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ID:(1037, 0)

brainhatch005

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![brainhatch005](showImage.php)
brainhatch005

ID:(8763, 0)

brainhatch007

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![brainhatch007](showImage.php)
brainhatch007

ID:(8890, 0)

brainhatch009

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![brainhatch009](showImage.php)
brainhatch009

ID:(8892, 0)

brainhatch011

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![brainhatch011](showImage.php)
brainhatch011

ID:(8894, 0)

brainhatch008

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![brainhatch008](showImage.php)
brainhatch008

ID:(8891, 0)

brainhatch010

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![brainhatch010](showImage.php)
brainhatch010

ID:(8893, 0)

brainhatch012

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![brainhatch012](showImage.php)
brainhatch012

ID:(8895, 0)

Control de Totales

Beschreibung

Las ecuaciones del sistema son para $c_1$ con

$s=0$:

$\displaystyle\frac{d}{dt}c_{1,0}=\displaystyle\frac{1}{2\pi Rd_1}j(j_{s0},c_{s0},c_0,c_{1,0})+\displaystyle\frac{2}{dl_2}D(c_{1,1}-c_{1,0})-\displaystyle\frac{1}{hd_1}D(c_{1,0}-c_{2,0})$

$\displaystyle\frac{d}{dt}c_{1,s}=\displaystyle\frac{1}{2\pi Rd_1}j(j_{s0},c_{s0},c_0,c_{1,s})-j(j_{s1},c_{s1},c_{1,s},c_{2,s}))+\displaystyle\frac{1}{d^2}D(c_{1,s+1}+c_{1,s-1}-2c_{1,s})$

$s=n$;

$\displaystyle\frac{d}{dt}c_{1,n}=\displaystyle\frac{1}{2\pi Rd_1}(j(j_{s0},c_{s0},c_0,c_{1,n})-j(j_{s1},c_{s1},c_{1,n},c_{2,n}))+\displaystyle\frac{1}{d^2}D(c_{1,n-1}-c_{1,n})$

y para $c_2$ con

$s=0$:

$\displaystyle\frac{d}{dt}c_{2,s}=\displaystyle\frac{1}{2\pi Rd_2}(j(j_{s0},c_{s0},c_{1,s},c_{2,s})-j(j_{s1},c_{s1},c_{2,s},c_3))+\displaystyle\frac{1}{d^2}D(c_{2,s+1}+c_{2,s-1}-2c_{2,s})$

$s=n$;

ID:(8896, 0)

Dependencia Espacial

Beschreibung

Para el caso estacionario las ecuaciones de la parte de la solución que dependen del perimetro se obtienen de

$\displaystyle\frac{\partial c_{1,s}}{\partial t}=\displaystyle\frac{1}{2\pi Rd_1}(D_{s0}(c_0-c_{1,s})-D_{s1}(c_{1,s}-c_{2,s}))+D\displaystyle\frac{\partial^2c_{1,s}}{\partial s^2}$

$\displaystyle\frac{\partial c_{2,s}}{\partial t}=\displaystyle\frac{1}{2\pi Rd_2}(D_{s0}(c_{1,s}-c_{2,s})-D_{s1}(c_{2,s}-c_3))+D\displaystyle\frac{\partial^2c_{2,s}}{\partial s^2}$

siendo$\displaystyle\frac{d^2}{ds^2}c_{1,s}=\displaystyle\frac{1}{2\pi Rd_1}\displaystyle\frac{(D_{s0}+D_{s1})}{D}c_{1,s}-\displaystyle\frac{1}{2\pi Rd_1}\displaystyle\frac{D_{s1}}{D}c_{2,s}$y$\displaystyle\frac{d^2}{ds^2}c_{2,s}=\displaystyle\frac{1}{2\pi Rd_2}\displaystyle\frac{(D_{s0}+D_{s1})}{D}c_{1,s}-\displaystyle\frac{1}{2\pi Rd_2}\displaystyle\frac{D_{s0}}{D}c_{1,s}$

ID:(8931, 0)

Solución Perimetral

Beschreibung

Como el vector de onda es positivo la solución es una suma de exponenciales $e^{ks}$ y $e^{-ks}$. Por otro lado, como debe ser simetrica respecto del origen en $s=0$ y continua en el otro extremo $s=\pi R$ se debe tener que la pendiente en dicho punto debe ser nula. Por ello la solución debe corresponder a un coseno hiperbolico. Si se considera el factor constante antes calculado será de la forma

