At this stage, we do not know whether things are running in
counter or co-current flow. Being chemical engineers with a firm
grasp of transport phenomena, however, we know that the
*rate* of heat exchange is proportional to the driving force
for the exchange (the temperature difference)! So, we might be
interested in looking at how the two temperatures vary along the
length of the exchanger.

In the previous example, we see that, while the two different
cases yield different results, both seem feasible. (As we will see
shortly, the only difference here would be the *area* need
for the heat exchange).

However, if we modify our example problem, so that instead of
cooling our fluid from 375*K* to 350*K*, we want to cool
it to 310*K*. We see that the total amount of heat being
transfer in the hot stream is:

$\displaystyle{q = -\left (\dot mc\Delta T \right)_H = \dot
m_H c_H (T_{H_{in}} - T_{H_{out}}) =
(0.5kg/s)(2100J/kgK)(375K-310K) = 68250 W}$

so that the water stream (which enters at 280*K* and 0.2
*kg*/*s* must also transfer that much, so its exit
temperature is given as:

$\displaystyle{q = \left (\dot mc\Delta T \right)_C= \dot m_C
c_C (T_{C_{out}} - T_{C_{in}}) = (0.2kg/s)(4200J/kgK)(T_{C_{out}}
-280K) = 68250 W}$

*T*_{Cout} = 361*K*

a result which makes perfect sense thermodynamically, but obviously
limits our exchanger configuration, when we consider transport!

Visualize the changing driving forces in co-current versus counter-current exchangers