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By Robert Rapier on Feb 3, 2009 with no responses

TDP Chemistry

While the most recent two posts (here and here) on Changing World Technologies’ thermal depolymerization (TDP) process were reposts, this one has not been previously posted here. Regular reader “Optimist” was initially optimistic about the TDP technology. During his investigation, he went through the chemistry of the process, and came away feeling a bit short-changed.

——————-

Perhaps the confusion about reasonable yields from TDP can be addressed by looking at the chemistry, as proposed by Dr. Adams at http://web.mit.edu/10.391J/www/0405SE05adams.pdf (RR: now hosted at) http://ergosphere.files.wordpress.com/2007/04/tdp_0405se05adams.pdf. As I see it, the conversion can be presented as three chemical reactions: hydrolysis, decarboxylation, and product degradation (for lack of a better term).

1. Hydrolysis (aka depolymerization)

H2-C-O-R
…….|
..H-C-O-R + 3H2O => C3H8O3 (glycerol) + 3 CH3-(CH2)14-COOH
…….|
H2-C-O-R

where R = CO-(CH2)14-CH3

Of course, there are many different fatty acids that occur, rather than the palmitic acid (C16) shown. However, for the sake of calculating the yield, the choice of fatty acid (typically in the C12 – C18 range) would have a fairly minor affect. Thus, palmitic acid will be used as an example.

2. Decarboxylation

CH3-(CH2)14-COOH => CH3-(CH2)13-CH3 +CO2

3. Product Degradation

CH3-(CH2)13-CH3 => 7C + 8CH4

Product degradation is necessary to explain the presence of carbon and low BTU gas in the products from TDP. While the actual reactions are likely to be very complex, the balanced reaction shown will suffice for the purpose of calculating yields.

Assumptions

1. All available fat/oil undergoes hydrolysis and all the fatty acid thus produced undergoes decarboxylation.

2. Deciding what fraction of the product undergoes degradation is more difficult. From Figure 5, http://ergosphere.files.wordpress.com/2007/04/cwt_genconflasvegas3_3_04.pdf it appears that CWT was originally expecting about a 10:1 mass ratio between oil produced and carbon produced. Using this ratio, it is estimated that about 20% of the product undergoes degradation. The calculation below will illustrate how this leads to the desired 10:1 ratio.

3. The entire feedstock consists of oil/fat.

4. All separations are perfect, i.e. all the fatty acid from the first stage proceeds to the second stage and all remaining product is recovered as TDP/TCP-40 oil.

From these assumptions, it would be obvious that the calculated yield would represent a maximum theoretical yield, as separations in a real plant can never be 100% perfect and no waste feedstock is going to be 100% lipid.

Calculations

The original fat (glyceryl tripalmitate) has a molecular mass of 806, the final oil product 212. Since each unit of fat produces three units of oil, the yield (before degradation) is:

3 x 212/806 = 78.9%

Factoring in degradation, the remaining yield (80%) would be: 78.9 x 0.8 = 63.1%

Proceeding in this manner the yield of all products can be calculated. The results are summarized below. Note that water is consumed during hydrolysis, hence the negative yield.

Product Yield w/o degradation Yield with degradation
Oil 78.9% 63.1%
Carbon dioxide (CO2) 16.4% 16.4%
Methane (CH4) 0 9.5%
Carbon (C) 0 6.3%
Glycerol 11.4% 11.4%
Water -6.7% -6.7%
TOTAL 100.0% 100.0%
Total gas production 16.4% 25.9%

Now let us do some comparison to Figure 5, which Dr. Adams is still holding up as a valid mass balance for the TDP process. On the left we have 92.9 t/d entering the process and on the right we have 69.8 t/d of oil produced, thus allowing for a yield of 69.8/92.9 = 75.1%! As our calculations, show that would only be possible with very limited product degradation.

Now look at the fuel gas production, listed as 7.5 t/d for a yield of 8.1%. As our calculations show, even with no product degradation you will produce twice as much gas as that. With product degradation, we expect to see THREE times as much gas. Did CWT forget to factor in the CO2 that would be produced? Note that a lipid feedstock would contain oxygen which would not be present in the hydrocarbon product, and would have to the report to the fuel gas. At the same time the oxygen would limit the fuel content of the fuel gas. If the oxygen is in the form of CO2, it would have NO fuel value. If it is in the form of CO, it takes valuable carbon from the oil product.

Keep in mind, these are theoretical maximum yields. Look at the four assumptions. Most importantly, we have not factored in the fact that any protein in the feedstock is effectively lost as amino acid fertilizer. According to the label on the turkey, the feedstock is 2/3 protein. If true, yields would significantly below what has been shown above. [Proteins are polymers constructed out of 20 different monomers, a.k.a. amino acids, as discussed at Wikipedia: http://en.wikipedia.org/wiki/Amino_acids. Of the 20, only 9 are nonpolar, i.e. hydrophobic or fat soluble. While this does not mean that only 45% of hydrolyzed turkey protein would dissolve in fat (turkey protein may contain a greater/smaller fraction of the fat soluble amino acids), it does point to a limited contribution of protein to the hydrocarbon product. Especially if one bears in mind that each amino acid contains at least one -COOH group (which would convert to CO2), at least one -NH2 group (which would be released as ammonia) and two contain sulfur which would be released as H2S.]

Let us now consider the actual plant performance (300 bbl/d from 270 t/d of waste). 300 bbl/d x 42 gal/bbl x 7.05 lb/gal / 2,000 lb/ton = 44.4 t/d oil. Assuming CWT got the make-up of the feedstock (left hand side of Figure 5) right, 270 t/d of feedstock would consist of 44% organics or 119.4 t/d. Based on the maximum theoretical yield (with degradation), the feedstock must include at least 44.4/0.631 = 70.4 t/d of fat. Thus, in spite of what the label claims it appears that the feedstock is at least 70.4/119.4 = 59% fat.

Bottom line: CWT’s original mass balance of the TDP process is a flight of fantasy, most likely because they forgot about the CO2 that the process produces. By continuing to present this mass balance as representative of plant performance, they are deliberately misleading people. Specifically, they are claiming a conversion efficiency that simple chemistry proves impossible to achieve. They should by now be able to factor in the correct gas production, carbon production and unconverted organics reporting to the “fertilizer” stream. The fact that they are reluctant to do so speaks volumes.

As for the actual conversion efficiency, I think most people will agree that the main product of interest is the oil. Yes, energy in the gas and carbon products can be recovered, but the main excitement is about the oil. Energy in the high-quality fertilizer (all water soluble monomers) are effectively lost, as this energy will not be recovered by the TDP facility. (The energy in that fertilizer may save a plant using the fertilizer some energy, thus increasing the yield of such a crop plant.) Assuming the reported conversion (300 bbl/d from 270 t/d of waste) is correct, that would mean 158 MM BTU/h in the feedstock and 63.3 MM BTU/h in the oil, for a conversion efficiency of 40%. MUCH below the original claim of 85%!