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THE ENGINEERING REFERENCE / GUIDE

LMTD Explained: Temperature Differences and Flow Arrangements

Short Answer

The log mean temperature difference (LMTD) is the mean driving force used in the heat exchanger sizing equation Q = UAΔTm, computed from the two end-of-exchanger temperature differences as ΔTlm = (ΔT1 − ΔT2)/ln(ΔT1/ΔT2). Because it correctly represents the exponential temperature profiles, it never exceeds the arithmetic mean. Flow arrangement controls how terminal temperatures pair: counter-flow gives the largest LMTD for the same four temperatures, parallel flow cannot heat the cold stream above the hot outlet, and multipass or cross-flow designs require a correction factor F.

Heat moves from hot to cold, and at every point inside a heat exchanger the local transfer rate depends on the local temperature difference between the two fluids. That difference is never constant: it is largest where the streams first meet and falls as heat is exchanged. The log mean temperature difference (LMTD) is the standard way to represent this continuously varying driving force with a single value for design work. Along with the flow arrangement inside the exchanger — parallel, counter-flow, cross-flow, or multipass — it decides how much heat transfer surface a given duty requires.

What the LMTD represents

Heat exchanger sizing starts from the basic relation Q = UAΔTm, where Q is the heat duty (the rate of heat transfer), U is the overall heat transfer coefficient, A is the heat transfer surface area, and ΔTm is the mean temperature difference between the hot and cold fluids (Ahmed, PROC 5071 lecture notes). For that mean, design practice uses the logarithmic mean temperature difference:

ΔTlm = (ΔT1 − ΔT2) / ln(ΔT1 / ΔT2)

Here ΔT1 and ΔT2 are the temperature differences between the hot and cold fluids at the two ends of the exchanger. Because these are differences, their values are identical whether the temperatures are stated in degrees Celsius or in kelvin, so the LMTD comes out the same in either unit system. The formula requires both terminal differences to be positive; a negative value signals that the requested duty is thermodynamically impossible for the chosen arrangement, as the worked example below shows.

Why a logarithmic mean?

Newton’s law of cooling ties the heat transfer rate to the instantaneous temperature difference between the hot and cold media, so in a heat transfer process the driving force varies with position. Under the idealized conditions used in hand calculation, the fluid temperature profiles along the exchanger are exponential curves rather than straight lines, and the average that correctly represents an exponentially varying quantity is the logarithmic mean (Engineering Toolbox).

A simpler choice — the arithmetic mean temperature difference, (ΔT1 + ΔT2)/2 — would be correct only if the driving force fell linearly from one end to the other. In reality it falls exponentially, so the arithmetic mean overstates the true average whenever the two terminal differences are unequal. The LMTD is always equal to or smaller than the arithmetic mean, and the gap widens as the ratio ΔT1/ΔT2 grows, which is why exchangers with very unequal end differences deserve careful treatment.

Two special cases are worth knowing. First, when ΔT1 equals ΔT2, direct substitution into the formula gives 0/0 because ln(1) = 0; the limiting value is simply the common difference, so an exchanger with 60 K at both ends has an LMTD of 60 K. Second, when one fluid changes phase — condensing steam is the classic case — its temperature stays essentially constant along the exchanger and its profile does not depend on the flow direction, so that constant temperature is simply used in both terminal differences (Engineering Toolbox).

How flow arrangements change the LMTD

The primary and secondary fluids in an exchanger may flow in the same direction (parallel or co-current flow), in opposite directions (counter-flow or counter-current flow), or perpendicular to each other (cross-flow) (Saari; Engineering Toolbox). The arrangement matters because it controls how the four terminal temperatures pair into ΔT1 and ΔT2 — and therefore what LMTD the same four temperatures produce.

Parallel (co-current) flow

Both streams enter at the same end. The driving force is greatest at the inlet and falls steadily toward the outlet as the temperature curves of the two fluids converge. Two consequences follow. The curves can approach but never cross, so the cold fluid cannot leave hotter than the hot fluid leaves — any duty that requires more is infeasible in this arrangement. And because ΔT1 and ΔT2 end up very unequal in most duties, the LMTD is penalized and more surface area is needed for the same heat load.