$c_1(s)=c_1^0+c_1(0)\displaystyl\frac{\cosh k(\pi R - s)}{\cosh k\pi R}$

y

$c_2(s)=c_2^0+c_2(0)\displaystyl\frac{\cosh k(\pi R - s)}{\cosh k\pi R}$

donde $c_1(0)$ y $c_2(0)$ son las concnetraciones frente al hatch. Para calcular ambas concentraciones basta con reemplazar las funciones en

$\displaystyle\frac{\partial c_{1,0}}{\partial t}=\displaystyle\frac{D_{s0}}{2\pi Rd_1}(c_0-c_{1,0})+\displaystyle\frac{2D}{l_2}\displaystyle\frac{\partial c_{1,s}}{\partial s}-\displaystyle\frac{D}{hd_1}(c_{1,0}-c_{2,0})$

$\displaystyle\frac{\partial c_{2,0}}{\partial t}=\displaystyle\frac{D}{hd_2}(c_{1,0}-c_{2,0})+\displaystyle\frac{2D}{l_2}\displaystyle\frac{\partial c_{2,s}}{\partial s}-\displaystyle\frac{D_{s2}}{2\pi Rd_2}(c_{2,0}-c_3)$

Como la derivada en el origen es$\displaystyle\frac{\partial}{\partial s}c_{1,s}|_{s=0}=-k\tanh(k\pi R)$y$\displaystyle\frac{\partial}{\partial s}c_{2,s}|_{s=0}=-k\tanh(k\pi R)$las ecuaciones se pueden reinterpretar como si la cocnetración en el capilar $c_0$ es$c_0\arrowright c_0+\displaystyle\frac{D}{D_{s0}\displaystyle\frac{4\piRd_1k}{l_2}\tanh(k\pi R)$y$c_3\arrowright c_3-\displaystyle\frac{D}{D_{s0}\displaystyle\frac{4\piRd_1k}{l_2}\tanh(k\pi R)$

ID:(8933, 0)

Simulator

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ID:(8457, 0)

Simulación

Modell

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Zuerst die Gleichung auswählen:

, dann die Variable auswählen:

Symbol

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Gleichung

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Gleichung

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Gleichungen

$\displaystyle\frac{d}{dt}c_{1,s}=\displaystyle\frac{1}{2\pi Rd_1}j(j_{s0},c_{s0},c_0,c_{1,s})-j(j_{s1},c_{s1},c_{1,s},c_{2,s}))+\displaystyle\frac{1}{d^2}D(c_{1,s+1}+c_{1,s-1}-2c_{1,s})$

dc_1,s/dt=(1/2piRd_1)(j(j_s0,c_s0,c_0,c_1)--j(j_s1,c_s1,c_1,s,c_2,s))+(1/d^2)D(c_1,s+1+c_1,s-1-2c_1,s)

(ID 8880)

$\displaystyle\frac{d}{dt}c_{2,s}=\displaystyle\frac{1}{2\pi Rd_2}(j(j_{s0},c_{s0},c_{1,s},c_{2,s})-j(j_{s1},c_{s1},c_{2,s},c_3))+\displaystyle\frac{1}{d^2}D(c_{2,s+1}+c_{2,s-1}-2c_{2,s})$

dc_1/dt=(1/2pi Rd_2)(j(j_s0,c_s0,c_1,s,c_2,s)-j(j_s1,c_s1,c_2,s,c_3))+(1/d^2)D(c_2,s+1+c_2.s-1-2c_2,s)

(ID 8881)

$\displaystyle\frac{d}{dt}c_{1,0}=\displaystyle\frac{1}{2\pi Rd_1}j(j_{s0},c_{s0},c_0,c_{1,0})+\displaystyle\frac{2}{dl_2}D(c_{1,1}-c_{1,0})-\displaystyle\frac{1}{hd_1}D(c_{1,0}-c_{2,0})$

dc_1,0/dt=(1/2pi Rd_1)j(j_s0,c_s0,c_0,c_1,0)+(2/dl_2)D(c_1,1-c_1,0)-(1/hd_1)D(c_1,0-c_2,0)

(ID 8882)

