Thermal Engineering
Log-Mean Temperature Difference (LMTD): Sizing a Heat Exchanger From Two End Gaps
Feed 90 °C hot oil against 25 °C cooling water and the temperature gap between them is not one number — it is 65 °C at one end and maybe 20 °C at the other, shrinking continuously along the tube. The log-mean temperature difference (LMTD) collapses that whole varying profile into a single effective driving temperature by taking the logarithmic average of the two end gaps, so a designer can size the exchanger with one clean equation instead of integrating along its length.
Formally, LMTD is the correct average temperature difference to use in the rate equation Q = U·A·ΔT_lm whenever the local ΔT between two fluids varies exponentially with position — the case for pure counterflow or parallel-flow single-phase heat exchange with constant U and specific heats. It converts a distributed thermal problem into the algebra of two boundary values, ΔT₁ and ΔT₂.
- TypeAveraging method for heat-exchanger thermal driving force
- Key equationΔT_lm = (ΔT₁ − ΔT₂) / ln(ΔT₁ / ΔT₂)
- Used inQ = U·A·F·ΔT_lm sizing of shell-and-tube, double-pipe, plate exchangers
- Correction factor F1.0 for pure counter/parallel flow; keep F ≥ 0.75–0.80 in design
- Valid whenConstant U, constant cp, single-phase or isothermal phase change
- Alternativeε-NTU method (preferred when outlet temperatures are unknown)
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What LMTD Is and Where Engineers Use It
The log-mean temperature difference is the single effective ΔT that makes the compact rate law Q = U·A·ΔT_lm exact for a two-fluid heat exchanger with pure counter- or parallel-flow. Here Q is the total heat duty (W), U the overall heat-transfer coefficient (W/m²·K), and A the heat-transfer surface area (m²). Because the two fluids' temperatures both change along the flow path, the local gap ΔT is not constant — LMTD supplies the mathematically correct average.
- Process plants: sizing shell-and-tube exchangers, reboilers, condensers, and crude-preheat trains per TEMA and API 660.
- HVAC and power: chilled-water coils, feedwater heaters, and steam condensers.
- Automotive/aerospace: radiators, oil coolers, intercoolers, and regenerators.
LMTD is the workhorse of the design problem: given a required duty and all four terminal temperatures, compute the surface area — and therefore the tube count, length, and cost — needed to deliver it. It is taught in every heat-transfer course alongside its rival, the ε-NTU method.
How the Log-Mean Average Is Derived
Consider a differential slice of a counterflow exchanger. The local heat transferred is dQ = U·(T_h − T_c)·dA = U·ΔT·dA. Energy balances on each stream give dQ = −C_h·dT_h = ±C_c·dT_c, where C = ṁ·cp is the heat-capacity rate (W/K). Combining these shows that d(ΔT)/ΔT = −U·(1/C_h ∓ 1/C_c)·dA, so ΔT decays exponentially along the surface.
Integrating from one end (ΔT₁) to the other (ΔT₂) over total area A and eliminating the capacity-rate terms with the overall balance yields the classic result:
- ΔT_lm = (ΔT₁ − ΔT₂) / ln(ΔT₁ / ΔT₂)
where ΔT₁ and ΔT₂ are the hot-minus-cold temperature differences at the two ends of the unit. The logarithm is the signature of the exponential temperature profile. The derivation assumes constant U, constant specific heats, no phase change (or a single isothermal one), negligible axial conduction, and no heat loss to surroundings — the same assumptions that make U a single lumped number.
Key Quantities and a Worked Sizing Example
Cool 2.0 kg/s of oil (cp ≈ 2.1 kJ/kg·K) from 120 °C to 60 °C using water entering at 25 °C and leaving at 55 °C, in a counterflow unit with U = 350 W/m²·K.
- Duty: Q = ṁ·cp·ΔT = 2.0 × 2100 × (120 − 60) = 252,000 W (252 kW).
- End gaps (counterflow): at the hot-oil-inlet end, ΔT₁ = 120 − 55 = 65 °C; at the oil-outlet end, ΔT₂ = 60 − 25 = 35 °C.
- LMTD: ΔT_lm = (65 − 35) / ln(65/35) = 30 / 0.619 = 48.5 °C. (The plain arithmetic mean, 50 °C, would over-predict the driving force and under-size the unit.)
- Area: A = Q / (U·ΔT_lm) = 252,000 / (350 × 48.5) = 14.8 m².
Run the same streams in parallel flow and the end gaps become 95 °C and 5 °C, giving ΔT_lm = 30.6 °C and demanding ~23.5 m² — roughly 60% more surface for identical duty. That gap is exactly why counterflow is the default.
Design in Practice: The F Correction Factor
Real shell-and-tube exchangers rarely run pure counterflow — a single shell pass typically has two, four, or more tube passes, mixing co- and counter-current regions. To keep using the convenient LMTD algebra, engineers multiply by a dimensionless correction factor F ≤ 1:
- Q = U·A·F·ΔT_lm, where ΔT_lm is computed as if the unit were pure counterflow.
F is read from TEMA/Bowman charts (or closed-form equations) as a function of two ratios: P = (t_out − t_in)/(T_in − t_in), the tube-side temperature effectiveness, and R = (T_in − T_out)/(t_out − t_in), the heat-capacity-rate ratio. Practical rules:
- Design to keep F ≥ 0.75–0.80; the F curves plunge steeply below that, so a small temperature error or fouling change can wreck performance.
