Log Mean Temperature Difference
Calculate the log mean temperature difference (LMTD) of a heat exchanger from the four terminal temperatures. Choose counterflow or parallel flow and enter the hot and cold inlet/outlet temperatures in °C, °F or K.
Find the Temperature Driving Force of a Heat Exchanger
LMTD Formula
Where:
- \(\Delta T_{lm}\) — log mean temperature difference (K or °C)
- \(\Delta T_1\) — temperature difference between the two fluids at one end of the exchanger
- \(\Delta T_2\) — temperature difference between the two fluids at the other end
- \(\ln\) — natural logarithm
The end differences depend on the flow arrangement. In counterflow, \(\Delta T_1 = T_{h,i} - T_{c,o}\) at the hot inlet end and \(\Delta T_2 = T_{h,o} - T_{c,i}\) at the hot outlet end. In parallel flow, \(\Delta T_1 = T_{h,i} - T_{c,i}\) at the common inlet end and \(\Delta T_2 = T_{h,o} - T_{c,o}\) at the common outlet end.
Because the temperature difference between the fluids varies along the exchanger, a simple average would misrepresent the true driving force. The logarithmic mean weights the smaller end difference more heavily, which is exactly what the heat-transfer equations require. When \(\Delta T_1 = \Delta T_2\), the formula reduces to \(\Delta T_{lm} = \Delta T_1\).
LMTD – Calculation Example
Example: a counterflow heat exchanger cools a hot stream from 150 °C to 100 °C while a cold stream is heated from 40 °C to 80 °C.
\[ \Delta T_1 = T_{h,i} - T_{c,o} = 150 - 80 = 70\ \text{K} \]
\[ \Delta T_2 = T_{h,o} - T_{c,i} = 100 - 40 = 60\ \text{K} \]
\[ \Delta T_{lm} = \frac{70 - 60}{\ln (70 / 60)} = \frac{10}{\ln 1.1667} = 64.86\ \text{K} \]
Result: \(\Delta T_{lm} \approx 64.86\) °C (identical in K, or 116.75 °F). Note that the logarithmic mean, 64.86, is slightly below the arithmetic mean of 65 K.
How to Use the Calculator
- Select the Flow arrangement: counterflow (fluids flow in opposite directions) or parallel flow (fluids flow in the same direction). The arrangement determines which end temperatures are paired.
- Enter the Hot fluid inlet temperature (Tₕᵢ) and pick its unit (°C, °F or K).
- Enter the Hot fluid outlet temperature (Tₕₒ) with its unit.
- Enter the Cold fluid inlet temperature (Tᶜᵢ) with its unit.
- Enter the Cold fluid outlet temperature (Tᶜᵒ) with its unit.
- Read the log mean temperature difference (ΔTₗₘ) together with the end differences ΔT₁ and ΔT₂, shown in °C, K and °F so you can compare the two arrangements directly.
Mixed units are allowed: each temperature can be entered in a different unit, because only temperature differences enter the LMTD. A difference of 1 K equals a difference of 1 °C and equals 9/5 °F.
What Is the Log Mean Temperature Difference?
The LMTD is the effective temperature driving force for heat transfer in a heat exchanger. The basic heat-transfer rate equation is \(Q = U \cdot A \cdot \Delta T_{lm}\), where \(Q\) is the heat duty, \(U\) the overall heat-transfer coefficient and \(A\) the heat-transfer area. The temperature difference between the hot and cold fluid is not constant along the exchanger — it is largest at one end and smallest at the other — so the LMTD replaces it with a single, physically correct average.
The reason a logarithmic (rather than arithmetic) mean is needed comes from integrating the local heat-transfer equation along the exchanger length. Where the two fluids run nearly parallel in temperature, the difference changes slowly and the two means nearly coincide. Where the difference changes rapidly — for example, in parallel flow near the inlet — the logarithmic mean correctly captures the fact that the heat transfer is dominated by the region where the temperature difference is small.
The flow arrangement matters a great deal. In counterflow the hot inlet meets the cold outlet, so the temperature difference stays relatively uniform and can even remain large over the whole length; in parallel flow the largest difference occurs at the common inlet and decays toward the outlet. For the same terminal temperatures, counterflow always gives a larger LMTD than parallel flow — in the example above, 64.86 °C versus 52.80 °C — which means a counterflow exchanger needs less heat-transfer area for the same duty.
The LMTD is central to exchanger sizing and rating. Designers use it, together with a correction factor for multi-pass and cross-flow shells, to determine the required area for a given duty. Once the LMTD is known, the analysis often continues with flow-side calculations: check the flow regime with our Reynolds Number calculator, estimate pipe pressure drop, or size the pump with the Pump Power calculator.
