Shell & Tube Heat Exchanger Sizing — LMTD Method

🌡️ Shell & Tube Heat Exchangers — Overview

Shell and tube heat exchangers (STHE) are the most widely used type of heat exchanger in the chemical process industry, oil & gas, power generation, and refining. They consist of a bundle of tubes enclosed in a cylindrical shell, with one fluid flowing through the tubes and another flowing over the outside of the tubes within the shell.

The LMTD (Log Mean Temperature Difference) method is the standard approach for sizing a new heat exchanger when both inlet and outlet temperatures of both streams are known.

Common TEMA Types

🔧

AEL / BEM (Fixed Tubesheet)

Simplest and cheapest. Tubes are fixed — no differential thermal expansion compensation. Best for small ΔT.

🔄

AES / BES (Floating Head)

Tube bundle can expand freely. Preferred for large ΔT services and where tube-side cleaning is required.

➿

AEU / BEU (U-Tube)

Tubes bent into U-shape. No rear header required — lower cost. Cannot clean tube interiors mechanically.

🌀

Kettle Reboiler (AKT)

Oversized shell with vapour disengagement space. Used for reboiling duties in distillation columns.

💡

TEMA standard: Heat exchanger types are designated by three letters — front head type, shell type, rear head type. E.g. BEM = Bonnet front head, single-pass shell, fixed tubesheet rear. Always specify TEMA class (R, C, or B) based on service severity.

🔥 Step 1 — Calculate Heat Duty (Q)

The heat duty is the rate of heat transfer between the two fluids. It is calculated from the energy balance on either the hot side or cold side:

// Heat Duty — Hot Side

Q = m_h × Cp_h × (T_h1 − T_h2)

m_h = Hot fluid mass flowrate (kg/s) | Cp_h = Specific heat capacity of hot fluid (kJ/kg·K) | T_h1 = Hot fluid inlet temperature (°C) | T_h2 = Hot fluid outlet temperature (°C)

// Heat Duty — Cold Side

Q = m_c × Cp_c × (T_c2 − T_c1)

m_c = Cold fluid mass flowrate (kg/s) | Cp_c = Specific heat of cold fluid (kJ/kg·K) | T_c1 = Cold inlet | T_c2 = Cold outlet

⚠️

Heat balance check: Always verify Q_hot ≈ Q_cold within ±5%. A large discrepancy indicates an error in stream data, a phase change that hasn't been accounted for, or significant heat losses to the environment.

// COUNTER-CURRENT FLOW (Preferred)

HOT FLUID →

T_h1 (inlet, high) ————————————————→ T_h2 (outlet, low)

← COLD FLUID

T_c2 (outlet, high) ←———————————————— T_c1 (inlet, low)

Shell inlet end Shell outlet end

📐 Step 2 — Calculate LMTD

The Log Mean Temperature Difference (LMTD) is the effective average temperature driving force across the heat exchanger. It is always calculated assuming pure counter-current flow first.

// LMTD — Counter-Current Flow

ΔT₁ = T_h1 − T_c2 (hot end temperature difference)

ΔT₂ = T_h2 − T_c1 (cold end temperature difference)

LMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁ / ΔT₂)

When ΔT₁ = ΔT₂, LMTD = ΔT₁ = ΔT₂ (isothermal case, e.g. condensers/reboilers)

✅

Counter-current is always preferred over co-current (parallel flow) because it gives a higher LMTD for the same terminal temperatures — meaning less heat transfer area is required. Counter-current also allows the cold fluid outlet to exceed the hot fluid outlet temperature, which co-current flow cannot achieve.

LMTD for Special Cases

Service Type	ΔT₁	ΔT₂	LMTD Note
Condenser (steam condensing)	T_steam − T_c1	T_steam − T_c2	T_h1 = T_h2 = T_steam (isothermal)
Reboiler (liquid boiling)	T_h1 − T_boil	T_h2 − T_boil	T_c1 = T_c2 = T_boil (isothermal)
Sensible heat exchange	T_h1 − T_c2	T_h2 − T_c1	Standard LMTD formula applies

📊 Step 3 — LMTD Correction Factor (F)

The LMTD formula assumes pure counter-current flow. Real shell and tube exchangers with multiple tube passes or shell passes deviate from this ideal. The F correction factor accounts for this:

// Corrected LMTD

ΔT_lm(corrected) = F × LMTD_countercurrent

F = LMTD correction factor (dimensionless, always ≤ 1.0)

F is a function of two dimensionless temperature ratios R and P:

// Temperature Ratios for F Charts

R = (T_h1 − T_h2) / (T_c2 − T_c1) [heat capacity ratio]

P = (T_c2 − T_c1) / (T_h1 − T_c1) [cold side effectiveness]

R and P are used to read F from TEMA F-factor charts for the specific shell configuration (1-2, 1-4, 2-4, etc.)

