What is LMTD in heat exchangers?

Under steady flow with constant and specific heats, heat exchanger duty is where is the log-mean temperature difference. For counterflow; cross/shell-and-tube use correction (RC Sachdeva / GATE HT).

What is LMTD? Write its expression for a double-pipe exchanger.

Log Mean Temperature Difference: where are the temperature differences between hot and cold fluids at the two ends.

Why use log mean rather than arithmetic mean ?

Local varies exponentially along the exchanger when and capacities are constant. Integrating yields the log mean, not the arithmetic mean.

Write the heat transfer rate using LMTD.

for pure counterflow or parallel flow (with appropriate end s). is the overall heat transfer coefficient.

Which arrangement gives higher LMTD for the same terminal temperatures — parallel or counterflow?

Counterflow generally gives a larger LMTD (and can achieve closer temperature approaches). Parallel flow LMTD is smaller; cold fluid cannot exceed hot outlet in the ideal limit.

What is the LMTD correction factor ?

For shell-and-tube, crossflow, and multipass exchangers, where is computed as if counterflow and from charts using and.

Define temperature effectiveness and capacity ratio used in charts.

(tube-side rise over inlet difference), (shell drop over tube rise). Notation varies by textbook; follow the chart’s definitions.

When is LMTD equal to the arithmetic mean temperature difference?

When (balanced counterflow with equal capacity rates, or condensers/evaporators with one fluid isothermal in a way that end differences match). Then.

(log mean reduces to the common value).

True/False: For identical terminal temperatures, parallel-flow LMTD counterflow LMTD.

False — counterflow LMTD is larger (or equal in special cases).

Correction factor for a multipass exchanger is typically:

Less than or equal to 1; only for pure counterflow equivalent behaviour.

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Study notes

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Qwestrum Engineering360 · Mechanical Engineering · Heat & Mass Transfer

Heat Exchanger LMTD

Under steady flow with constant

U

and specific heats, heat exchanger duty is

q=UA\,\Delta T_{\mathrm{lm}}\,F

where

\Delta T_{\mathrm{lm}}

is the log-mean temperature difference. For counterflow

\Delta T_{\mathrm{lm}}=(\Delta T_a-\Delta T_b)/\ln(\Delta T_a/\Delta T_b)

; cross/shell-and-tube use correction

F

(RC Sachdeva / GATE HT).

Exam tip: never replace LMTD with an arithmetic mean of end ΔT unless ΔT_a ≈ ΔT_b; keep U and A in consistent SI units.

Key formulas & points

Skim these first — then read the full notes below.

Energy balance: $q=\dot{m}_h c_{p,h}(T_{h,i}-T_{h,o})=\dot{m}_c c_{p,c}(T_{c,o}-T_{c,i})$ .
Rate equation: $q=UA\Delta T_{\mathrm{lm}}$ (counter/parallel with proper end ΔT).
LMTD: $\Delta T_{\mathrm{lm}}=\dfrac{\Delta T_a-\Delta T_b}{\ln(\Delta T_a/\Delta T_b)}$ .
Counterflow: usually larger LMTD than parallel for same terminal temperatures.
Shell-and-tube / crossflow: $q=UA\,F\,\Delta T_{\mathrm{lm,cf}}$ with $F$ from charts ( $F\gtrsim 0.75$ preferred).
Overall $U$ : $\dfrac{1}{UA}=\sum R$ including films, wall, fouling.
NTU–effectiveness is alternative when outlet temperatures are unknown.

Topic details

Definition and physical meaning

A heat exchanger transfers heat between two fluids without mixing (in recuperators). The LMTD method sizes or rates an exchanger when inlet/outlet temperatures are known (or assumed).

Log-mean temperature difference

\Delta T_{\mathrm{lm}}=\frac{\Delta T_a-\Delta T_b}{\ln(\Delta T_a/\Delta T_b)}

where

\Delta T_a

and

\Delta T_b

are the temperature differences between hot and cold fluids at the two ends of the exchanger.

Heat transfer rate

q=UA\,\Delta T_{\mathrm{lm}}

(for pure parallel or counterflow). For more complex flow arrangements:

q=UA\,F\,\Delta T_{\mathrm{lm,cf}}

with

\Delta T_{\mathrm{lm,cf}}

based on counterflow end differences and

F\le 1

a configuration correction factor.

