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习题 6.3

题目

某二元精馏塔分离苯-甲苯混合物。已知:F=100F = 100 kmol/h,zF=0.5z_F = 0.5xD=0.95x_D = 0.95xW=0.05x_W = 0.05α=2.5\alpha = 2.5,进料为泡点液体(q = 1)。

若操作回流比 R=1.2R = 1.2,求实际理论板数(用 Gilliland 关联式估算)。


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计算过程

1. Fenske 方程(最小理论板数)

Nmin=log(0.950.050.950.05)log2.5=log361log2.5=2.5580.398=6.43N_{\min} = \frac{\log\left(\frac{0.95}{0.05} \cdot \frac{0.95}{0.05}\right)}{\log 2.5} = \frac{\log 361}{\log 2.5} = \frac{2.558}{0.398} = 6.43

2. Underwood 方程(最小回流比)

2.5×0.52.5θ+1×0.51θ=1\frac{2.5 \times 0.5}{2.5 - \theta} + \frac{1 \times 0.5}{1 - \theta} = 1

解得 θ=1.75\theta = 1.75

Rmin=2.5×0.951.751=0.357R_{\min} = \frac{2.5 \times 0.95}{1.75} - 1 = 0.357

3. Gilliland 关联式

X=RRminR+1=1.20.3571.2+1=0.8432.2=0.383X = \frac{R - R_{\min}}{R + 1} = \frac{1.2 - 0.357}{1.2 + 1} = \frac{0.843}{2.2} = 0.383

Y=0.5660.566(0.3830.5660.383+0.566)0.6=0.5660.566×(0.193)0.6Y = 0.566 - 0.566\left(\frac{0.383 - 0.566}{0.383 + 0.566}\right)^{0.6} = 0.566 - 0.566 \times (-0.193)^{0.6}

Y=0.566+0.566×0.1930.6=0.566+0.566×0.381=0.782Y = 0.566 + 0.566 \times 0.193^{0.6} = 0.566 + 0.566 \times 0.381 = 0.782

N=Y+Nmin1Y=0.782+6.4310.782=7.2120.218=33.1N = \frac{Y + N_{\min}}{1 - Y} = \frac{0.782 + 6.43}{1 - 0.782} = \frac{7.212}{0.218} = 33.1

答案

实际理论板数 NN \approx 33 块(含塔釜)。


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