Heat Exchangers & Coils · Study · PE Mechanical: HVAC and Refrigeration · FE → PE Prep
Heat Exchangers & Coils
7% of exam
LMTD and effectiveness-NTU methods, shell-and-tube and plate-and-frame exchangers, correction factors, and cooling/heating coil duty.
5 concepts
C. Heat Exchangers
The LMTD Method and the F Correction Factor
Counterflow vs parallel flow, the log-mean temperature difference, and the F factor that lets Q = U·A·LMTD·F size shell-and-tube and cross-flow exchangers.
Almost every piece of HVAC and refrigeration equipment is a heat exchanger wearing a different costume: a chilled-water coil, a shell-and-tube condenser, a plate heat exchanger on a glycol loop, an economizer. The PE exam tests whether you can size one or rate one, and the workhorse for that is the log-mean temperature difference (LMTD) method built around Q=UAΔTlmF
. Master this one equation family and you have the spine of the entire Heat Exchangers knowledge area. The handbook lays it out in the NCEES PE Mechanical Reference Handbook — §5.5 Heat Exchangers, and the single most common way candidates lose points here is using an arithmetic average temperature difference where a logarithmic one is required, or forgetting the
F
factor entirely on a multipass unit.
Why a log-mean, not an arithmetic mean
As two streams exchange heat along the length of an exchanger, their temperature difference ΔT changes from one end to the other — and the local heat flux is proportional to that local ΔT. Because the driving force decays, the correct average that reproduces the total duty is the logarithmic mean of the two end differences, not their arithmetic average. The arithmetic mean always overstates the effective driving force (the two are equal only when the ends are identical), so using it undersizes the area and you fail the problem. The LMTD is defined from the temperature difference at each terminal of the exchanger.
ΔTlm=ln(ΔT1/ΔT2)ΔT1−ΔT2
Counterflow vs parallel flow
Direction matters. In parallel (cocurrent) flow the two streams enter at the same end, so the largest ΔT is at the inlet and the streams converge — the cold outlet can never exceed the hot outlet. In counterflow the streams enter at opposite ends, the temperature difference is more uniform, and the LMTD is always larger for the same terminal temperatures. A larger LMTD means less area for the same duty, so counterflow is the preferred arrangement and is what you assume unless told otherwise. Counterflow also permits a temperature cross (cold outlet hotter than hot outlet), which parallel flow physically cannot reach.
Writing the terminal differences correctly
For counterflow, pair each end's hot and cold streams: ΔT1=THi−TCo at the hot-inlet end and ΔT2=THo−TCi at the hot-outlet end. For parallel flow, both fluids enter together, so ΔT1=THi−TCi and ΔT2=THo−TCo. A special case to memorize: when ΔT1=ΔT2 the logarithm is indeterminate, and the LMTD simply equals that common difference — a phase-change stream (condenser, evaporator, steam coil) on one side often approaches this because its temperature is nearly constant. The handbook directs you to treat a condenser as a counterflow device.
ΔT1=THi−TCo,ΔT2=THo−TCi(counterflow)
The F correction factor
Real shell-and-tube and cross-flow exchangers are neither pure counterflow nor pure parallel flow — a 1-shell-pass, 2-tube-pass unit runs partly each way. The fix is to compute the LMTD as if the unit were ideal counterflow, then multiply by a correction factor F≤1 that captures the geometric penalty. F is read from charts (or computed) as a function of two dimensionless ratios: the thermal effectiveness P of the tube-side stream and the capacity-rate ratio R. Pure counterflow gives F=1; a well-designed multipass exchanger keeps F≥0.80, and a value below about 0.75 signals a near-vertical region of the chart where small temperature errors blow up the answer — a cue to add a shell pass.
P=Ti−tito−ti,R=to−tiTi−To
Putting it together: the rate equation
With LMTD and F in hand, the heat-transfer rate equation closes the sizing problem. You typically get Q from a first-law energy balance on whichever stream is fully specified, Q=m˙cpΔT, then solve for the area A given an overall coefficient U (a separate concept). Keep U and A on the same reference surface — if U is an outside-area coefficient, A must be the outside area. Note ΔTlm is a temperature difference, so it has the same numeric value in °F or °R; never convert it.
Q=UAΔTlmF
Exam strategy
Run a fixed sequence: (1) sketch the device and label all four terminal temperatures; (2) get Q from the stream with both temperatures and a known flow, Q=m˙cpΔT; (3) compute the counterflow LMTD even for a multipass unit; (4) if it is shell-and-tube or cross-flow, compute P and R and read F; (5) solve Q=UAΔTlmF for the unknown. Two reflexes save points: use the logarithmic mean (the arithmetic mean is a distractor answer), and never apply F to a single-pass counterflow or to a pure condenser/evaporator where F=1. If ΔT1≈ΔT2 within a few percent, the arithmetic and log means agree and either is acceptable.
Key equations
Heat-exchanger rate equationQ=UAΔTlmF
Duty Q (Btu/h); U overall coefficient (Btu/h·ft²·°F) on area A (ft²); ΔTlm counterflow LMTD; F correction factor (=1 for pure counter/parallel flow and for phase-change streams).
