51 - Heat Exchangers - LMTD Method

\[Q = UA \Delta T_m \tag*{(1)}\] For an elemental area \(dA\), \[dQ = U (\Delta T) (dA) \tag*{(2)}\] where \(\Delta T = T_h-T_c\).

From heat capacity relations, for the cold and hot fluids, we have \[\begin{align*} dQ &= -C_c\; dT_c \tag*{(3a)} \\ dQ &= -C_h\; dT_h \tag*{(3b)} \end{align*}\] where \(C_c=\dot{m_c}C_{P,c}\), and \(C_h=\dot{m_h}C_{P,h}\)

\[\begin{aligned} \Delta T &= T_h - T_c \\ d(\Delta T) &= dT_h - dT_c\\ \text{          Substituting for $dT_h$ and $dT_c$ from Eqn.(3), we get}\\ d(\Delta T)&= -\frac{dQ}{C_h} + \frac{dQ}{C_c} =dQ\left(\frac{1}{C_c} - \frac{1}{C_h} \right) \end{aligned}\] Substituting for \(dQ\) from Eqn.(2), we get \[d(\Delta T) = U (\Delta T) (dA)\left(\frac{1}{C_c} - \frac{1}{C_h} \right)\] Rearranging, \[\frac{d(\Delta T)}{\Delta T} = U (dA)\left(\frac{1}{C_c} - \frac{1}{C_h} \right)\]

By continuing the derivation similar to co-current, we get the final relation as \[Q = UA\Delta T_{\text{lm}}\] where \[\boxed{\Delta T_m = \frac{\Delta T_2 - \Delta T_1}{\ln \dfrac{\Delta T_2}{\Delta T_1}} = \Delta T_{\text{lm}} = \text{LMTD}}\] LMTD = Log Mean Temperature Difference

This final relation is same as that cocurrent. But \(\Delta T_1\) and \(\Delta T_2\) are not the same.