Hi,
I want to calculate the amount of feedstock that is needed to produce 1500m^3 biogas per day with anaerobic digestion. I have tried so many different ways to calculate but can't seem to get is quite right. The BMP is 0,204 m^3 CH4/kg VS, the VS content is 90% and TS is 12%. Now I am thinking about this approach:
Total VS = 1500 * 0,6 / 0,204
Biomass [kg] = Total VS / (VS[%] * TS[%])
This gives me 21 tons feedstock per day, which I think sounds a bit too much?
Would be vary thankful for some input and guidance in how to calculate!
Dear Hanna Svensson ,
Your calculation formulas are correct but the result is 4412 kg VS/d and in 4412/(12% x 90%) = 41 tons/d of wet biomass.
These high VS and biomass loads are due to the very low 0,204 Nm³ CH4/kgVS from the BMP test.
In our full scale AD(+) plants we get 0,75 Nm³ CH4/kgVS-removed. At 60% VS removal the BMP should be 0,75/60% = 1,25 Nm³ CH4/kgVS which is 6 times higher! Hence it seems that the BMP test has not been performed correctly.
Thank you so much for the help!
Do you know how the methane yield is related to the methane concentration? Since the methane yield doesn't seem to be correct I would like to do a sensitivity analysis to see how different values of the yield affect the required biomass. Can I then vary just the methane yield or is it correlated to the methane concentration? In calculations above I multiplied the biogas volume with 0,6 (60% methane concentraiton) to get the methane gas volume.
My pleasure Hanna.
The methane concentration in the biogas depends primarily on the composition of the feedstock (sugars, proteins, ...) as well as on the operating pH. Higher pH results in less CO2 in the biogas and hence higher methane concentrations. In some cases we get up to 85% methane.
- Badges
- Science topic
- Similar topics
- Agricultural Science
- Agronomy