Counter-flow (counter-current)

The streams enter at opposite ends, so the hot inlet sits next to the cold outlet and the hot outlet next to the cold inlet. The terminal pairings become ΔT1 = hot inlet − cold outlet and ΔT2 = hot outlet − cold inlet. The driving force is much more uniform along the length, and the temperature curves may pass one another: the cold fluid can leave hotter than the hot fluid leaves. For the same four terminal temperatures, counter-flow always yields an equal or larger LMTD than parallel flow, and therefore a smaller required area, which is why it is the preferred arrangement wherever piping allows.

Cross-flow and multipass arrangements

Cross-flow units (common in compact exchangers and air-cooled equipment) and multipass shell-and-tube designs do not match the one-dimensional ideal the plain formula assumes. A 1-shell-pass, 2-tube-pass exchanger, for instance, begins like counter-flow and turns back like parallel flow. The standard remedy is an LMTD correction factor: Q = UA·F·ΔTlm, where F is read from charts for the specific geometry and terminal temperatures (Saari, Heat Exchanger Dimensioning). A low F indicates that much of the surface contributes little driving force; as a widely used rule of thumb, designers aim for F of roughly 0.75 or higher and reconsider the geometry when it falls below.

A hypothetical worked example

Suppose a hot process stream is to be cooled from 120°C to 80°C while a cold stream is heated from 20°C to 60°C (hypothetical numbers; all differences are the same in °C and K). The four temperatures are identical; only the arrangement changes.

ArrangementΔT1ΔT2LMTD
Parallel (co-current)120 − 20 = 100 K80 − 60 = 20 K(100 − 20)/ln(100/20) = 80/ln 5 ≈ 49.7 K
Counter-flow120 − 60 = 60 K80 − 20 = 60 KTerminal differences equal, so 60 K

With identical temperatures, the counter-flow LMTD is about 20% larger. Since A = Q/(U·ΔTlm) and U is similar in both cases, the parallel-flow design would need roughly 20% more surface to deliver the same duty. Note also a common trap: the parallel-flow arithmetic mean here is (100 + 20)/2 = 60 K, which coincidentally matches the counter-flow LMTD and would wrongly suggest the two designs perform alike — the logarithmic mean is what exposes the penalty.

Now make the duty harder: the cold stream must reach 90°C while the hot stream leaves at 80°C. In parallel flow, the outlet-end difference becomes 80 − 90 = −10 K, and the logarithm of a negative ratio is undefined — the mathematics is flagging a physical impossibility. Counter-flow remains feasible: ΔT1 = 120 − 90 = 30 K, ΔT2 = 80 − 20 = 60 K, giving an LMTD of (60 − 30)/ln 2 ≈ 43 K. Heating the cold stream above the hot outlet is something only counter-flow can achieve.

Assumptions behind the method

The LMTD expression is derived under simplifying assumptions, and a design should be checked against them (Ahmed):

  • Steady operation with constant mass flow rates.
  • Each fluid is at a uniform temperature over any cross-section, so only position along the exchanger matters.
  • Specific heat capacities are constant, or phase change occurs at essentially constant temperature.
  • The overall heat transfer coefficient U is constant along the exchanger length.
  • Heat losses to the surroundings are negligible — all heat leaving the hot fluid enters the cold fluid.
  • The flow arrangement follows the ideal pattern assumed in pairing the terminal temperatures; departures are handled with the correction factor F.

Real equipment meets these conditions only approximately — U changes with temperature-dependent properties and fouling, for example — but they are generally acceptable for design estimates, with appropriate margins applied.

