Short Answer
When all four terminal temperatures of a heat exchanger are known, the fastest route to its heat duty runs through one number: the log mean temperature difference, or LMTD. For a counterflow exchanger — one in which the hot and cold fluids move in opposite directions — cooling the hot stream from 150 °C to 90 °C while heating the cold stream from 30 °C to 70 °C produces end temperature differences of 80 K and 60 K and an LMTD of about 69.5 K. With an overall heat-transfer coefficient of 500 W/m²·K acting across 10 m² of surface area, that corresponds to a duty of roughly 348 kW [1]. The sections below define the quantity, work this example through in full, show the same method applied to a Fahrenheit problem, and flag the mistakes that most often derail the calculation.
What the log mean temperature difference represents
A heat exchanger transfers heat between two fluids without letting them mix, using a conducting wall between them. The driving force for that transfer at any point along the exchanger is the local difference between the hot stream temperature and the cold stream temperature. This difference is not constant: the hot fluid cools and the cold fluid warms as they flow, so the gap between the streams shifts continuously from one end of the exchanger to the other.
The LMTD condenses that varying gap into a single equivalent difference. Multiplied by the overall heat-transfer coefficient U and the surface area A, it reproduces the actual duty Q — the rate at which heat is transferred, in watts — through the relation Q = U × A × LMTD. The coefficient U bundles wall conduction and the thermal resistance of the fluid films on both sides into one conductance; a value of 500 W/m²·K means 500 watts pass through each square metre for every kelvin of driving force.
The logarithmic form is what makes the average faithful. When the two end differences differ, the local difference decays exponentially along the length, and the appropriate average is the logarithmic mean: LMTD = (ΔT1 − ΔT2) ÷ ln(ΔT1 ÷ ΔT2), where ln denotes the natural logarithm. Only when the two end differences are equal does the logarithmic mean coincide exactly with the simple arithmetic mean; as the end differences spread apart, the arithmetic mean increasingly overstates the true driving force [1].
Pairing the ends correctly in countercurrent flow
In a countercurrent (counterflow) arrangement, the two fluids travel in opposite directions: the hot stream enters one end while the cold stream enters the other. Consequently, the hottest point of the hot stream faces the coldest point of the cold stream, and the two end differences are formed by crossing the streams:
- ΔT1 at the hot-inlet end: hot inlet minus cold outlet.
- ΔT2 at the hot-outlet end: hot outlet minus cold inlet.
In parallel flow, by contrast, both streams enter the same end and leave together, so the differences pair inlets with inlets (ΔT1 = hot inlet − cold inlet) and outlets with outlets (ΔT2 = hot outlet − cold outlet). For the same four temperatures, this parallel pairing yields a smaller LMTD than the countercurrent pairing [1], one reason designers favour counterflow. A common textbook convention labels the hot stream T and the cold stream t, with end 1 at the hot inlet; in countercurrent arrangements, end 2 is then the cold-fluid inlet, so ΔT1 = T1 − t1 and ΔT2 = T2 − t2 [3].
Calculating the LMTD step by step
- Collect the four terminal temperatures: hot inlet and outlet, cold inlet and outlet.
- Pair the ends according to the flow arrangement. For countercurrent flow, subtract the cold outlet from the hot inlet to get ΔT1, and the cold inlet from the hot outlet to get ΔT2.
- Apply the logarithmic mean: LMTD = (ΔT1 − ΔT2) ÷ ln(ΔT1 ÷ ΔT2).
- Sanity-check the result: the LMTD must fall between ΔT1 and ΔT2.
- Compute the duty: Q = U × A × LMTD, keeping all units consistent.
Worked example: a counterflow exchanger cooling 150 °C to 90 °C
Consider the reference problem: hot fluid cooling from 150 °C to 90 °C against cold fluid heating from 30 °C to 70 °C, with U = 500 W/m²·K and A = 10 m² in counterflow arrangement [1]. The end pairing works out as follows:
| End of exchanger | Hot fluid | Cold fluid | Difference |
|---|---|---|---|
| Hot inlet meets cold outlet | 150 °C | 70 °C | ΔT1 = 80 K |
| Hot outlet meets cold inlet | 90 °C | 30 °C | ΔT2 = 60 K |
Substituting into the logarithmic mean:
LMTD = (80 − 60) ÷ ln(80 ÷ 60) = 20 ÷ 0.2877 ≈ 69.5 K [1].
The duty follows immediately:
Q = U × A × LMTD = 500 × 10 × 69.5 ≈ 347,600 W, or about 348 kW [1].
Cross-checking the arithmetic
Two quick checks build confidence in the result. First, 69.5 K sits between the end differences of 60 K and 80 K, exactly where a logarithmic mean must land. Second, comparing with the parallel-flow pairing for the same temperatures shows why the crossing matters: parallel flow pairs the inlets (150 − 30 = 120 K) and the outlets (90 − 70 = 20 K), and the same formula then gives (120 − 20) ÷ ln(120 ÷ 20) = 100 ÷ 1.7918 ≈ 55.8 K. That is markedly lower than the correct countercurrent value, and an exchanger sized on it would be larger than necessary for the actual duty — the direct consequence of pairing the wrong ends [1]. The arithmetic mean, at (80 + 60) ÷ 2 = 70 K, happens to sit close to the true LMTD here because the ratio 80 ÷ 60 = 1.33 is modest; the gap between the two means widens as the end differences spread [1].
