$7$:$19$ PM Wed $28$ Feb

From the error function values for Ficker's second law table, select the error function for the argument value of $0\mathrm{.}128$ as $0\mathrm{.}144$

$\frac{1-c_x}{0\mathrm{.}9}=0\mathrm{.}144$

$c_x=1-0\mathrm{.}1296$

$c_x=0\mathrm{.}87\%C$

$\frac{?}{\text{b}\text{a}\text{r}}(11)$

I couldn't find

from it

Hence, the diffusing concentration of carbon content $c_x$ at a distance of $0\mathrm{.}01cm$ is $0\mathrm{.}87\%C.$

Calculate the diffusing concentration of carbon content $($c$)$ at a distance of $0\mathrm{.}05cm.$

Substitute ${4\mathrm{.}210}^{-7}\frac{c{m}^2}{(s)}$ for D$,1\%$ C for C$,1$hr for $t,0\mathrm{.}10\%$ C for Co$,$ and $0\mathrm{.}01cm$ for $x$ in Equation $(1).$

$\frac{1\%-c_x}{1\%-0\mathrm{.}10\%}=$eff$(\frac{0\mathrm{.}05((cm))}{2\sqrt[\phantom{2}]{(42{10}^{-7}c\frac{m^2}{s})(1hr)(\frac{3,600((s))}{1hr})}})$

$\frac{1-c_x}{0\mathrm{.}9}=$erf$(0\mathrm{.}643)$

From the error function values for Flicker's second law table, select the error function for the argument value of $0\mathrm{.}643$ as $0\mathrm{.}63.$

$]=[0\mathrm{.}63$

Hence, the diffusing concentration of carbon content $c_x$ at distance of $0\mathrm{.}05cm$ is $0\mathrm{.}43\%C$

What would you like to do next?

Send to an expert $\frac{20}{20}$

You have $20$ expert questions left. Renews on Mar $18$