Now calculate the specific enthalpies of the vapor $($V$)$ and liquid $($L$)$ streams of the evaporator with respect to the state of the feed $F$ to the evaporator using the following data for the properties of the evaporator contents.

$)=100C,P=(1(atm)|$

Then use those specific enthalpies to recalculate the mass flow rate of the required steam.

$S^{}=1,k\frac{g}{h}$

Evaporation is commonly used to concentrate dissolved solids in a liquid feed stream and produce pure water vapor. In the process below, heat for the water vaporization is supplied by condensation of steam in a heating coil passing through in the evaporator.

An aqueous slurry at $40\mathrm{.}0C$ and $1\mathrm{.}01325$ bar containing $5\mathrm{.}00wt\%$ dissolved salt is fed to the evaporator, and a liquid containing $25\mathrm{.}0$ wt$\%$ dissolved salt leaves as product together with a water vapor stream. The internal pressure in the evaporator is $600mmHg$ absolute$,$ and the vapor $($V$)$ and liquid $($L$)$ streams leaving the evaporator are at the same temperature as the evaporator contents.

Saturated steam at $10\mathrm{.}0$ bar absolute enters the heating coil and saturated liquid at the same pressure exits the heating coil, with the latent heat evaporating water in the process.