Comparing two functions and determining coefficients. I get 1/1.

[Mamed]

The objective is to find some coefficients by comparing two equations.


$T_2 \cdot \frac{R_1}{R_1+R_2} + T_1 \cdot \frac{R_2}{R_1+R_2}$

and

$T_2 \cdot k_1 + T_1 \cdot k_2$

I compare and set


$k_1 = \frac{R_1}{R_1+R_2} (1)$

$k_2 = \frac{R_2}{R_1+R_2} (2)$

I expand the equations and throw them around

$(1) -> R_1 = R_2 \cdot \frac{k_1}{1-k_1} (3)$
$(2) -> R_2 = R_1 \cdot \frac{k_2}{1-k_2} (4)$

I put in (3) in (4) and get


$R_2 = R_2 \cdot \frac{k_1}{1-k_1} \cdot \frac{k_2}{1-k_2} -> 1 = \frac{k_1}{1-k_1} \cdot \frac{k_2}{1-k_2}$

So i don't know how to get the values of $R_1$ and $R_2$ as a function of $k_1$ and $k_2.$

The solution for this is given and is correct when I put it in the equations above and it fulfills everything but i just don't know how he has gotten it.

$k_x$ is a long function based on parameters that i just rebrand for simplicity.

 

[Stephen Tashi]

The objective is to find some coefficients by comparing two equations.


$T_2 \cdot \frac{R_1}{R_1+R_2} + T_1 \cdot \frac{R_2}{R_1+R_2}$

and

$T_2 \cdot k_1 + T_1 \cdot k_2$

If those are two equations then where are the "=" signs? Do you mean "functions" instead of "equations"? Are the two functions assumed to be equal?

 

[Mamed]

No they have it all expressed as

$T_m = T_2 \underbrace{\left( \frac{\displaystyle r_2^2}{\displaystyle r_2^2-r_1^2}-\frac{\displaystyle 1}{ \displaystyle 2 \ln{\left( \displaystyle r_2 / r_1 \right) }} \right)}_{k_1} + T_1 \underbrace{\left( \frac{ \displaystyle 1}{\displaystyle 2 \ln{\left(\displaystyle r_2 / r_1 \right) }}-\frac{\displaystyle r_1^2}{\displaystyle r_2^2-r_1^2} \right)}_{k_2}$

and then they have the equation as i have given above

$T_m = T_2 \cdot \frac{\displaystyle R_1}{\displaystyle R_1+R_2} + T_1 \cdot \frac{\displaystyle R_2}{\displaystyle R_1+R_2}$

And then they say "Comparing the coefficients of $T_1$ and $T_2$ in the two equations we get"

$R_1 = \frac{1}{\displaystyle 4\pi kL} \left( \frac{\displaystyle 2r_2^2\ln{\left(r_2/r_1\right)} }{\displaystyle r_2^2-r_1^2}-1 \right)$

$R_2 = \frac{1}{\displaystyle 4\pi kL} \left( 1- \frac{\displaystyle 2r_1^2\ln{\left(r_2/r_1\right)} }{\displaystyle r_2^2-r_1^2} \right)$

What i want to know is how they arrive at the values for $R_1$ and $R_2$

 

[gsal]

There must be something else you are not telling because the equations for R1 and R2 contain 1/(4.pi.k.L) ...which is nowhere in the coefficients k1 and k2...

 

[Mamed]

Those are for thermal conductivity and and the length, when you put them in the equation they are canceled out.

Tm is a mean temperature in an lumped parameter cylinder with no internal heat generation. And that's how they arrive at the solution. No further explaining done. And i want to know how they arrive at the solution rather than just look at the solution. Not holding back anything.

 

[gsal]

Then, maybe you need to reveal the entire problem at hand...we are driving blind here...without knowing the context...maybe if we knew more...for example, you just revealed that we are talking about a cylinder without internal heat generation...what can you tell about T1 and T2, where are they supposed to be?

Sparing us from the details is not working...tell us everything, now.

 

[Mamed]

lol ok, the r2 is outer radius at temperature T2 and r1 is inner with T1. L is the length.

And the resistance is set so that

T1----R1-----Tm------R2------T2--->Q

If we use Ohms law here, then Tm = T2+Q*R2

We know that Q = (T1-T2)/(R1+R2) -> Tm = T2 + R2*(T1-T2)/(R1+R2)

Tm = T2*R1/(R1+R2) + T1*R2/(R1+R2)

Thats the equation form the beginning and then he has the other equation that he compares with which i have already given. And then he magically arrives at that solution.

That specific equation he derives form a cylinder with the equationd^2T/dr^2 + 1/r* dT/dr + g/k = 0 but here g = 0 as there is no internal heat generation.

 

[gsal]

Hhhmmm..

So, if R1 and R2 add up to the total thermal resistance in the radial direction of the cylinder (tube, rather)...then

R1/(R1+R2) is a fraction of it (a number less than 1.0), and

R2/(R1+R2) is another fraction ( <1.0) and complementary...

meaning, together, they add up to 1, correct?

R1/(R1+R2) + R2/(R1+R2) = (R1+R2)/(R1+R2) = 1.0

that means that also k1+k2=1 ...the knowledge of this may help you in your calculations

Also, those formulas about the thermal resistance look too complicated...why do you have r² in there? Make sure that when calculating the thermal resistance you do not mix what the cross-sectional area available for heat flow is...in other words, if the heat is flowing from the inside wall of the cylinder to the outside wall...the area for flow at a given radius is 2∏rL ...this appears in the denominator of the formula for resistance:

dR = dr/(2∏rLκ)

when you integrate this over a given range, you do not have r2, you just get a difference of ln:

(ln(R2)-ln(R1))

Now, if the heat is flowing in the axial direction of the cylinder, well, then the area available for heat flow does include r²...so, just make sure you have all your ducks in a row...