Why Does the Cylindrical Wall Heat Equation Solution Include Logarithms?

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SUMMARY

The discussion centers on the solution to the cylindrical wall heat equation, specifically the derivation of the temperature distribution T(r) = C1 * ln(r) + C2. The equation is derived from the differential equation 1/r d/dr (r dT/dr) = 0, leading to the integration of d/dr (r dT/dr) = 0. This results in the expression (r dT/dr) = C1, which upon further integration yields the logarithmic solution for temperature as a function of the radial coordinate r.

PREREQUISITES
  • Differential equations, specifically the method of separation of variables
  • Understanding of heat transfer principles in cylindrical coordinates
  • Basic integration techniques, including logarithmic integration
  • Familiarity with boundary conditions in thermal analysis
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  • Study the derivation of the heat equation in cylindrical coordinates
  • Learn about boundary value problems in heat transfer
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Students and professionals in engineering, particularly those specializing in thermal dynamics, as well as anyone interested in solving differential equations related to heat transfer in cylindrical systems.

clava345
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I'm a little bit rusty with my differential equations, and can't seem to see how solving for 1/r d/dr (r dT/dr)=0 has the solution T(r)=C_1*ln⁡(r)+C_2
 
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integration of d/dr (r dT/dr)=0 leads to (r dT/dr) = C1
dT = C1 dr/r
integration of dT=C1 dr/r leads to t = C1 ln(r) + C2
 
Thanks
 

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