Our project aims to use Shewanella bacteria for phosphorus accumulation and electricity generation in space. Our current idea involves three cycles:
In the following discussion, we will quantitatively calculate the flow of phosphorus elements and corresponding energy in each path, and finally draw a conclusion that introducing Shevanella into space system can save 70.58% of phosphate fertilizer in space phosphorus cycling.
(a) The bacterial metabolic rate is calculated based on the amount of substrate consumed per hour or per minute.
(b) The metabolic rate of Shewanella bacteria is similar to that of common heterotrophic bacteria.
(c) Shewanella 's metabolic electricity generation mechanism.[1]
Suppose we have:
Then we get:
In summary, each lactose molecule can produce a maximum of 12 electrons and a minimum of 4 electrons.
The detailed derivation process can be found in ODE MODEL 1.2.3.
Each bacterium consumes approximately 10-11 moles of lactate per hour.
Each Shewanella bacterium consumes about 10⁻¹¹ moles of lactate per hour. Converted to a per-minute consumption rate:
$$ \frac{\mathrm{1}\mathrm{0}^{\mathrm{-11}}\mathrm{mol/h} }{\mathrm{60min{/}h}}\mathrm{=1.67}\times\mathrm{1}\mathrm{0}^{\mathrm{-13}}\mathrm{mol/min} \tag{1.1} $$
1. Total bacterial number: Assume the total number of bacteria is 108 based on experimental data.
2. Total lactate consumption
$$ \mathrm{1.67}\times\mathrm{1}\mathrm{0}^{\mathrm{-13}}\mathrm{mol/min}{\times}\mathrm{10^8}=1.67*10^{-5}\mathrm{mol/min} \tag{1.2} $$
Under this assumption, $10^8$ Shewanella bacteria consume approximately $1.67*10^{-5}$ moles of lactate per minute, which is about $0.167$ micromoles of lactate.
Coulombic efficiency represents the efficiency with which microorganisms convert the chemical energy in substrates into electrical energy. Coulombic efficiency formula is:
$$ CE=\frac{I\cdot t}{n\cdot F\cdot\Delta C} \tag{1.3} $$
(a). Calculation of Theoretical Charge Equivalent of Electrons: Based on the previous calculation (1.2), assume that consuming one mole of lactose produces 4 moles of electrons. Thus, the electron equivalent of lactose is:
$$ n=1.67\times\mathrm{1}\mathrm{0}^{\mathrm{-5}}\mathrm{mol}\times\mathrm{4=6.68}\times\mathrm{1}\mathrm{0}^{\mathrm{-5}}\mathrm{mol\ }\mathrm{e}^- \tag{1.4} $$
Then, using the Faraday constant to convert the number of moles of electrons into charge:
$$ electrons\ into\ charge =n\times F\mathrm{=6.68}\times\mathrm{1}\mathrm{0}^{\mathrm{-5}}\mathrm{mol}\times\mathrm{96485}C/\mathrm{mol=6.44}C \tag{1.5} $$
(b). Calculation of Actual Charge: Actual charge comes from the measured current. According to our team’s experimental results, the actual current is $I = 1500 μA/cm²$. Assuming an electrode area of 2 cm², the total current is $300 μA$.
$$ actual\mathrm{\ c}harge=I\times\mathrm{t=3000}\times\mathrm{1}\mathrm{0}^{\mathrm{-6}}A\times\mathrm{60s=1.8}\times\mathrm{1}\mathrm{0}^{\mathrm{-1}}C \tag{1.6} $$
(c). Calculation of Coulombic Efficiency: $$ CE=\frac{1.8\times10^{-1}C}{6.44C}\times100%=2.8% \tag{1.7} $$
Under these conditions, the Coulombic efficiency of the electricity-generating bacteria is approximately $2.8%$. This means that only $2.8%$ of the lactate metabolic energy is converted into electrical energy, with the rest likely used for bacterial growth and other metabolic activities.
In a space station cycle using potato cultivation as an example, every $100g$ of potatoes contains $40mg$ of phosphorus, while an adult needs 700mg of phosphorus per day ($70%$ can be absorbed, the rest excreted in feces). This means an intake of $1g$ of phosphorus is needed, requiring about $2500g$ of potatoes, or ten potatoes.
According to our team's experimental results, the phosphorus recovery efficiency of Shewanella is about $30%$, meaning that about $30%$ of the phosphorus in astronauts' urine and feces can be recycled.
For potato cultivation, each $100g$ of tuber requires $0.2g $ of phosphate fertilizer, of which phosphorus constitutes about $8.5%$. Therefore, every $100g$ of potatoes requires $17mg$ of phosphorus (most of the phosphorus comes from the original soil).
For a typical space station with three people, $3g$ of phosphorus is needed daily (equivalent to $7500g$ of potatoes), of which Shewanella bacteria can recover about $0.9g$ of phosphorus. Growing $7500g$ of potatoes requires $1.275g$ of phosphorus aside from what is provided by the soil.
$$ \frac{0.9g}{1.275g}\times100%=70.58% \tag{2.1} $$
In conclusion, the phosphorus recovery efficiency of Shewanella bacteria is about $30%$, which can save $70.58%$ of phosphate fertilizer in space phosphorus cycling.
Furthermore, according to experimental results, $10^8$ bacteria can accumulate $0.2mg$ of phosphorus, so to handle the daily waste of three astronauts, about $15$ liters of bacterial solution ( at $10^8 cells/ml$ ) would be required.
[1]: Kouzuma A: Molecular mechanisms regulating the catabolic and electrochemical activities of Shewanella oneidensis MR-1. Biosci Biotechnol Biochem 2021, 85:1572-1581.