Now that we have looked at the *global* or
*macroscopic* view of a heat exchanger, it is useful to look
into more detail as to what goes on *inside* the black
box...

Before, we did a balance around the entire heat exchanger, if
instead, we now look at a *microscopic* balance around a
small portion of the exchanger (the following analysis will only be
applicable (as it is) for a single-pass exchanger (either parallel
of counter-current flow)):

If our small section has a length, $\Delta x$, and a perimeter
$P$, we can say that the amount of heat transfered from the hot
stream (at $T_H$) to the cold stream (at $T_C$), is given by:

$\displaystyle{q = UP\Delta x (T_H-T_C)}$

which is equal to (from a "macro" view of the section):

$\displaystyle{q = \dot m_H c_H (T_{H_{in}}-T_{H_{out}}) = \dot m_H c_H (T_{H_{x}}-T_{H_{x+\Delta x}}) }$

so that

$\displaystyle{UP\Delta x (T_H-T_C) = \dot m_H c_H (T_H{_{x}}-T_H{_{x+\Delta x}}) }$

dividing through by $\Delta x$ and letting $\Delta x \rightarrow \infty$

$\displaystyle{-UP(T_H-T_C) = \dot m_H c_H \frac{dT_H}{dx} }$

we also note that $q = \dot m_C c_C (T_C{_{out}}-T_C{_{in}})$, so that we can write (by the same logic as above)

$\displaystyle{UP(T_H-T_C) = \dot m_C c_C \frac{dT_C}{dx} }$

dividing through by $\dot m c$ we get

$\displaystyle{-\frac{UP}{\dot m_Hc_H}(T_H-T_C) = \frac{dT_H}{dx} }$

and

$\displaystyle{\frac{UP}{\dot m_Cc_C}(T_H-T_C) = \frac{dT_C}{dx} }$

if we subtract the "hot-stream" equation from the "cold-stream" equation, we get

$\displaystyle{\frac{UP}{\dot m_Cc_C}(T_H-T_C) + \frac{UP}{\dot m_Hc_H}(T_H-T_C)= \frac{dT_C}{dx} - \frac{dT_H}{dx}}$

which we can rearrange to be

$\displaystyle{UP(T_H-T_C)\left ( \frac{1}{\dot m_Cc_C} + \frac{1}{\dot m_HcH} \right )= \frac{d(T_C-T_H)}{dx}}$

dividing through by
(*T*_{H}-*T*_{C}), moving
the *dx* to the other side, and multiplying both sides by a
-1

$\displaystyle{\frac{d(T_H-T_C)}{(T_H-T_C)} = -UP\left ( \frac{1}{\dot m_Cc_C} + \frac{1}{\dot m_Hc_H} \right ) dx}$

if we integrate (Note: here, we are assuming the $U \neq U(x)$, but only a minor modification is necessary if $U=U(x)$) from 0 to $L$ (and denote the temperatures at those points as $T_0$ and $T_L$), we get

$\displaystyle{ln\left (\frac{(T_{H,L}-T_{C,L})}{(T_{H,0}-T_{C,0})} \right ) = -UPL \left ( \frac{1}{\dot m_Cc_C} + \frac{1}{\dot m_HcH} \right )}$

going *way* back to the beginning, you should recall that
$q = C_H(T_{H,in}-T_{H,out}) = C_C(T_{C,out}-T_{C,in})$, so
plugging these two relations in for $C_H$ and $C_C$, we get

$\displaystyle{ln\left (\frac{(T_{H,L}-T_{C,L})}{(T_{H,0}-T_{C,0})} \right ) = -UPL \left ( \frac{T_{C,L}-T_{C,0}}{q} - \frac{T_{H,L}-T_{H,0}}{q} \right )}$

solving for *q* gives

$\displaystyle{q = -UPL \left ( \frac{(T_{C,L}-T_{C,0}) - (T_{H,L}-T_{H,0})}{ln\left (\frac{(T_{H,L}-T_{C,L})}{(T_{H,0}-T_{C,0})} \right )} \right )}$

which we can rearrange one last time to be

$\displaystyle{q = UPL \left ( \frac{(T_{H,L}-T_{C,L}) - (T_{H,0}-T_{C,0})}{ln\left (\frac{(T_{H,L}-T_{C,L})}{(T_{H,0}-T_{C,0})} \right )} \right )}$

if we define a log mean temperature, $\Delta T_{LM}$, as

$\displaystyle{\Delta T_{LM} = \frac{(T_{H,L}-T_{C,L}) - (T_{H,0}-T_{C,0})}{ln\left (\frac{(T_{H,L}-T_{C,L})}{(T_{H,0}-T_{C,0})} \right )} }$

we can write our final result as

$\displaystyle{q = UPL \Delta T_{LM}}$

Calculate the "average" driving force from the heat flow and resistance