CONVECTION HEAT TRANSFERHeat TransferQk_{Dr}A_{lm} (T_{2}T_{3})Qh_{2}A_{2} (T_{3}T_{4})DTo can be expressed as the sum of the DT of the three individual processes.DTo(T_{1}T_{2})(T_{2}T_{3})(T_{3}T_{4})If the basic relationship for each process is solved for its associated temperature difference andsubstituted into the expression for DT_{o} above, the following relationship results.DToQ1h_{1}A_{1}Drk A_{lm}1h_{2}A_{2}This relationship can be modified by selecting a reference cross-sectional area A_{o}.DToQA_{o}A_{o}h_{1}A_{1}Dr Aok A_{lm}A_{o}h_{2}A_{2}Solving for results in an equation in the form .QQU_{o} A_{o D}T_{o}Q1A_{o}h_{1}A_{1}Dr Aok A_{lm}A_{o}h_{2}A_{2}A_{o D}T_{o}where:(2-10)U_{o}1A_{o}h_{1}A_{1}Dr Aok A_{lm}A_{o}h_{2}A_{2}Equation 2-10 for the overall heat transfer coefficient in cylindrical geometry is relativelydifficult to work with. The equation can be simplified without losing much accuracy if the tubethat is being analyzed is thin-walled, that is the tube wall thickness is small compared to the tubediameter. For a thin-walled tube, the inner surface area (A_{1}), outer surface area (A_{2}), and logmean surface area (A_{1m}), are all very close to being equal. Assuming that A_{1}, A_{2}, and A_{1m} areequal to each other and also equal to A_{o} allows us to cancel out all the area terms in thedenominator of Equation 2-11.HT-02 Page 22 Rev. 0