Sign in to save

Bookmark this page so you can find it later.

Sign in to save

Bookmark this page so you can find it later.

The log mean temperature difference method is a core tool for sizing and analyzing heat exchangers when two fluids exchange thermal energy through a wall. It converts a changing temperature difference along the exchanger into one effective average driving force. This matters because the local heat-transfer rate is larger where the fluid temperatures are farther apart and smaller where they are closer together.

Engineers use the method to estimate heat-transfer area, heat duty, and outlet temperatures in devices such as condensers, boilers, radiators, and process heat exchangers.

The method combines the heat-transfer design equation Q = UAΔTlm with an energy balance on the hot and cold streams. For a heat exchanger with temperature differences ΔT1 and ΔT2 at the two ends, the log mean temperature difference is ΔTlm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2). Counterflow usually gives a larger LMTD than parallel flow for the same inlet and outlet temperatures, so it often needs less surface area for the same heat duty.

Real multi-pass and crossflow exchangers often require a correction factor F, giving Q = UAFΔTlm.

Understanding Engineering: Heat Exchanger LMTD Method

The overall heat transfer coefficient is often the hardest part of a real design. It represents every barrier between the two fluids. Heat must move from the hot fluid into the wall, conduct through the wall material, then pass from the wall into the cold fluid.

Each step adds thermal resistance. A thin copper wall offers little resistance, while a thick stainless steel wall offers more. In many exchangers, the main resistance comes from slow-moving fluid next to the wall, not from the metal itself.

Higher fluid speed can improve transfer by disturbing this boundary layer. It can also raise pumping power and pressure loss. Good design balances these effects rather than simply choosing the fastest flow.

Before finding area, engineers determine the required heat duty from each stream. The heat lost by the hot side should closely match the heat gained by the cold side. Small differences can occur because of heat loss to surroundings or rounded data.

A large mismatch usually signals a problem with units, flow rate, temperature measurement, or assumed heat capacity. Students should keep mass flow rate units consistent with the heat capacity units.

For example, using kilograms per second with heat capacity in joules per kilogram per degree gives heat rate in watts. This basic energy check prevents many design mistakes before the temperature driving force is calculated.

The temperature pattern reveals whether a proposed exchanger is physically possible. A temperature difference must remain positive throughout ordinary heat exchange. If the calculated profiles cross, the requested outlet temperatures cannot be reached with that flow arrangement.

Counterflow is especially useful when the cold outlet needs to become nearly as hot as the hot outlet. In parallel flow, both streams leave at the same end and their temperatures tend to move closer together quickly.

In counterflow, each fluid meets a different part of the other stream, which preserves a useful temperature difference over more of the length. This is why arrangement affects required area even when the inlet temperatures and flow rates are unchanged.

Real equipment becomes less effective over time. Scale from hard water, corrosion products, oil films, and biological growth can coat the surfaces. These deposits add resistance and reduce heat transfer.

Designers commonly allow extra area for expected fouling, especially in cooling-water service. They must consider pressure drop, cleaning access, material compatibility, and safe operating pressure as well.

A calculation may predict a compact exchanger, yet the final unit may need wider passages so it can be cleaned or so particles do not block it. When learning this method, sketch the flow directions first, label every inlet and outlet temperature clearly, and check that the two end temperature differences use the correct paired locations.

Key Facts

  • Heat exchanger design equation: Q = UAΔTlm.
  • Log mean temperature difference: ΔTlm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2).
  • Hot-stream energy balance: Q = mh ch (Th,in - Th,out).
  • Cold-stream energy balance: Q = mc cc (Tc,out - Tc,in).
  • For parallel flow, ΔT1 = Th,in - Tc,in and ΔT2 = Th,out - Tc,out.
  • For counterflow, ΔT1 = Th,in - Tc,out and ΔT2 = Th,out - Tc,in.

Vocabulary

LMTD
The log mean temperature difference is the effective average temperature difference that drives heat transfer along a heat exchanger.
Overall heat-transfer coefficient
The overall heat-transfer coefficient U measures the combined ability of convection, wall conduction, and fouling layers to transfer heat.
Heat duty
Heat duty Q is the rate of thermal energy transferred from one fluid to the other.
Counterflow
Counterflow is a heat exchanger arrangement where the hot and cold fluids move in opposite directions.
Correction factor
The correction factor F adjusts the ideal LMTD result for exchanger arrangements that are not simple parallel flow or counterflow.

Common Mistakes to Avoid

  • Using the arithmetic average temperature difference instead of LMTD is wrong because the temperature driving force usually changes nonlinearly along the exchanger.
  • Mixing up ΔT1 and ΔT2 for parallel and counterflow is wrong because each flow arrangement pairs different end temperatures.
  • Forgetting the correction factor F for crossflow or multi-pass exchangers is wrong because Q = UAΔTlm applies directly only to ideal one-pass parallel or counterflow cases.
  • Using inconsistent units for U, A, and Q is wrong because W, m2, and W/(m2 K) must combine so the temperature difference is in K or °C.

Practice Questions

  1. 1 A counterflow heat exchanger has Th,in = 120 °C, Th,out = 70 °C, Tc,in = 25 °C, and Tc,out = 60 °C. Calculate ΔT1, ΔT2, and ΔTlm.
  2. 2 A heat exchanger has U = 500 W/(m2 K), A = 12 m2, and ΔTlm = 35 K. Calculate the heat-transfer rate Q.
  3. 3 For the same inlet and outlet temperatures, explain why a counterflow heat exchanger often requires less heat-transfer area than a parallel-flow heat exchanger.