$8\mathrm{.}4-3.$ Effect of Evaporator Pressure on Capacity and Product Composition. Recalculate Example $8\mathrm{.}4-1$ but use an evaporator pressure of $41\mathrm{.}4$kPa instead of $101\mathrm{.}32$kPa abs$.$ Use the same steam pressure, area $A,$ and heat$-$transfer coefficient $U$ in the calculations.

$($a$)$ Do this to obtain the new capacity or feed rate under these new conditions. The composition of the liquid product will be the same as before.

$($b$)$ Do this to obtain the new product composition if the feed rate is increased to $18144k\frac{g}{h}.$

EXAMPLE $8\mathrm{.}4-1.$ Heat$-$Transfer Area in Single$-$Effect Evaporator

A continuous single$-$effect evaporator concentrates $9072k\frac{g}{h}$ of a $1\mathrm{.}0wt\%$ salt solution entering at $311\mathrm{.}0K(37\mathrm{.}8C)$ to a final concentration of $1\mathrm{.}5wt\%.$ The vapor space of the evaporator is at $101\mathrm{.}325$kPa $(1\mathrm{.}0$atm abs$)$ and the steam supplied is saturated at $143\mathrm{.}3$kPa. The overall coefficient $U=1704\frac{W}{m^2}*K.$ Calculate the amounts of vapor and liquid product and the heat$-$transfer area required. Assume that, since it is dilute, the solution has the same boiling point as water.