- A low or undefined F signals a temperature cross (cold outlet exceeds hot outlet) — split the duty across multiple shells in series (1-2, 2-4 arrangements) to raise F.
- For an isothermal phase change (pure condensation or evaporation), F = 1 because one stream stays at constant temperature.
LMTD Versus the ε-NTU Method
The two standard methods answer different questions. The LMTD/F-factor method shines for sizing: when all four terminal temperatures are specified, area falls out in a single step, A = Q/(U·F·ΔT_lm). But if you only know the inlet temperatures — a rating problem, e.g. checking an existing unit at a new flow — LMTD forces iteration, because you cannot compute ΔT_lm without the very outlet temperatures you are solving for.
The ε-NTU method sidesteps this. It defines effectiveness ε = Q/Q_max and the number of transfer units NTU = U·A/C_min, then uses geometry-specific relations — for counterflow, ε = (1 − exp[−NTU(1−C_r)]) / (1 − C_r·exp[−NTU(1−C_r)]), with C_r = C_min/C_max. Outlet temperatures come out directly, no iteration.
- Use LMTD when outlet temps are known (design/sizing).
- Use ε-NTU when only inlet temps are known (rating/performance).
The two are mathematically equivalent and interconvertible; most software carries both. LMTD also has a well-known removable singularity at ΔT₁ = ΔT₂, where you simply set ΔT_lm = ΔT₁.
Limits, Failure Modes, and Why It Matters
LMTD is exact only under its assumptions, and each broken assumption is a trap:
- Variable U: viscous oils and near-critical fluids make U change severely along the length. Split the exchanger into zones, compute a local LMTD per zone, and sum the areas — a single overall LMTD lies.
- Multi-phase duty: a condenser with desuperheating → condensing → subcooling regions must be broken into a zone analysis; each region has its own LMTD and U.
- Fouling: scale and biofilm add a resistance R_f (m²·K/W) that lowers U over time; designers pad A with a fouling factor, but an oversized clean exchanger can run too cold and worsen fouling.
- Temperature cross: pushing a multi-pass unit past its F limit yields a physically impossible cross — the algebra warns you before the hardware fails to perform.
Its significance endures: LMTD turns a distributed differential-equation problem into two subtractions, a logarithm, and a divide — the compact link between a specified thermal duty and the physical steel a plant must buy. Get the LMTD wrong by treating it as an arithmetic mean and you under-size the surface, and the exchanger never makes its rated duty.
| Aspect | LMTD / F-factor method | ε-NTU method |
|---|---|---|
| Best when | All four terminal temperatures are known — a sizing (design) problem | Only inlet temperatures known — a rating (performance) problem |
| Core quantity | ΔT_lm = (ΔT₁−ΔT₂)/ln(ΔT₁/ΔT₂) | ε = Q/Q_max, NTU = U·A/C_min |
| To find area A | A = Q / (U·F·ΔT_lm) — direct, one step | Requires iteration: guess A → NTU → ε → Q → repeat |
| To find outlet temps | Requires iteration (ΔT_lm depends on unknown outlets) | Direct from ε-NTU relation, no iteration |
| Multi-pass geometry | Apply chart/analytic correction factor F(P,R) < 1 | Use geometry-specific ε(NTU, C_r) relation |
| Failure case | Undefined when ΔT₁ = ΔT₂ (use ΔT_lm = ΔT₁) | No singularity; well-behaved throughout |
Frequently asked questions
Why use a logarithmic mean instead of a simple arithmetic average of the two end gaps?
Because the temperature difference between the two fluids decays exponentially along the flow path, not linearly. The logarithm is the exact result of integrating that exponential profile. The arithmetic mean always over-estimates the true average driving force (except when the two end gaps are equal, where they coincide), so using it under-sizes the exchanger.
What is ΔT₁ and ΔT₂ — how do I pick the two end gaps?
They are the hot-fluid-minus-cold-fluid temperature differences evaluated at the two physical ends of the exchanger. For counterflow, one end pairs the hot inlet with the cold outlet and the other pairs the hot outlet with the cold inlet. The LMTD formula is symmetric, so it does not matter which end you label 1 or 2.
What happens when ΔT₁ equals ΔT₂?
The formula gives 0/ln(1) = 0/0, which is indeterminate. This occurs, for example, when both heat-capacity rates are equal in counterflow, giving a constant ΔT everywhere. The limit is simply ΔT_lm = ΔT₁ = ΔT₂, so you use that common value directly. Software guards this case to avoid a divide-by-zero.
What is the F correction factor and when do I need it?
F (≤ 1) accounts for the fact that multi-pass and cross-flow exchangers are not pure counterflow. You compute ΔT_lm as if the unit were counterflow, then use Q = U·A·F·ΔT_lm. F is read from charts as a function of the ratios P and R. Design so F stays above about 0.75–0.80, because the curves fall off a cliff below that.
When should I use ε-NTU instead of LMTD?
Use ε-NTU when only the inlet temperatures are known — a rating or performance-check problem. LMTD requires the outlet temperatures to form ΔT₁ and ΔT₂, so with unknown outlets it forces iteration, whereas ε-NTU gives outlet temperatures and duty directly. For sizing with all four temperatures known, LMTD is faster.
Is LMTD valid when one fluid is boiling or condensing?
Yes, provided the phase change is isothermal (a pure single-component fluid at constant pressure). That stream sits at constant temperature, F = 1, and the LMTD formula still applies with one of the two stream temperatures held fixed. For non-isothermal changes — desuperheating, subcooling, or multi-component condensation — you must split the exchanger into zones and compute a separate LMTD per zone.