Physical parameters

Symbol	Meaning	SI unit
$q$	Heat transfer rate	$\mathrm{W}$
$U$	Overall heat-transfer coeff.	$\mathrm{W/(m^2\cdot K)}$
$A$	Heat-transfer area	$\mathrm{m^2}$
$\Delta T_{\mathrm{lm}}$	Log-mean temp. difference	$\mathrm{K}$
$F$	LMTD correction factor	—
$\dot{m}$	Mass flow rate	$\mathrm{kg/s}$
$c_{p}$	Specific heat	$\mathrm{J/(kg\cdot K)}$
$T_{h,i},T_{h,o}$	Hot in / out	$\mathrm{K}$ or °C
$T_{c,i},T_{c,o}$	Cold in / out	$\mathrm{K}$ or °C

Fig — Shell-and-tube counter-flow (schematic)

Schematic diagram for study — aligned with standard B.Tech / GATE syllabus.

Core assumptions (LMTD method)

1. Steady flow and heat transfer.
2. Negligible heat loss to surroundings (well insulated).
3. Negligible kinetic/potential energy changes; PE/KE ignored in energy balance.
4. Constant specific heats

c_{p}

(or use mean values).
5. Constant overall coefficient

U

along the exchanger (or suitable average).
6. Single-phase fluids (or isothermal phase-change on one side — then LMTD still applies with care).
7. No axial conduction in the wall that smears the ΔT profile excessively.

U

c_{p}

vary strongly, integrate locally or use numerical methods / ε–NTU with care.

Energy balance and end temperature differences

Capacity rates:

C_h=\dot{m}_h c_{p,h}

C_c=\dot{m}_c c_{p,c}

Formulas (Indian textbook notation)

$q = C_{h}(T_{h,i}-T_{h,o}) = C_{c}(T_{c,o}-T_{c,i})$

Parallel flow (both fluids enter at same end):

\Delta T_a=T_{h,i}-T_{c,i},\qquad\Delta T_b=T_{h,o}-T_{c,o}

Counterflow (fluids enter at opposite ends):

\Delta T_a=T_{h,i}-T_{c,o},\qquad\Delta T_b=T_{h,o}-T_{c,i}

Why log-mean? Local

dq = U(T_{h}-T_{c})dA

. With linear relation between

\Delta T

and

q

from energy balances, integrating yields LMTD — not the arithmetic mean

(\Delta T_a+\Delta T_b)/2

, except when

\Delta T_a\approx\Delta T_b

Overall U and fouling

For a plane wall or thin tube (approximate):

\frac{1}{U}=\frac{1}{h_i}+R_{f,i}+\frac{t}{k}+R_{f,o}+\frac{1}{h_o}

(area basis must be consistent —

U_{i} A_{i} = U_{o} A_{o}

Fouling factors

R_{f}

account for scale/deposits; design

U

is lower than clean

U

Shell-and-tube: use

F

charts with parameters

P=\frac{T_{t,o}-T_{t,i}}{T_{s,i}-T_{t,i}},\qquad R=\frac{T_{s,i}-T_{s,o}}{T_{t,o}-T_{t,i}}

(tube and shell fluid temperatures). Avoid designs with

F<0.75

(temperature-cross sensitivity).

Counterflow vs parallel flow

For the same four terminal temperatures, counterflow LMTD

\ge

parallel-flow LMTD, so required

A=q/(U\,\mathrm{LMTD})

is smaller for counterflow.

Counterflow can also achieve

T_{c,o}>T_{h,o}

(cold outlet hotter than hot outlet) — impossible in simple parallel flow without violating second-law ΔT direction along the device.

Special case: if one fluid is isothermal (evaporation/condensation), parallel and counterflow LMTD expressions coincide.

Step-by-step problem approach

1. Sketch flow arrangement; label all four terminal temperatures.
2. From energy balance, find unknown outlet (if one unknown) using

q=C\Delta T

.
3. Form

\Delta T_a

\Delta T_b

for parallel or counterflow.
4. Compute

\Delta T_{\mathrm{lm}}=(\Delta T_a-\Delta T_b)/\ln(\Delta T_a/\Delta T_b)

.
5.

q=UA\Delta T_{\mathrm{lm}}

(or include

F

).
6. If sizing:

A=q/(U F\Delta T_{\mathrm{lm}})

.
7. Check

\Delta T_a

and

\Delta T_b

same sign and nonzero; if equal, LMTD = that value.
8. Units: °C differences equal Kelvin differences.

