MHE: Heat Exchanger Analysis (the "Thermo" Problem)

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 $\displaystyle{\hat H}$ = cpT, we write this in its final form:

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


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


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 Tout-Tin (hence the negative for the hot stream).


Solve the "thermodynamic" problem


A Light oil (cp = 2100 J/kg K) is cooled by water in a heat exchanger. The oil flows at 0.5 kg/s and is cooled from 375K to 350K. If the water starts out at 280K 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:
$\displaystyle{q = \left (\dot mc\Delta T \right)_C= \dot m_C c_C (T_{C_{out}} - T_{C_{in}}) = (0.2kg/s)(4200J/kgK)(T_{C_{out}} -280K) = 26250 W}$

TCout = 311K