Solve the following ODE $$\frac{dy}{dt} = ye^{-t}$$ to $t=6$. Plot the resulting function $y(t)$. The initial condition is $y_0=1$.
Solve the following rate equation for a coal particle in a furnace. $$ \frac{dT}{dt} = \left(\frac{hA}{m c_p}(T_f - T) + \frac{\sigma A}{m c_p}(T_f^4-T^4)\right).$$
The initial particle temperature is $T_0=500$ K.
Use D=100 $\mu$m, tend = 0.05 s.
The following data is given:
Variable | Value | Units |
---|---|---|
$\rho_p$ | 1000 | kg/m$^3$ |
$c_p$ | 1380 | J/kg$\cdot$K |
$k$ | 0.1 | W/m$\cdot$K |
$Nu$ | 2 | -- |
$\sigma$ | 5.67E-8 | W/m$^2$K$^4$ |
$T_f$ | 1500 | K |
Also, the area, mass and Nusselt number $Nu$ are given, respectively, by: $$ A = \pi D^2, $$ $$ m = \frac{\pi}{6}D^3\rho_p,$$ $$ Nu = \frac{hD}{k}.$$
Plot the particle temperature as a function of time. Label and format your plot (include units).
We are performing a chemical reaction as follows.
$$\mbox{Rxn 1: }\phantom{xxx} A+B\rightarrow C $$$$\mbox{Rxn 2: }\phantom{xxx} B+C\rightarrow D $$Here, symbols $A$, $B$, $C$, $D$ denote species concentrations in mol/L. The initial concentrations are $A_0=1$, $B_0=1$, $C_0=0$, $D_0=0$. Also, $k_1=1\,L/mol*s$, and $k_2=1.5\,L/mol*s$. The concentrations obey the following rate equations: $$ \frac{dA}{dt} = -k_1AB,$$ $$ \frac{dB}{dt} = -k_1AB - k_2BC,$$ $$ \frac{dC}{dt} = k_1AB - k_2BC,$$ $$ \frac{dD}{dt} = k_2BC.$$
Solve for the concentrations of $A$, $B$, $C$, and $D$ as functions of time to $t=3$ s.
Also, once the species concentrations are found, compute the selectivity defined as $S=C/(C+D)$ as a function of time. (S is initially undefined (since C and D are zero), but you can set it equal to 1 at t=0 since C forms before D.
Plot the concentrations of A, B, C, and D, and also selectivity S, as functions of time on the same plot. Label the axes as "time (s)" and “concentration (mol/L)”. Include a legend so we can see what the curves correspond to.
What can you do to maximize your selectivity? Consider $k_1$, $k_2$ and initial concentrations.