The $10\%$ NaOH solution is fed to the first unit of a three$-$effect evaporator at a temperature of $300$ K with a flow rate of $10$ kg$/$s$.$ the pressure in the third effect is $13\mathrm{.}3$ kN$/m^2$ however, the NaOH taken from this effect is $40\%.$ The heating temperature of the solution is $4\mathrm{.}18$ kJ$/$kg$.$K can be taken. The boiling point increase will not be neglected.The steam is saturated water vapor at a pressure of $205$ kN$/m^2$ and is fed to the first effect and the condensate is saturated. it is separated as a liquid. If the heating areas are equal, calculate the areas, temperature differences and steam consumption.

$(U_1=5k\frac{W}{m^2}.K)(U_2=3\mathrm{.}2k\frac{W}{m^2}.K)(U_3=1\mathrm{.}4k\frac{W}{m^2}.K)$