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.

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

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