SOLUTION AT BOTTOM!!! Can someone explain to me how they got B$=800$ kg$/$hr in the yellow highlighted portion?? Please for the love of God do NOT copy another solution on here. They're all wrong.

D$3.*$ A distillation column separating ethanol from water is shown. Pressure is $1k\frac{g}{c}{m}^2.$ Instead of

having a condenser, a stream of pure liquid ethanol is added directly to the column to serve as the

reflux. This stream is a saturated liquid. The feed is $40wt\%$ ethanol and is at $-20C.$ Feed flow

rate is $2000k\frac{g}{h}.$ We desire a distillate concentration of $80wt\%$ ethanol and a bottoms

composition of $5$ wt $\%$ ethanol. A total reboiler is used, and the boilup is a saturated vapor. The

cooling stream is input at $C=1000k\frac{g}{h}.$ Find the external boilup rate, $\frac{?}{\text{b}\text{a}\text{r}}(V).$ Note: Set up the

equations, solve in equation form for $\frac{?}{\text{b}\text{a}\text{r}}(V)$ including explicit equations for all required terms, read

off all required enthalpies from the enthalpy composition diagram $($Figure $2-4),$ and then calculate

a numerical answer.

$3.$D$3.$ External balances: $F+C=B+D$

For a total reboiler: $x_N=x_B,y_N=x_N=x_B,h_N=h_B=92$

Solve Eqs. $(1)$ and $(2)$ for B: $B=\frac{Fz+C{x}_C-F{y}_D-C{y}_D}{x_B-y_D}$

Thus

$B=\frac{800+1000-1600-800}{(\mathrm{.}05-\mathrm{.}8)}=800$ and $D=2200$

From Eq$.(3),,Q_R=B{h}_B+D{H}_D-F{h}_F-C{h}_C$

$Q_R=(800)(92)+(2200)(327)-(2000)(-30\mathrm{.}75)-(1000)(50)=804,500ca\frac{l}{h}\frac{?}{b}$

$ar(V)=\frac{Q_R}{H_{\text{r}\text{e}\text{b}}-h_N}=\frac{804500}{617-92}=1532\mathrm{.}4k\frac{g}{h}$