Heat Exchanger Types & the LMTD Method

From flow arrangements to a practical design equation: Q = U·A·ΔT_lm

coreheat transfersophomore
⏱️ About 18 min

Why do two exchangers with the same inlet temperatures deliver different heat duties? The answer is hidden in how the temperature difference changes along the length.

💡
The big idea: The driving force for heat exchange is the temperature difference between streams, and the LMTD captures the correct average ΔT when it varies along the exchanger.
🎯 By the end, you'll be able to
  • Distinguish parallel-flow vs counter-flow temperature profiles
  • Define and compute the log-mean temperature difference (LMTD)
  • Use Q = U·A·LMTD to relate duty, area, and overall coefficient
  • Explain why counter-flow often gives a larger LMTD than parallel-flow
📎 Helpful to know first
  • Convection Correlations for External Flow

What a heat exchanger does

A heat exchanger transfers energy between a hot stream and a cold stream across a wall. In many undergraduate problems we treat the exchanger as steady-state and ignore axial conduction, so the key question becomes: what is the average temperature driving force along the length?

Because the hot stream cools while the cold stream warms, the temperature difference is usually not constant—and that is why we use the log-mean temperature difference (LMTD).

🔑 Common exchanger types (quick map)

Double-pipe: one fluid in an inner tube, the other in an annulus; simple and good for teaching.

Shell-and-tube: many tubes inside a shell; robust and widely used in industry.

Plate: corrugated plates form narrow channels; high U and compact for clean fluids.

Parallel-flow vs counter-flow

Parallel-flow (co-current): both fluids enter the same end and flow in the same direction. The temperature difference is largest at the inlet and shrinks rapidly as the temperatures approach each other.

Counter-flow: fluids enter from opposite ends. The temperature difference can remain more uniform, often giving a larger average driving force for the same inlet/outlet temperatures.

Parallel-flowCounter-flowHot inCold inHot outCold outT_hT_cx →Hot inHot outCold inCold outT_hT_cx →
Parallel-flow tends to lose driving force quickly (profiles converge), while counter-flow keeps a more uniform temperature difference along the length.
\[ \Delta T_{\mathrm{lm}}=\frac{\Delta T_1-\Delta T_2}{\ln\!\left(\Delta T_1/\Delta T_2\right)} \]
Log-mean temperature difference (valid when ΔT varies approximately exponentially along the exchanger).

End temperature differences for counter-flow

For a counter-flow exchanger, define the end temperature differences as:

ΔT1 = Th,in − Tc,out and ΔT2 = Th,out − Tc,in.

Then LMTD is computed from the log-mean formula. (For parallel-flow the end differences are paired at the same end.)

\[ Q = U\,A\,\Delta T_{\mathrm{lm}} \]
LMTD sizing/analysis equation (overall coefficient U and area A).
📝 Worked example: A counter-flow heat exchanger cools a hot stream from 150°C to 100°C while heating a cold stream from 20°C to 90°C. Compute the LMTD.
  1. Counter-flow pairing: one end has Th,in facing Tc,out, the other has Th,out facing Tc,in.
  2. ΔT1 = Th,in − Tc,out = 150 − 90 = 60 K.
  3. ΔT2 = Th,out − Tc,in = 100 − 20 = 80 K.
  4. Compute ln(ΔT1/ΔT2) = ln(60/80) = ln(0.75) = −0.287682.
  5. LMTD = (ΔT1 − ΔT2)/ln(ΔT1/ΔT2) = (60 − 80)/(−0.287682) = 69.52 K.
✓ 69.52 K
✏️ Practice: Counter-flow exchanger: Th,in = 150°C, Th,out = 90°C, Tc,in = 40°C, Tc,out = 110°C. Compute LMTD.
K
Solution
  1. Compute end differences (counter-flow): ΔT1 = Th,in − Tc,out = 150 − 110 = 40 K.
  2. ΔT2 = Th,out − Tc,in = 90 − 40 = 50 K.
  3. Compute ln(ΔT1/ΔT2) = ln(40/50) = ln(0.8) = −0.223143551.
  4. LMTD = (ΔT1 − ΔT2) / ln(ΔT1/ΔT2) = (40 − 50)/(−0.223143551) = 44.814 K.

Check your understanding

1. In the LMTD method, why do we use a log-mean instead of an arithmetic mean of ΔT?
In standard 1D exchanger analysis (steady, negligible axial conduction), the temperature difference often changes approximately exponentially, yielding a log-mean form.
2. For the same inlet/outlet temperatures, counter-flow generally gives a larger LMTD than parallel-flow because:
Counter-flow tends to maintain a higher ΔT over more of the length, increasing the effective average driving force.
3. If ΔT1 equals ΔT2 exactly, the LMTD is:
When ΔT is constant, the correct mean is that constant value. LMTD approaches ΔT1 as ΔT2 → ΔT1.
🔎 Related lessons
Next: LMTD Correction Factors
✅ Key takeaways
  • Heat exchangers come in many hardware forms; the temperature driving force is the unifying concept.
  • Parallel-flow profiles converge quickly; counter-flow tends to preserve a larger average ΔT.
  • LMTD provides the correct mean ΔT when the driving force varies along the length.
  • The core sizing relationship is Q = U·A·ΔT_lm (with U lumping the resistances).
➡️ Next, we fix a common real-world gap: many exchangers are not true counter-flow, so we correct LMTD using a factor F from standard charts.
Want to test yourself on this? Try the Chemical Aptitude test →