The main costs of a wastewater treatment plant are sludge disposal, chemical consumption, power, maintenance, and personnel. This article will detail cost budgeting methods based on the actual operation of wastewater treatment plants.

 

 

0 1

Sludge disposal cost

0 1 Sludge yield coefficient per 10,000 tons

(Estimate this way if no specific data is given in the feasibility study)

SS less than 100, coefficient 1.0;

SS between 100 and 200, coefficient 1.3;

SS greater than 200, coefficient 1.6

That is: 10,000 tons of water produces 1.6 tons of 20% absolutely dry sludge, which is converted to sludge with 80% water content as 1.6/0.2=8 tons. Converted to sludge with 60% water content is 4 tons.

02 Mud and water conversion relationship example

(1) 1 ton of sludge with 80% water content, how to calculate the water content after drying to 0.5 tons of sludge?

Answer: After drying, there is 0.5 tons of sludge, which means 0.5 tons of water has evaporated.

The original water content is 80%, which means there is 0.8 tons of water, and 0.3 tons of water remains.

The water content is 0.3÷0.5=60%

(2) The relationship between the amount of absolutely dry sludge and the amount of treated water and concentration. The amount of absolutely dry sludge is equal to the amount of water treated by the equipment multiplied by the sludge concentration:

For example, if the equipment treats 10 m3/h of water, and the sludge water content is 99.5%, the relative sludge concentration is 0.5%. The amount of absolutely dry sludge of the equipment is 10*0.005*1000=50kg. The amount of treated water multiplied by the sludge concentration and then multiplied by 1000 is equal to the number of kilograms processed per hour. This calculation results in 50kg/h.

Knowing the amount of absolutely dry sludge to calculate the amount of water treated, the calculation method of the equipment is: the amount of absolutely dry sludge divided by the sludge concentration/(1-water content), for example, if the equipment treats 60kg/h of absolutely dry sludge, and the water content is 99%, the amount of water treated is 60/(1-0.99)=6000kg, one ton is equal to one thousand kilograms, so it is equal to 6T/h.

03 Relationship between sludge water content and density

04 Unit of absolutely dry sludge

The unit of absolutely dry sludge is kgDS/d, and the empirical value of PAM cationic dosing is 4g/kgDS.

05 Sludge disposal fee

The above is the calculation of sludge quantity. The sludge disposal fee is generally charged by the ton, and the price is determined according to the local industry situation. In addition, the sludge disposal fee also includes the sludge transportation fee. Some disposal companies are responsible for transportation, so the transportation fee is included in the disposal fee.

 

 

0 2

Reagent addition cost

0 1 PAC and PAM anion addition

(1) PAC and PAM preparation concentration (taken from the reagent packaging bag and added to the dissolving tank)

According to experience, the reagent preparation concentration is: 5%-10% for PAC in the dissolving tank, and 0.1%-0.3% for PAM. The above data is based on mass ratio, which means that 50-100kg of PAC and 1-3kg of PAM are added per cubic meter of water. This concentration is relatively high, and the solubility of PAM is limited. A medium-speed stirrer needs to be fully stirred to completely dissolve it. The dissolution concentration can be appropriately increased to 0.3-0.5% in summer. Taking the PAC dissolution concentration as 10% and the PAM dissolution concentration as 0.5%, 100kg of PAC and 5kg of PAM are dissolved per cubic meter of water respectively. Adjust the flow rate of the diaphragm metering pump, calculate according to 1 cubic meter/24 hours, that is, Q=42 liters per hour, to achieve the ideal flocculation effect.

(2) PAC and PAM addition

The concentration of chemical agents added to wastewater is generally 50-100ppm for PAC and 1-2ppm for PAM. The unit ppm is one millionth, so the conversion is 50-100g of PAC and 1-2g of PAM per ton of wastewater. Manufacturers generally recommend trial use according to this addition amount. If the daily wastewater treatment volume is 2000 cubic meters, and the PAC addition concentration is 50ppm and the PAM addition concentration is 2ppm, then the daily PAC usage is 100kg and the PAM usage is 4kg.

