2 Answers

Paljoshi · Answer 1 · 2020-02-07T00:00:09+0000

Assumptions:

Step-by-step explanation:

Use the appropriated heat transfer equation (cylindrical coordinates),

$\frac {1}{r} \frac {d}{dr} (\ r \frac {dT}{dr}) + \frac {q}{k} \ = \ 0$

After separating variables and integrating we obtained the following expression,

$r(dT)/(dr) \ = \ - (q)/(2k) \ r^(2) \ + \ C_(1)$

After repeating the procedure state above we obtained the following expression,

$T(r) \ = \ - (q)/(2k) \ r^(2) \ + \ C_(1) ln(r) \ + \ C_(2)$

According to the image attached, the corresponding boundary values for the system are

$\left \{ {{(dT)/(dr)=0 \ , \ r={0}} \atop {T(r_(0))=T_(s)}} \right.$

Then, the temperature distribution in the rod along r axis corresponds to,

$T(r) \ = \ (qr_(0)^(2))/(4k) (1 \ - (r^2)/(r_(0)^(2)) ) \ + \ T_(s)$

After deriving the function, we obtained the expression that follows,

$(dT)/(dr) \ = \ - (qr)/(2k)$

Evaluating the previous expression at the surface of the rod,

$(dT)/(dr) \lvert_{r = r_(0)}} \ = (- (qr_(0))/(2k))$

Answer:

The heat transferred by the finite extended rod corresponds to,

$q_(r) = -k(2 \pi L) (dT)/(dr)$

Replacing the derivate term, we obtained the expression that follows,

$q_(r)  = -k(2 \pi L) \ (- (qr_(0))/(2k))\\q_(r) \ = \ \pi \ L \ {q \ r_(0)$

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