Assumptions, Limitations and Practical Tips
- Constant properties: the classical LMTD assumes constant specific heats, constant overall heat-transfer coefficient \(U\), and no phase change along the exchanger. Condensers and evaporators, where one fluid stays at a nearly constant temperature, still work with the formula but need care.
- No heat loss: the exchanger is assumed adiabatic with respect to the surroundings, and both fluids are assumed perfectly mixed across each cross-section.
- Positive end differences: both \(\Delta T_1\) and \(\Delta T_2\) must be positive. A zero or negative difference means a temperature cross, which cannot occur in a single-pass exchanger — the calculator flags this rather than returning a misleading number.
- Equal end differences: when \(\Delta T_1 = \Delta T_2\) the logarithm diverges, but the mathematical limit is simply \(\Delta T_{lm} = \Delta T_1\). The calculator applies this limit automatically.
- Correction factors: for multi-pass shell-and-tube and cross-flow arrangements, the true driving force is \(F \cdot \Delta T_{lm}\) with a correction factor \(F < 1\) taken from exchanger design charts. This calculator covers the ideal single-pass counterflow and parallel-flow cases.
- Special case — one fluid condensing or evaporating: its inlet and outlet temperatures are equal (e.g. \(T_{h,i} = T_{h,o}\)), and the formula still applies directly.
Frequently Asked Questions
How do I calculate the LMTD by hand?
First find the two end temperature differences. In counterflow, \(\Delta T_1 = T_{h,i} - T_{c,o}\) and \(\Delta T_2 = T_{h,o} - T_{c,i}\); in parallel flow use the inlet end and outlet end pairs instead. Then apply \( \Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} \). Always subtract in the same order, keep both differences positive, and use the natural logarithm (not log base 10). If the two differences are equal, the LMTD equals that value.
What is a good or typical LMTD value?
There is no universal target — the LMTD depends entirely on your process temperatures. As a rough guide, water-to-water exchangers often land in the 5–20 °C range, while gas streams or refrigeration duties may show 20–50 °C. A small LMTD means a large heat-transfer area is needed for the same duty, so designers trade it against capital cost. Very small values (below a few degrees) usually signal an oversized or uneconomical exchanger.
Why is my result an error or 'temperature cross'?
The calculator flags a temperature cross when either end difference is zero or negative — that is, the cold fluid is not colder than the hot fluid everywhere along the exchanger. Check that the hot outlet is above the cold inlet (counterflow) or above the cold outlet (parallel flow). If your data truly require the cold outlet to exceed the hot outlet, a single-pass exchanger cannot achieve it; you need multiple passes, a different arrangement, or the temperatures are simply mixed up.
Can I mix units, and why does the LMTD come out in °C or K?
Yes, each of the four temperatures can be entered in a different unit because the LMTD depends only on temperature differences, not absolute temperatures. A difference of 1 K equals a difference of 1 °C, so the LMTD is numerically identical in K and °C. To express it in °F, multiply by \(9/5\). What you must never do is subtract raw readings in mixed units — the calculator converts each temperature first.
Why not just use the arithmetic average of the two end differences?
The arithmetic mean always overestimates the true driving force. Heat transfer concentrates where the temperature difference is smallest, so the correct average — obtained by integrating the local rate equation along the exchanger — weights the smaller end difference more heavily. When the two end differences are close (within about 20%), the error is minor, but with a large ratio such as 70 K versus 10 K the arithmetic mean can overstate the LMTD by 30% or more, leading to a seriously undersized exchanger.
How is the LMTD used to size a heat exchanger?
Combine it with the rate equation \( Q = U \cdot A \cdot \Delta T_{lm} \). Determine the heat duty \(Q\) from an energy balance on one fluid, \( Q = \dot{m} c_p (T_{in} - T_{out}) \), estimate the overall coefficient \(U\) from correlations or typical values, then solve for the required area \( A = Q / (U \Delta T_{lm}) \). For multi-pass shell-and-tube or cross-flow exchangers, multiply by a correction factor \(F < 1\) from standard design charts before computing the area.
When does the LMTD formula not apply?
The classical formula assumes constant specific heats, a constant overall heat-transfer coefficient, and no phase change. It still works directly when one fluid condenses or evaporates at a constant temperature (enter its inlet and outlet temperatures as equal). It breaks down when properties vary strongly with temperature, when \(U\) changes significantly along the exchanger, or when the flow arrangement is not simple single-pass counterflow or parallel flow — those cases need a correction factor or a stepwise (zonal) calculation.