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在循环系统中给药

發(fā)布時(shí)間:2025/3/21 windows 24 豆豆
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Drug delivery in circulatory system

生物醫(yī)學(xué)研究中心的一組科學(xué)家被要求研究藥物在循環(huán)系統(tǒng)中的轉(zhuǎn)運(yùn)。在初始階段,他們?cè)O(shè)計(jì)了一個(gè)簡(jiǎn)單的場(chǎng)景來(lái)校準(zhǔn)他們的初始設(shè)計(jì)。他們雇用你來(lái)協(xié)助他們的研究通過(guò)對(duì)一個(gè)二維板進(jìn)行數(shù)值研究。你知道,要研究這種輸運(yùn),你必須解下列對(duì)流-擴(kuò)散方程

其中為藥物的濃度,為速度,為藥物在血液中的擴(kuò)散系數(shù)常數(shù)。血流是單向的(在上式中只存在分量),其速度由下式模擬


結(jié)束時(shí),你的工作,你需要提供一個(gè)壓縮文件夾包含所有您已經(jīng)實(shí)現(xiàn)的功能除了主要腳本運(yùn)行整個(gè)仿真的接口(你必須加載變量直接在主腳本,不允許用戶動(dòng)態(tài)修復(fù)它們)。所有的編碼都需要用適當(dāng)?shù)淖⑨寔?lái)解釋。此外,您必須提供一個(gè)PDF文件,其中包含以下結(jié)果和討論。

利用MATLAB的contourf命令繪制以下時(shí)間點(diǎn)的藥物沿平板分布:t=0,0.5,1,1.5,2,2.5,3s。討論結(jié)果和主要特征的觀察物理行為的濃度。

MATLAB求解代碼

clc close all clear %% initalisation lx=4*pi; y_max=pi; y_m=0.5*y_max; D=0.1; dt=0.1; nx=51; ny=21; tmax=3; %% intitalisation of x,y,t dx=lx/(nx-1); x=0:dx:lx; dy=y_max/(ny-1); y=0:dy:y_max; nt=tmax/dt+1; t=0:dt:tmax;%% Processing Velocity u=zeros(nx,ny,nt); for i=1:nxfor j=1:nyfor k=1:ntu(i,j,k)=3*(1-((x(i)-2*pi)^2/(2*pi^2))*cos(pi*t(k))*sin(x(i)))*1-((y(j)-y_m)^2/y_m^2);endend end %% concentration at inital condition c=zeros(nx,ny,nt); for i=1:4nxfor j=1:nyc(i,j,1)=1-(y(j)-y_m)^2/(y_m^2);end end %% Values for AP, AN, AS, AW, AE from discretisationas=-D*(dt)/(dy^2); ap= 1+(2*D*dt/dy^2)+(2*D*dt/dx^2); an=-D*(dt)/dy^2;b=reshape(c(:,2:ny-1,1),[nx*(ny-2),1]); f(:,1)=b; %stores the value of concentration A = zeros(nx*(ny-2),nx*(ny-2)); %creates a specific size of matrix Afor k=2:nt for i=1:nxfor j=2:ny-1ae(i,j,k)=u(i,j,k)*dt/(2*dx)-D*dt/(dx^2); %depends on the U velocity vectoraw(i,j,k)=-u(i,j,k)*dt/(2*dx)-D*dt/(dx^2); %''endend%%Types%T1 centrefor i=2:(nx-1)for j=3:ny-2pointer=(j-2)*nx+i; A(pointer,pointer)=ap;A(pointer,pointer+1)=ae(i,j,k);A(pointer,pointer-1)=aw(i,j,k);A(pointer,pointer+nx)=an;A(pointer,pointer-nx)=as; endend%T2 leftfor j=3:ny-2i=1;pointer=(j-2)*nx+i;A(pointer,pointer)=ap;A(pointer,pointer+1)=ae(i,j,k);A(pointer,pointer+nx-2)=aw(i,j,k);A(pointer,pointer+nx)=an;A(pointer,pointer-nx)=as; end%T3 rightfor j=3:ny-2i=nx;pointer=(j-2)*nx+i;A(pointer,pointer)=ap;A(pointer,pointer-nx+2)=ae(i,j,k);A(pointer,pointer-1)=aw(i,j,k);A(pointer,pointer+nx)=an;A(pointer,pointer-nx)=as; end%T4 bottomfor i=2:nx-1j=2;pointer=(j-2)*nx+i;A(pointer,pointer)=ap;A(pointer,pointer+1)=ae(i,j,k);A(pointer,pointer-1)=aw(i,j,k);A(pointer,pointer+nx)=an;end%T5 topfor i=2:nx-1j=ny-1;pointer=(j-2)*nx+i;A(pointer,pointer)=ap;A(pointer,pointer+1)=ae(i,j,k);A(pointer,pointer-1)=aw(i,j,k);A(pointer,pointer-nx)=as; end%T6 bottom lefti=1;j=2;pointer=(j-2)*nx+i;A(pointer,pointer)=ap;A(pointer,pointer+1)=ae(i,j,k);A(pointer,pointer+nx-2)=aw(i,j,k);A(pointer,pointer+nx)=an;%T7 bottom righti=nx;j=2;pointer=(j-2)*nx+i;A(pointer,pointer)=ap;A(pointer,pointer-nx+2)=ae(i,j,k);A(pointer,pointer-1)=aw(i,j,k);A(pointer,pointer+nx)=an;%T8 top lefti=1;j=ny-1;pointer=(j-2)*nx+i;A(pointer,pointer)=ap;A(pointer,pointer+1)=ae(i,j,k);A(pointer,pointer+nx-2)=aw(i,j,k);A(pointer,pointer-nx)=as;%T9 top righti=nx;j=ny-1;pointer=(j-2)*nx+i;A(pointer,pointer)=ap;A(pointer,pointer-nx+2)=ae(i,j,k);A(pointer,pointer-1)=aw(i,j,k);A(pointer,pointer-nx)=as;%solvingZ=A\b; f(:,k)=Z; b=Z; end%% Resultsc(:,2:ny-1,:)=reshape(f,[nx,ny-2,nt]);%reshapes the concentration values ffor h=1:5:31 figurecontourf(c(:,:,h)')xlabel('x (m)')ylabel('y (m)')time=(h-1)*dt;title([ 'Concentration over time' , num2str(time),'s']) end

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