1.11 – Practice Problem 1

Question

In wastewater treatment, an important step in the process is sedimentation. Sedimentation is the process of allowing particulate matter in a fluid to settle to the bottom of a tank. In the sedimentation step, a coagulant can be used to coagulate bacteria and fine particles to allow the sedimentation to occur faster. The most common molecular species used for coagulation is alum, $A l_{2} (S O_{4})_{3} \cdot 14 H_{2} O$ .

You are a chemical engineer working in a wastewater treatment plant. You are asked to make a material balance on the sedimentation process. Wastewater flows into a sedimentation tank at 200 kg/s. The wastewater contains 0.01 wt% of bacteria, 2 wt% particulate matter, and the balance pure water. Pure alum is added to the sedimentation tank in a separate stream. Alum allows 80% of the bacteria and 100 % of the particulate matter to settle. Just enough alum is added for this ratio of settlement. The sludge at the bottom of the tank exits the tank with 20 wt% of water and 40 wt% of particulate matter. The semi-treated water exits the tank in a separate stream.

a) Draw and label a block flow diagram (BFD).

b) Perform a degree of freedom analysis.

c) Fully balance the remainder of the block flow diagram.

Answer

a) Draw and label a block flow diagram (BFD)

First, let's draw the basic layout of the BFD:

Attribution: Said Zaid-Alkailani & UBC

Using the information given, let's fill out as much of the BFD as we can.

$x_{(1, H_{2} O)} = 1 - x_{(1, P a r t i c u l a t e)} - x_{(1, B a c t e r i a)} = 1 - 0.02 - 0.0001 = 0.9799$

$x_{(2, A l_{2} (S O_{4})_{3} \cdot 14 H_{2} O)} = 1$

$x_{(4, H_{2} O)} = 0.20$

$x_{(4, P a r t i c u l a t e)} = 0.40$

Attribution: Said Zaid-Alkailani & UBC

b) Perform a degree of freedom analysis

We have 7 unknowns, so we need to find 7 equations. First lets do all of the mass balances:

${\dot{m}}_{1} x_{(1, H_{2} O)} = {\dot{m}}_{3} x_{(3, H_{2} O)} + {\dot{m}}_{4} x_{(4, H_{2} O)}$

${\dot{m}}_{1} x_{(1, P a r t i c u l a t e)} = {\dot{m}}_{4} x_{(4, P a r t i c u l a t e)}$

${\dot{m}}_{1} x_{(1, B a c t e r i a)} = {\dot{m}}_{3} x_{(3, B a c t e r i a)} + {\dot{m}}_{4} x_{(4, B a c t e r i a)}$

${\dot{m}}_{2} x_{(2, A l_{2} (S O_{4})_{3} \cdot 14 H_{2} O)} = {\dot{m}}_{4} x_{(4, A l_{2} (S O_{4})_{3} \cdot 14 H_{2} O)}$

Now lets do the mass fraction balance:

$x_{(3, H_{2} O)} = 1 - x_{(3, B a c t e r i a)}$

$x_{(4, P a r t i c u l a t e)} = 1 - x_{(4, H_{2} O)} - x_{(4, B a c t e r i a)} - x_{(4, A l_{2} (S O_{4})_{3} \cdot 14 H_{2} O)}$

Next lets write our the extra equation:

$0.80 \cdot {\dot{m}}_{1} x_{(1, B a c t e r i a)} = {\dot{m}}_{4} x_{(4, B a c t e r i a)}$

Finally we can solve for the degrees of freedom:

$DOF = 7 - 7 = 0$

c) Fully balance the remainder of the block flow diagram

To fully we balance the diagram we must first solve for all of the unknowns.