Common mistakes in exams

• Using arithmetic mean

(\Delta T_a+\Delta T_b)/2

instead of LMTD.
• Swapping parallel vs counterflow end pairings.
• Applying

F = 1

to multipass shell-and-tube without checking charts.
• Mixing

U

based on inner area with outer area

A

.
• Allowing a temperature cross in parallel-flow calculations.
• Forgetting fouling when computing design

U

Calculator

LMTD

\mathrm{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)}

ΔT₁KΔT₂K

Result

28.8539K

LMTD = (40 − 20) / ln(40/20) = 28.85 K

Worked examples

Try the problem first — open the solution when you are ready to check.

Counterflow LMTD and area

Problem

Hot oil cools from

T_{h,i}=120^\circ\mathrm{C}

T_{h,o}=80^\circ\mathrm{C}

. Cold water heats from

T_{c,i}=30^\circ\mathrm{C}

T_{c,o}=50^\circ\mathrm{C}

q=50\,\mathrm{kW}

U=400\,\mathrm{W/(m^2\cdot K)}

. Find LMTD (counterflow) and required

A

Solution

Formulas (Indian textbook notation)

$\Delta T_a=T_{h,i}-T_{c,o}=120-50=70\,\mathrm{K}$

Formulas (Indian textbook notation)

$\Delta T_b=T_{h,o}-T_{c,i}=80-30=50\,\mathrm{K}$

Formulas (Indian textbook notation)

$\Delta T_{\mathrm{lm}}=\frac{70-50}{\ln(70/50)}=\frac{20}{\ln(1.4)}=59.44\,\mathrm{K}$

Formulas (Indian textbook notation)

$A=\frac{q}{U\Delta T_{\mathrm{lm}}}=\frac{50000}{400\times 59.44}=2.103\,\mathrm{m^2}$

Parallel vs counterflow LMTD

Problem

Same terminal temperatures as above. Compute parallel-flow LMTD and compare.

Find outlet from energy balance then LMTD

Problem

Hot fluid:

\dot{m}_h=0.5\,\mathrm{kg/s}

c_{p,h}=2.0\,\mathrm{kJ/(kg\cdot K)}

T_{h,i}=150^\circ\mathrm{C}

. Cold:

