Team:Nanjing-China/Model
Model
Bacterial growth model
Model introduction:
In this experiment, this model can be approximated that the bacterial concentration is directly proportional to the OD value, and 1 OD corresponds to \(1.44×10^8\) cells/ml.The phosphorus concentration of the sewage in the device is 8mg/l. After operating the equipment for a period of time, the original sewage phosphorus in the device is removed. At this time, new sewage is introduced for further treatment, and the phosphorus concentration of the sewage is still 8mg/l. But in other experiments, we changed the water inlet speed to explore the optimal water inlet speed.
Model content:
We assume that the bacterial growth follows the Logistic Model, setting the bacterial concentration to x, and the blocking function as r(x).
Then there is \(\frac{dx}{dt}={r(x)x}\),\({x(0)}={x_0}\)
Assume that \({r(x)}={r(1-\frac{x}{x_m})}\)
Then \({x(t)}=\frac{x_m}{1+(\frac{x_m}{x_0}-1)e^{s(t)}}\)
\({s(t)}={s_0+s_1t+s_2t^2+s_3t^3}\)
When the water inlet speed is 5 L/h, \({x_m}={167.7437}\) (unit: \(10^6\) cells/ml);\({s_0}={0.1632}\);\({s_1}={-0.5126}\);\({s_2}={0.0875}\);\({s_3}={-0.0060}\).
When the water inlet speed is 10 L/h, \({x_m}={345.1465}\);\({s_0}={-0.0483}\);\({s_1}={-0.1042}\);\({s_2}={-0.0333}\);\({s_3}={0.0014}\).
We comprehensively analyzed the relationship between the influent velocity and the growth function, trying to construct a function of the growth curve parameters with respect to the influent velocity.
Since there are too many parameters in \({s(t)}={s_0+s_1t+s_2t^2+s_3t^3}\),we simplify it and only consider \({s(t)}={s_0+s_1t}\).
Assume that the water inlet speed is u (unit L/h),then \({s_0}={0.28+0.02u}\),\({s_1}={-0.35}\),\({x_m}={34u}\).
Usually, the initial bacterial concentration is about \(25×10^6\) cell / mL.
\({x(t)}={\frac{34u}{1+(1.36u-1)e^{0.28+0.02u-0.35t}}}\)
We selected five values of u=5, 8, 10, 12, and 15 to build five growth curves.
Figure 1: Growth curve of bacteria
Experimental device absorption phosphorus model
Model introduction:
According to the operation scheme of the device, there are 25L of sewage (phosphorus concentration of 8mg/L) in the device at the beginning, and within 4 hours, the bacteria in the device can absorb all the phosphorus. After the fourth hour, the sewage was introduced into the device (phosphorus concentration was still 8mg/L), and the bacteria continued to absorb the phosphorus in the newly-introduced sewage. At the beginning, due to the bacteria have strong ability to absorb phosphorus, the phosphorus is always in the state of "supply in short supply" means that the concentration of phosphorus in the water in the device is generally zero. However, after a certain period of time, due to the weakening of the ability of bacteria to absorb phosphorus, the phenomenon of "phosphorus penetration" occurs, that is, the phosphorus cannot be completely absorbed, and the working efficiency of the device will decrease at this time. We believe that the total amount of phosphorus absorbed by bacteria in per unit time is gradually decreasing (the efficiency of absorbing phosphorus is getting worse), but the rate of decline is constant. Therefore, calculating the absorption rate requires integrating the product of the bacterial concentration and the absorption efficiency.
Model content:
(1) First, we need to explore when the original phosphorus in the sewage can be absorbed completely. Since the ability of bacteria to absorb phosphorus is kept at a high level in the first few hours, it can be approximated as a fixed value.Assuming that the amount of phosphorus absorbed by bacteria in the T time is y(T)mg, then:\({y(T)}={1.525×\int_{0}^{T}x(t)dt}\)
In the calculation we assume \({x(t)}={\frac{345}{1+13e^{-0.0483-0.1042t-0.0333t^2+0.0014t^3}}}\)
Figure 2: Bacterial absorption of phosphorus before the device enters water
The curve formed by (t, x, y) passes through the y = 200 plane at t = 3.7, indicating that the bacteria can absorb the original phosphorus in the apparatus approximately 3.7 hours after the start of the reaction. It is possible to use 3.7 hours (3.5 hours or 4 hours) as the starting point of the influent time.
