Heat Exchangers — Fundamentals and Types

Heat exchangers transfer thermal energy between fluids separated by a solid wall or by direct contact. They are essential in power plants, chemical processes, HVAC, refrigeration, automotive, and electronics cooling. This page covers classifications, common designs, governing equations, sizing methods, and practical selection tips.

Classification

Flow Arrangements

Parallel Flow

Both fluids enter at the same end and move in the same direction. Hot–cold temperature difference is large at the inlet and diminishes toward the outlet; outlet cold temperature cannot exceed the outlet hot temperature.

Counterflow

Fluids move in opposite directions, maximizing the average driving temperature difference. Counterflow provides higher effectiveness for given area and can achieve cold outlet temperatures approaching hot inlet temperature.

Crossflow

Fluids move perpendicular to each other. Behavior depends on mixing: one or both streams may be unmixed (confined by fins or plates) or mixed (plenum/headers allow lateral mixing). Crossflow needs a correction factor when using LMTD.

Multi-pass and Shell Arrangements

Shell-and-tube units often use multiple tube passes and/or multiple shell passes to enhance heat transfer and control pressure drop. Baffles direct shell-side flow across tube bundles (increasing turbulence and area utilization).

Common Heat Exchanger Types

Double-Pipe Heat Exchanger

Shell-and-Tube Heat Exchanger

Plate Heat Exchanger (PHE)

Finned-Tube (Air-Cooled) Exchanger

Plate-Fin and Compact Exchangers

Spiral Heat Exchanger

Two-Phase Devices

Direct-Contact Heat Exchangers

Thermal Design Fundamentals

Energy Balance

For single-phase on both sides:

\[ Q = \dot{m}_h c_{p,h}\,(T_{h,in} - T_{h,out}) = \dot{m}_c c_{p,c}\,(T_{c,out} - T_{c,in}) \]

Define capacity rates and ratios:

\[ C_h = \dot{m}_h c_{p,h},\quad C_c = \dot{m}_c c_{p,c},\quad C_{\min} = \min(C_h, C_c),\quad C_{\max} = \max(C_h, C_c),\quad C_r = \frac{C_{\min}}{C_{\max}} \]

Overall Heat Transfer Coefficient

The overall conductance combines film, wall, and fouling resistances. For a cylindrical wall (tube side “i”, shell side “o”):

\[ \frac{1}{U_o A_o} = \frac{1}{h_o A_o} + R_{f,o} + \frac{\ln(r_o/r_i)}{2\pi k L} + \frac{R_{f,i}}{A_i} + \frac{1}{h_i A_i} \]

Use a consistent reference area (inner, outer, or mean) when relating \( U \), \( A \), and \( Q \).

LMTD Method

The log-mean temperature difference (LMTD) for a single shell/tube pass counterflow or parallel-flow exchanger:

\[ \Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln\!\left(\dfrac{\Delta T_1}{\Delta T_2}\right)}, \quad Q = U\,A\,F\,\Delta T_{lm} \]

\(\Delta T_1\) and \(\Delta T_2\) are end temperature differences. The correction factor \( F \le 1 \) accounts for crossflow or multipass arrangements.

Effectiveness–NTU Method

Define effectiveness \( \varepsilon \) and NTU:

\[ \varepsilon = \frac{Q}{C_{\min}(T_{h,in} - T_{c,in})},\qquad \mathrm{NTU} = \frac{U A}{C_{\min}} \]

Common formulas:

Worked Example — Sizing by LMTD

Design a counterflow exchanger. Water is cooled by water.

Heat duty:

\[ Q = \dot{m}_c c_{p,c}(T_{c,out}-T_{c,in}) = 1.5\times 4180\times (60-20) = 2.508\times 10^5\,\text{W} \]

Hot outlet:

\[ T_{h,out} = T_{h,in} - \frac{Q}{\dot{m}_h c_{p,h}} = 90 - \frac{2.508\times 10^5}{2\times 4180} \approx 60^\circ\text{C} \]

End temperature differences (counterflow):

\[ \Delta T_1 = T_{h,in} - T_{c,out} = 90 - 60 = 30\,\text{K},\quad \Delta T_2 = T_{h,out} - T_{c,in} = 60 - 20 = 40\,\text{K} \]

LMTD and required area (\(F=1\) for ideal counterflow):

\[ \Delta T_{lm} = \frac{40-30}{\ln(40/30)} \approx 34.75\,\text{K},\qquad A = \frac{Q}{U\,\Delta T_{lm}} \approx \frac{2.508\times 10^5}{800\times 34.75} \approx 9.0\,\text{m}^2 \]

Check with \( \varepsilon\text{–NTU} \):

\[ C_h=8360,\; C_c=6270\,\text{W/K}\Rightarrow C_{\min}=6270,\; C_r=0.75,\quad \mathrm{NTU}=\frac{UA}{C_{\min}}\approx \frac{800\times 9.0}{6270}\approx 1.15 \]

Counterflow effectiveness:

\[ \varepsilon=\frac{1 - e^{-\mathrm{NTU}(1 - C_r)}}{1 - C_r e^{-\mathrm{NTU}(1 - C_r)}} = \frac{1 - e^{-1.15\times 0.25}}{1 - 0.75\,e^{-1.15\times 0.25}} \approx 0.571 \]

Then \(Q=\varepsilon\,C_{\min}(T_{h,in}-T_{c,in})\approx 0.571\times 6270\times 70\approx 2.51\times 10^5\,\text{W}\) (consistent).

Selection Guidelines

Fouling and Maintenance

Pressure Drop and Mechanical Constraints

Quick Reference