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Heat Transfer · Chapter 9 of 10 · Advanced

Heat Exchangers

Two streams, one wall, no mixing. Size one with the log-mean temperature difference; rate one with effectiveness and NTU.

01

Readiness check

From Chapters 2 and 6. Tick only what you can do closed-notes.

  • Add thermal resistances in series and find an overall value.
  • Use q = ṁ cp ΔT on a flowing stream.
  • Find a convection coefficient from a correlation.
  • Work with natural logs and exponentials.
  • Track which stream is hot and which is cold.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsReview series resistances in Chapter 2; U is built the same way.
3 or more weak itemsRevisit the resistance network of Chapter 2 first.
02

The core idea

A heat exchanger is a wall with a convective resistance on each side; combine them into one coefficient U, then use the right temperature difference.

q = UA·ΔTlmΔTlm = (ΔT₁−ΔT₂)/ln(ΔT₁/ΔT₂)ε = q/qmax, NTU = UA/Cmin

Because the stream temperatures change along the exchanger, the driving difference is not constant; the log-mean temperature difference is its correct average. When you know all four terminal temperatures you size the area with LMTD; when you do not, effectiveness-NTU gives the performance without iteration.

The skill works when: you combine both film coefficients (and fouling) into U, and pick LMTD for sizing or NTU for rating.
The skill breaks down when: a plain arithmetic average ΔT is used, or LMTD is forced on a problem with two unknown outlet temperatures.
The concept. In parallel flow the two streams converge and the driving difference collapses; in counterflow they stay apart, keeping ΔT large along the whole length, which is why counterflow is the more effective arrangement.
03

The skills, taught in order

Heat exchangers reuse the resistance network of Chapter 2 and the coefficients of Chapter 6, then add two analysis methods. Six skills take you from configuration to a sized or rated unit.

9.1 What a heat exchanger does

It transfers heat between two fluids through a separating wall without mixing them. Common types are the double-pipe, the shell-and-tube, the compact plate, and the crossflow (as in a car radiator), each a variation on the same wall-between-streams idea.

9.2 The overall coefficient U

The two convective resistances and the wall resistance act in series: 1/UA = 1/(hA)hot + Rwall + 1/(hA)cold, plus fouling once the surfaces foul. U is dominated by the smallest coefficient, so a gas side usually controls the design.

9.3 Parallel versus counterflow

In parallel flow both streams enter the same end and their temperatures converge, capping performance. In counterflow they enter opposite ends, keeping the difference large throughout; for the same area and U it transfers more heat. Counterflow is the benchmark.

9.4 The LMTD method

For sizing, when the inlet and outlet temperatures are known, q = UA·F·ΔTlm, where ΔTlm = (ΔT₁−ΔT₂)/ln(ΔT₁/ΔT₂) and F is a correction factor (F = 1 for pure counterflow). Solve for the area A.

9.5 The effectiveness-NTU method

For rating, when outlet temperatures are unknown, define effectiveness ε = q/qmax with qmax = Cmin(Th,in−Tc,in), the number of transfer units NTU = UA/Cmin, and the capacity ratio Cr = Cmin/Cmax. Each configuration has an ε(NTU, Cr) relation that gives performance directly.

MethodUse whenYou solve for
LMTDall four terminal temperatures knownarea A (sizing)
Effectiveness-NTUoutlet temperatures unknownq and outlets (rating)

9.6 Choosing and using a method

Sizing a new unit to hit known temperatures is an LMTD job; predicting how an existing unit performs with given inlets is an NTU job. The two are equivalent, just rearranged for what is known, and many problems can be cross-checked with both.

Engineering connection: radiators, condensers, evaporators, oil coolers, HVAC coils, and process plant; the film coefficients come from Chapters 6 and 8.

04

Worked example 1: sizing by LMTD

A counterflow exchanger cools oil from 100 °C to 60 °C while heating water at 0.5 kg/s (cp = 4180 J/kg·K) from 20 °C to 40 °C. With an overall coefficient U = 300 W/m²·K, find the heat duty and the required area.