Log-mean temperature differenceΔTlm=ln(ΔT1/ΔT2)ΔT1−ΔT2
Problem. Hot water enters a double-pipe exchanger at 200∘F and leaves at 150∘F; cold water enters at 80∘F and leaves at 120∘F. Find the LMTD for (a) counterflow and (b) parallel flow.
Problem. A 1-shell-pass, 2-tube-pass exchanger uses 20,000lbm/h of hot water (shell side) entering at 250∘F and leaving at 175∘F to heat a tube-side stream from 100∘F
Common pitfalls
•Using the arithmetic mean 21(ΔT1+ΔT2) instead of the log mean. It always overstates the driving force and undersizes area; the exam offers it as a distractor.
•Pairing terminals wrong. Counterflow pairs hot-inlet with cold-outlet; mixing up the ends silently swaps parallel and counterflow values.
•Forgetting the F factor on a multipass shell-and-tube or cross-flow unit — or, conversely, applying F to a single-pass counterflow or to a condenser/evaporator where F=1.
•Converting ΔTlm to absolute temperature. It is a temperature difference: 74.9∘F=74.9∘R. Convert only individual temperatures when an absolute value is genuinely needed.
•Reading F in the steep, near-vertical region of the chart (typically F<0.75) where the answer is hypersensitive — that is a design signal to add a shell pass, not to trust the number.
•Mismatching U and A reference surfaces — using an outside-area U with an inside area, which the overall-U concept warns against.
•Computing duty from the wrong stream. Use the stream with both temperatures and a known flow; both streams carry the same Q, so do not add them.
References
NCEES PE Mechanical Reference Handbook — §5.5 Heat Exchangers
ASHRAE Handbook—Fundamentals, Heat Transfer chapter — LMTD, F-factor charts for shell-and-tube and cross-flow geometries
Incropera & DeWitt, Fundamentals of Heat and Mass Transfer — Derivation of LMTD and the correction-factor relations
The Effectiveness-NTU Method
Capacity-rate ratio, NTU = UA/Cmin, and the effectiveness relations that rate a heat exchanger when outlet temperatures are unknown.
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Overall U, Fouling and the Resistance Network
The series convective, conductive and fouling resistances that build the overall U, why U depends on the reference area, and how fouling factors size real surface.
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Shell-and-Tube vs Plate-and-Frame Selection
Shell-and-tube configurations and passes, plate-and-frame compactness and close approach, gasketed vs brazed plates, and the trade-offs that drive HVAC selection.
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F. Cooling and Heating Coils
Cooling and Heating Coil Duty (Sensible/Latent Split, Rows)
Total, sensible and latent coil duty from the 1.10/4,840/4.5 standard-air coefficients, the sensible heat ratio and coil condition line, rows and fins, and the air/water balance.
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ΔT1,ΔT2 are the terminal temperature differences (°F or °R — same value). Reduces to ΔT1 when the two ends are equal.
Hot inlet pairs with cold outlet, hot outlet with cold inlet. Use for the LMTD baseline even on multipass units.
Both streams enter the same end. Always gives a smaller LMTD than counterflow for identical temperatures.
Energy balance on one stream; for water m˙cp≈500(gpm) Btu/h·°F. Both streams carry the same Q (adiabatic shell).
Tube-side temperature rise over the maximum available difference; t = tube (usually cold) stream, T = shell stream. 0≤P≤1.
Ratio of the two stream temperature changes; equals m˙cp of tube side over that of shell side. R may be above or below 1.
Closed form for the most common shell-and-tube geometry (valid R=1). The exam usually supplies F from a chart; know how P and R feed it.
Avoids the 0/0 logarithm; common when one side is condensing or evaporating at constant temperature.
is on the same reference surface as
U
(inside or outside).
,
ΔT2=THo−TCi=150−80=70∘F
.
ΔTlm=ln(80/70)80−70=0.1335310=74.9∘F.
(b) Parallel:
ΔT1=200−80=120∘F
,
ΔT2=150−120=30∘F
.
ΔTlm=ln(120/30)120−30=1.386390=64.9∘F.
**Answers: 74.9 °F counterflow, 64.9 °F parallel.** Sanity check: counterflow exceeds parallel for the same temperatures (as it must), and both lie below the arithmetic mean of
75∘F
— the log mean is always the smaller, more conservative driving force.
to
160∘F
. With
U=120Btu/h⋅ft2⋅∘F
, find the required surface area.
Solution. Duty from the shell (hot) stream, cp=1.0: Q=20,000(1.0)(250−175)=1.50×106Btu/h.
Counterflow LMTD: ΔT1=250−160=90, ΔT2=175−100=75.
ΔTlm=ln(90/75)90−75=0.1823215=82.3∘F.
Correction factor: P=250−100160−100=0.400, R=160−100250−175=1.25. From the 1-2 chart (or the closed form) F=0.877.
Area: A=UΔTlmFQ=120(82.3)(0.877)1.50×106=173ft2.
**Answer: A = 173 ft² (with F = 0.877).** Sanity check: F sits comfortably above 0.80, so the geometry is reasonable; ignoring F would have given 152ft2, a 12% undersize.