Heat exchanger

The local heat transfer rate for both streams read $\begin{aligned} c \dot V \frac{dT_h}{dx} dx &= -d\dot Q = -k w (T_h(x) - T_c(x)) dx \\ \pm c \dot V \frac{dT_c}{dx} dx &= d\dot Q = k w (T_h(x) - T_c(x)) dx \\ \end{aligned}$ where the plus sign and the minus sign in the latter equation respectively denoting the parallel and countercurrent flow. With the parameter $\lambda = 2 k w/c \dot V$ we get the coupled differential equations $\left( \begin{array}{c} -T_h'(x) \\ \pm T_c'(x) \end{array} \right) = \frac{\lambda}{2} \left( \begin{array}{cc} 1 & - 1\\ 1 & -1 \end{array} \right) \left( \begin{array}{c} T_h(x) \\ \pm T_c(x) \end{array} \right)$

Countercurrent Flow

$\frac{dT_h}{dx} = \frac{dT_c}{dx}$ applies, so that the gradient $T_h(x) - T_c(x) = \Delta T$ results a constant. Therefore, the functions show a linear course $\begin{aligned} T_h(x) &= T_{h1} - a x \\ T_c(x) &= T_{c1} + a (L - x) \end{aligned}$ with $\Delta T = T_{h1} - T_{c1} - a L = - (2/\lambda) \cdot T_h'(x) = 2 a/\lambda$ , so that $\begin{aligned} a &= \frac{T_{h1} - T_{c1}}{ \frac{2}{\lambda} + L }\\ T_{c2} &= T_c(0) = T_{c1} + \frac{L}{\frac{2}{\lambda} + L} (T_{h1} - T_{c1}) \end{aligned}$ in the limiting case $L \to \infty$ we get $T_{c2} = T_{h1}$ , so that the cold water stream can get as hot as the incoming hot water.

Parallel Flow

$\frac{dT_h}{dx} + \frac{dT_c}{dx} = 0$ applies, so that the mean temperature $T_m = (T_{h} + T_{c})/2 = \text{const}$ stays constant. Therefore, only the temperature difference $\Delta T(x) = T_h(x) - T_c(x)$ must be considered, so that differential equation becomes onedimensional $\Delta T'(x) = - \lambda \Delta T(x)$ which is solved by $\Delta T(x) = (T_{h1} - T_{c1}) \exp(-\lambda x)$ . The final temperature of the cold stream results to $T_{c2} = \frac{T_{h1} + T_{c1}}{2} - \frac{T_{h1} - T_{c1}}{2} e^{-\lambda L}$ in the limit $L \to \infty$ we get $T_{c2} = T_m$ , so that cold water stream can never exceed the mean temperature.

Plotting both final temperature as function of the length $L$ results $T_{c2}^\text{countercurrent}(L) > T_{c2}^\text{parallel}(L)$ for all $L > 0$ , so that the countercurrent configuration is always more effective.