For a heat exchanger analysis, the first step is to do an
overall macroscopic heat balance.

We are actually doing an energy balance that is simplified (we will
see this soon). In other words, we assume that our exchanger is
insulated, that none of our thermal energy is converted from or to
other forms of energy, and that all of the heat that goes
*in* must come *out* (just hopefully not in the same
places!). We will tend to call this the *thermodynamic* part
of the problem.

Using the average velocities (or assuming a plug flow) across all inlets and outlets and recalling that we (generally) have two of each, we can write this in a much simpler form as:

$\displaystyle{\left (\rho \hat H v A \right)_{H_{in}} - \left (\rho \hat H v A \right)_{H_{out}} = \left (\rho \hat H v A \right)_{C_{out}} - \left (\rho \hat H v A \right)_{C_{in}}}$

recalling that the mass flow rate is the product of density,
average velocity, and cross sectional area, as well as the fact
that $\displaystyle{\hat H}$ = c_{p}T, we write this in
its final form:

$\displaystyle{\left (\dot mcT \right)_{H_{in}} - \left (\dot mcT \right)_{H_{out}} = \left (\dot mcT \right)_{C_{out}} - \left (\dot mcT \right)_{C_{in}}}$

or

$\displaystyle{q = -\left (\dot mc\Delta T \right)_H = \left (\dot mc\Delta T \right)_C}$

This type of equation is probably familiar from your
Thermodynamics class. Here, we have defined the **heat duty** as
the total amount of heat that is transferred from one stream to the
other. Also, we have used the convention that the change in
temperature refers to T_{out}-T_{in} (hence the
negative for the hot stream).

Solve the "thermodynamic" problem

A Light oil (*c*_{p} = 2100 *J*/*kg
K*) is cooled by water in a heat exchanger. The oil flows at 0.5
*kg*/*s* and is cooled from 375*K* to 350*K*.
If the water starts out at 280*K* and flows at 0.2
*kg*/*s*. What is the water outlet temperature?

Our macroscopic energy balance yields:

$\displaystyle{q = -\left (\dot mc\Delta T \right)_H = \dot
m_H c_H (T_{H_{in}} - T_{H_{out}}) =
(0.5kg/s)(2100J/kgK)(375K-350K) = 26250 W}$

Our macroscopic energy balance yields:

$\displaystyle{q = \left (\dot mc\Delta T \right)_C= \dot m_C
c_C (T_{C_{out}} - T_{C_{in}}) = (0.2kg/s)(4200J/kgK)(T_{C_{out}}
-280K) = 26250 W}$

*T*_{Cout} = 311*K*