Figure 1. Counterflow sizing. The duty is set by the water stream; the log-mean temperature difference is 49.3 °C, giving a required area of about 2.8 m².
  1. ProblemFind the heat duty and the area for the counterflow exchanger in Figure 1.
  2. Given / findOil 100 → 60 °C, water 20 → 40 °C at ṁcp = 0.5 × 4180, U = 300, counterflow. Find q and A.
  3. AssumptionsSteady operation, constant properties, negligible losses, F = 1 (pure counterflow).
  4. ModelGet q from the water stream, form ΔTlm from the end differences, then A = q/(U·ΔTlm).
  5. Equationsq = ṁcpΔT ΔTlm = (ΔT₁−ΔT₂)/ln(ΔT₁/ΔT₂) A = q/(U·ΔTlm)
  6. Solveq = 0.5 × 4180 × (40−20) = 41 800 W. End differences: ΔT₁ = 100−40 = 60 °C, ΔT₂ = 60−20 = 40 °C. ΔTlm = (60−40)/ln(60/40) = 20/0.405 = 49.3 °C. A = 41 800/(300 × 49.3) = 2.8 m².
  7. CheckThe log-mean (49.3 °C) lies between the two end differences (40 and 60) and below their arithmetic mean (50), as it must. Using the arithmetic mean would have undersized the area slightly.
  8. ConclusionA 2.8 m² counterflow surface meets the duty. A parallel-flow unit would need more area for the same job, because its driving difference collapses toward the outlet, which is why counterflow is preferred when layout allows.
Result. q = 41.8 kW, ΔTlm = 49.3 °C, required area A ≈ 2.8 m².
05

Worked example 2: rating by effectiveness-NTU

Take that same exchanger (UA = 847 W/K, counterflow) but now feed it hot oil at 120 °C with capacity rate Ch = 1000 W/K and water at 20 °C with Cc = 2090 W/K. The outlet temperatures are unknown. Find the heat rate and both outlet temperatures.

Figure 2. Rating the same unit with new inlets. NTU = 0.85 and Cr = 0.48 give ε = 0.52, so the exchanger delivers about 51.6 kW and the outlets follow directly, no iteration needed.
  1. ProblemFind q and the outlet temperatures for the exchanger in Figure 2, given only the inlets.
  2. Given / findUA = 847 W/K, Ch = 1000, Cc = 2090, Th,in = 120 °C, Tc,in = 20 °C, counterflow. Find q, Th,out, Tc,out.
  3. AssumptionsSteady operation, constant properties, counterflow ε relation applies.
  4. ModelForm Cmin, Cr, and NTU; use the counterflow effectiveness relation; then q = εCminΔTmax and energy balances for the outlets.
  5. EquationsNTU = UA/Cmin, Cr = Cmin/Cmax ε = [1−e−NTU(1−Cr)]/[1−Cre−NTU(1−Cr)] q = εCmin(Th,in−Tc,in)
  6. SolveCmin = 1000, Cr = 1000/2090 = 0.48, NTU = 847/1000 = 0.85. The counterflow relation gives ε = 0.52. q = 0.52 × 1000 × (120−20) = 51 600 W. Then Th,out = 120 − 51 600/1000 = 68.4 °C and Tc,out = 20 + 51 600/2090 = 44.7 °C.
  7. CheckBoth outlets lie between the two inlet temperatures, and the hot stream (smaller C) changes more than the cold, as Cmin demands. No iteration was needed, the advantage of NTU when outlets are unknown.
  8. ConclusionThe same hardware, given different inlets, is rated directly by effectiveness-NTU. Sizing (LMTD) and rating (NTU) are two views of one exchanger; choose the method that matches what you already know.
Result. ε = 0.52, q ≈ 51.6 kW, Th,out = 68.4 °C, Tc,out = 44.7 °C.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Arithmetic-mean ΔTArea slightly off, worse at large ΔT spread"Did I use the log-mean?"Use ΔTlm; the driving difference varies along the length.
LMTD with unknown outletsStuck, or guessing outlet temperatures"Do I know all four temperatures?"If outlets are unknown, switch to effectiveness-NTU; no iteration needed.
Wrong Cminqmax and ε computed from the wrong stream"Which stream has the smaller ṁcp?"qmax uses Cmin; the Cmin stream undergoes the larger temperature change.
Forgetting foulingClean-surface U overpredicts long-term duty"Is fouling resistance included?"Add fouling factors to 1/UA; fouling can dominate after service.
07

Practice ladder

Level 1 · Direct skill

A counterflow unit has end differences of 50 °C and 20 °C. Find the log-mean temperature difference.

