Team:ETH Zurich/bubbelingModel

The first step of AROMA is to take up air samples and dissolve them in the medium. We decided to use a bubble column, where air is bubbled into the medium for a short period of time. This will lead to the transfer of our compound of interest from air into the medium. In order to see if this is a feasible and fast approach, a model was constructed to see what kind of concentrations of our compound in the medium could be achieved and to look at the time-scale required for the bubbling process. Our model showed that the bubble column is a feasible choice for AROMA. Using a pump speed of 1*10^-3 $m^3/s $ results in the maximum concentration of dissolved 2,4-DNT, a compound found in land mines, within minutes.

Schematic representation of the bubble column

Mass Transfer Model

The aim of the bubbling model is to calculate the transfer of our molecule of interest from air into medium. We therefore start by having a look at the equation for mass transfer across a surface:

\begin{equation} \label{eq:1} \ N \cdot A = k \cdot (c_0 -c_m) \cdot A \end{equation}

Here, $N$ describes the rate of mass transferred per area at the interface, $k$ is the mass transfer coefficient and $c_{0}$ and $c_{m}$ are the concentrations at the medium-to-gas interface and the concentration in medium, respectively. $A$ represents the total surface area.
Note that here it's assumed that all substances are well mixed except near the medium-to-gas interface. (reference 1)

In order to solve equation (\ref{eq:1}), we first have to find the mass transfer coefficient $k$. When we assume that there is only a diffusive flux and no free convection, we can use Fick's 1st law in spherical coordinates to express $k$ as \begin{equation} \label{eq:2} \ k = \frac{2D}{d} . \end{equation}

$D$ represents the diffusion coefficient of our molecule of interest in the medium, whereas $d$ stands for the diameter of the air bubble.

In our case, we can say that the transfer of mass into the medium over a period of time is equal to the dissolution of our compound over time and therefore we replace $N*A$ by

\begin{equation} \label{eq:3} \ - \frac{\partial c_i \cdot v_b}{\partial t} \end{equation}

with $c_i$ representing the concentration in the air bubble and $v_{b}$ the volume of the bubble

Combing equations (\ref{eq:1}) , (\ref{eq:2}) and (\ref{eq:3}) will give us

\begin{equation} \frac{\partial c_i \cdot v_b}{\partial t} = -k \cdot A \cdot (c_o - c_m) \end{equation}

Using the ideal gas law, we can replace $c_i$ by $\frac{p}{R \cdot T}$ where $p$, $R$ and $T$ represent the pressure in the bubble, the ideal gas constant and the absolute temperature, respectively. This will give us the following mass balance that we need to solve:

\begin{equation} \label{eq:5} \frac{\partial p_i}{\partial t} = -k \cdot a \cdot R \cdot T \cdot (c_o - c_m) \end{equation}

where $a = \frac{A}{v_{b}}$ .

Solving the mass balance

Henry's law

In order to solve this mass balance, we need to introduce Henry's ideal gas law. This law states that the ''amount of dissolved gas is proportional to it's partial pressure in the air'', which is represented by the equation:

\begin{equation} \ H = \frac{c}{p}\ \end{equation}

where H is the characteristic Henry value for a molecule.

When solving the mass balance in (\ref{eq:5}), we have to distinguish between two different situations. Situation 1 applies when $H \cdot p_1 < c_s $, meaning that the amount of dissolved gas does not exceed the maximum solubility of this compound in water.

Situation 2 represents $H \cdot p_1> c_s $, meaning that the partial pressure will cause a concentration of compound in the medium that is higher that it's maximum solubility. As this is not possible, we will in this case say that the amount of dissolved gas is equal to the solubility limit of this compound in water.

The timepoint where situation 1 turns into situation 2, e.g. where $H \cdot p_1 = c_s $, is referred to as $t_{switch}$

Situation 1

For situation 1, we set $c_o = H \cdot p_i$, according to Henry's law. Substituting this in equation (\ref{eq:5}) and subsequently integrating the formula gives us:

\begin{equation} \ p_i = K_1 \cdot e^{-k \cdot a \cdot R \cdot T \cdot H \cdot t} + \frac{c_m}{H} \end{equation}

Applying the boundary condition $p_i(tswitch) = c_s / H $ results into:

\begin{equation} \ p_i(t) = \frac{c_m}{H} +\frac{c_s - c_m}{H} \cdot e^{-k \cdot a \cdot R \cdot T \cdot H \cdot (t-tswitch)} \end{equation}

Situation 2

For situation 2, we will set $c_o = c_s$ , for the reason described above. Integrating equation (5) with this information gives:

\begin{equation} \ p_i = K_2 \cdot -k \cdot a\cdot R \cdot T \cdot (c_s - c_m) \cdot t \end{equation}

Applying the boundary condition $p(0) = p_\infty $ leads to

\begin{equation} \ p_i(t) = p_\infty -k \cdot a \cdot R \cdot T \cdot (c_s - c_m) \cdot t \end{equation}

$ p_\infty $ represents the pressure in the air.

