LOOO 200 l2oD200 T200180

4000 = 4000 OK

A balance around hold tank 1 gives

A balance around hold tank 2 gives


Adsorption processes are often used as a follow-up to chemical wastewater treatment to remove organic reaction products that cause taste, odor, color, and toxicity problems. The equilibrium relationship between adsorbents (solid materials that adsorb organic matter, e.g., activated carbon) and adsorbates (substances that are bound to the adsorbates, e.g., benzene) may be simply expressed as q = Kc"

where q — amount of organic matter adsorbed per amount of adsorbent c = concentration of organic matter in water n = experimentally determined constant K = equilibrium distribution constant

Based upon the data given in the table below that were obtained from a laboratory sorption experiment, a college student who worked for a company as an intern was asked to evaluate the K and n values for a certain type of activated carbon to be used to remove undesirable by-products formed during chemical treatment. What are the values of K and n that the intern should have generated?

Sorption Data Collected in Laboratory Experiments

c (mg/L)

<7 0*g/g)












The equilibrium relationship given in the problem statement can be linearized by taking the log of both sides of the equation. This yields the following equation:

A plot of log(</) versus log(c) yields a straight line if this relationship can be used to represent the experimental data. The slope of this line is equal to n, while the

Figure 100. Log-log plot of sorption data.


intercept is log(A'). The experimental data analyzed using the equation above are plotted in Figure 100, showing that the data fit this linearized isotherm quite well. The equation generated from a regression analysis indicates that n = 1.48

provided the units of c and q are mg/L and jug/q, respectively.


A fixed-bed ion exchanger is to be used to soften 20,000 gal of water between regenerations. Hardness in the water averages 400 mg/L as CaCOj. If the capacity to breakthrough of the exchange material is 25 kilograms (kg) (as CaC03)/ft3, and the time interval between regenerations is 8h, determine the dimensions of the exchanger.


The exchange capacity required is

The tentative volume of exchange material required is


Assuming a bed depth of 6 ft, the surface area will be

For a surface area of 6.23 ft2, the superficial velocity of flow through the bed will be



A large wastewater treatment facility currently uses chlorine for disinfection. The average chlorine dosage is 6.0 mg/L at an average flowrate of 70 MGD. Chlorine (Cl2) costs $1.00/lb. Sulfur dioxide (S02) is used to dechlorinate the effluent before it is discharged (a requirement placed on the facility to protect the fish inhabiting the receiving water) and is consumed at an average dose of 2.0 mg/L. Its cost is $1.20/lb.

Concerned about operating costs and the risks of Cl2 usage, the utility has completed a pilot study showing that ultraviolet (UV) light disinfection could be used to replace Cl2 and S02. The UV system will have a capital cost of $12,000,000, will cost $500,000 a year to operate with a 20-year service life. The utility will draw from bank savings accounts currently earning 8.0% simple interest per year to pay for the UV system's capital cost. Does the savings in operating costs justify the capital cost? Assume straight-line depreciation of the UV system, with $0 salvage value at the end of its service life.

Additional Information

Conversion factor: A dose of 1 mg/L = 8.34 lb/106 gal (MG) Straight-line depreciation equation:

Service life

Rate of return on investment (ROI) equation in %/yr:

Scenario A operating costs — Scenario B operating costs

Investment where scenario A = estimated cost for continuing the status quo scenario B = estimated cost for the proposed alternative


Calculate the daily Cl2 and S02 dosage rates as follows:

Cl2 dosage rate = (70MGD)(6.0mg/L)(8.34L • lb/MG • mg) = 3500 lb/day

S02 dosage rate = (70MGD)(2.0mg/L)(8.34L • lb/MG • mg) = 1170 lb/day

CHM.10 ULTRAVIOLET VS. CHLORINE DISINFECTION These daily rates are then converted to annual rates:

Annual Cl2 dosage rate = (daily Cl2 rate)(day/yr)

Annual S02 dosage rate = (daily S02 rate)(day/yr)

The annual costs for Cl2 and S02 addition are calculated as follows:

Annual Cl2 cost = (annual Cl2 dosage rate)(Cl2 cost in $/lb) = (1,278,000 lb/yr)($l .00/lb) = $l,278,000/yr

Annual S02 cost = (annual S02 dosage rate)(S02 cost in lb) = (427,000 lb/yr)($1.20/lb) = $512,000/yr

Total operating cost = Annual Cl2 cost + Annual S02 cost = $l,278,000/yr + $512,000/yr = $l,790,000/yr

The straight-line depreciation of the system is then calculated as follows:

Depreciation = (initial value — salvage value)/service life = ($12,000,000 - $0)/(20 yr) = $600,000/yr

The total annual cost of the UV system is then calculated as:

Total annual cost = Straight-line depreciation + Operating cost = $600,000/yr + $500,000/yr = $l,100,000/yr

The annual cost savings for switching to the UV system is determined as follows:

Annual cost savings = Scenario A operating cost — Scenario B operating cost where scenario A is the total annual Cl2 cost and scenario B is the total annual UV cost.

Annual cost savings = Annual Cl2 cost — Annual UV cost = $ 1,790,000/yr -$1,100,000/yr = $690,000/yr

The expected ROI on the investment in the UV system is determined as follows:

ROI = 100(Annual cost savings)/(investment) = 100($690,000/yr)/(2,000,000) = 100(0.0575) = 5.8%

This is less than the 8.0% simple interest the utility could make from cash in its bank account (at the time of preparation of this problem). Thus, based on operating cost considerations alone, the utility should not invest in the UV system. However, other considerations may make UV more attractive. There may be hidden costs to continued Cl2 and S02 usage, such as maintaining safety equipment and training, hazardous materials planning, worker and public concerns, stockholder support and, perhaps, capital costs to maintain or upgrade the Cl2 and S02 storage and dosing systems. Additional calculations should be made that include these as well as potentially other hidden costs before a final decision is made.

Renewable Energy 101

Renewable Energy 101

Renewable energy is energy that is generated from sunlight, rain, tides, geothermal heat and wind. These sources are naturally and constantly replenished, which is why they are deemed as renewable. The usage of renewable energy sources is very important when considering the sustainability of the existing energy usage of the world. While there is currently an abundance of non-renewable energy sources, such as nuclear fuels, these energy sources are depleting. In addition to being a non-renewable supply, the non-renewable energy sources release emissions into the air, which has an adverse effect on the environment.

Get My Free Ebook

Post a comment