CHE-25

Log-mean temperature difference

ΔTlm = (ΔT1−ΔT2)/ln(ΔT1/ΔT2). Driving force of a two-stream exchanger.

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Heat exchangersLMTD

Governing equation

\Delta T_{\mathrm{lm}}=\dfrac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}

$\Delta T_2 \in \left[1, 200\right]$

where

\Delta T_1: End difference 1 (K)
\Delta T_2: End difference 2 (K)
\Delta T_{\mathrm{lm}}: LMTD (K)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

\Delta T_{\mathrm{lm}}=\dfrac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}

$\Delta T_2 \in \left[1, 200\right]$

$\Delta T_2 = 1$ $\Delta T_{\mathrm{lm}} = 10.57$

Along the curve

②

Domain of the variable — blue

\Delta T_2_{\min} = 1,\quad \Delta T_2_{\max} = 200

$\Delta T_2 \in \left[1, 200\right]$

$\Delta T_2 = 1$ $\Delta T_2 = 1$

Along the curve

③

Response curve — green

\Delta T_{\mathrm{lm}}=\dfrac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}

$\Delta T_2 \in \left[1, 200\right]$

$\Delta T_2 = 1$ $\Delta T_{\mathrm{lm}} = 10.57$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}\Delta T_{\mathrm{lm}}}{\mathrm{d}\Delta T_2}

$\Delta T_2 \in \left[1, 200\right]$

$\Delta T_2 = 2.252$ $\Delta T_{\mathrm{lm}}^{\prime} = 1.764$

Along the curve

Lecture brief

Historical brief

From CSTR/PFR mole balances and Arrhenius rates to McCabe–Thiele stages and NTU exchangers, chemical engineering is conservation plus equilibrium. The lab is that design arithmetic. This sheet (CHE-25 — Log-mean temperature difference) is the form associated with LMTD. Working symbols: $\Delta T_1$ , $\Delta T_2$ $\rightarrow$ $\Delta T_{\mathrm{lm}}$ . Integrating dQ = U (T_h−T_c) dA with linear T vs Q on each side yields the log-mean of the two end differences.

Purpose

Purpose: compute

\Delta T_{\mathrm{lm}}

from

\Delta T_1

\Delta T_2

in Chemical engineering via

\Delta T_{\mathrm{lm}}=\dfrac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}

ΔTlm = (ΔT1−ΔT2)/ln(ΔT1/ΔT2). Driving force of a two-stream exchanger. Use it when a real chemical engineering question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

\Delta T_1 = 40.000\,\mathrm{K}

\Delta T_2 = 15.000\,\mathrm{K}

, the governing relation

\Delta T_{\mathrm{lm}}=\dfrac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}

yields

\Delta T_{\mathrm{lm}} = 25.49\,\mathrm{K}

. Two T profiles, two end gaps, an LMTD bar. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

End difference 1 \Delta T_140 KEnd difference 2 \Delta T_215 K

Outputs

LMTD \Delta T_{\mathrm{lm}}25.49 K

Reading speed

Watch on YouTube

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Narration of this film

Two T profiles, two end gaps, an LMTD bar.

Integrating dQ = U (T_h−T_c) dA with linear T vs Q on each side yields the log-mean of the two end differences.

Reading speed

Watch on YouTube