(3) Accurate calculation of PAC addition:

The concentration of phosphorus to be removed in the deep treatment section is 1.5mg/L, multiplied by the water volume is the amount of phosphorus to be removed: 1.5*20000=30kg/d. It is known that 27g of aluminum is needed to remove 31g of phosphorus, and the addition coefficient is 2, so the addition amount of Al is 30/31*27*2=52.258kg/d.

Al ₂ O ₃ Al accounts for 54/102 in ₂ O ₃ The addition amount of Al is 52.258*102/54=98.8kg/d, and the effective content is 10% Al ₂ O ₃ The required amount of PAC is 98.8/10%=988kg/d, and the converted addition amount is 988/20000=49.4mg/L. After simplification: (102*effective content*x/31) mg/L, where x is the concentration of phosphorus to be removed.

Empirical values:

Removing 1.5mg/L of phosphorus requires a PAC addition of 50mg/L.

Removing 2.0mg/L of phosphorus requires a PAC addition of 66mg/L.

(4) Accurate calculation of PAM addition:

PAM (polyacrylamide) is divided into three categories according to its structure: cationic (CPAM), anionic (APAM), and non-ionic (NPAM). Anionic PAM is mainly used in the coagulation and sedimentation of industrial wastewater (containing more metal cationic charges), cationic PAM is widely used in the sludge dewatering section of domestic sewage, flocculating negatively charged colloids; non-ionic PAM is mostly used with inorganic flocculants such as polyaluminum and polyferric, and is more suitable for acidic wastewater.

Cationic PAM is generally prepared at a concentration of 3‰ for dewatering of domestic sewage sludge. When the dosage is 0.8mg/L~1mg/L, a sludge cake with lower water content can be obtained. Generally, about 4kg of solid cationic PAM is needed for every 1m3 of wet sludge with a water content of about 80%. (The specific chemical dosage is related to the sludge concentration in the thickening tank).

Anionic PAM dosage: 1~2mg/L.

02 Sodium hypochlorite dosage

(1) Disinfection:

Generally, wastewater treatment plants use industrial-grade sodium hypochlorite with an effective chlorine content of 10%. The density of this solution is 1.18g/mL. Design specifications require that the concentration of NaClO added to wastewater treatment plants is 6mg~15mg/L (in terms of effective chlorine). The specific dosage needs to be determined in combination with the actual effluent water quality, through the determination of fecal coliform count and residual chlorine.

(2) Breakpoint chlorination:

To remove 1mg/L of ammonia nitrogen from wastewater, according to the ratio of 1:0.088, 100 tons of wastewater requires 8.8 kg of 10% sodium hypochlorite, which is converted to a dosage of 8800g/1.18=7457ml≈7.5L. (88mg/L)

03 Cationic PAM dosage

0.8~1mg/L or 4g/kgDS

04 Defoamer dosage

The dosage of Al salt dephosphorizing agent is shown in the PAC dosage. The calculation of Fe salt dephosphorizing agent is similar to that of Al salt. The calculation is as follows, taking ferrous sulfate (FeSO ₄ ) as an example:

The concentration of phosphorus to be removed in the deep treatment section is 1.5mg/L, multiplied by the water volume is the amount of phosphorus to be removed: 1.5*20000=30kg/d. It is known that 56g of iron is needed to remove 31g of phosphorus, and the dosage coefficient is 2, then the dosage of Fe is 30*56/31*2=108.39kg/d.

FeSO ₄ Fe accounts for 56/152 in ₄ , then the dosage of FeSO ₄ is 108.39*152/56=294.2kg/d. For FeSO with an effective content of 90%

05 Carbon source dosage

(1) COD equivalent

The COD equivalent of a carbon source can be understood as the number of milligrams of oxygen required when a unit volume or unit mass of the carbon source is completely oxidized, with units of mg/L, mg/g or mg/kg.

