CEE 370

Fall 2010

Homework #8

0.5 pts each

Drinking Water Problems

 

1.      Do Problem 10-4 in the D&M text

The following minearl analysis was reported for Michigan State Well water. Determine the following  in units of mg/L as CaCO3

a. total hardness,

b. carbonate hardness and

c. noncarbonate hardness

 

Mineral

Conc (mg/L)

Mineral

Conc (mg/L)

Fluoride

1.1

Silica (as SiO2)

13.4

Chloride

4.0

Bicarbonate

318.0

Nitrate

0.0

Sulfate

0.5

Sodium

14.0

Iron

0.5

Potassium

1.6

Manganese

0.07

Calcium

96.8

Zinc

0.27

Magnesium

30.4

Barium

0.2

 

Solution: 

 

a. Begin by converting to mg/L as CaCO3

 

 

            Compound      mg/L as ion    EW/EW     mg/L as CaCO3

 

                  Ca                    96.8              2.50              242.0

                  Mg                   30.4              4.12              125.3

                HCO3              318.0              0.82              260.9

 

b. Calculate TH, CH and NCH

 

            TH = 242.0 + 125.3 = 367.3 mg/L as CaCO3

 

            CH = 260.9 mg/L as CaCO3 

 

            NCH = TH ‑ CH = 367.3 ‑ 260.9 = 106.4 mg/L as CaCO3

 

 

2.      Do Problem 10-6 in the D&M text.

Two parallel flocculation basins are to be used to treat a water flow of 0.150 m3/s.  If the design detention time is 20 min, what is the volume of each tank?

 

Solution:

 

V = Qt0 = (0.150 m3/s)(20 min)(60 s) = 180 m3

 

                                                    180

Volume for each tank = ‑‑‑‑‑‑ = 90 m3

                                                      2

 

 

 

3.      Do Problem 10-8 in the D&M text.

Assuming a conservative value for an overflow rate, determine the suface area ( in m2) of each of two sedimentation tanks that together must handle a flow of 0.05162 m3/s of lime softening floc.  Use an overflow rate of 57 m3/day/m2.

 

Solution: 

 

a. With 57 m3/d‑m2 as a conservative overflow rate, i.e. one that will yield the larger and, hence, more conservative surface area.

 

b. Since two tanks (assume in parallel):

 

                     0.05162 m3/s

            Q = ‑‑‑‑‑-----‑‑‑‑‑‑‑‑‑ = 0.02581 m3/s

                             2

 

c. And surface area of each tank

 

                      (0.02581 m3/s)( 86,400 s/d)

            As = ‑‑‑------‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑ = 39.123 or 39 m2

                               57 m3/d‑m2

 

 

4.      Do Problem 10-12 in the D&M text.

The water flow meter at the Westwood water plant is malfunctioning.  The plant superintendent tells you the four dual-media filters (each 5.00 m x 10.0 m) are loaded at a velocity of 280 m/day.  What is the flow rate through the filters (in m3/s)?

 

Solution: 

 

a. Note that 280 m/d = 280 m3/d‑m2 

 

b. Compute flow in m3/d

 

            Q = (280 m3/d‑m2)(4)(5.00 m)(10.00 m)

 

            Q = 56,000 m3/d

 

c. Convert to m3/s

 

                    56,000 m3/d

            Q = ‑‑‑‑‑----‑‑‑‑‑‑‑‑ =  0.648 m3/s

                    86,400 s/d

 

 

 

 

Wastewater Treatment Problems

 

5.      Do Problem 11-2 in the D&M text.

If the terminal settling velocity of a particle falling in aquiescent water having a temperature of 15°C is 0.0950 cm/s, what is its diameter?  Assume a particle density of 2.05 g/cm3 and a density of water equal to 1000 kg/m3.  Assume Stokes’ law applies.

 

 

 

 

d = 0.004349 cm  = 0.04343 mm

 

 

 

6.      Microbial Growth Problem (not in book)

You’re operating a batch reactor.  At the start (time =0) you have 0.85 mg/L of biomass.  After two full weeks of operation you find that the biomass concentration is 286 mg/L.

a. What is the specific growth rate assuming simple exponential growth throughout?

b. Is the value you calculated in part “a” equal to the mmax ?  Why or why not?  Explain.

 

 

 

From equation on Slide #8 of Lecture #16:

 

a.    Is the value you calculated in part “a” equal to the mmax ?  Why or why not?  Explain.

 

It is not necessarily equal to the mmax because the growth rate may or may not be restricted due to limited substrate concentration or other environmental factors.  Information on growth limitation is not provided, so we cannot assume that such limitation does not exist.  In other words, the growth is exponential, but not necessarily “unlimited”.  With better conditions it might be able to grow at a faster exponential rate.

 

 

 

 

7.      Do Problem 11-20 in the D&M text.

The 500- bed Lotta Hart Hospital has a small activated sludge plant to treat its wastewater.  The average daily hospital discharge is 1200 L/day per bed, and the average soluble BOD5 after primary settling is 500 mg/L.  the aeration tank has effective liquid dimensions of 10.0 m wide x 10.0 m long x 4.5 m deep.  The plant operating parameters are as follows: MLVSS = 2000 mg/L, MLSS = 1.2 x MLVSS, and return sludge concentration = 12,000 mg/L  (VSS).  Determine:

a. Aeration Period in hrs

 

 

Ɵ = 18 hours

 

b. F/M ratio

 

F/M = 0.33 d-1

 

If you use MLSS for X instead of MLVSS (which is more common), you would have gotten 0.278

 

 

8.      Do Problem 11-23 in the D&M text.

Using the following assumptions, determine:

a. the solids retention time (days), and the

b. cell wastage flow rate

 for the Lotta Hart Hospital problem above.

 

Assumptions:

Suspended solids in effluent = 30 mg/L

Bacterial decay rate = 0.060 day-1

Wastage is from the return sludge line

Inert fraction of suspended solids = 66.67%

Yield coefficient = 0.60

Allowable BOD in effluent = 30.0 mg/L

 

 

a., the SRT

 

 

And rearranging

and

 

 

 

 

 

If you use MLSS for X instead of MLVSS (which is more common), you would have gotten 10.3 days

 

 

 

b. the waste sludge flow

 

 

Qw = 12,240 L/d

 

 

If you use MLSS for X instead of MLVSS (which is more common), you would have gotten 10,400 L/d