F Factor Guidelines

F Value	Interpretation	Action
F = 1.0	Pure counter-current or isothermal	Ideal — no correction needed
F = 0.85–1.0	Acceptable deviation	Normal design range
F = 0.75–0.85	Marginal	Consider adding shell pass or redesign
F < 0.75	Unacceptable — large area penalty	Redesign: split into two shells, add shell pass

⚠️

Design rule: Never accept F below 0.75 in a design. The heat transfer area penalty becomes excessive and the design becomes very sensitive to small changes in operating conditions. Split into two shells in series instead.

⚡ Step 4 — Overall Heat Transfer Coefficient (U)

The overall heat transfer coefficient U combines all resistances to heat transfer between the two fluids — shell-side film, tube wall conduction, tube-side film, and fouling on both sides.

💧

Shell-side film 1/h_o

+

🔩

Shell fouling R_fo

+

🪨

Tube wall t/k_w

+

🔩

Tube fouling R_fi

+

💧

Tube-side film 1/h_i

// Overall Heat Transfer Coefficient (simplified for thin-walled tubes)

1/U = 1/h_o + R_fo + t/k_w + R_fi + 1/h_i

h_o = Shell-side film coefficient (W/m²·K) | h_i = Tube-side film coefficient (W/m²·K) | R_fo, R_fi = Shell & tube fouling resistances (m²·K/W) | t = Tube wall thickness (m) | k_w = Tube wall thermal conductivity (W/m·K)

Typical Overall U Values — Quick Reference

Service	U (W/m²·K)	Notes
Water-to-water	800 – 1500	Common cooling duty
Steam condenser (water coolant)	1000 – 3000	High h_o for condensing steam
Gas-to-gas	15 – 50	Low film coefficients both sides
Gas-to-liquid	20 – 300	Gas side controls
Organic liquid-to-organic liquid	100 – 400	Depends on viscosity
Boiling liquid (reboiler)	500 – 2000	High boiling-side coefficient
Viscous oil-to-water	50 – 200	Oil viscosity controls

🔩 Step 5 — Fouling Resistances

Fouling is the accumulation of unwanted deposits (scale, biofilm, corrosion products, particulates) on heat transfer surfaces over time. It increases thermal resistance and reduces effective U — heat exchangers must be overdesigned to account for end-of-run fouling conditions.

🚨

Plant reality: Fouling is the single biggest cause of heat exchanger underperformance in process plants. A badly fouled heat exchanger can lose 30–60% of its clean heat transfer area effectiveness. Always include fouling resistances in your design — and never assume clean conditions for a new exchanger in service.

Typical TEMA Fouling Resistances

Fluid / Service	R_f (m²·K/W)	TEMA Class
Cooling water (treated)	0.0001 – 0.0002	TEMA R/C
Cooling water (river/untreated)	0.0002 – 0.0003	TEMA B
Steam (clean)	0.00009	TEMA R
Process organic liquids	0.0002	Typical
Process aqueous solutions	0.0002	Typical
Crude oil / heavy hydrocarbons	0.0005 – 0.001	Refinery service
Seawater	0.0001	With anti-fouling treatment

Cleanliness Factor (CF)

// Cleanliness Factor

CF = U_fouled / U_clean × 100%

Typical design target: CF = 75–85%. Means the fouled U is 75–85% of the clean U. This directly sets your overdesign margin.

📏 Step 6 — Heat Transfer Area & Tube Count

Once Q, LMTD, F, and U are known, the required heat transfer area is calculated directly:

// Required Heat Transfer Area

A = Q / (U × F × LMTD)

A = Required heat transfer area (m²) | Q = Heat duty (W) | U = Overall coefficient (W/m²·K) | F = LMTD correction factor | LMTD = Log mean temperature difference (K)

Number of Tubes Required

// Tube Count

A_per_tube = π × d_o × L

N_tubes = A / A_per_tube

d_o = Tube outside diameter (m) | L = Effective tube length (m) | Standard tube lengths: 1.83, 2.44, 3.66, 4.88, 6.10 m

Standard Tube Sizes (TEMA)

OD (mm)	Common BWG	Wall (mm)	ID (mm)	Typical Use
19.05 (¾")	16 BWG	1.65	15.75	Most common general service
25.4 (1")	14 BWG	2.11	21.18	Fouling or viscous services
12.7 (½")	18 BWG	1.24	10.22	High-pressure services
38.1 (1½")	12 BWG	2.77	32.56	Very fouling services

🧮 Complete Worked Example

Duty: Cool 10,000 kg/h of process water from 80°C to 40°C using cooling water inlet at 30°C, outlet at 40°C. Design a 1-2 shell and tube heat exchanger.

