Heat Transfer by Conduction Resistance through a Cylindrical Wall

Written by Jerry Ratzlaff on . Posted in Thermodynamics

Heat Transfer by Conduction resistance through a Cylindrical Wall Formula

\(\large{ R_t =  \frac   {  ln \;  \left(   \frac  {r_2 }  { r_1 } \right) }   { 2 \; \pi \; k \; l  }   }\)   

Where:

 Units English Metric
\(\large{ R_t }\) = thermal resistance \(\large{\frac{hr-F}{Btu}}\) \(\large{\frac{K}{W}}\)
\(\large{ l }\) = length of material \(\large{ft}\) \(\large{m}\)
\(\large{ ln }\) = natural logarithm \(\large{dimensionless}\)
\(\large{ \pi }\) = Pi \(\large{3.141 592 653 ...}\)
\(\large{ r_1 }\) = radius inside diameter (ID) \(\large{in}\) \(\large{mm}\)
\(\large{ r_2}\) = radius outside diameter (OD) \(\large{in}\) \(\large{mm}\)
\(\large{ k }\) = thermal conductivity \(\large{\frac{Btu-ft}{hr-ft^2-F}}\) \(\large{\frac{W}{m-K}}\)

 

P D Logo 1

Tags: Heat Transfer Equations Heat Equations