# Background

In thermal equilibrium without phase change, the heat transfer $$Q$$ between two fluids with parallel flow or counter flow configuration is expressed as \begin{align} Q \; = \; U A \; LMTD \end{align} where $$U$$ is the heat transfer coefficients, $$A$$ is the contact area for heat exchange, and $$LMTD$$ is the log mean temperature difference given by \begin{align} LMTD \; = \; \frac{\Delta T_{A} \; - \Delta T_{B}}{\ln\big(\frac{\Delta T_{A}}{\Delta T_{B}}\big)} \end{align} $$\Delta T_{A}$$ and $$\Delta T_{B}$$ is the temperature difference between two fluid at terminal $$A$$ and $$B$$ respectively

To extend to other flow configurations, it is said that a correction factor is multiplied to $$LMTD$$, such that this equation hold \begin{align} Q \; = \; UA \; \big(F \times LMTD \big) \end{align} where $$F$$ is temperature dependent.

# Question

It is interesting to see that for any flow configuration, $$Q \; = \; UA \; \big(F \times LMTD \big)$$ still holds. It looks like $$F$$ alone capture the flow configuration, and the $$LMTD$$ is factorised out from the integral. Is there any mathematical derivation?

• For any particular flow configuration, you have to derive F separately, or have its value pre-tabulated in a table. Commented Jul 27 at 9:19
• FWIW, ISO-80000-2 (paywalled, unfortunately), recommends using single letters for variables. Things like $LMTD$ look to be the product for 4 variables, rather than some function of temperature. It probably would be better to write it as a function like $f(T_a,T_b)=(T_a-T_b)/\log(T_a/T_b)$ or similar. Commented 2 days ago