Basic Models:

Treatment area:

where | |

C = |
Wetland effluent concentration (mg/L). |

Co = |
Wetland influent concentration (mg/L). |

C* = |
Background concentration (mg/L). |

hlra= |
Annual hydraulic loading rate (m/yr). |

Kt = |
Rate constant at temperature T (m/yr). |

K20 = |
Rate constant at 20°C (m/yr). |

e = |
Temperature coefficient. |

a, = |
Treatment area of wetland (m2). |

Qo = |
Annual influent wastewater flow rate (m3/yr) |

z = |
Safety factor. |

Note: (1) These are areal-based models written in terms of hydraulic loading per unit area as compared to a detention time base for the model in Table 6.16. Detention time (HRT) and hydraulic loading (HLR) are directly related for a specific set of wetland conditions. All of the models should, therefore, produce similar results, but they do not. The difference is due to their derivation from different data sets and to the internal position of C* and a safety factor (z) in the models in this table. The other models treat the background concentration and a safety factor as external boundary conditions. (2) The Q0 and the HLR as used in the above models are the influent flow rate only and do not include adjustment for water gains or losses through precipitation, evapotranspiration, or seepage. (3) The porosity (n),water depth (y), and detention time (t) in the wetland are not considered in these areal-based design models. (4) The safety factor (z) is the ratio of the annual average concentration to the maximum monthly concentration for the pollutant of concern as derived from the database used by Kadlec and Knight (1996). (5) Because of the internal position of the C* and z in this treatment area model, it is not possible to design a system large enough to achieve an effluent with background concentrations. As the required effluent concentration (Ce) approaches background (C*), the required wetland area approaches infinity.

BOD5 Removal:

C* (mg/L) = 3.5 + 0.053(C0). ez (for C*> = 1.00. z = 0.59.

Note: The treatment area model cannot be solved for BOD5 effluent values approaching background levels. For example, if an effluent of 7 mg/L is desired, the influent cannot exceed 12 mg/L. An effluent of 6 mg/L would be impossible to achieve.

TSS Removal:

ez (for C*> = [CT* = C20* (e>(T-20> ] = 1.065.

Note: The treatment area model cannot be solved for low effluent (Ce) values. For example, if an effluent TSS concentration of 15 mg/L is required, the influent (C0) cannot exceed 17 mg/L if the area model is to be solved.

Organic Nitrogen Removal:

K20 (m/yr) = 17. 0 = 1.05. C* (mg/L) 1.5. z = 0.555.

Ammonia Removal:

Nitrate Removal:

Total Nitrogen Removal:

K20 (m/yr) = 22. 0 = 1.09. C* (mg/L) = 1.5. z = 0.625.

Total Phosphorus Removal:

Fecal Coliform Removal:

K20 (m/yr) = 75. 0 = 1.00. C* (mg/L) = 300. z = 0.333.

Background Concentration:

The background concentration (C*) is included internally in each design model. Wetland Sizing:

The parameter (BOD5, etc.) that requires the largest treatment area for removal is the limiting design factor, and that area should be selected for the intended project. The wetland should then provide acceptable treatment for all other parameters of concern.

Safety Factor:

The safety factor (z) is included internally in the logarithmic portion of the design model for treatment area.

Source: Kadlec, R.H. and Knight, R., Treatment Wetlands, Lewis Publishers, Boca Raton, FL, 1996. With permission.

A FWS wetland is to be designed to reduce the BOD from 100 mg/L to 15 mg/L. The flow is 0.9 mgd and the wastewater temperature is 68°F. Compare the land areas needed using the volumetric/detention time approach and the areal loading rate approach.

1. Using Equation 6.23, solve for the land area for the volumetric/detention time approach. Use a depth of 2 ft and a porosity of 0.8. Use KT = 0.68: As = ajln(Cfl/Ce)/KT<y)(n)]

As = (0.9)(3.069 ac-ft/mil gal)[ln100/15]/(0.68)(2)(0.8) As = (2.76)(1.897)/(1.088) As = 4.81 ac

2. Calculate the land area using the hydraulic loading rate method. Using Equation 6.25, calculate the land area. Convert flow into m3/yr; use Kt = 34; and set z = 1, C = 8.8:

As = (1,243,372 m3/yr)/(34)[(15 - 8.8)/(100 - 8.8)]

Comment

The major differences between these models are the C* values and the internal position of the safety factor (z) in the logarithmic portion of this area model; for example, if safety factor z is 0.59 instead of 1, the calculated area in step 2 would be 67 acres.

Was this article helpful?

## Post a comment