Show answer

ΔTlm = (50−20)/ln(50/20) = 30/ln(2.5) = 30/0.916 = 32.7 °C, below the arithmetic mean of 35 °C as expected.

Level 2 · Mixed concept

Build U for a tube with inside coefficient hi = 8000 and outside ho = 100 W/m²·K (thin wall, equal areas, no fouling). Which side controls?

Show answer

1/U = 1/8000 + 1/100 = 0.000125 + 0.010 = 0.010125, so U ≈ 99 W/m²·K. The outside (gas) side controls almost entirely; improving the water side barely helps. This is why gas-side surfaces are finned.

Level 3 · Independent problem

For the Worked Example 2 exchanger, what would ε approach if the area were made very large (NTU → ∞)? What limits it?

Show answer

For counterflow, ε → 1 only if Cr → 0; with Cr = 0.48, ε approaches (1−e−∞)/(1−0.48e−∞) = 1/1 ... in the limit εmax = 1 is unreachable, and effectiveness saturates near 1 slowly. Infinite area cannot beat the second-law limit set by Cr; beyond NTU ≈ 3 to 5 added area buys little.

Level 4 · Transfer to real engineering

Pick a real exchanger (car radiator, home boiler, AC condenser). Identify its type and flow arrangement, estimate U and the duty, and state whether you would analyse it by LMTD or NTU and why.

What good work looks like

Type and flow arrangement named, U built from both film coefficients, the method chosen to match the known data, and a duty or area with a sanity check.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Here is what I know about the exchanger. Tell me whether LMTD or NTU is the right method."
"Check that I identified Cmin and the flow arrangement correctly before I compute ε."
"Size this exchanger." Choosing the method and building U is the judgment being trained.
"What is the effectiveness?" Forming NTU and Cr yourself is the skill.

Portfolio task

Analyse one real exchanger both ways: size it by LMTD for a target duty, then rate it by NTU at off-design inlets, and confirm the two are consistent.

Must include: U built from both sides, ΔTlm with correct end differences, and an ε-NTU rating that reproduces the duty.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. How is the overall coefficient U built?

From the series convection and wall resistances (plus fouling): 1/UA = 1/(hA)hot + Rwall + 1/(hA)cold.

2. Why is counterflow better than parallel flow?

It keeps the temperature difference large along the whole length, so for the same U and area it transfers more heat.

3. Write the log-mean temperature difference.

ΔTlm = (ΔT₁−ΔT₂)/ln(ΔT₁/ΔT₂), using the two end differences.

4. Define effectiveness and NTU.

ε = q/qmax with qmax = Cmin(Th,in−Tc,in); NTU = UA/Cmin.

5. When do you use NTU rather than LMTD?

When outlet temperatures are unknown (rating an existing unit); LMTD suits sizing with all temperatures known.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-solve the LMTD sizing from a blank page.
+3 daysOne NTU rating with new inlets.
+7 daysMixed set: build U from Chapter 6 coefficients, then size.
+30 daysBring it together with the radiation of Chapter 10.
10

Textbook mapping

ItemMapping
Primary sourceLienhard and Lienhard, A Heat Transfer Textbook (6th ed), Chapter 3 (heat exchanger design)
Cross-referenceIncropera, Ch. 11 (heat exchangers) · Çengel and Ghajar, Ch. 11
Core topics9.1 Types · 9.2 Overall coefficient U · 9.3 Parallel vs counterflow · 9.4 LMTD · 9.5 Effectiveness-NTU · 9.6 Choosing a method
Engineering connectionRadiators, condensers, evaporators, and HVAC coils; film coefficients from Chapters 6 and 8.
Read nextChapter 10: Thermal Radiation.