$\displaystyle\frac{d}{dt}c_{2,0}=\displaystyle\frac{1}{hd_2}D(c_{1,0}-c_{2,0})+\displaystyle\frac{2}{dl_2}D(c_{2,1}-c_{2,0})-\displaystyle\frac{1}{2\pi Rd_2}j(j_{s2},c_{2s},c_{2,0},c_3)$

d_2l_2dc_2,0/dt=(l_2/h)D(c_1,0-c_2,0)+(2d_2/d)D(c_2,1-c_2,0)-(l_2/2pi R)j(j_s2,c_2s,c_2,0,c_3)

(ID 8883)

$\displaystyle\frac{d}{dt}c_{1,n}=\displaystyle\frac{1}{2\pi Rd_1}(j(j_{s0},c_{s0},c_0,c_{1,n})-j(j_{s1},c_{s1},c_{1,n},c_{2,n}))+\displaystyle\frac{1}{d^2}D(c_{1,n-1}-c_{1,n})$

dc_1,n/dt=(1/2piRd_1)(j(j_s0,c_s0,c_0,c_1)--j(j_s1,c_s1,c_1,n,c_2,n))+(1/d^2)D(c_1,n-1-c_1,n)

(ID 8884)

c_{1,0}=\displaystyle\frac{16D^2\pi^2R^2c_1^0d_1d_2hk^2\tanh(k\pi R)^2+ ((8D^2R^2c_2^0d_2+8D^2R^2c_1^0d_1)kl_2\pi^2+(4DD_{s0}Rc_0d_2+4DD_{s2}Rc_1^0d_1)hkl_2\pi)\tanh(k\pi R)+(2DD_{s2}Rc_3+2DD_{s0}Rc_0)l_2^2\pi+D_{s0}D_{s2}c_0hl_2^2}{16D^2\pi^2R^2d_1d_2hk^2tanh(k\pi R)^2+((8D^2R^2d_2+8D^2R^2d_1)kl_2\pi^2+(4DD_{s0}Rd_2+4DD_{s2}Rd_1)hkl_2\pi)\tanh(k\pi R)+(2DD_{s2}+2DD_{s0})\pi Rl_2^2+D_{s0}D_{s2}hl_2^2}

(ID 8885)

$\displaystyle\frac{\partial c_{1,s}}{\partial t}=\displaystyle\frac{1}{2\pi Rd_1}(D_{s0}(c_0-c_{1,s})-D_{s1}(c_{1,s}-c_{2,s}))+D\displaystyle\frac{\partial^2c_{1,s}}{\partial s^2}$

\displaystyle\frac{\partial c_{1,s}}{\partial t}=\displaystyle\frac{1}{2\pi Rd_1}(D_{s0}(c_0-c_{1,s})-D_{s1}(c_{1,s}-c_{2,s}))+D\displaystyle\frac{\partial^2c_{1,s}}{\partial s^2}

(ID 8923)

$\displaystyle\frac{\partial c_{2,s}}{\partial t}=\displaystyle\frac{1}{2\pi Rd_2}(D_{s0}(c_{1,s}-c_{2,s})-D_{s1}(c_{2,s}-c_3))+D\displaystyle\frac{\partial^2c_{2,s}}{\partial s^2}$

\displaystyle\frac{\partial c_{2,s}}{\partial t}=\displaystyle\frac{1}{2\pi Rd_2}(D_{s0}(c_{1,s}-c_{2,s})-D_{s1}(c_{2,s}-c_3))+D\displaystyle\frac{\partial^2c_{2,s}}{\partial s^2}

(ID 8924)

\displaystyle\frac{\partial c_{1,0}}{\partial t}=\displaystyle\frac{D_{s0}}{2\pi Rd_1}(c_0-c_{1,0})+\displaystyle\frac{2D}{l_2}\displaystyle\frac{\partial c_{1,s}}{\partial s}-\displaystyle\frac{D}{hd_1}(c_{1,0}-c_{2,0})

(ID 8925)

\displaystyle\frac{\partial c_{2,0}}{\partial t}=\displaystyle\frac{D}{hd_2}(c_{1,0}-c_{2,0})+\displaystyle\frac{2D}{l_2}\displaystyle\frac{\partial c_{2,s}}{\partial s}-\displaystyle\frac{D_{s2}}{2\pi Rd_2}(c_{2,0}-c_3)

(ID 8926)