1. Solve for ${\dot{m}}_{4}$ using particulate balance:

${\dot{m}}_{1} x_{(1, P a r t i c u l a t e)} = {\dot{m}}_{4} x_{(4, P a r t i c u l a t e)}$

${\dot{m}}_{4} = \frac{{\dot{m}}_{1} x_{(1, P a r t i c u l a t e)}}{x_{(4, P a r t i c u l a t e)}} = \frac{(200) (0.02)}{0.4} \frac{k g}{s} = 10 \frac{k g}{s}$

2. Solve for $x_{(4, B a c t e r i a)}$ using the extra equation:

$0.80 \cdot {\dot{m}}_{1} x_{(1, B a c t e r i a)} = {\dot{m}}_{4} x_{(4, B a c t e r i a)}$

$x_{(4, B a c t e r i a)} = \frac{0.80 \cdot {\dot{m}}_{1} x_{(1, B a c t e r i a)}}{{\dot{m}}_{4}} = \frac{(0.80) (200 \frac{k g}{s}) (0.0001)}{10 \frac{k g}{s}} = 0.0016$

==== 3. Solve for $x_{(4, A l_{2} (S O_{4})_{3} \cdot 14 H_{2} O)}$ using the fractional mass balance:

$x_{(4, A l_{2} (S O_{4})_{3} \cdot 14 H_{2} O)} = 1 - x_{(4, H_{2} O)} - x_{(4, B a c t e r i a)} - x_{(4, P a r t i c u l a t e)} = 1 - 0.20 - 0.0016 - 0.40 = 0.3984$

4. Solve for ${\dot{m}}_{2}$ using the alum balance:

${\dot{m}}_{2} x_{(2, A l_{2} (S O_{4})_{3} \cdot 14 H_{2} O)} = {\dot{m}}_{4} x_{(4, A l_{2} (S O_{4})_{3} \cdot 14 H_{2} O)} = (10 \frac{k g}{s}) (0.3984) = 3.984 \frac{k g}{s}$

5. Solve for ${\dot{m}}_{3}$ using a combination of the water balance and bacteria balance:

${\dot{m}}_{1} x_{(1, B a c t e r i a)} = {\dot{m}}_{3} x_{(3, B a c t e r i a)} + {\dot{m}}_{4} x_{(4, B a c t e r i a)}$

${\dot{m}}_{3} = \frac{{\dot{m}}_{1} x_{(1, B a c t e r i a)} - {\dot{m}}_{4} x_{(4, B a c t e r i a)}}{x_{(3, B a c t e r i a)}}$

${\dot{m}}_{1} x_{(1, H_{2} O)} = {\dot{m}}_{3} x_{(3, H_{2} O)} + {\dot{m}}_{4} x_{(4, H_{2} O)}$

${\dot{m}}_{3} = \frac{{\dot{m}}_{1} x_{(1, H_{2} O)} - {\dot{m}}_{4} x_{(4, H_{2} O)}}{x_{(3, H_{2} O)}}$

We can equate these two equations to get:

$\frac{{\dot{m}}_{1} x_{(1, B a c t e r i a)} - {\dot{m}}_{4} x_{(4, B a c t e r i a)}}{x_{(3, B a c t e r i a)}} = \frac{{\dot{m}}_{1} x_{(1, H_{2} O)} - {\dot{m}}_{4} x_{(4, H_{2} O)}}{x_{(3, H_{2} O)}}$

and using the fractional mass balance

$x_{(3, H_{2} O)} = 1 - x_{(3, B a c t e r i a)}$

we can solve for $x_{(3, B a c t e r i a)}$

$\frac{{\dot{m}}_{1} x_{(1, B a c t e r i a)} - {\dot{m}}_{4} x_{(4, B a c t e r i a)}}{x_{(3, B a c t e r i a)}} = \frac{{\dot{m}}_{1} x_{(1, H_{2} O)} - {\dot{m}}_{4} x_{(4, H_{2} O)}}{1 - x_{(3, B a c t e r i a)}}$

$\frac{0.004}{x_{(3, B a c t e r i a)}} = \frac{193.98}{1 - x_{(3, B a c t e r i a)}}$

$x_{(3, B a c t e r i a)} = 0.00002$

$∴ x_{(3, H_{2} O)} = 1 - 0.00002 = 0.99998$

$∴ {\dot{m}}_{3} = \frac{{\dot{m}}_{1} x_{(1, H_{2} O)} - {\dot{m}}_{4} x_{(4, H_{2} O)}}{x_{(3, H_{2} O)}} = \frac{193.98}{0.99998} \frac{k g}{s} = 193.98 \frac{k g}{s}$

Now that we have completely solved the material balance, we can complete our BFD:

Attribution: Said Zaid-Alkailani & UBC