纤维状颗粒的空气动力学直径误差传递公式

基于Prodi et al (1982)的纤维颗粒空气动力学等效直径计算公式推导出以纤维颗粒的长度、直径为直接测量变量的等效直径误差传递公式。

Prodi et al¹给出了纤维状颗粒的空气动力学等效直径计算公式： $$ D=\frac{3W}{2}\sqrt{\frac{\rho}{\frac{0.385}{\ln{2\beta}-0.5}+\frac{1.23}{\ln{2\beta}+0.5}}} $$ 其中$W$是纤维的直径，$\beta$是基于投影尺寸的纵横比（长/宽，$L/W$）²。在做文献调研时，需要计算一些已发表实验中采用纤维颗粒的等效直径及误差，但是论文中通常只会给出纤维颗粒的直径和长度测量数据及误差，即$W±\sigma_W,L±\sigma_L$。这时虽然可以很方便地计算出$D=f(W,L)$，却不能直接知道$D$的误差是多少。本文给出从长度、直径误差计算出等效直径误差的推导过程和误差传递公式。

以偏导的形式表达等效直径误差， $$ \sigma_D^2=\sigma_W^2\left(\frac{\partial D}{\partial W}\right)^2_{\overline W}+\sigma_L^2\left(\frac{\partial D}{\partial L}\right)^2_{\overline L} $$ 接下来开始对两个偏导项逐个击破。

对$W$的偏导

$$ \frac{\partial D}{\partial W}=\frac32\cdot\sqrt{\frac{\rho}{\frac{0.385}{\ln{2\beta}-0.5}+\frac{1.23}{\ln{2\beta}+0.5}}} +\frac{3W}{2}\cdot\frac{\partial g(W,L)}{\partial W} $$

$$ \begin{aligned} {\rm where~~} g(W,L)&=\left[ \frac1\rho \left( \frac{0.385}{\ln(L/W)+\ln2-0.5}+\frac{1.23}{\ln(L/W)+\ln2+0.5} \right) \right]^{-\frac12}\\ &=\sqrt\rho\cdot h(W,L)^{-\frac 12}\\ {\rm where~~}h(W,L)&= \frac{0.385}{\ln(L/W)+\ln2-0.5}+\frac{1.23}{\ln(L/W)+\ln2+0.5} \end{aligned} $$

$$ \therefore~ \frac{\partial g(W,L)}{\partial W}=-\frac{\sqrt\rho}2\cdot h(W,L)^{-\frac32}\cdot\frac{\partial h(W,L)}{\partial W} $$

$$ \begin{aligned} \frac{\partial h(W,L)}{\partial W}&=\frac{\partial}{\partial W}\left[ \frac{0.385}{\ln(L/W)+\ln 2-0.5}+\frac{1.23}{\ln(L/W)+\ln2+0.5} \right]\\ &=\frac{0.385}{\left[\ln(L/W)+\ln 2-0.5\right]^2}\cdot\frac1W +\frac{1.23}{[\ln(L/W)+\ln2+0.5]^2}\cdot\frac1W\\ &=\frac{0.385}{\left[\ln(2\beta)-0.5\right]^2\cdot W} +\frac{1.23}{[\ln(2\beta)+0.5]^2\cdot W} \end{aligned} $$

$$ \begin{aligned} \therefore~\frac{\partial g(W,L)}{\partial W}&=-\frac{\sqrt\rho}{2W}\cdot \left( \frac{0.385}{\ln(2\beta)-0.5}+\frac{1.23}{\ln(2\beta)+0.5} \right)^{-\frac32}\cdot\left( \frac{0.385}{\left[\ln(2\beta)-0.5\right]^2\cdot W} +\frac{1.23}{[\ln(2\beta)+0.5]^2\cdot W} \right)\\ &=-\frac{1}{2W}\sqrt{ \frac\rho {\left[ \frac{0.385}{\ln(2\beta)-0.5}+\frac{1.23}{\ln(2\beta)+0.5} \right]^3} } \times\left( \frac{0.385}{\left[\ln(2\beta)-0.5\right]^2} +\frac{1.23}{[\ln(2\beta)+0.5]^2} \right) \end{aligned} $$

$$ \begin{aligned} \therefore~\frac{\partial D}{\partial W}&=\frac32\cdot\sqrt{\frac{\rho}{\frac{0.385}{\ln{2\beta}-0.5}+\frac{1.23}{\ln{2\beta}+0.5}}} +\frac{3W}{2}\cdot\frac{\partial g(W,L)}{\partial W}\\ &=\frac32\cdot\sqrt{\frac{\rho}{\frac{0.385}{\ln{2\beta}-0.5}+\frac{1.23}{\ln{2\beta}+0.5}}}-\frac34\cdot\sqrt{ \frac\rho {\left[ \frac{0.385}{\ln(2\beta)-0.5}+\frac{1.23}{\ln(2\beta)+0.5} \right]^3} } \times\left( \frac{0.385}{\left[\ln(2\beta)-0.5\right]^2} +\frac{1.23}{[\ln(2\beta)+0.5]^2} \right)\\ &=\frac32\sqrt{\frac{\rho}{\frac{0.385}{\ln{2\beta}-0.5}+\frac{1.23}{\ln{2\beta}+0.5}}} \left( 1-\frac12\frac{ \frac{0.385}{\left[\ln(2\beta)-0.5\right]^2} +\frac{1.23}{[\ln(2\beta)+0.5]^2} } {\frac{0.385}{\ln(2\beta)-0.5}+\frac{1.23}{\ln(2\beta)+0.5}} \right) \end{aligned} $$

对$L$的偏导

$$ \begin{aligned} \frac{\partial D}{\partial L}&=\frac{3W}{2}\times\frac{\partial g(W,L)}{\partial L}\\ \frac{\partial g(W,L)}{\partial L}&=-\frac{\sqrt \rho}{2}h(W,L)^{-\frac32}\frac{\partial h(W,L)}{\partial L} \end{aligned} $$

由于$W$和$-L$在函数$h(W,L)$里的地位是相同的，所以根据前一节对$h(W,L)$求偏导的计算可以很快得到

$$ \begin{aligned} \frac{\partial h(W,L)}{\partial L}&=-\frac{\partial h(W,L)}{\partial W}\\ &=-\frac{0.385}{[\ln (2\beta)-0.5]^2\cdot L}-\frac{1.23}{[\ln(2\beta)+0.5]^2\cdot L} \end{aligned} $$

$$ \begin{aligned} \therefore~\frac{\partial D}{\partial L}&=\frac{3}{4\beta} \sqrt{ \frac{\rho}{\frac{0.385}{\ln2\beta-0.5}+\frac{1.23}{\ln2\beta+0.5}} } \left(\frac{\frac{0.385}{[\ln(2\beta)-0.5]^2}+\frac{1.23}{[\ln(2\beta)+0.5]^2}}{\frac{0.385}{\ln2\beta-0.5}+\frac{1.23}{\ln2\beta+0.5}}\right) \end{aligned} $$