Bubble Residence time

The last parameter we need to determine is the bubble residence time. This will help us to find out for which time period we need to solve the mass balance. As we will implement a short bubble column on our robot, we can assume that the residence time of the bubble is smaller than the total dissolution time. Hence, we assume a nearly constant bubble radius along the column. [reference 2]

Using Stoke's law we first determine the bubble's rising speed.

\begin{equation} \ v = \frac{d^2}{18} * (\rho_l - \rho_g) * \frac{g}{\mu} \end{equation}

$ \rho $ stands for the density of either the medium ($ \rho_l $) or the air $ \rho_g $, $g$ is the standard gravity, and $\mu$ represents the viscosity of water.

Then, implementing the length of our column will give us a residence time of:

\begin{equation} \ \tau = \frac{L}{v}\ \end{equation}

From partial pressure to a concentration in the medium

In order to plot the concentration of our compound in the medium, the $c_{m}$ at different time points is defined as:

\begin{equation} \ c_m = c_m(0) + \frac{time \cdot Q \cdot T }{V_l \cdot R} \cdot (p(0) - p(t)) \end{equation}

where $Q$ represents the volumetric gas flow rate, time the time period we set for updating the current medium concentration, $V_{l}$ the volume of the fluid in the bubble column, $c_{m}(0)$ the concentration in the medium at timepoint 0, and p(0) and p(time) the pressure at timepoint 0 and the current timepoint of simulation.

This means that for each defined period of time we update the initial medium concentration with the concentration transferred from the bubble to the medium. Here we need to take into account how much air is pumped into the bubble column (Q) and the total volume in the bubble column ($V_{l}$). The formula tells us that the larger the total medium volume in the column, the more air needs to be dissolved in order to obtain the same concentration of compound in the medium. It's therefore important that we keep the liquid volume small in the robot and that we implement a high flow rate, as this will facilitate the uptake speed.

Results

Incorporating the parameters from our hardware, e.g. the power of the pump, the bubble size obtained by our pump and the size of our bubble column, we obtain figure 2.1 of the concentration of the molecule 2,4-DNT, a compound in land mines, in medium over time. The graph is based on the concentration directly above a land mine. This means that this is the graph we would obtain when the robot is at the source.

Concentration of 2,4-DNT in the medium over time. An increase in concentration is the result of mass transfer of the compound of interest from air to the medium using a bubble column.

The partial pressure within a bubble decreases upon increasing residence time in the bubble column, due to mass transfer of the air molecules into the medium. The first bubbles that are introduced into the column will have a stronger decrease in the partial pressure than air bubbles that enter the column at a later stage. This corresponds to equation (1).

Measurement

Figure 2.1 shows that it takes around 1 hour before the medium is saturated. In the ideal case we would like to obtain the $c_s$ at the source, as this means that we achieve the upper limit of the biosensor at the source. For this to be achieved, we need to improve one of our hardware parts. Looking at the influence of our pump speed (Q) on the bubbling time that's needed to reach $c_s$, we can see in figure 3.1 that increasing the pump speed will give a $c_s$ that is reached within 5 minutes.

Increasing the air flow rate into the bubble column, by increasing the speed of the pump, improves the time needed to obtain maximum solubility.

Increasing the air flow rate into the bubble column leads to higher volume of bubbled air per second. This plot is obtained by multiplying the bubbling time with the flow rate of the pump.

TODO

Conclusion

When implementing a pump with a speed of 1*10^-3 $m^3/s $, the bubble column serves as a fast and efficient method to transfer our molecule of interest from air to the medium. Within minutes, a concentration of 2,4-DNT in the medium is obtained that matches the almost with the upper range of our biosensors, meaning that the maximum output of our biosensors is achieved at the source. Hence,further away from the source there is room for lower outputs of our sensors.

References

[1] Cussler, Edward Lansing. Diffusion: mass transfer in fluid systems. Cambridge university press, 2009.
[2] Martínez, I., and P. A. Casas. "Simple model for CO2 absorption in a bubbling water column." Brazilian Journal of Chemical Engineering 29.1 (2012): 107-111.

[1] [1] Cussler, Edward Lansing. Diffusion: mass transfer in fluid systems. Cambridge university press, 2009.

[2] [2] Martínez, I., and P. A. Casas. "Simple model for CO2 absorption in a bubbling water column." Brazilian Journal of Chemical Engineering 29.1 (2012): 107-111.