Question

algae is initially grown in a pond where it is fed an inoculum (a substance introduced to the body to increase immunity to a disease. i.e. heat, contamination). After the inoculum pond, the algae is fed to a second pond where growth will continue to take place. Only consider the second pond for the mass balance. 130 g/s of algae and water slurry mixture is fed to a second cultivation pond with dimensions of 2000 m^2. Evaporative losses are on average at a rate of 0.84 m^3/(m^2*yr). The resulting mixture leaving the second pond is 80 wt% algae. Recall 1 m^3 = 1000 L & the density of water = 1 g/cm^3

a. Draw a process flowchart

b. Perform a degree of freedom analysis

c. Calculate P1 (evaporative stream loss) and P2

d. Write mass balances for total mass and figure out compositions of algae and water in each stream

Answer 1

Solution:

(a) Pocess chart:

(b) Degree of freedom analysis:

Degree of freedom (DOF) = number of unknowns - number of equations

Here unknowns are = P2 ; Input (Algae and water composition) = 3

Equations are = overall mass balance ; Algae balance and water balance = 3

So, DOF = 3-3 = 0 (Problem is well-defined)

(C) Calculation of P1 (evaporative stream loss) and P2:

Evaporation rate =

Overall mass banace in second pond gives:

I = P1 + P2

130 = 53.27 +P2

P2 = 76.73 g/s

(d) Total mass balane is shown in above part (C), now we will do component balance:

Algae balance:

130*X_A1 = P₂ * 0.8

X_A1 = (76.73 *0.8/130) = 0.4722 or 47.22 %

X_A1 = 47.22 %

Water balance:

(130 * X_W1 )= (P₁ * 1 )+ (P₂ * 0.2)

X_W1 = (53.27+(76.73*0.2)/130) = 0.5278

X_W1 = 52.78%