(2) From the whole process of bacteria absorption of phosphorus in the device, the ability of bacteria to absorb phosphorus is not suitable for the fixed value. Considering the changes of the environment in the device, such as the decrease of bacterial vigor, we believe that the rate of phosphorus absorption by bacteria A relationship with time as a function.
Assume \({y(T)}={\int_{0}^{T}v(t)x(t)dt}\)
Then v(t) is a linear function, and the result is fitted according to the experimental data v(t)=-0.14t+1.85,which is \({y(T)}={\int_{0}^{T}(-0.14t+1.85)x(t)dt}\)
In the bacterial growth curve section, we have obtained the relationship between x(t) and the influent velocity u:\({x(t)}={\frac{34u}{1+(1.36u-1)e^{0.28+0.02u-0.35t}}}\)
Next we explore when phosphorus breakthrough will occur. It is not difficult to obtain that the total phosphorus content (both absorbed and unabsorbed) in the plant should be the sum of the phosphorus content in the raw sewage and the phosphorus content in the new sewage, ie:\({P(y)}={\begin{cases}200,&\text{t≤4}\\\\{200+8u(t-4)},&\text{t＞4}\end{cases}}\)
According to the research needs, the main consideration is t>4.The next step is to find the solution of P(T)=y(T):\({200+8u(T-4)}={\int_{0}^{T}(-0.14t+1.85)\frac{34u}{1+(1.36u-1)e^{0.28+0.02u-0.35t}}}\)
Figure 3: Comparison of bacteria absorption phosphorus curve and total phosphorus content in the device after water is introduced into the device
Figure 4: Relationship between maximum phosphorus uptake, phosphorus breakthrough time and water inflow rate after the device is flooded
According to this equation, we can calculate the relationship between the influent velocity, the total phosphorus uptake time, and the amount of phosphorus absorbed. We have found that when the influent velocity reaches a higher value, the phosphorus in the sewage is always in a state where it cannot be removed, which is inefficient. (Figure 3, u = 15L / h)
The most ideal situation is that the water inlet speed is in the range of 8~12L/h, and the amount of phosphorus absorbed by the bacteria is the highest when the phosphorus penetrates, about 960mg. When the water inlet speed is 10 L/h, it takes 13.5 hours to reach the penetration, and when the water inlet speed is 12 L/h, it takes only 10.9 hours to reach the penetration. When the water inlet speed is higher, the time taken to achieve penetration is shorter, but the total amount of phosphorus absorbed will also decrease, that is, the efficiency will decrease. Based on this equation, we plot the relationship between the influent velocity and the total phosphorus absorption (Figure 5).
The optimum water inlet speed is 10.4L/h, and the total phosphorus absorption when reaching the phosphorus penetration is 967mg. Therefore, the water inlet speed is preferably controlled at about 10L/h in practical application.
Figure 5: Relationship between maximum phosphorus uptake and water inflow rate after the device is flooded
Condensed bottle absorption phosphorus model and comparison
Model introduction:
Similarly, we believe that the rate at which bacteria absorb phosphorus is a function of time. And we scaled the conical flask to the volume of the device and then compared the difference between the conical flask and the device.
Model content:
First, we got the bacterial growth equation:\({x(t)}={\frac{74.223}{1+3.282e^{-0.3889t-0.2543t^2+0.0234t^3}}}\)
Assume \({y(T)}={\int_{0}^{T}v(t)x(t)dt}\)
Then v(t) is a linear function, and the result is fitted according to the experimental data v(t)=-0.0725t+0.97,ie:\({y(T)}={\int_{0}^{T}(-0.0725t+0.97)x(t)dt}\)
Conical flask release phosphorus model
We combined the Eckenfelder model's reaction model at low organic concentrations to construct a phosphorus and phosphorus release model. According to the basic content of this model, it is not difficult to get:\({\frac{dP}{dt}}={KP(P_{max}-P)}\)
Therefore \({P(t)}={\frac{P_{max}}{1+e^{-k(t-t_0)P_{max}}(P_{max}-P_0)/P_0}}\)
Figure 6: Relationship between released phosphorus concentration and release time in an Erlenmeyer flask experiment
\(P_0\) is the initial concentration of the reaction. \(P_{max}\) is the maximum phosphorus concentration during phosphorus release.k is the release rate parameter and we get a fitted value of -0.0018.