Given Data

Parameter	Hot Side (Process Water)	Cold Side (Cooling Water)
Fluid	Process water	Cooling water
Inlet temperature	80°C	30°C
Outlet temperature	40°C	40°C
Mass flowrate	10,000 kg/h	To be calculated
Cp (kJ/kg·K)	4.18	4.18
Fouling resistance	0.0002 m²·K/W	0.0002 m²·K/W

Step 1 — Heat Duty

// Q Calculation

Q = 10000/3600 × 4180 × (80−40)

Q = 2.778 × 4180 × 40 = 464,400 W ≈ 464.4 kW

Cooling water flowrate = Q / (Cp × ΔT) = 464,400 / (4180 × 10) = 11.11 kg/s = 40,000 kg/h

Step 2 — LMTD (Counter-Current)

// LMTD

ΔT₁ = T_h1 − T_c2 = 80 − 40 = 40°C (hot end)

ΔT₂ = T_h2 − T_c1 = 40 − 30 = 10°C (cold end)

LMTD = (40 − 10) / ln(40/10) = 30 / 1.386 = 21.6°C

Step 3 — F Correction Factor (1-2 exchanger)

// R and P

R = (80−40) / (40−30) = 40/10 = 4.0

P = (40−30) / (80−30) = 10/50 = 0.20

From TEMA F-chart for 1-2 exchanger at R=4.0, P=0.20 → F ≈ 0.88

Step 4 — U Value (assumed)

// U with Fouling

U_clean (water-water) ≈ 1200 W/m²·K (assumed)

1/U_fouled = 1/1200 + 0.0002 + 0.0002 = 0.00083 + 0.0004

U_fouled = 1 / 0.00123 ≈ 813 W/m²·K

Step 5 — Required Heat Transfer Area

// Area Calculation

A = Q / (U × F × LMTD)

A = 464,400 / (813 × 0.88 × 21.6)

A = 464,400 / 15,474 = 30.0 m²

Step 6 — Tube Count

// Number of Tubes (19.05mm OD, 3.66m length)

A_per_tube = π × 0.01905 × 3.66 = 0.219 m²

N_tubes = 30.0 / 0.219 = 137 tubes → use 140 tubes

Summary of Results

Parameter	Result	Unit
Heat duty (Q)	464.4	kW
LMTD (counter-current)	21.6	°C
F correction factor	0.88	—
U fouled	813	W/m²·K
Required heat transfer area	30.0	m²
Number of tubes	140	tubes
Tube size	19.05mm OD × 3.66m	—
Shell type	1-2 TEMA E	—

📋 LMTD Design Procedure — Summary

1

Calculate heat duty Q

From energy balance on hot or cold stream. Verify Q_hot ≈ Q_cold.
2

Calculate LMTD (counter-current)

Using terminal temperature differences ΔT₁ and ΔT₂. Use simplified formula when ΔT₁ = ΔT₂.
3

Find F correction factor

Calculate R and P, read F from TEMA chart for your shell-and-tube pass configuration. Ensure F ≥ 0.75.
4

Estimate U value

Use literature/experience values or calculate from individual film coefficients. Apply fouling resistances to get U_fouled.
5

Calculate required area A

A = Q / (U × F × LMTD). Add 10–20% overdesign margin for uncertainty.
6

Select tube size and calculate tube count

Choose standard TEMA tube OD and length. Calculate N_tubes = A / (π × d_o × L). Round up to even number.
7

Select shell diameter

From TEMA tube-count tables for selected tube pitch and layout (triangular or square). Verify shell-side velocity and pressure drop.

Shell & Tube Heat Exchanger Sizing — LMTD Method with Worked Example

🌡️ Shell & Tube Heat Exchangers — Overview

Common TEMA Types

AEL / BEM (Fixed Tubesheet)

AES / BES (Floating Head)

AEU / BEU (U-Tube)

Kettle Reboiler (AKT)

🔥 Step 1 — Calculate Heat Duty (Q)

// COUNTER-CURRENT FLOW (Preferred)

📐 Step 2 — Calculate LMTD

LMTD for Special Cases

📊 Step 3 — LMTD Correction Factor (F)

F Factor Guidelines

⚡ Step 4 — Overall Heat Transfer Coefficient (U)

Typical Overall U Values — Quick Reference

🔩 Step 5 — Fouling Resistances

Typical TEMA Fouling Resistances

Cleanliness Factor (CF)

📏 Step 6 — Heat Transfer Area & Tube Count

Number of Tubes Required

Standard Tube Sizes (TEMA)

🧮 Complete Worked Example

Given Data

Step 1 — Heat Duty

Step 2 — LMTD (Counter-Current)

Step 3 — F Correction Factor (1-2 exchanger)

Step 4 — U Value (assumed)

Step 5 — Required Heat Transfer Area

Step 6 — Tube Count

Summary of Results

📋 LMTD Design Procedure — Summary

Calculate heat duty Q

Calculate LMTD (counter-current)

Find F correction factor

Estimate U value

Calculate required area A

Select tube size and calculate tube count

Select shell diameter

Calculate Heat Exchanger Area Instantly