$c_1^0=\displaystyle\frac{D_{s0}(D_{s0}+D_{s1})}{D_{s1}^2+D_{s0}D_{s1}+D_{s0}^2}c_0+\displaystyle\frac{D_{s1}^2}{D_{s1}^2+D_{s0}D_{s1}+D_{s0}^2}c_3$

c_1^0=D_s0(D_s0+D_s1)c_0/(D_s1^2+D_s0D_s1+D_s0^2)+D_s1^2c_3/(D_s1^2+D_s0D_s1+D_s0^2)

(ID 8927)

$c_2^0=\displaystyle\frac{D_{s1}^2}{D_{s1}^2+D_{s0}D_{s1}+D_{s0}^2}c_0+\displaystyle\frac{D_{s1}(D_{s0}+D_{s1})}{D_{s1}^2+D_{s0}D_{s1}+D_{s0}^2}c_3$

c_2^0=c_0D_s1^2/(D_s1^2+D_s0D_s1+D_s0^2)+c_3D_s1(D_s0+D_s1)/(D_s1^2+D_s0D_s1+D_s0^2)

(ID 8928)

$c_{1,0}=\displaystyle\frac{1+\displaystyle\frac{h}{2\pi R}\displaystyle\frac{D_{s2}}{D}}{1+\displaystyle\frac{D_{s2}}{D_{s0}}+\displaystyle\frac{h}{2\pi R}\displaystyle\frac{D_{s2}}{D}}c_0+\displaystyle\frac{\displaystyle\frac{D_{s2}}{D_{s0}}}{1+\displaystyle\frac{D_{2s}}{D_{s0}}+\displaystyle\frac{h}{2\pi R}\displaystyle\frac{D_{s2}}{D}}c_3$

c_{1,0}=\displaystyle\frac{1+\displaystyle\frac{h}{2\pi R}\displaystyle\frac{D_{s2}}{D}}{1+\displaystyle\frac{D_{s2}}{D_{s0}}+\displaystyle\frac{h}{2\pi R}\displaystyle\frac{D_{s2}}{D}}c_0+\displaystyle\frac{\displaystyle\frac{D_{s2}}{D_{s0}}}{1+\displaystyle\frac{D_{2s}}{D_{s0}}+\displaystyle\frac{h}{2\pi R}\displaystyle\frac{D_{s2}}{D}}c_3

(ID 8929)

$c_{2,0}=\displaystyle\frac{1}{1+\displaystyle\frac{D_{s2}}{D_{s0}}+\displaystyle\frac{h}{2\pi R}\displaystyle\frac{D_{s2}}{D}}c_0+\displaystyle\frac{\displaystyle\frac{D_{s2}}{D_{s0}}+\displaystyle\frac{h}{2\pi R}\displaystyle\frac{D_{s2}}{D}}{1+\displaystyle\frac{D_{2s}}{D_{s0}}+\displaystyle\frac{h}{2\pi R}\displaystyle\frac{D_{s2}}{D}}c_3$

c_{2,0}=\displaystyle\frac{1}{1+\displaystyle\frac{D_{s2}}{D_{s0}}+\displaystyle\frac{h}{2\pi R}\displaystyle\frac{D_{s2}}{D}}c_0+\displaystyle\frac{\displaystyle\frac{D_{s2}}{D_{s0}}+\displaystyle\frac{h}{2\pi R}\displaystyle\frac{D_{s2}}{D}}{1+\displaystyle\frac{D_{2s}}{D_{s0}}+\displaystyle\frac{h}{2\pi R}\displaystyle\frac{D_{s2}}{D}}c_3

(ID 8930)

$k^2=\displaystyle\frac{1}{4\pi R}\left(\displaystyle\frac{(D_{s0}+D_{s1})}{D}\left(\displaystyle\frac{1}{d_1}+\displaystyle\frac{1}{d_2}\right)\pm\sqrt{4\displaystyle\frac{D_{s0}D_{s1}}{D^2d_1d_2}+\displaystyle\frac{(D_{s0}+D_{s1})^2}{D^2}\left(\displaystyle\frac{1}{d_1}-\displaystyle\frac{1}{d_2}\right)^2}\right)$

(ID 8932)

(ID 8934)

Beispiele

brainhatch005

![brainhatch005](showImage.php)
brainhatch005

(ID 8763)

brainhatch007

![brainhatch007](showImage.php)
brainhatch007

(ID 8890)

brainhatch009

![brainhatch009](showImage.php)
brainhatch009

(ID 8892)

brainhatch011

![brainhatch011](showImage.php)
brainhatch011

(ID 8894)

brainhatch008

![brainhatch008](showImage.php)
brainhatch008

(ID 8891)

brainhatch010

![brainhatch010](showImage.php)
brainhatch010

(ID 8893)

brainhatch012

![brainhatch012](showImage.php)
brainhatch012

(ID 8895)

