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, $Al_{2}(SO_{4})_{3}\cdot 14H_{2}O$ .

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.

$x_{(1,{\text{ }}H_{2}O)}=1-x_{(1,{\text{ }}Particulate)}-x_{(1,{\text{ }}Bacteria)}=1-0.02-0.0001=0.9799$

$x_{(2,{\text{ }}Al_{2}(SO_{4})_{3}\cdot 14H_{2}O)}=1$

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

$x_{(4,{\text{ }}Particulate)}=0.40$

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:

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

${\dot {m}}_{1}x_{(1,Particulate)}={\dot {m}}_{4}x_{(4,Particulate)}$

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

${\dot {m}}_{2}x_{(2,{\text{ }}Al_{2}(SO_{4})_{3}\cdot 14H_{2}O)}={\dot {m}}_{4}x_{(4,{\text{ }}Al_{2}(SO_{4})_{3}\cdot 14H_{2}O)}$

Now lets do the mass fraction balance:

$x_{(3,{\text{ }}H_{2}O)}=1-x_{(3,Bacteria)}$

$x_{(4,{\text{ }}Particulate)}=1-x_{(4,H_{2}O)}-x_{(4,Bacteria)}-x_{(4,{\text{ }}Al_{2}(SO_{4})_{3}\cdot 14H_{2}O)}$

Next lets write our the extra equation:

$0.80\cdot {\dot {m}}_{1}x_{(1,Bacteria)}={\dot {m}}_{4}x_{(4,Bacteria)}$

Finally we can solve for the degrees of freedom:

${\text{DOF}}=7-7=0$

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:

${\dot {m}}_{1}x_{(1,Particulate)}={\dot {m}}_{4}x_{(4,Particulate)}$

${\dot {m}}_{4}={\frac {{\dot {m}}_{1}x_{(1,Particulate)}}{x_{(4,Particulate)}}}={\frac {(200)(0.02)}{0.4}}{\text{ }}{\frac {kg}{s}}=10{\text{ }}{\frac {kg}{s}}$

2. Solve for $x_{(4,Bacteria)}$ using the extra equation:

$0.80\cdot {\dot {m}}_{1}x_{(1,Bacteria)}={\dot {m}}_{4}x_{(4,Bacteria)}$

$x_{(4,Bacteria)}={\frac {0.80\cdot {\dot {m}}_{1}x_{(1,Bacteria)}}{{\dot {m}}_{4}}}={\frac {{\big (}0.80{\big )}{\big (}200{\text{ }}{\frac {kg}{s}}{\big )}{\big (}0.0001{\big )}}{10{\text{ }}{\frac {kg}{s}}}}=0.0016$

==== 3. Solve for $x_{(4,{\text{ }}Al_{2}(SO_{4})_{3}\cdot 14H_{2}O)}$ using the fractional mass balance:

$x_{(4,{\text{ }}Al_{2}(SO_{4})_{3}\cdot 14H_{2}O)}=1-x_{(4,H_{2}O)}-x_{(4,Bacteria)}-x_{(4,{\text{ }}Particulate)}=1-0.20-0.0016-0.40=0.3984$

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

${\dot {m}}_{2}x_{(2,{\text{ }}Al_{2}(SO_{4})_{3}\cdot 14H_{2}O)}={\dot {m}}_{4}x_{(4,{\text{ }}Al_{2}(SO_{4})_{3}\cdot 14H_{2}O)}={\bigg (}10{\text{ }}{\frac {kg}{s}}{\bigg )}{\bigg (}0.3984{\bigg )}=3.984{\text{ }}{\frac {kg}{s}}$

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

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

${\dot {m}}_{3}={\frac {{\dot {m}}_{1}x_{(1,{\text{ }}Bacteria)}-{\dot {m}}_{4}x_{(4,{\text{ }}Bacteria)}}{x_{(3,{\text{ }}Bacteria)}}}$

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

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

We can equate these two equations to get:

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

and using the fractional mass balance

$x_{(3,{\text{ }}H_{2}O)}=1-x_{(3,{\text{ }}Bacteria)}$

we can solve for $x_{(3,{\text{ }}Bacteria)}$

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

${\frac {0.004}{x_{(3,{\text{ }}Bacteria)}}}={\frac {193.98}{1-x_{(3,{\text{ }}Bacteria)}}}$

$x_{(3,{\text{ }}Bacteria)}=0.00002$

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

$\therefore {\text{ }}{\dot {m}}_{3}={\frac {{\dot {m}}_{1}x_{(1,H_{2}O)}-{\dot {m}}_{4}x_{(4,H_{2}O)}}{x_{(3,H_{2}O)}}}={\frac {193.98}{0.99998}}{\text{ }}{\frac {kg}{s}}=193.98{\text{ }}{\frac {kg}{s}}$

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

Attribution: Said Zaid-Alkailani & UBC