A solution containing $3\%$ solids at a temperature of $18\mathrm{.}5$ C will be concentrated to $18\%$ in a single$-$stage evaporator with a heating area of $42$ m$^2$ and a total heat transfer coefficient of $2000$ w$/$m$^2.$ The heat used is $200$ lpa and the saturation temperature of the water is $120\mathrm{.}2$ C$.$ The boiling temperature in the evaporator is $60\mathrm{.}1$ C$.$ Answer using the following information: vapor enthalpy of the steam $=65$ kj$/$kg condensate enthalpy of the steam $=505$ kj$/$kg atmospheric pressure $=101\mathrm{.}3$ kpa enthalpy of the feed $65$ kj$/$kg solvent vapor enthalpy $=2619$ kj$/$kg concentrated enthalpy at the evaporation temperature $=272$ kj$/$kg $1-$ draw a schematic evaporator shape and show the data on the figure $2-$ calculate the required steam amount in kg$/$second $($ignore the boiling increase $3-$ find the amounts of the dilute solution fed to the evaporator, the evaporated solvent and the resulting concentrated solution in kg$/$second