Carbon sources are divided into dry agents and liquids. For dry agent carbon sources, we generally calculate by unit mass, generally in grams or kilograms. For example, the COD equivalent of 1Kg of dry glucose powder is 1Kg, which is converted to the commonly used mg unit as 1 million mg/Kg. For liquid carbon sources, the industry uses liters (L) as the unit. Many carbon source manufacturers claim that the COD equivalent of liquid carbon sources is hundreds of thousands, which is the amount of COD of 1L of liquid carbon source. We feel it is amazing, but after conversion, it is not as high as the COD equivalent of 98% glucose (density is calculated as 1kg/L).

Of course, the COD equivalent is not the only evaluation index for the quality of carbon sources. Utilization rate, cost performance, denitrification efficiency, carbon source residue, etc., should also be considered. Therefore, low COD equivalent does not necessarily mean that it is not a good carbon source! For example, sodium acetate is a recognized good carbon source, but its dry agent COD equivalent is only 780,000 mg/Kg, and many liquid sodium acetates are only around 300,000 mg/L. The COD equivalent is not as good as glucose, and the price is several times that of glucose, but it is indeed a better carbon source than glucose!

(2) Sodium acetate dosage calculation

5 times the TN to be removed * influent volume / 0.78

0 6 Reverse osmosis chemical dosage

Reverse osmosis antiscalant dosage: 3mg/L

Reverse osmosis reductant dosage: 3-5mg/L

0 7 Power cost calculation

Project	Daily power consumption (kw.h/d)	Electricity price (yuan/kw.h)	Daily electricity cost (yuan/d)	Daily water treatment volume (T/d)	Electricity cost per ton of water (yuan/T)	Power consumption per ton of water
Electricity cost	8000	0.8	6400	20000.00	0.3200	0.4000
Basic electricity charge	Transformer capacity KV.A	Basic charge yuan/month.KVA	Monthly charge (yuan/month)	Daily treatment capacity (T/d)	Electricity cost per ton of water (yuan/T)
	2000	22	44000	20000.00	0.0723
Total					0.3923

0 8 Maintenance cost calculation

The annual routine maintenance rate is calculated as 1% of the total equipment investment, and the annual overhaul rate is calculated as 0.5% of the total equipment investment.

 

 

0 5

Labor cost calculation

1 plant manager, 1 deputy plant manager, 3 department managers, 1 finance staff, 2 office staff, 1 procurement staff, 1 warehouse manager, 3*3=9 main control staff, 6 maintenance staff, 3 laboratory staff, 3 canteen staff, 2 cleaning staff, 3 landscaping staff, 4 sludge pressing staff, totaling 40 people. (Increase or decrease according to the actual situation of each wastewater treatment plant, and wages are calculated specifically according to the local salary level of the same industry).

 

 

0 6

Outsourcing testing and laboratory fees, etc.

1. Standard wastewater treatment plants should have a dedicated laboratory capable of testing various conventional pollutants in the influent, effluent, and process segments. Costs should include not only infrastructure and equipment but also consumables and laboratory glassware.

2. Almost all wastewater treatment plants currently require online monitoring equipment for influent and effluent. This cost includes equipment purchase and maintenance.

3. Based on the wastewater treatment plant's discharge permit requirements, the plant needs to implement a self-monitoring program. Items that the plant's laboratory cannot test must be outsourced to qualified third-party organizations for testing and reporting. This cost is the third-party testing fee.

4. Online monitoring equipment requires quarterly calibration and a calibration report to verify the accuracy of the monitoring data. This cost is the online monitoring calibration fee.

 

 

0 7

Tap water fee

The main use of tap water in wastewater treatment plants is in the chemical dosing area, where it is used for preparing chemicals. The amount of water depends mainly on the plant's design scale and process, and employee water usage may also be considered.

 

 

0 8

Other expenses

Other expenses include heating, food, landscaping, management, and employee welfare. For outsourced operations, profit and taxes must also be considered.