Control de Totales

Las ecuaciones del sistema son para $c_1$ con

$s=0$:

$\displaystyle\frac{d}{dt}c_{1,0}=\displaystyle\frac{1}{2\pi Rd_1}j(j_{s0},c_{s0},c_0,c_{1,0})+\displaystyle\frac{2}{dl_2}D(c_{1,1}-c_{1,0})-\displaystyle\frac{1}{hd_1}D(c_{1,0}-c_{2,0})$

$\displaystyle\frac{d}{dt}c_{1,s}=\displaystyle\frac{1}{2\pi Rd_1}j(j_{s0},c_{s0},c_0,c_{1,s})-j(j_{s1},c_{s1},c_{1,s},c_{2,s}))+\displaystyle\frac{1}{d^2}D(c_{1,s+1}+c_{1,s-1}-2c_{1,s})$

$s=n$;

$\displaystyle\frac{d}{dt}c_{1,n}=\displaystyle\frac{1}{2\pi Rd_1}(j(j_{s0},c_{s0},c_0,c_{1,n})-j(j_{s1},c_{s1},c_{1,n},c_{2,n}))+\displaystyle\frac{1}{d^2}D(c_{1,n-1}-c_{1,n})$

y para $c_2$ con

$s=0$:

$\displaystyle\frac{d}{dt}c_{2,s}=\displaystyle\frac{1}{2\pi Rd_2}(j(j_{s0},c_{s0},c_{1,s},c_{2,s})-j(j_{s1},c_{s1},c_{2,s},c_3))+\displaystyle\frac{1}{d^2}D(c_{2,s+1}+c_{2,s-1}-2c_{2,s})$

$s=n$;

(ID 8896)

Dependencia Espacial

Para el caso estacionario las ecuaciones de la parte de la soluci n que dependen del perimetro se obtienen de

$\displaystyle\frac{\partial c_{1,s}}{\partial t}=\displaystyle\frac{1}{2\pi Rd_1}(D_{s0}(c_0-c_{1,s})-D_{s1}(c_{1,s}-c_{2,s}))+D\displaystyle\frac{\partial^2c_{1,s}}{\partial s^2}$

$\displaystyle\frac{\partial c_{2,s}}{\partial t}=\displaystyle\frac{1}{2\pi Rd_2}(D_{s0}(c_{1,s}-c_{2,s})-D_{s1}(c_{2,s}-c_3))+D\displaystyle\frac{\partial^2c_{2,s}}{\partial s^2}$

(ID 8931)

Soluci n Perimetral

Como el vector de onda es positivo la soluci n es una suma de exponenciales $e^{ks}$ y $e^{-ks}$. Por otro lado, como debe ser simetrica respecto del origen en $s=0$ y continua en el otro extremo $s=\pi R$ se debe tener que la pendiente en dicho punto debe ser nula. Por ello la soluci n debe corresponder a un coseno hiperbolico. Si se considera el factor constante antes calculado ser de la forma

$c_1(s)=c_1^0+c_1(0)\displaystyl\frac{\cosh k(\pi R - s)}{\cosh k\pi R}$

y

$c_2(s)=c_2^0+c_2(0)\displaystyl\frac{\cosh k(\pi R - s)}{\cosh k\pi R}$

donde $c_1(0)$ y $c_2(0)$ son las concnetraciones frente al hatch. Para calcular ambas concentraciones basta con reemplazar las funciones en

Como la derivada en el origen es$\displaystyle\frac{\partial}{\partial s}c_{1,s}|_{s=0}=-k\tanh(k\pi R)$y$\displaystyle\frac{\partial}{\partial s}c_{2,s}|_{s=0}=-k\tanh(k\pi R)$las ecuaciones se pueden reinterpretar como si la cocnetraci n en el capilar $c_0$ es$c_0\arrowright c_0+\displaystyle\frac{D}{D_{s0}\displaystyle\frac{4\piRd_1k}{l_2}\tanh(k\pi R)$y$c_3\arrowright c_3-\displaystyle\frac{D}{D_{s0}\displaystyle\frac{4\piRd_1k}{l_2}\tanh(k\pi R)$

(ID 8933)

Simulator

(ID 8457)

ID:(1037, 0)