Question

A triple-effect evaporator is a process that removes pure water from a salt water feed. The process consists of three units connected in series. The feed to the process contains 3.43 wt % NaCl in water. In the first effect, sufficient water is evaporated to concentrate the solution to 3.89 wt % NaCl. The second effect concentrates the solution further to 4.22 wt %, and the brine discharged from the process contains 5.02 wt % NaCl. a. Draw a diagram of the process, then do a degrees of freedom analysis on each effect and on the entire process to determine which of these choices of system are appropriate starting point(s) for the calculation. b. Determine the total mass of pure water provided by the process per 100 kg of feed. c. Solve for the flow rates of all streams based on 100 kg feed.

Answer 1

Degree of freedom on first effect

Number of unknowns = 2 (L1 and W1)

Number of independent equations = 3 (total balance , NaCl balance, water balance)

Degree of freedom = Number of unknowns - Number of independent equations

= 2 - 3 = - 1

Degree of freedom on second effect

Number of unknowns = 2 (L2 and W2)

Number of independent equations = 3 (total balance , NaCl balance, water balance)

Degree of freedom = Number of unknowns - Number of independent equations

= 2 - 3 = - 1

Degree of freedom on third effect

Number of unknowns = 2 (L3 and W3)

Number of independent equations = 3 (total balance , NaCl balance, water balance)

Degree of freedom = Number of unknowns - Number of independent equations

= 2 - 3 = - 1

Degree of freedom on entire process

Number of unknowns = 4 (L3, W1, W2, W3)

Number of independent equations = 3 (total balance , NaCl balance, water balance)

Degree of freedom = Number of unknowns - Number of independent equations

= 4 - 3 = 1

Total feed F = 100 kg

NaCl in feed = 100 x 3.43/100 = 3.43 kg

Water in feed = 100 - 3.43 = 96.57 kg

First effect

Total balance

Feed in = water out + concentrated liquid out

F = W1 + L1

100 = W1 + L1

NaCl balance

NaCl in feed = W1*0 + L1*0.0389

3.43 = L1*0.0389

L1 = 88.17 kg

W1 = 100 - 88.17 = 11.83 kg

Second effect

Total balance

Concentrated from first effect in = water out + concentrated liquid out

L1 = W2 + L2

88.17 = W2 + L2

NaCl balance

NaCl in = NaCl out

L1*0.0389 = L2*0.0422

88.17*0.0389 = L2*0.0422

L2 = 81.28 kg

W2 = 88.17 - 81.28 = 6.89 kg

Third effect

Total balance

Concentrated from second effect in = water out + concentrated liquid out

L2 = W3 + L3

81.28 = W3 + L3

NaCl balance

NaCl in = NaCl out

L2*0.0422 = L3*0.0502

81.28*0.0389 = L3*0.0422

L3 = 74.92 kg

W3 = 81.28 - 74.92 = 6.36 kg

Total water out from the process = W1 + W2 + W3

= 11.83 + 6.89 + 6.36

= 25.08 kg

Now we can check our calculations

Overall balance

F = L3 + W1 + W2 + W3

100 = 74.92 + 11.83 + 6.89 + 6.36

100 = 100

Our calculation is correct