Tutor profile: Aadhil B.

Given:$$ \frac{5}{2} \cos^{-1}x +\sin^{-1}x=\frac{4}{3}[\tan^{-1}(2)+\tan^{-1}(3)]$$ Find the value of x.

First, we will compute the right hand side of the equation. Using the identity: $$\tan^{-1}(A)+\tan^{-1}(B)=\tan^{-1}(\frac{A+B}{1-A.B})$$, we can write, $$\frac{4}{3}[\tan^{-1}(2)+\tan^{-1}(3)]=\frac{4}{3}[\tan^{-1}(\frac{2+3}{1-2.3})]$$ $$=\frac{4}{3}[\tan^{-1}(-1)]$$ We know that $$\tan (\frac{\pi}{2}+\theta)$$ is negative in the second quadrant with a value of $$-\tan \theta$$ That is, $$\tan (\frac{\pi}{2}+\frac{\pi}{4})=-\tan \frac{\pi}{4}=-1$$ Therefore, $$\tan^{-1}(-1)=\tan^{-1}(\tan (\frac{\pi}{2}+\frac{\pi}{4}))=\frac{3\pi}{4}$$ $$\implies \frac{4}{3}[\tan^{-1}(-1)]= \frac{4}{3} \times \frac{3\pi}{4}=\pi$$ Let us now compute the left hand side of the equation. $$ \frac{5}{2} \cos^{-1}x +\sin^{-1}x$$ can be written as $$\frac{3}{2} \cos^{-1}x +\cos^{-1}x+\sin^{-1}x$$ (This step is done to bring the expression in terms of $$\cos^{-1}x +\sin^{-1}x$$) Using the identity: $$\cos^{-1}x+\sin^{-1}x=\frac{\pi}{2}$$, we will get, $$\frac{3}{2} \cos^{-1}x +\cos^{-1}x+\sin^{-1}x=\frac{3}{2} \cos^{-1}x +\frac{\pi}{2}$$ Now ,equate the left hand side and right hand side of the equation. That is, $$\frac{3}{2} \cos^{-1}x +\frac{\pi}{2}=\pi$$ $$\implies \frac{3}{2} \cos^{-1}x=\frac{\pi}{2}$$ $$\implies \cos^{-1}x=\frac{\pi}{3}$$ $$\implies x=\cos \frac{\pi}{3}=\frac{1}{2}$$ $$\therefore, x=\frac{1}{2} $$

A first order liquid phase reaction, $$A \to B$$ is carried out isothermally in a mixed flow reactor at a steady state. A conversion of 75% is achieved. If the mixed flow reactor is replaced by five smaller and identical isothermal mixed flow reactors in series (assuming the same inlet conditions and the same overall conversion), what is the percentage reduction in total volume of the reactor? Also, obtain the space time $$\tau_f$$ in terms of the initial and final concentration of the reactant, if the mixed flow reactor is replaced by infinite smaller, identical and isothermal reactors in series? Given: the rate constant of this first order reaction=$$k$$, initial concentration of the reactant=$$C_0$$ and the final concentration of the reactant in the case of infinite reactors in series =$$C_l$$.

