Département : Energies et Génie des Procédés. Première année Master: Génie Chimique. Matière : Réacteurs non-idéaux et bioréacteurs Devoir à domicile (à faire par groupes de 3 étudiants et à remettre par mail le lundi 18 mars 2024). Exercice. Une impulsion d'une solution de \( \mathrm{NaCl} \) concentrée est introduite dans un récipient de volume \( V=1 \mathrm{~m}^{3} \). Le débit volumique de liquide est gardé constant \( Q=1 \mathrm{~m}^{3} / \mathrm{min} \). La mesure de la concentration de ce traceur en sortie du récipient a permis de tracer l'évolution ci-dessous. Cette sortie peut être assimilée à deux impulsions d'aires respectifs 16 et 24 sortant respectivement aux temps 0,25 minutes et 1,5 minutes. Un des modèles d'écoulement possibles pour ce récipient serait celui constitué de deux zones piston placées en parallèle. 1) Tracez un schéma de ce modèle et reportez dessus les diffërentes grandeurs qui le caractérisent. 2) Déterminez la fonction de transfert de ce modèle. 3) Déterminez la fonction qui représente la concentration de traceur en sortie du modèle dans le domaine du temps résultant d'une injection impulsion à l'entrée. 4) Déterminez les paramètres de ce modèle (les volumes et les débits volumiques dans chacun de deux pistons) par comparaison avec les mesures expérimentales reportés sur le schéma. 5) Existe-il un volume mort dans ce réacteur réel ? Justifiez votre réponse.
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Les grandeurs qui caractérisent ce modèle sont les volumes de chaque piston (V1 et V2), les débits volumiques dans chaque piston (Q1 et Q2), et les temps de résidence dans chaque piston (t1 et t2). Show more…
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In chemical vapor deposition (CVD), a semiconducting or insulating solid material is formed in a reaction between a gaseous species and a species adsorbed on the surface of silicon wafers (disks about $10 \mathrm{cm}$ in diameter and $1 \mathrm{mm}$ thick). The coated wafers are subjected to further processing to produce the microelectronic chips in computers and most other electronic devices in use today. In one such process, silicon dioxide (MW $=60.06, \mathrm{SG}=2.67$ ) is formed in the reaction between gaseous dichlorosilane (DCS) and adsorbed nitrous oxide: $$\mathrm{SiH}_{2} \mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{N}_{2} \mathrm{O}(\mathrm{ads}) \rightarrow \mathrm{SiO}_{2}(\mathrm{s})+2 \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{HCl}(\mathrm{g})$$ A mixture of DCS and $\mathrm{N}_{2} \mathrm{O}$ flows through a "boat reactor" - a horizontal pipe in which 50 to 100 silicon wafers about $12 \mathrm{cm}$ in diameter and $1 \mathrm{mm}$ thick are set upright along the reactor length, with about $20 \mathrm{mm}$ separation between each wafer. A side view of the reactor is shown below: The feed gas enters the reactor at a rate of 3.74 SCMM (standard cubic meters per minute) and contains 22.0 mole\% DCS and the balance $\mathrm{N}_{2} \mathrm{O}$. In the reactor, the gas flows around the wafers, DCS and $\mathrm{N}_{2} \mathrm{O}$ diffuse into the spaces between the wafers, $\mathrm{N}_{2} \mathrm{O}$ is adsorbed on the wafer surfaces, and the adsorbed $\mathrm{N}_{2} \mathrm{O}$ reacts with gascous DCS. The silicon dioxide formed remains on the surface, and the nitrogen and hydrogen chloride go into the gas phase and eventually leave the reactor with the unconsumed reactants. The temperature and absolute pressure in the reactor are constant at $900^{\circ} \mathrm{C}$ and 604 millitorr. (a) The percentage conversion of DCS at a certain axial position (distance along the length of the reactor) is 60\%. Calculate the volumetric flow rate (m $^{3} / \mathrm{min}$ ) of gas at this axial position. (b) The rate of deposition of silicon dioxide per unit area of wafer surface is given by the formula $$r\left(\frac{\mathrm{mol} \mathrm{SiO}_{2}}{\mathrm{m}^{2} \cdot \mathrm{s}}\right)=3.16 \times 10^{-8} \mathrm{p}_{\mathrm{DCS}} p_{\mathrm{N}_{2} \mathrm{O}}^{0.65}$$ where $p_{\mathrm{DCS}}$ and $p_{\mathrm{N}, \mathrm{o}}$ are the partial pressures of $\mathrm{DCS}$ and $\mathrm{N}_{2} \mathrm{O}$ in millitorr. What is $r$ at the axial position in the reactor where the DCS conversion is $60 \% ?$ (c) Consider a wafer located at the axial position determined in Part (b). How thick is the silicon dioxide layer on that wafer after two hours of reactor operation, assuming that gas diffusion is rapid enough at the low reactor pressure for the composition of the gas (and hence the component partial pressures) to be uniform over the wafer surface? Express your answer in angstroms, where 1 $\AA=1.0 \times 10^{-10} \mathrm{m}$. (Hint: You can calculate the rate of growth of the $\mathrm{Si} \mathrm{O}_{2}$ layer in $\mathrm{A} / \mathrm{min}$ from $r$ and propertics of $\mathrm{SiO}_{2}$ given in the problem statement.) Would the thickness be greater or less than this value at an axial position closer to the reactor entrance? Briefly explain your answer.
