Problem 4.41 Seawater Evaporation Seawater containing 3.50 wt% salt passes throu

Question

Problem 4.41 Seawater Evaporation Seawater containing 3.50 wt% salt passes through a series of 9 evaporators. Roughly equal quantities of water are vaporized in each of the 9 units and then condensed and combined to obtain a product stream of fresh water. The brine leaving each evaporator but the 9th is fed to the next evaporator. The brine leaving the gen evaporator contains 4.70 wt% salt. It is desired to produce 1 x 104 L/h of fresh water. Feed Rate How much seawater must be fed to the process? the tolerance is +/-2% Outlet Brine What is the mass flow rate of concentrated brine out of the process? i kg/h the tolerance is +12%

Explanation / Answer

Let the mass feed rate of input sea water be X,

Assuming the density of water being produced be 1, so the mass flow rate of pure water will be 1000kg/hr.

the mass flow rate of brine(concentrated salt) going out will be X-1000

Applying salt material balance between inlet and outlet stream we get,

X*3.5%=1000*0%+(X-1000)*4.

Solving for X we get X=39166kg/h<Ans>

So the inlet mass flow rate of sea water/ Feed rate=39166kg/h

Outlet brine=X-10000

=>39166-10000=29166kg/h<Ans>

Part B

Outlet salt% from fourth evaporator

The water outlet % from fourth evaporator will be 4/9*10000=4444Kg/h

Now the outlet brine mass flowrate =39166-4444=34721kg/h

Balancing salt between inlet and outlet streams we get(Let outlet salt concentration beY wt%)

36166*3.5=34721*Y

=>Y=3.64wt%<Ans>

Part C

The Fractional Yeild

Yeild = Water recovered/ Water fed

=> Yeild=10000/3916=0.255<Ans>

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