Plases use MATLAB A $3{}_{1}kW$ electric mixer with a $0\mathrm{.}2{}_{2}m$ diameter impeller is used to mix $1000kg$ liquid mixture. The

change in turbulent kinetic energy $($TKE$)$ per unit mass can be expressed as

$d\frac{3\frac{u^2}{2}}{d}t=$ TKE production rate $/$ mass $-$ TKE dissipation rate $/$ mass.

If all energy input to the electric mixer goes into mixing, we have

TKE production rate $=3{}_{1}kW=3{}_1,000W.$

The dissipation or decay rate of TKE may be estimated as TKE dissipation rate ~~$\frac{u^3}{L}.$ The size of the

mixing eddies may be assumed to scale with the impeller diameter. Doing so$,$ we get TKE dissipation

rate ~~$\frac{u^3}{0\mathrm{.}2}{}_2.$ Therefore, for $1000kg$ of liquid, the energy balance is

Differentiating $3\frac{u^2}{2}$ gives

$d\frac{3\frac{u^2}{2}}{d}t=0\mathrm{.}03{}_1-\frac{u^3}{2{}_{2}0}.$

This can be rearranged into

$3ud\frac{u}{d}t=0\mathrm{.}03{}_1-\frac{u^3}{2{}_{2}0}.$

$\{3\frac{u}{0\mathrm{.}03{}_1-\frac{u^3}{2{}_{2}0}}\}du=dt.$

Estimate the time it takes to reach $7a_3\%$ of the maximum mixing intensity.

Here, ${}_1,{}_2,{}_3$ are the smallest, second smallest, and third smallest non$-$zero digits, respectively, of

your student ID$.$ E$.$g$.,$ ID $=110985247$ gives ${}_1=1,{}_2=2,$ and ${}_3=4,$ i$.$e$.,\{3\frac{u}{0\mathrm{.}031-\frac{u^3}{220}}\}du$

$=dt$ can Places use MATLAB, also the answer is supposed to be $1\mathrm{.}3$can