The nuclear pore complex (NPC) creates a barrier to the free exchange of molecules between the nucleus and cytosol, but in a way that remains mysterious. In yeast, for example, the central pore of the NPC has a diameter of 35 $\mathrm{nm}$ and is $30 \mathrm{nm}$ long, which is somewhat smaller than its vertebrate counterpart. Even so, it is large enough to accommodate virtually all components of the cytosol. Yet the pore allows passive diffusion of molecules only up to about 40 kd; entry of anything larger requires help from a nuclear import receptor. Selective permeability is controlled by protein components of the NPC that have unstructured, polar tails extending into the central pore. These tails are characterized by periodic repeats of the hydrophobic amino acids phenylalanine (F) and glycine (G). At high enough concentration $(\sim 50 \mathrm{mM})$, the FG-repeat domains of these proteins can form a gel, with a meshwork of interactions between the hydrophobic $\mathrm{FG}$ repeats (Figure $\mathrm{Q} 12-2 \mathrm{A}$ ). These gels allow passive diffusion of small molecules, but they prevent entry of larger proteins such as the fluorescent protein mCherry fused to maltose binding protein (MBP) (Figure Q12-2B). (The fusion to MBP makes mCherry too large to enter the nucleus by passive diffusion.) However, if the nuclear import receptor, importin, is fused to a similar protein, MBP-GFP, the importin-MBP-GFP fusion readily enters the gel (Figure $\mathrm{Q} 12-2 \mathrm{B}$ ). A. $\quad$ FG-repeats only form gels in vitro at relatively high concentration $(50 \mathrm{mM}) .$ Is this concentration reasonable for FG repeats in the NPC core? In yeast, there are about 5000 FG-repeats in each NPC. Given the dimensions of the yeast nuclear pore $(35 \mathrm{nm} \text { diameter and } 30 \mathrm{nm} \text { length })$ calculate the concentration of FG-repeats in the cylindrical volume of the pore. Is this concentration comparable to the one used in vitro? B. A second question is whether the diffusion of importin-MBP-GFP through the FG-repeat gel is fast enough to account for the efficient flow of materials between the nucleus and cytosol. From experiments of the type shown in Figure $\mathrm{Q} 12-2 \mathrm{B}$, the diffusion coefficient $(D)$ of importin-MBP-GFP through the FG-repeat gel was determined to be about $0.1 \mu \mathrm{m}^{2} / \mathrm{s}$. The equation for diffusion is $t=x^{2} / 2 D,$ where $t$ is time and $x$ is distance. Calculate the time it would take importin-MBP-GFP to diffuse through a yeast nuclear pore $(30 \mathrm{nm})$ if the pore consisted of a gel of FG-repeats. Does this time seem fast enough for the needs of a eukaryotic cell?
The patch clamp technique is widely used in electrophysiology to study the gating properties of ion channels. A glass micropipette with an open tip is brought into contact with a cell or a vesicle containing the ion channel of interest. As a result of the strong adhesion between the glass and the lipid membrane, the membrane gets pulled into the pipette as shown in Figure 11.49. Figure 11.49 : Membrane patch and pipette. Schematic showing the membrane patch shape (grey) at zero applied pressure. The pipette has a radius of $R_{p}$ and as a result of adhesion, the membrane is pulled up the pipette by a distance $L_{0}$ Although the glass-membrane adhesion is crucial to avoiding spurious leakage currents not resulting from the channels themselves, this adhesion can also have negative side effects. For example, as the membrane adheres to the pipette, it induces a tension that is different from the resting tension in the membrane patch, and this could lead to errors in the estimates of gating tension of mechanosensitive channels, for example. In this problem, we use our understanding of elasticity of membranes to estimate the size of this adhesion-induced tension. (a) Consider the geometry shown in Figure 11.49 , where the membrane patch has a cylindrical adhered domain and a circular free domain. The elastic energy associated with the patch comes from two contributions, namely, the stretch energy and the glass-bilayer adhesion energy. The stretch energy results from the fact that the membrane is stretched when it is pulled into the pipette, This contribution to the free energy has a quadratic dependence on the areal strain $\phi=\left(A-A_{0}\right) / A_{1},$ and is given by $$G_{\text {stretch }}=\frac{1}{2} K_{\mathrm{A}} \phi^{2} A_{0}$$, where $A_{0}$ is the unstretched area of the membrane patch and $K_{\mathrm{A}} \approx 50 \mathrm{k}_{\mathrm{B}} \mathrm{T} / \mathrm{nm}^{2}$ is the stretch modulus. The adhesion energy is proportional to the area of the contact domain $A_{\text {adh }}$ and is given by $G_{\text {adh }}=-\gamma A_{\text {adh }},$ where $\gamma$ is the adhesion energy (that is, the energy per unit area). Using these facts and the geometry shown in Figure 11.49 obtain the expression for the total elastic energy of the membrane patch in terms of the areal strain $\phi$. (b) Minimize the elastic energy with respect to $\phi$ to obtain the expression for the equilibrium areal strain. Compute the tension $\tau$ in the membrane patch in equilibrium using the relation $\tau=K_{A} \phi .$ For $\gamma \sim 0.5 k_{B} T / n m^{2},$ a typical value for glass-bilayer interaction, estimate the equilibrium areal strain and the tension in the patch. (c) Assuming that the unstretched area of the patch is $A_{0}=2 \times \pi R_{\mathrm{P}}^{2},$ what is the length of membrane tube $L$ that lines the sides of the pipette? How does this length compare with the length computed from simple geometrical considerations, namely, by assuming that the membrane patch does not get stretched when it is drawn into the pipette. (d) Show that membrane bending does not contribute significantly to the energy budget of the membrane patch. Do this by computing the bending energy of cylindrical membrane material within the pipette and compare the resulting energy with the stretch energy and adhesion energy you already obtained.
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