Introduction & Context

The Logarithmic Mean Temperature Difference (LMTD) is the representative temperature driving force for heat exchange between two process streams. It is used in the design and rating of shell‑and‑tube, plate, and double‑pipe heat exchangers because the local temperature difference between hot and cold fluids varies continuously along the heat‑transfer surface. Replacing the true, position‑dependent ΔT with a single, rigorously averaged value allows the overall heat‑transfer coefficient U to be treated as constant, yielding the classic design equation \( Q = U A \Delta T_{\text{lm}} \). LMTD is therefore central to sizing new exchangers (finding area A) or checking whether an existing exchanger can deliver a specified duty Q, as explained in our heat‑exchanger duty calculation guide.

Methodology & Formulas

Energy balance
The heat duty is first fixed by the hot stream (or cold stream) enthalpy change: \[ Q = \dot m_{\text{h}}\,c_{p,\text{h}}\,(T_{\text{h,in}} - T_{\text{h,out}}) \] The cold-stream outlet temperature follows from the same duty: \[ T_{\text{c,out}} = T_{\text{c,in}} + \frac{Q}{\dot m_{\text{c}}\,c_{p,\text{c}}} \]
End temperature differences
Define the two terminal ΔT values: \[ \Delta T_1 = T_{\text{h,in}} - T_{\text{c,out}} \] \[ \Delta T_2 = T_{\text{h,out}} - T_{\text{c,in}} \]
LMTD
The logarithmic mean is: \[ \Delta T_{\text{lm}} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)} \] A small positive lower limit (≈1 × 10⁻⁹ °C) is imposed on the denominator to avoid division by zero when the two ends are nominally equal.
Required heat-transfer area
Rearranging the design equation gives: \[ A = \frac{Q}{U\,\Delta T_{\text{lm}}} \]

Validity regimes for LMTD
Parameter	Lower limit	Upper limit	Comment
\(\Delta T_1/\Delta T_2\)	> 0	< 100	Outside this range the arithmetic mean is often adequate
\(\Delta T_1,\;\Delta T_2\)	> 0	—	Negative or zero values indicate temperature crossover; LMTD is physically invalid

LMTD is the constant temperature driving force that, when multiplied by overall heat-transfer coefficient and area, gives the same heat duty as the actual varying temperature difference in a heat exchanger. It is always smaller than the arithmetic mean and therefore yields a conservative, physically realistic duty. Engineers use LMTD because heat-transfer rate is proportional to the logarithmic average of the two terminal differences, not their simple average.

Use the same equation for both configurations, but take the terminal differences correctly:

Counter-flow: ΔT₁ = hot_in − cold_out, ΔT₂ = hot_out − cold_in
Parallel-flow: ΔT₁ = hot_in − cold_in, ΔT₂ = hot_out − cold_out
LMTD = (ΔT₁ − ΔT₂) ⁄ ln(ΔT₁ ⁄ ΔT₂)

Counter-flow almost always gives a higher LMTD for the same inlet/outlet temperatures, so less area is required.

If either ΔT₁ or ΔT₂ equals zero, the log term becomes undefined and LMTD collapses to zero, implying infinite area is required—clearly impractical. A negative terminal difference indicates a temperature cross; multistage or counter-current exchangers are then needed. In practice, keep the smallest terminal difference ≥ 3 °C to avoid oversized units and ensure controllability.

For shell-and-tube or plate exchangers with multiple passes, the simple LMTD is multiplied by an F factor:

Q = U A F LMTD_counter
F depends on dimensionless parameters P and R obtained from T-sheets or charts (TEMA).
Keep F ≥ 0.75 to avoid steep local temperature gradients and poor distribution.

If F drops below 0.75, consider increasing the number of shell passes or switching to pure counter-flow.

Yes, provided the phase-change side remains isothermal (e.g., condensing steam). One terminal difference becomes constant, simplifying the LMTD calculation. If both sides undergo phase change or if there is significant sub-cooling/super-heating, break the exchanger into zones and calculate LMTD separately for each zone, then sum the duties.

Worked Example – LMTD for a Crude-Oil/Water Heat Exchanger

A small topping unit pre-heats 2 kg s⁻¹ of light crude oil from 20 °C to 52 °C before it enters the flash column. Hot heavy-naphtha rundown (2 kg s⁻¹) is available at 90 °C and may be cooled to 50 °C. A single-pass shell-and-tube exchanger with an overall heat-transfer coefficient of 800 W m⁻² K⁻¹ is proposed. Determine the required heat-transfer area.

Hot fluid mass flow rate, m_h = 2.0 kg s⁻¹
Hot fluid specific heat, c_p,h = 4180 J kg⁻¹ K⁻¹
Hot inlet temperature, T_h,in = 90 °C
Hot outlet temperature, T_h,out = 50 °C
Cold fluid mass flow rate, m_c = 2.5 kg s⁻¹
Cold fluid specific heat, c_p,c = 4180 J kg⁻¹ K⁻¹
Cold inlet temperature, T_c,in = 20 °C
Overall heat-transfer coefficient, U = 800 W m⁻² K⁻¹

Calculate the duty Q from the hot side: \[ Q = m_h\,c_{p,h}(T_{h,\text{in}} - T_{h,\text{out}}) \] \[ Q = 2.0 \times 4180 \times (90 - 50) = 334\,400\ \text{W} \]
Determine the cold outlet temperature: \[ Q = m_c\,c_{p,c}(T_{c,\text{out}} - T_{c,\text{in}}) \Rightarrow T_{c,\text{out}} = 20 + \frac{334\,400}{2.5 \times 4180} = 52\ \text{°C} \]
Find the two end temperature differences: \[ \Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}} = 90 - 52 = 38\ \text{K} \] \[ \Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}} = 50 - 20 = 30\ \text{K} \]
Compute the logarithmic mean temperature difference: \[ \Delta T_{\text{lm}} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)} \] \[ \Delta T_{\text{lm}} = \frac{38 - 30}{\ln(38/30)} = 33.8\ \text{K} \]
Calculate the required area: \[ A_s = \frac{Q}{U\,\Delta T_{\text{lm}}} \] \[ A_s = \frac{334\,400}{800 \times 33.8} = 12.4\ \text{m}^2 \]

Final Answer: A heat-transfer area of 12.4 m² is required for the proposed naphtha/crude-oil exchanger.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle

Logarithmic Mean Temperature Difference (LMTD)

Introduction & Context

🚀 Skip the Manual Math!

Methodology & Formulas

Worked Example – LMTD for a Crude-Oil/Water Heat Exchanger