Help please. : ) The electric potential surrounding a charged spherical nanopart

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Help please. : )

The electric potential surrounding a charged spherical nanoparticle immersed in an electrolyte solution is governed by the following equation: In the above, phi is the potential inside the electrolyte solution; epsilon f is the permittivity of the solution (epsilon f = 7.08 times 10-10 F/m); F is the Faraday constant (F = 96490 C/mol); R is the universal gas constant (R = 8.31 J/(K-mol)); T is the fluid temperature (T=300K); zi and Ci0 are, respectively, the valance and bulk concentration of the ith ionic species. There are four species, H+ (i=l and z1=l), K+ (i=2 and Z2=l), CI- (i=3 and Z3= -1), and OH+ (i=4 and Z4= -1), in the electrolyte solution, and their corresponding bulk concentrations are: When pH LE 7: In the above, CKCI represents the concentration of the background salt KC1; pH is the pH value of the electrolyte solution, which is adjusted by KOH or HC1. The boundary conditions for the governed equation (1) are: On the nanoparticle surface (r=Rp): In the above, represents the thickness of the electric double layer. The surface charge density in Equation (4) is determined by where NTotal = 1.328 times 10-5 mol/m2, KA = 10-.6+3, KB = 10-1.9+3, and the concentration of H+ on the particle's surface is Note that phi s in equation (7) represents the electric potential on the surface of the particle. Use the BVP4c to solve the boundary value problem and show the following results: Plot the surface charge density, sigma, as a function of pH ranging from 3 to 9 using step size of 0.2 when CKCL=10 mM and Rp=40 times l0-9 m. Plot the surface charge density, sigma, as a function of CKCL ranging from 1 mM to 100 mM using step size of 1 mM when pH=6 and Rp=40 times l0-9 m. Plot the surface charge density, sigma, as a function of Rp ranging from 10-8 m to 10-7 m using step size of 10-9 m when pH=6 and CKCL = 10 mM.

Explanation / Answer

plot(t,y(:,1),'-',t,y(:,2),'--') title('Solution of van der Pol Equation, mu = 1'); xlabel('time t'); ylabel('solution y'); legend('y_1','y_2') function dydt = vdp1000(t,y) dydt = [y(2); 1000*(1-y(1)^2)*y(2)-y(1)]; [t,y] = ode15s(@vdp1000,[0 3000],[2; 0]); plot(t,y(:,1),'-'); title('Solution of van der Pol Equation, mu = 1000'); xlabel('time t'); ylabel('solution y_1'); function [t,y] = solve_vdp(mu) tspan = [0 max(20, 3*mu)]; y0 = [2; 0]; % Call the ODE solver ode15s. [t,y] = ode15s(@vdp,tspan,y0); % Define the ODE function as nested function, % using the parameter mu. function dydt = vdp(t,y) dydt = [y(2); mu*(1-y(1)^2)*y(2)-y(1)]; end end function rigidode %RIGIDODE Euler equations: rigid body without external forces tspan = [0 12]; y0 = [0; 1; 1]; % Solve the problem using ode45 ode45(@f,tspan,y0); % ------------------------------------------------------------ function dydt = f(t,y) dydt = [ y(2)*y(3) -y(1)*y(3) -0.51*y(1)*y(2) ]; function fem1ode(N) %FEM1ODE Stiff problem with a time-dependent mass matrix if nargin < 1 N = 19; end h = pi/(N+1); y0 = sin(h*(1:N)'); tspan = [0; pi]; % The Jacobian is constant. e = repmat(1/h,N,1); % e=[(1/h) ... (1/h)]; d = repmat(-2/h,N,1); % d=[(-2/h) ... (-2/h)]; % J is shared with the derivative function. J = spdiags([e d e], -1:1, N, N); d = repmat(h/6,N,1); % M is shared with the mass matrix function. M = spdiags([d 4*d d], -1:1, N, N); options = odeset('Mass',@mass,'MStateDep','none', ... 'Jacobian',J); [t,y] = ode15s(@f,tspan,y0,options); figure; surf((1:N)/(N+1),t,y); set(gca,'ZLim',[0 1]); view(142.5,30); title(['Finite element problem with time-dependent mass ' ... 'matrix, solved by ODE15S']); xlabel('space ( x/pi )'); ylabel('time'); zlabel('solution'); %-------------------------------------------------------------- function yp = f(t,y) % Derivative function. yp = J*y; % Constant Jacobian provided by outer function end % End nested function f %-------------------------------------------------------------- function Mt = mass(t) % Mass matrix function. Mt = exp(-t)*M; % M is provided by outer function end % End nested function mass %-------------------------------------------------------------- end

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