Practical notes and common pitfalls

  • Pair the terminals to match the real arrangement. Reusing a counter-flow pairing for a parallel-flow design inflates the predicted LMTD and undersizes the exchanger.
  • Check feasibility first. If a terminal difference comes out negative for the chosen arrangement, the duty cannot be achieved; switch to counter-flow or revisit the target temperatures.
  • Watch tight approaches. As the smaller terminal difference shrinks toward zero, the LMTD collapses with it and the required area grows sharply, so realistic approach temperatures matter in specification.
  • Apply the correction factor for multipass and cross-flow geometries, and treat a low F as a warning sign rather than a number to accept.
  • Know when LMTD is the wrong tool. The method needs all four terminal temperatures. When sizing a new exchanger with known temperatures, that is fine; when predicting the outlets of an existing exchanger of known area, the outlets are unknown and an iterative LMTD calculation or the ε-NTU (effectiveness) method is more convenient (Ezgi, 2017; Saari).

Key takeaways

  • The LMTD compresses a varying temperature driving force into one value for the sizing equation Q = UAΔTlm.
  • The logarithmic — not arithmetic — mean reflects the exponential temperature profiles; the two agree only when the terminal differences are equal.
  • Flow arrangement determines how terminal temperatures pair up: counter-flow maximizes the LMTD, parallel flow cannot heat the cold stream above the hot outlet, and multipass or cross-flow designs need a correction factor F.

Concepts that sit alongside the LMTD in heat exchanger work include the overall heat transfer coefficient U and the alternative ε-NTU analysis method; each is treated in more depth in its own article.

FAQ

Why is a logarithmic mean used instead of the arithmetic mean?

The temperature difference between the fluids varies non-linearly (exponentially) along the exchanger, and the log mean represents that variation correctly. The arithmetic mean, which would only be right for a linear profile, overstates the true average driving force whenever the two terminal differences are unequal. The two means coincide only when the terminal differences are equal.

Can the cold fluid leave a heat exchanger hotter than the hot fluid's outlet?

Yes, but only in counter-flow, where the temperature curves of the two fluids are allowed to pass one another. In parallel flow the curves converge but cannot cross, so the cold outlet must remain below the hot outlet. A duty violating this produces a negative terminal difference, which signals an infeasible design for that arrangement.

What happens if one terminal temperature difference is zero?

As the smaller terminal difference approaches zero, the LMTD approaches zero and the required area grows without bound, so exchangers are designed with a finite approach temperature. If the terminal differences are exactly equal, the formula reads 0/0 because ln(1) = 0, and the LMTD is taken as the limiting value, which is that common difference.

What does the LMTD correction factor F account for?

It corrects the ideal counter-flow LMTD for geometries that do not follow the one-dimensional pattern, such as shell-and-tube exchangers with multiple tube passes and cross-flow units: Q = UA·F·ΔTlm. F is read from charts for the specific arrangement. A low F means much of the surface contributes little driving force, so a widely used rule of thumb is to keep F at roughly 0.75 or higher.

When should the ε-NTU method be used instead of the LMTD method?

The LMTD method needs all four terminal temperatures, which suits sizing a new exchanger for a known duty. When predicting the performance of an existing exchanger of known area, the outlet temperatures are unknown, so a direct LMTD solution requires iteration; the ε-NTU (effectiveness–number of transfer units) method then gives the outlet temperatures directly.

References

  1. Ahmed, S. Log Mean Temperature Difference (LMTD). PROC 5071: Process Equipment Design I, Memorial University. https://www.mun.ca/engineering/crise/media/production/memorial/academic/faculty-of-engineering/c-rise-engineering/people/PROC5071_HO_7_1_LMTD.pdf
  2. Saari, J. Heat Exchanger Dimensioning. Lappeenranta University of Technology, LUT Energy. https://sistemas.eel.usp.br/docentes/arquivos/5817712/LOQ4086/saari__heat_exchanger_dimensioning.pdf
  3. The Engineering ToolBox. Arithmetic and Logarithmic Mean Temperature Difference. https://www.engineeringtoolbox.com/arithmetic-logarithmic-mean-temperature-d_436.html
  4. Ezgi, C. (2017). Basic Design Methods of Heat Exchanger. InTech. https://doi.org/10.5772/67888

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