\dot{m}_c=0.8\,\mathrm{kg/s}

c_{p,c}=4.18\,\mathrm{kJ/(kg\cdot K)}

T_{c,i}=25^\circ\mathrm{C}

T_{c,o}=45^\circ\mathrm{C}

. Counterflow,

U=500\,\mathrm{W/(m^2\cdot K)}

. Find

A

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

1
What is LMTD? Write its expression for a double-pipe exchanger.
Model answer
Log Mean Temperature Difference: $\mathrm{LMTD}=\dfrac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}$ where $\Delta T_1,\Delta T_2$ are the temperature differences between hot and cold fluids at the two ends.
2
Why use log mean rather than arithmetic mean $\Delta T$ ?
Model answer
Local $\Delta T$ varies exponentially along the exchanger when $U$ and capacities are constant. Integrating $dq=U\,dA\,\Delta T$ yields the log mean, not the arithmetic mean.
3
Write the heat transfer rate using LMTD.
Model answer
$q=UA\,\mathrm{LMTD}$ for pure counterflow or parallel flow (with appropriate end $\Delta T$ s). $U$ is the overall heat transfer coefficient.
4
Define $\Delta T_1$ and $\Delta T_2$ for parallel flow and counterflow.
Model answer
Parallel: $\Delta T_1=T_{h,\mathrm{in}}-T_{c,\mathrm{in}}$ , $\Delta T_2=T_{h,\mathrm{out}}-T_{c,\mathrm{out}}$ . Counterflow: $\Delta T_1=T_{h,\mathrm{in}}-T_{c,\mathrm{out}}$ , $\Delta T_2=T_{h,\mathrm{out}}-T_{c,\mathrm{in}}$ .
5
Which arrangement gives higher LMTD for the same terminal temperatures — parallel or counterflow?
Model answer
Counterflow generally gives a larger LMTD (and can achieve closer temperature approaches). Parallel flow LMTD is smaller; cold fluid cannot exceed hot outlet in the ideal limit.
6
What is the LMTD correction factor $F$ ?
Model answer
For shell-and-tube, crossflow, and multipass exchangers, $q=UAF\,\mathrm{LMTD}_{\mathrm{cf}}$ where $\mathrm{LMTD}_{\mathrm{cf}}$ is computed as if counterflow and $F\le 1$ from charts using $P$ and $R$ .
7
Define temperature effectiveness $P$ and capacity ratio $R$ used in $F$ charts.
Model answer
$P=\dfrac{t_2-t_1}{T_1-t_1}$ (tube-side rise over inlet difference), $R=\dfrac{T_1-T_2}{t_2-t_1}$ (shell drop over tube rise). Notation varies by textbook; follow the chart’s definitions.
8
When is LMTD equal to the arithmetic mean temperature difference?
Model answer
When $\Delta T_1=\Delta T_2$ (balanced counterflow with equal capacity rates, or condensers/evaporators with one fluid isothermal in a way that end differences match). Then $\mathrm{LMTD}=\Delta T_{\mathrm{am}}$ .
9
How do you treat a condenser or evaporator in LMTD analysis?
Model answer
One fluid stays at saturation temperature. End differences use that constant $T_{\mathrm{sat}}$ . Energy balance: $q=\dot{m}h_{fg}$ (phase change) equals $\dot{m}c_p\Delta T$ of the single-phase stream.
10
State assumptions behind the classical LMTD derivation.
Model answer
Steady flow; constant $U$ ; constant specific heats (or use enthalpy); negligible heat loss to surroundings; single-pass pure parallel/counterflow; kinetic/potential changes neglected; no axial conduction in fluids.
11
Compare LMTD and NTU–effectiveness methods — when is each convenient?
Model answer
LMTD: when all four terminal temperatures (or enough to get both end $\Delta T$ s) are known — sizing $A$ . NTU– $\varepsilon$ : when outlet temperatures are unknown — rating problems with known $UA$ and inlet conditions.
12
What happens to required area if LMTD decreases for fixed $q$ and $U$ ?
Model answer
From $q=UA\,\mathrm{LMTD}$ , $A=q/(U\,\mathrm{LMTD})$ rises — closer temperature approaches need more area.
13
Can cold fluid exit hotter than hot fluid exit in counterflow? In parallel flow?
Model answer
Counterflow: yes, possible (temperature cross of exit streams). Parallel flow: no — cold outlet cannot exceed hot outlet for the same exchanger.
14
How is fouling incorporated with LMTD design?
Model answer
Fouling factors add to $\frac{1}{U}$ : $1/U_o=1/h_o+R_{f,o}+\cdots+R_{\mathrm{wall}}+R_{f,i}(A_o/A_i)+\cdots$ . Design $U$ is lower than clean $U$ , increasing required $A$ .
15
If $\Delta T_1$ and $\Delta T_2$ differ greatly, why is arithmetic mean unsafe?
Model answer
Arithmetic mean overpredicts the effective driving force when $\Delta T_1/\Delta T_2$ is large, underestimating required area. Always use LMTD (or $F\cdot\mathrm{LMTD}$ ).

Exams & GATE

1
Textbook: RC Sachdeva (heat exchangers); Incropera Ch. 11.
2
Never average end temperature differences arithmetically when $\Delta T$ varies — use LMTD.
3
GATE favourites: counter vs parallel comparison, $F$ correction concept, and fouling factors in $U$ .

📖 Standard books (India)

Fundamentals of Engineering Heat & Mass Transfer — RC Sachdeva
Read: Ch. 11–12
Heat transfer and heat exchangers

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📚Key formulas & points

Topic details

Definition and physical meaning

Core assumptions (LMTD method)

Energy balance and end temperature differences

Overall U and fouling

Counterflow vs parallel flow

Step-by-step problem approach

Common mistakes in exams

LMTD

Worked examples

Counterflow LMTD and area

Parallel vs counterflow LMTD

Find outlet from energy balance then LMTD

Practice questions

📝Exams & GATE

📖 Standard books (India)

Explore related topics

See real mechanical engineering careers

Key formulas & points

Exams & GATE