Seawater containing 3.50 wt% salt passes through a series of 9 evaporators. Roug

Question

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 9th evaporator contains 4.70 wt% salt.

It is desired to produce 1 x 104 L/h of fresh water.

1)How much seawater must be fed to the process?(in kg/h)

2)What is the mass flow rate of concentrated brine out of the process?(in kg/h)

3)What is the weight percent of salt in the outlet from the 4th evaporator?

4)
What is the fractional yield of fresh water from the process (kg H2O recovered/kg H2O in process feed)?

I am having a lot of trouble converting stuff as i have forgotten a lot of the basic conversions so please notate where these conversion units are coming from

thank you!

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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