Primer - Working With Total Solids, Fixed Solids and Volatile Solids in Sludge and Biosolids Michael D. Doran, DEE Professor of Practice Department of Civil & Environmental Engineering University of Wisconsin-Madison August 2013 Often as environmental engineers we must perform calculations and analysis of data that involve volumetric and mass flows of sludge and biosolids streams of varying concentrations of solids (raw, digested, thickened, dewatered, etc.). This can be a challenge since it involves terms used in practice that many of us may not be familiar with. First Let s Define some Terms the most important terms used with these concepts are: Term Meaning Significance Sludge A suspension of residuals from a treatment process in water. This may be chemical solids and settled and/or filtered suspended solids from a water treatment plant, or settled primary solids, waste activated sludge, or filter backwash from a wastewater treatment plant. Sludge normally has to processed further so that it can be managed in an economical and environmentally sound fashion. Sludge is frequently dilute (high proportion of water to total mass) and needs concentration to reduce its volume for economic management. From a wastewater treatment plant, sludge normally contains a high fraction of unstable organic matter that requires treatment prior to management of the material in an environmentally sound Biosolids Total Solids (%) Abbreviated TS Wastewater treatment sludge that has been treated biologically or by other means (such as high temperature drying) to render it more stable (lower organic concentration) and to reduce pathogen levels. The percentage of the original mass of a sample that remains after water is evaporated from the sample in a laboratory drying oven at 103 C. So, if 20 g of sample were placed in an evaporating dish of known weight and, after drying, the sample weight 1.0 g more than the dish, the sample would have had 5.0% TS. (1.0/20)100%. manner. Wastewater biosolids can be applied to land as part of a beneficial reuse program. Wastewater sludge normally cannot. Important when the total mass of solids in a stream is to be computed or when calculations involving the total solids are to be performed. 1
Term Meaning Significance Fixed Solids (%) Abbreviated FS Volatile Solids (%) Abbreviated VS Volatile Solids Destroyed (%) Specific Gravity Abbreviated as sp gr Thickening The fraction as a percent of the total solids (TS) that remain in a sample after combustion in a laboratory furnace at 600 C. So, if the 1.0 g of solids from the TS along with the dish from the above example were placed in a 600% furnace until all of the combustible material was removed from the sample, and 0.25 g of material remained (above the dish weight), then the sample would have 25% fixed solids (25% FS). (0.25/1.0)100%. The fraction as a percent of the total solids (TS) that are removed by the sample from combustion in a laboratory furnace at 600 C. This is determined by subtracting the FS (from above) from 100%. So, for above, if the FS were 25% the VS would be 75%. During digestion or other process for stabilizing organic sludge or other material (usually wastewater treatment or solid waste management) the fraction of the VS present that are converted to methane, carbon dioxide, or other gases). The VS destroyed is determined by comparing TS, VS and FS values before and after the biological process. The mass of a volume of material compared to the same volume of water. For example if 20 ml of a sludge sample was weighed in an evaporating dish (prior to evaporation) and the mass was found to be 21 g greater than the weight of the dish, the specific gravity would be 1.05 (20 ml of water would weigh 20 g). (21/20=1.05) A process to increase the TS in sludge or biosolids. Normally unthickened sludge would have a TS of 0.5% - 3.0% whereas a thickened sludge would have a TS of 4% to 6% Representative of the solids that can be considered inert in that they are not degraded or broken down in digestion or would not be available as food for microorganisms or other life forms. Representative of the solids that can be degraded or broken down by biological activity and may provide food for microorganisms and other life forms. Representative of the performance of a sludge stabilization process, the higher the VS destroyed the more efficient the process. Representative of the stability of the treated material (the lower the remaining VS the more stable). Important factor to be more precise with calculations when solids concentrations exceed several percent. This is because volumes of sludge and biosolids weigh more than the same volume of water (ratio is the specific gravity) and this needs to be accounted for when performing calculations involving mass flow or volumetric flow. Can dramatically reduce the volumetric flow rate of sludge or biosolids that must be handled in further treatment or removed from a treatment plant for beneficial reuse. Care must be taken that flows and loadings returning to the treatment processes from the supernatant, overflow or filtrate from the thickening process are accounted for in the design flows and loadings of the treatment processes, as they can be a significant fraction of the process loadings. 2
