MHE: Heat Exchanger Analysis (the driving force)

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 375K to 350K, we want to cool it to 310K. 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 280K and 0.2 kg/s must also transfer that much, so its exit temperature is given as:
$\displaystyle{q = \left (\dot mc\Delta T \right)_C= \dot m_C c_C (T_{C_{out}} - T_{C_{in}}) = (0.2kg/s)(4200J/kgK)(T_{C_{out}} -280K) = 68250 W}$

TCout = 361K

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