f}}}} \right)}^{1/2}}} $ 8.89 King && Harrison[17] $ {\frac{{{U_{ms}}}}{{\sqrt {2gH} }} = \left( {\frac{{{\rho _p}}}{{{\rho _p}}}} \right)\frac{{{d_p}}}{{{D_c}}}{{\left( {\frac{{{D_i}}}{{{D_c}}}} \right)}^{1/3}}{{\left( {\frac{{{\rho _p} - {\rho _f}}}{{{\rho _f}}}} \right)}^{1/2}}}$ 2.14 Fane && Mitchell[18] $ {\frac{{{U_{ms}}}}{{\sqrt {2gH} }} = \left( {0.2D_c^*} \right)\frac{{{d_p}}}{{{D_c}}}{{\left( {\frac{{{D_i}}}{{{D_c}}}} \right)}^{1/3}}{{\left( {\frac{{{\rho _p} - {\rho _f}}}{{{\rho _f}}}} \right)}^{1/2}}}$ 0.42 Wen-Yu[10] ${\frac{{{d_p}{u_{mf}}{\rho _g}}}{{{\mu _g}}} = {\rm{ }}{{\left[ {C^2_1 + {C_2}\frac{{d^3_p{\rho _g}\left( {{\rho _p} - {\rho _g}} \right)g}}{{{\mu ^2}}}} \right]}^{1/2}}} $ 7.86
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表 3 最大喷动高度
Table 3. Maximum spoutable bed height
研究者 经验计算公式 计算结果/m Mcnab&&Bridgwaterti[19] ${{H_m} = \frac{{D^2_c}}{{{d_P}}}{{\left( {\frac{{{D_c}}}{{{D_i}}}} \right)}^{2/3}}\frac{{568{b^2}}}{{Ar}}{{\left( {\sqrt {1 + 35.9 \times {{10}^{ - 6}}Ar} - 1} \right)}^2}} $ 0.45 Littman[20] $ {\frac{{{H_m}{d_p}}}{{{D^2} - d^2_p}} = 0.345{{\left( {\frac{{{d_p}}}{D}} \right)}^{ - 0.384}}}$ 0.373 Cecen[21] ${\frac{{{H_m}{d_p}}}{{{D^2}}} = 0.99{{\left( {\frac{{{D_i}}}{D}} \right)}^{0.856}}{{\left( {\frac{{{\rho _p} - \rho }}{\rho }\frac{{g{D_i}}}{{{u_{mf}}{u_t}}}} \right)}^{0.144}}} $ 0.4
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