Ứng dụng Gom Rác GRAC

F/M ratio (Food to Microorganism)

  • Thread starter Hải Anh
  • Ngày gửi
H

Hải Anh

Guest
#1
Bộ ứng cứu tràn đổ hóa chất và môi trường
Mình tìm thấy cái này khá hay dàn h cho phần tính toán khi vận hành hệ thống xử lý nước thải, các bạn tham khảo nhé:
20 • MARCH 1999• FLORIDA WATER RESOURCES JOURNAL
Any one of a number of factors can affect overall CBOD and TSS removal efficiencies of an activated sludge process. Some factors are the responsibility of the design engineers and are essentially uncontrollable as far as operations go. These include aeration tankage, settling tanks, flow equalization, return activated sludge, aeration supply,waste sludge handing, and digested sludge disposal.
Other factors affecting efficiencies are functions of the connected system, such as daily flow volume and and raw sewage CBOD, and are again uncontrollable by the operator. These
factors may vary seasonally, on weekdays, or on weekends.
But there are other factors affecting efficiency that are controllable by the operator. These include the food to microorganism (F/M) ratio (which is controlled by the wasting of biological growth), proper raw sewage flow equalization, return activated sludge rates, digester supernatant return, digested sludge disposal, and control of residual dissolved oxygen in aeration tanks (valving of centrifugal blowers, proper sheaves on positive displacement blowers, turning on or off blowers,
time switch operation of blowers).
Of these factors that can be controlled by the operator, the most important is the F/M ratio. The “food” in the ratio is the CBOD entering the process. The “microorganisms” are the activated sludge solids in the aeration tanks, which are measured as ppm or mg/L of MLTSS.
To establish and maintain a consistent CBOD and TSS secondary waste removal from raw sewage, an activated sludge process must maintain the weight of food to weight of microorganisms
under aeration within the limits of the type of treatment being provided by the facility design parameters.
Many operators and engineers have never grasped the importance of F/M ratio control. Part of the reason is because of the many different methods of calculating it. The three main methods of controlling sludge wasting include the F/M ratio, the Sludge Age method {SA}, and the Mean
Cell Residence Time method {MCRT}. In reality, these methods all measure the same thing, and, in fact, the F/M ratio is merely the reciprocal of the SA. The MCRT is somewhat more complex.
In the final ana-lysis, all three methods do the same thing: measure the ratio of food to microorganisms by weight in the aeration tanks.
I find the F/M ratio to be the easiest to calculate and the easiest to understand. MCRT and SA can be confusing because they are expressed in days, which many people confuse with calendar days; they are really only indexes.
The key to understanding the F/M ratio is that it is not just concentration versus concentration, it is the weight of food (CBOD) compared to the weight of microorganisms (MLTSS).
Volume Formulas
To calculate an F/M ratio, you must know the volume of water being aerated. The volume of water in a rectangular tank in millions of gallons is:
V = (L x W x H x 7.48)/1,000,000 (Equation 1)
where L = the interior length and width of the tank in feet,
W = the interior length and width of the tank in feet,
H = water depth in feet,
7.48 = the number of gallons in a cubic foot,
1,000,000 = the number of gallons in a million gallons.
The volume of water in a round tank in millions of gallons is:
V = p x (D/2)2 x H x 7.48)/1,000,000 (Equation 2)
Or simplifying, V = (5.87 x D2 x H)/1,000,000
Where p = 3.142
D = interior diameter of the tank in feet,
H = water depth in feet,
7.48 = the number of gallons in a cubic foot,
1,000,000 = the number of gallons in a million gallons.
The Pounds Formula
To calculate the F/M ratio, both the food and the microorganisms must be compared in the form of mass or weight. In the U.S. the measure usually is in weight, i.e., pounds. Realizing that one
gallon of water weighs 8.34 pounds, the following formula converts a flowrate in MGD with a concentration in parts per million (which, for our purposes, can be assumed to be the same
as mg/L) to a weight in pounds per day:
W = ppm x 8.34 lb/gallon x MGD (Equation 3)
where W = weight in pounds
For example, assume that a complete-mix activated sludge treatment plant has an average daily flow into aeration of 1.0 MGD, a CBOD into aeration of 200 ppm, and aeration volume of
0.25 million gallons, and an MLTSS of 2500 ppm. The weight of food entering the process, from Equation 3, is 200 x 8.34 x 1.0 = 1670 pounds per day.
The weight of microorganisms under aeration, again from Equation 3, is 2500 x 8.34 x 0.25 = 5210 pounds.
The F/M ratio is then 1670/5210 = 0.32. Since the standard process F/M ratio (see accompanying
table) is from 0.25 to 0.50, the example process is within proper efficient operating parameters.
For another example, assume that the same process during a seasonal low-flow period has an average daily flow into aeration of 0.7 MGD, a CBOD into aeration of 170 ppm, and the same
MLTSS as before.
The weight of food into the activated sludge process, from Equation 3,
is 170 x8.34 x 0.7 = 992 lb/day.
The weight of microorganisms under aeration, from F/M Ratio and the Operation of an Activated Sludge Process Activated Sludge Process Ranges for F/M Ratio Control Process Range Names Common SWT ASP Names F/M Range
Extended Aeration Extended Aeration 0.05-0.15 Lb CBOD5/1 Lb MLTSS
Sequencing Batch Reactors
Race Track or Orbital Ditch
Standard Activated Sludge Conventional Activated Sludge 0.25-0.5 Lb CBOD5/1 Lb MLTSS
Contact Stabilization
Step Aeration
Complete (or Homogenous) Mix
Others used with nutrient removal
Hi-Rate Activated Sludge HRAS based on desired removal 1.0-10 Lb CBOD5/1 Lb MLTSS
{75 to 60% efficiency)

Equation 3, is 2500 x 8.34 x 0.25 = 5210 lb.
The F/M ratio is then 992/5210 = 0.19, which is below the standard range of 0.25 to 0.50. The
process has too much sludge (weight of microorganisms) under aeration for the food being
provided. The operator should waste sludge to lower the MLTSS in aeration.Determining Optimal MLTSS
The optimal MLTSS is based on the ratioproportion formula, R1:R2::p1:p2, where the product of the means is equal to the product, or: (R1 x P2) = (R2 x P1) (Equation 4)
where R1 = the desired F/M ratio
= 0.25 for the above example.
R2 = the MLTSS corresponding to the desired F/M ratio, or 1.0 pounds for an F/M ratio of 0.25.
P2 = optimal MLTSS
P1 = weight of food in aeration from
Equation 3, or 992 lb/day in the above example.
Therefore, for the example plant:
(0.25 x P2) = (1.0 x 992)
or, P2 = (1.0 x 992)/0.25 = 3968 lbs MLTSS.
The optimal aeration MLTSS concentration can be found by rewriting Equation 3 as:
ppm = (pounds per day MLTSS)/(8.34 x MG)
ppm = (3968 lb/day)/(8.34 x 0.25 MG) = 1900.
Therefore, 1900 ppm is the maximum concentration of MLTSS to achieve a F/M ratio of
0.25.

...........
Bài khá dài, chỉ cắt một đoạn cần thôi.
Còn dưới đây là chương trình tính tự động chỉ số F/M(online)
 
Sửa lần cuối bởi điều hành viên:
Top