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U 3 = a 2 4 π R e r f c R 4 t / a 2 , t > 0 R = ( x - x o ) 2 + ( y - y o ) 2 + ( z - z o ) 2 1 / 2
The a 2 .factor has the units of time / L 2 . If time is made dimensionless with respect to a 2 / R o 2 and R with respect to R o , then the factor will disappear from the argument of the erfc.

Assignment 7.4

Plot the profiles of the response to a continuous source in 1, 2, and 3 dimensions using the MATLAB code contins.m and continf.m in the diffuse subdirectory. From the integral of the profiles as a function of time, determine the magnitude, spatial and time dependence of the source. Note: The exponential integral function, expint will give error messages for extreme values of the argument. It still computes the correct values of the function.

Convective-diffusion equation

The convective-diffusion equation in one dimension will be expressed in terms of velocity and dispersion,

u t + v u x = K 2 u x 2 u ( x , ) = 0 , x > 0 u ( 0 , t ) = 1 , t > 0

The independent variables can be transformed from ( x , t ) to a spatial coordinate that translates with the velocity of the wave in the absence of dispersion, ( y , t ) .

y = x - v t

This transforms the equation to the diffusion equation in the transformed coordinates.

u t = K 2 u y 2

To see this, we will transform the differentials from x to y .

y t = - v y x = 1

The total differentials expressed as a function of ( x , t ) or ( y , t ) are equal to each other.

d u = u t x d t + u x t d x d u = u t y d t + u y t d y

The total differentials expressed either way are equal. The partial derivatives in t and x can be expressed in terms of partial derivatives in t and y by equating the total differentials with either d t or d x equal to zero and dividing by the non-zero differential.

u t x = u t y + u y t y t x = u t y - v u y t and u x t = u y t y x t = u y t 2 u x 2 t = 2 u y 2 t

Substitution into the original equation results in the transformed equation. This result could have been derived in fewer steps by using the chain rule but would not have been as enlightening.

The boundary condition at x = 0 is now at changing values of y . We will seek an approximate solution that has the boundary condition u ( y - ) = 1 . A simple solution can be found for the following initial and boundary conditions.

u ( y , 0 ) = 1 , y < 0 1 / 2 , y = 0 0 , y > 0 u ( y - , t ) = 1 u ( y , t ) = 0

This system is a step with no dispersion at t = 0 . Dispersion occurs for t > 0 as the wave propagates through the system. The solution can be found with a similarity transform, which we will discuss later. For now, the approximate solution is given as

u = 1 2 erfc y 4 K t = 1 2 erfc x - v t 4 K t

The boundary condition at x = 0 will be approximately satisfied after a small time unless the Peclet number is very small.

Similarity transformation

In some cases a partial differential equation and its boundary conditions (and initial condition) can be transformed to an ordinary differential equation with boundary conditions by combining two independent variables into a single independent variable. We will illustrate the approach here with the diffusion equation. It will be used later for hyperbolic PDEs and for the boundary layer problems.

The method will be illustrated for the solution of the one-dimensional diffusion equation with the following initial and boundary conditions. The approach will follow that of the Hellums-Churchill method.

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
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I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
is it a question of log
Commplementary angles
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what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
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ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
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Smarajit Reply
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Source:  OpenStax, Transport phenomena. OpenStax CNX. May 24, 2010 Download for free at http://cnx.org/content/col11205/1.1
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