We know that, for a mixed flow reactor, $$\tau = \frac{V}{v_0}=\frac{C_0-C_1}{-r_A}$$ , where, $$\tau$$= Space time $$V$$= Total volume of the reactor, $$v_0$$= Volumetric flow rate (constant), $$C_1$$=The outlet concentration, $$-r_A$$=The rate of the reaction=$$kC_1$$(since, this is a first order reaction). It is given that the fraction of the conversion achieved, $$X$$=0.75 $$\therefore C_1=C_0 X=0.75 C_0$$ Hence, $$ \frac{V}{v_0}=\frac{C_0-C_1}{-r_A}=\frac{C_0-C_1}{kC_1}=\frac{C_0-0.75 C_0}{0.75 k C_0}=\frac {1}{3k}$$ $$\therefore V=0.333\frac{v_0}{k}$$ After the replacement, let $$\tau_i$$ and $$V_i$$ be the space time and the volume of the reactor $$i$$ respectively. Since, the reactors are in series, the volumetric flow rate is constant. $$\therefore$$, we can write the performance equation for a mixed flow reactor $$i$$ as follows: $$\tau_i=\frac{V_i}{v_0}=\frac{C_{i-1}-C_i}{kC_i}$$ $$\implies \frac{C_{i-1}}{C_i}=1+k\tau_i$$ There are 5 identical reactors and given that the overall conversion remains the same. Hence, $$X=0.75=1-\frac{C_0}{C_5}$$ and, $$C_5=0.75C_0$$ $$\therefore \frac{C_0}{C_5}=\frac{1}{1-X}=\frac{C_0}{C_1}\times \frac{C_1}{C_2}\times \frac{C_2}{C_3}\times \frac{C_3}{C_4}\times \frac{C_4}{C_5}=(1+k\tau_1)(1+k\tau_2)(1+k\tau_3)(1+k\tau_4)(1+k\tau_5)$$ Here, the space time is the same in all equal size mixed flow reactors of equal volumes. That is, $$\tau_1=\tau_2=\tau_3=\tau_4=\tau_5=\tau_m$$ $$\therefore, \frac{C_0}{C_5}={(1+k\tau_m)}^{5}$$ $$\implies \tau_m=\frac{1}{k} [\frac{C_0}{C_5}^\frac{1}{5}-1]$$ The overall space time for the 5 reactors = $$\tau_{new}=5 \tau_m=5 \frac{V_m}{v_0}=\frac{5}{k} [\frac{C_0}{C_5}^\frac{1}{5}-1]$$ where, $$V_m$$, the reduced volume= $$\frac{5v_0}{k} [\frac{C_0}{C_5}^\frac{1}{5}-1]=\frac{5v_0}{k} [\frac{C_0}{0.75C_0}^\frac{1}{5}-1]=0.296\frac{v_0}{k}$$ $$\therefore,$$ the percentage reduction in volume = $$\frac{V-V_m}{V} \times 100=\frac{0.333-0.296}{0.333} \times 100=11.11$$ percentage When the mixed flow reactor is replaced by infinite smaller reactors in series, the equation of the space time reduces to that of a plug flow reactor. $$\therefore, \lim\limits_{n \to \infty} \tau_m=\lim\limits_{n \to \infty} \frac{n}{k} [\frac{C_0}{C_l}^\frac{1}{n}-1] = \frac{1}{k} \ln(\frac{C_0}{C_l})$$ , where, $$n$$ is the number of reactors. $$\implies\lim\limits_{n \to \infty} \tau_m=\tau_f=\frac{1}{k} \ln(\frac{C_0}{C_l})$$

Consider the following linear ordinary differential equation in one-dimension at steady state: $$\frac{d^{2} u(x)}{dx^{2}} + f(x) =0$$, in $$(0,L)$$ where, $$x$$ is the spatial coordinate, $$u(x)$$ is the temperature field as a function of $$x$$, $$L$$ is the length of the domain and $$f(x)$$ is the forcing function. It is given that $$f(x)=\sin x$$, $$u(0)=u_0$$, and $$\frac{du(L)}{dx}=u_l$$. Find the temperature field $$u(x)$$.

Given: $$\frac{d^{2} u(x)}{dx^{2}} + f(x) =0$$, in $$(0,L)$$ $$\implies\frac{d^{2} u(x)}{dx^{2}} + \sin x =0$$ (Since, $$f(x)=\sin x$$) $$\implies\frac{d^{2} u(x)}{dx^{2}} =-\sin x$$ We integrate the above equation with respect to $$x$$, $$\implies\int \frac{d^{2} u(x)}{dx^{2}}\mathrm{d}x =-\int \sin x\mathrm{d}x$$ $$\implies\frac{du(x)}{dx}=\cos x + C_1$$ , where $$C_1$$ is a constant of integration. Now, integrate the above equation with respect to $$x$$ again, $$\implies\int \frac{du(x)}{dx}\mathrm{d}x =\int \cos x\mathrm{d}x+\int C_1\mathrm{d}x$$ $$\implies u(x)=\sin x+C_1x+C_2$$, where $$C_2$$ is another constant of integration. We need to use the boundary conditions to evaluate the constants $$C_1$$ and $$C_2$$. Given: $$u(0)=u_0$$ (this is a Dirichlet boundary condition), and $$\frac{du(L)}{dx}=u_l$$ (this is a Neumann boundary condition). Let us evaluate $$C_2$$: We know that, $$u(x)=\sin x+C_1x+C_2$$ $$\therefore u(0)=\sin 0+C_1\times 0+C_2=C_2$$ $$\implies C_2=u(0)=u_0$$, using the Dirichlet boundary condition. $$\therefore u(x)=\sin x+C_1x+u_0$$ Now, let us evaluate $$C_1$$: We need to apply the Neumann boundary condition to obtain $$C_1$$. $$\therefore$$ we need to evaluate the derivative of $$u(x)$$ with respect to $$x$$. $$ \frac{du(x)}{dx}=\cos x+C_1$$ $$\implies \frac{du(L)}{dx}=\cos L+C_1$$ $$\implies u_l=\cos L+C_1$$, using the Neumann boundary condition. That is, $$C_1=u_l-\cos L$$ $$\therefore$$, the solution for the temperature field, $$u(x)=\sin x+(u_l-\cos L)x+u_0$$