The same method in Fahrenheit
The formula is independent of the temperature scale, as a liquid-to-liquid counterflow exchanger example from a nuclear facility shows. There, the cold fluid is heated from 120 °F to 310 °F while the hot fluid enters at 500 °F and leaves at 400 °F [2]. Pairing the ends in counterflow gives ΔT1 = 500 − 310 = 190 °F at the hot-inlet end and ΔT2 = 400 − 120 = 280 °F at the hot-outlet end, so LMTD = (280 − 190) ÷ ln(280 ÷ 190) = 90 ÷ 0.3878 ≈ 232 °F. Note that here the larger difference sits at the hot-outlet end; the formula is symmetric in ΔT1 and ΔT2, since ln(ΔT1 ÷ ΔT2) = −ln(ΔT2 ÷ ΔT1), so labeling the ends consistently is what matters. If the duty were then required in SI units, the difference scale converts as 232 °F × 5/9 ≈ 129 K, the size of the same difference expressed in kelvin.
Common mistakes and how to avoid them
- Pairing the wrong ends. Applying parallel-flow pairing to a counterflow problem shrinks the LMTD — about 55.8 K instead of 69.5 K in the worked example above — and understates the driving force [1]. Confirm the flow arrangement before forming ΔT1 and ΔT2.
- Substituting the arithmetic mean. (ΔT1 + ΔT2) ÷ 2 is convenient but overstates the driving force once the end differences differ substantially; treat it only as a rough check when the two differences are similar [1].
- Mixing unit systems. A difference of one degree Celsius equals exactly one kelvin, so ΔT1 and ΔT2 are interchangeable between those two scales. Fahrenheit differences are on a different scale and must be multiplied by 5/9 before combining with SI values of U and A.
- Using the uncorrected LMTD on multipass exchangers. Shell-and-tube units with more than one pass are neither pure counterflow nor pure parallel flow, so the countercurrent LMTD used directly misstates their driving force, as discussed next.
When a plain countercurrent LMTD needs adjusting
Real process exchangers often deviate from the ideal single-pass counterflow model. In a 1-2 shell-and-tube unit — one shell pass and two tube passes — the flow pattern is a compromise, so designers first calculate the LMTD as if the unit were single-pass counterflow and then apply a correction factor to obtain the corrected LMTD used in the duty equation [3]. A worked design problem of exactly this kind subcools methanol condensate from 95 °C to 40 °C against brackish water rising from 25 °C to 40 °C: the countercurrent end differences come out as ΔT1 = 95 − 40 = 55 °C and ΔT2 = 40 − 25 = 25 °C before the correction is applied [3]. Correction factors are obtained from standard design charts keyed to the pass arrangement and the temperature approach.
The LMTD method also presupposes that all four terminal temperatures are known, whether measured or found from energy balances of the form Q = m·cp·ΔT, where m is the mass flow rate and cp the specific heat capacity. When only inlet temperatures are known — often the case when rating an exchanger against uncertain loads — the effectiveness–NTU method is the standard alternative; both routes were invoked, for instance, to determine the overall heat-transfer coefficient of a double-pipe oil cooler in an Engineering Stack Exchange problem [4]. Phase-change duties such as condensation add further subtlety, since a condensing stream holds a nearly constant temperature while releasing heat, and such cases are usually handled with dedicated design procedures.
Practical takeaways
- Identify the flow arrangement first, then pair the ends: hot inlet with cold outlet for ΔT1, hot outlet with cold inlet for ΔT2 in counterflow.
- Apply LMTD = (ΔT1 − ΔT2) ÷ ln(ΔT1 ÷ ΔT2) and confirm the result lies between the two end differences.
- Multiply by U and A to obtain the duty Q = U × A × LMTD, keeping units consistent throughout.
- Apply a correction factor for multipass shell-and-tube geometries, and switch to the effectiveness–NTU method when outlet temperatures are unknown.
FAQ
Why not just use the simple average of the end temperature differences?
The local temperature difference varies exponentially along the exchanger length. The arithmetic mean of the end differences overstates the true driving force once the two end differences spread apart, while the logarithmic mean tracks that decaying difference. When the end differences are nearly equal the two means agree, which is why the arithmetic mean can serve only as a rough check.
What happens if the two end temperature differences are equal?
When ΔT1 equals ΔT2, the term ln(ΔT1 ÷ ΔT2) approaches zero and the limit of the logarithmic-mean formula equals that common difference, so the LMTD is simply ΔT1. The formula does not break down; it reduces to the arithmetic mean, which is identical in this case.
Does it matter which end I call ΔT1 and which ΔT2?
No. The logarithmic mean is symmetric: ln(ΔT1 ÷ ΔT2) = −ln(ΔT2 ÷ ΔT1), so swapping the two end differences leaves the LMTD unchanged. In the Fahrenheit example the larger difference sat at the hot-outlet end; consistent labeling is what matters, not which end is larger.
When should I use the effectiveness–NTU method instead of the LMTD method?
The LMTD method requires all four terminal temperatures — hot inlet and outlet, cold inlet and outlet — whether measured or obtained from energy balances. When only inlet temperatures are known, for example when rating an exchanger against uncertain flows, the effectiveness–NTU method works directly from inlet data and is the standard alternative.
Do shell-and-tube exchangers with multiple passes use the same LMTD?
Not directly. A unit such as a 1-2 shell-and-tube exchanger is neither pure counterflow nor pure parallel flow, so designers compute the countercurrent LMTD first and then multiply it by a correction factor from standard charts to obtain the corrected LMTD used in the duty equation.
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