A $6wt.\%$NaOH solution is to be concentrated to $44wt\%$ using a double effect reverse feed evaporator $($see figure below$).$ Saturated steam is available at $0\mathrm{.}370$ barg, and the condenser which condenses the vapour from the second effect can maintain a vacuum at $-0\mathrm{.}89$ barg.

Other details:

The two effects are identical.

Boiling Point Rise $($BPR$),$ BPR $(K)=200x^2($where $x=$ mass fraction of NaOH $).$

Feed flow: $21$ tons$/$day$,$ feed temperature is $50C.$

Heat transfer coefficients: $U_1=1\mathrm{.}5k\frac{W}{m^2}K$ and $U_2=3\mathrm{.}0k\frac{W}{m^2}K$

Specific heat capacity of NaOH solution,

$C_p^L=4\mathrm{.}22-4x(k\frac{J}{k}g*K)$;$C_p^V=1\mathrm{.}96k\frac{J}{k}g*K$

$1\mathrm{.}1$ Perform material balance calculations aimed at evaluating all stream flow rates.

$1\mathrm{.}2$ Calculate the boiling point elevation in effect one and two, and $T_2.$

$1\mathrm{.}3$ Calculate the temperature and pressure in effect one.

$1\mathrm{.}4$ Calculate the mass flow rate of steam.

$1\mathrm{.}5$ Calculate the area of heat transfer.