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Qwestrum Engineering360 · Chemical Engineering · Heat Transfer

Heat Exchanger Design

Exchanger duty is q = UA·ΔT_lm, where the log-mean temperature difference accounts for the varying gap along the length and U bundles all film, wall and fouling resistances in series; when outlet temperatures are unknown you switch to the effectiveness-NTU method.
Exam tip: keep SI units consistent end-to-end, write the governing relation symbolically before substituting, and sanity-check magnitude and sign.

Key formulas & points

Skim these first — then read the full notes below.

  • Fcorrectionfactorformulti−passshell−and−tube:q=UAFΔTlmF correction factor for multi-pass shell-and-tube: q = U A F \Delta T_{lm}
  • Effectiveness-NTU method when outlet T unknown
  • Fouling resistances R_f degrade performance over time

Topic details

Introduction

This topic ties the whole heat-transfer section together into equipment sizing. You compute the overall coefficient U from the individual film and fouling resistances, evaluate the log-mean temperature difference for the flow arrangement, apply an F-correction factor for multipass shell-and-tube units, and size the area A. When only inlet temperatures are known, the ε-NTU method avoids trial and error.

Key relations & formulas

q=UAΔTlmq = U A \Delta T_{lm}
(overall heat transfer)
ΔTlm=(ΔT1−ΔT2)ln(ΔT1ΔT2)\Delta T_{lm} = \frac{(\Delta T_{1} - \Delta T_{2})}{ln}(\frac{\Delta T_{1}}{\Delta T_{2}})
(counter-current LMTD)
1U=1hh+Rf,h+Lkwall+Rf,c+1hc\frac{1}{U} = \frac{1}{h_{h}} + R_{f},h + \frac{L}{k_{wall}} + R_{f},c + \frac{1}{h_{c}}
(overall resistance)
Hot fluid →← Cold fluid
Fig — Shell-and-tube counter-flow (schematic)

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

Counter-flow heat exchanger. LMTD method — RC Sachdeva / standard heat transfer texts used in Indian universities.

Notation and sign conventions

Relation 1 —
q=UAΔTlmq = U A \Delta T_{lm}
q=UAΔTlmq = U A \Delta T_{lm}
(overall heat transfer)
Write this relation with symbols exactly as in Process Heat Transfer — Kern before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
ΔTlm=\Delta T_{lm} =
ΔTlm=(ΔT1−ΔT2)ln(ΔT1ΔT2)\Delta T_{lm} = \frac{(\Delta T_{1} - \Delta T_{2})}{ln}(\frac{\Delta T_{1}}{\Delta T_{2}})
(counter-current LMTD)
Write this relation with symbols exactly as in Process Heat Transfer — Kern before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
1U=1hh+Rf,h+Lkwall+Rf,c+1hc\frac{1}{U} = \frac{1}{h_{h}} + R_{f},h + \frac{L}{k_{wall}} + R_{f},c + \frac{1}{h_{c}}
1U=1hh+Rf,h+Lkwall+Rf,c+1hc\frac{1}{U} = \frac{1}{h_{h}} + R_{f},h + \frac{L}{k_{wall}} + R_{f},c + \frac{1}{h_{c}}
(overall resistance)
Write this relation with symbols exactly as in Process Heat Transfer — Kern before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Concept in depth

The temperature difference driving heat transfer changes continuously along the exchanger, so a single representative value is needed — the log-mean captures this exactly for constant-property, constant-U service. Counter-current flow gives a larger and more uniform driving force than co-current, which is why it is preferred. The overall coefficient U is a series combination of the hot film, wall conduction, cold film and two fouling resistances; the smallest coefficient (usually the gas side) dominates U. LMTD works when outlet temperatures are known; the effectiveness-NTU approach, using ε = q/q_max, is the natural choice for rating an existing exchanger.

Assumptions and validity limits

State assumptions explicitly before using any relation for heat exchanger design — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Heat Transfer (Chemical) viva and GATE descriptive questions, listing valid assumptions often earns separate marks.

Step-by-step problem approach

1. Read the question and list given data with SI units (common in Heat Transfer (Chemical) papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to heat exchanger design.
4. Use equation 1:
q=UAΔTlmq = U A \Delta T_{lm}
.
5. Use equation 2:
ΔTlm=\Delta T_{lm} =
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.

Applications & exam relevance

Heat Exchanger Design appears in heat exchangers and reactors. In Indian chemical curricula this topic is tested because it connects theory to heat exchange in process equipment.
GATE and semester exams often combine heat exchanger design with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use heat exchanger design?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

Students apply the counter-current LMTD to a co-current arrangement, forget the F-correction for multipass units, and omit fouling resistances that can halve U in service. Using arithmetic-mean instead of log-mean ΔT when the terminal differences are very unequal also introduces error.

Quick revision checklist

Before attempting heat exchanger design problems, confirm you can:
1.
Fcorrectionfactorformulti−passshell−and−tube:q=UAFΔTlmF correction factor for multi-pass shell-and-tube: q = U A F \Delta T_{lm}

2. Effectiveness-NTU method when outlet T unknown
3. Fouling resistances R_f degrade performance over time
Revise the solved examples in Process Heat Transfer — Kern and one previous-year GATE or university paper for this unit.

Worked examples

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

Counter-current LMTD and area

Problem

Hot fluid cools 120→80 °C while cold fluid heats 30→70 °C (counter-current). Duty is 200 kW, U = 500 W/m²·K. Find the area.

Solution

ΔT₁ = 120−70 = 50, ΔT₂ = 80−30 = 50, so LMTD = 50 °C (equal ends). A = q/(U·LMTD) = 200000/(500×50) = 8 m².

Conceptual check — Heat Exchanger Design

Problem

In a Heat Transfer (Chemical) semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of heat exchanger design." What should a complete answer include?

Exams & GATE

Kern Ch. 6 — LMTD requires both outlet temperatures or use ε-NTU.

📖 Standard books (India)

  • Process Heat Transfer — Kern

    Read: Syllabus unit

    Heat exchangers and process heat transfer

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