Term Meaning Significance Dewatering A process to increase the TS in sludge or biosolids. Normally dewatered sludge or biosolids would have a TS of 20% to 30%. Can dramatically reduce the volumetric flow rate of sludge or biosolids that must be handled in further treatment or removed from a treatment plant for beneficial reuse. Care must be taken that flows and loadings returning to the treatment processes from the supernatant, overflow or filtrate from the thickening process are accounted for in the design flows and loadings of the treatment processes, as they can be a significant fraction of the process loadings. Where did some of these terms come from? It might help to think about how the parameters are measured (and the limitations of that) to better understand why the terms above are used. It starts in the laboratory where the samples are analyzed using an accurate lab balance, and evaporating dishes that are of known mass (they are weighed after washing and drying), and by placing known volumes of sample in evaporating dishes for treatment. The samples can be weighed to determine the weight of the dish and the known volume of sample. Subtracting the dish weight gives the weight of the sample, and the weight per unit volume can be compared with water (1.0 mg/ml) to arrive at the sample specific gravity. After drying at 103 C, following evaporation of water, weighing the sample and subtracting the dish weight allows the TS to be determined (mass of dried material divided by original mass of material). After combustion in a lab furnace at 600 C the fixed (non-volatile) solids percentage (FS) of the original dry mass can be determined by weighing the sample, subtracting the dish weight and comparing with the weight of sample after evaporation. This, expressed as a percentage, can be subtracted from 100% to arrive at the volatile fraction percentage (VS). This graphic may help explain how this is done. 3
A note on suspended solids vs. total solids: Suspended matter is captured on a 0.45 µ (0.45 micro meter) filter. Total solids therefore includes suspended (larger than 0.45 µ) matter plus smaller particles (colloids for example) and all dissolved matter. For sludge, the ratio of suspended to total matter is so high that 100% is usually assumed for calculations. For example, if suspended solids removed in a primary clarifier amounted to 5,000 kg/d at 3.0% solids, the concentration of the suspended solids (neglecting sp gr) would be about 30,000 mg/l. because 1.0 L weighs 1 x 10 6 mg and 3.0% of that would be 30 x 10 3 mg. Now, if there were 500 mg/l of dissolved solids (typical value for domestic wastewater) you can see that only a very small error results in Solids are normally given in total solids (TS) rather than total suspended solids (TSS) for sludge and biomass. treating the settled suspended matter as TS (500/30,000)(100%) = 1.7%. Usually, therefore, in sludge or biosolids streams the distinction between suspended solids (TSS) and total solids (TS) is ignored. Another problem is that you cannot effectively measure TSS in a sludge or biosolids sample because a filter must be used that will quickly blind so that inadequate sample can be filtered to arrive at a reasonable result. Computing Mass Flowrate from Volumetric Flowrate and Concentration this is a simple exercise as long as you keep track of your units and think carefully about what you are trying to achieve in the calculation. Here is an example: How many kg/d of TS are contained in 100 m 3 /d of sludge of 5.0% TS having a sp gr of 1.01? = QC = (100 m 3 ) (1,000 kg Water (1.01 kg Sludge) ( 0.05 kg TS) = 5,000 kg/d of TS (d) (m 3 water) (kg Water) (kg Sludge) Note that this is the dry matter (not including water) contained in 100 m 3 /d of the sludge. Computing Volumetric Flowrates from Mass Flowrates and Concentration similar to above, here is an example: How many m 3 /d must be pumped for a mass flow of 5,000 kg/d of TS at 5.0% TS and a sp gr of 1.01? Q = = (5,000 kg TS)(m 3 Water)(kg Water) (kg Sludge) = 99 m 3 /d ~ 100 m 3 /d C (d) (1,000 kg)(1.01 kg Sludge)(0.05 kg TS) Computing Concentration from Mass Flowrate and Volumetric Flowrate similar to above, here is an example: If 100 m 3 /d of sludge was pumped and it was known that this contained 5,000 kg/d of TS at a sp gr of 1.01, what is the TS concentration in %? C = = (5,000 kg TS) (d) (m 3 Water) (1.01 kg Sludge)(100%) = 5.0% Q (d) (100 m 3 )(1,000 kg)(kg Water) (1.0) Working with Digestion, VS, FS and VS Destruction I have seen a lot of errors made in my career with this type of calculation, so do not be hard on yourself if you find it a bit elusive at first. All becomes pretty clear when you think about the fact that the FS (in kg/d of fixed solids) is not affected by digestion since they are not broken down. Then, if a certain percentage of the VS in the feed material (say 60% of the kg/d of VS in the feed) is converted to biogas (it is not really destroyed), the TS after digestion (in kg/d) is the FS (in kg/d) plus the VS that has not been destroyed (n kg/d). Here are some examples: 4
5,000 kg/d of undigested sludge (mass of dry solids basis) is feed at 5.0% TS and 75% VS to an anaerobic digester. If 65% of the VS are destroyed what mass/day of VS is that, what is the mass/day of TS in the digested biosolids, and what is the %TS concentration in the digested sludge? Dest = f( ) = (0.75 VS)(0.65 Dest)(5,000 kg/d) = 2,400 kg/d VS Dest = - Dest = 5,000 kg/d 2,400 kg/d = 2,600 kg/d of Digested Biosolids (mass of dry solids basis) %TS Biosolids = (%TSRaw)(kg/d Digested) = (5.0%)(2,600)/(5,000) = 2.6% (kg/d Raw) A digested sludge contains 62% VS and the influent VS was 75%. What percent VS was destroyed during digestion? One way to think about this is to track 100 kg of feed sludge. That would contain 75 kg of VS and 25 kg of FS if the sludge were 75% VS (75% of TS are VS). After digestion the 25 kg of FS would remain, and this would amount to 38% of the TS after digestion (62% are VS). The TS (from the original 100 kg must therefore be 25/0.38 = 66 kg. Since 62% of that is VS, the VS remaining would be 41 kg (0.62)(66)kg. On this basis we can compute the percentage of the original VS destroyed as [(75-41)/75]100=45%. Many folks asked this question would simply say: %VS destroyed must be just 75%-62%=13% VS based on the original information. This is of course wrong you need to think about this on a mass balance basis! 5