Can you please solve this question with clear steps and answer, its a very important question for my exam. A double$-$effect evaporator is used to concentrate $10,000$ kg$/$h of caustic soda solution from $9\%$ to

$47\%$ by wt$.$ NaOH. For this purpose, a backward feed arrangement is used. The feed enters the

evaporator at $309$ K$.$ Process steam at $787\mathrm{.}941$ kPa $(442\mathrm{.}7$K$,2048\mathrm{.}7$ kJ$/$kg$)$ is available and in the

second effect a pressure of $14\mathrm{.}665$ kPa $(326\mathrm{.}3$K$,2377\mathrm{.}8$ kJ$/$kg$)$ is maintained. First evaporator is

maintained at $401\mathrm{.}303$kPa$(143\mathrm{.}73$C$,2132\mathrm{.}98$ kJ$/$kg$).$ Calculate the heating surface area in both the

effects. Calculate the steam consumption and evaporation in each effect. Neglect boiling point rise.

The overall heat transfer coefficients in the first and second effects are $2326$ and $1744\mathrm{.}5$ W$/($m$2$

$$K$),$

respectively. Take a specific heat of $3\mathrm{.}77$ kJ$/($kg$$K$),$ for all caustic streams.