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

Exam areas

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 = U A\,\Delta T_{lm} F

. 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

\Delta T

changes from one end to the other — and the local heat flux is proportional to that local

\Delta T

. 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.

\Delta T_{lm} = \dfrac{\Delta T_1 - \Delta T_2}{\ln\!\left(\Delta T_1/\Delta T_2\right)}

Counterflow vs parallel flow

Direction matters. In parallel (cocurrent) flow the two streams enter at the same end, so the largest

\Delta T

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 for the same terminal temperatures its LMTD is never smaller — larger whenever both streams change temperature, equal when one of them condenses or boils at constant temperature. 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:

\Delta T_1 = T_{Hi} - T_{Co}

at the hot-inlet end and

\Delta T_2 = T_{Ho} - T_{Ci}

at the hot-outlet end. For parallel flow, both fluids enter together, so

\Delta T_1 = T_{Hi} - T_{Ci}

and

\Delta T_2 = T_{Ho} - T_{Co}

. A special case to memorize: when

\Delta T_1 = \Delta T_2

the logarithm is indeterminate, and the LMTD simply equals that common difference — this happens in a balanced counterflow exchanger (equal

\dot m c_p

on both sides), where

\Delta T

stays constant along the length. It is NOT the condenser case: a constant-temperature phase-change side gives unequal ends whenever the other stream changes temperature. The handbook directs you to treat a condenser as a counterflow device.

\Delta T_1 = T_{Hi}-T_{Co}, \qquad \Delta T_2 = T_{Ho}-T_{Ci} \quad\text{(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 \le 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 \ge 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 = \dfrac{t_o - t_i}{T_i - t_i}, \qquad R = \dfrac{T_i - T_o}{t_o - t_i}

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 = \dot{m}c_p\,\Delta T

, 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

\Delta T_{lm}

is a temperature difference, so it has the same numeric value in °F or °R; never convert it.

Q = U A\,\Delta T_{lm}\,F

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=\dot mc_p\Delta T

; (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=UA\,\Delta T_{lm}F

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

\Delta T_1 \approx \Delta T_2

within a few percent, the arithmetic and log means agree and either is acceptable.

Key equations

Heat-exchanger rate equation

Q = U A\,\Delta T_{lm}\,F

Duty $Q$ (Btu/h); $U$ overall coefficient (Btu/h·ft²·°F) on area $A$ (ft²); $\Delta T_{lm}$ counterflow LMTD; $F$ correction factor ( $=1$ for pure counter/parallel flow and for phase-change streams).

Log-mean temperature difference

\Delta T_{lm} = \dfrac{\Delta T_1 - \Delta T_2}{\ln\!\left(\Delta T_1/\Delta T_2\right)}

$\Delta T_1,\Delta T_2$ are the terminal temperature differences (°F or °R — same value). Reduces to $\Delta T_1$ when the two ends are equal.

Counterflow terminal differences

\Delta T_1 = T_{Hi}-T_{Co}, \quad \Delta T_2 = T_{Ho}-T_{Ci}

Hot inlet pairs with cold outlet, hot outlet with cold inlet. Use for the LMTD baseline even on multipass units.

Parallel-flow terminal differences

\Delta T_1 = T_{Hi}-T_{Ci}, \quad \Delta T_2 = T_{Ho}-T_{Co}

Both streams enter the same end. Gives a smaller LMTD than counterflow for identical temperatures (equal only when one stream is isothermal).

First-law duty

Q = \dot{m}\,c_p\,(T_{in}-T_{out})

Energy balance on one stream; for water $\dot m c_p \approx 500\,(\text{gpm})$ Btu/h·°F. Both streams carry the same $Q$ (adiabatic shell).

F-factor parameter P (effectiveness)

P = \dfrac{t_o - t_i}{T_i - t_i}

Tube-side temperature rise over the maximum available difference; $t$ = tube (usually cold) stream, $T$ = shell stream. $0\le P\le 1$ .

F-factor parameter R (capacity ratio)

R = \dfrac{T_i - T_o}{t_o - t_i}

Ratio of the two stream temperature changes; equals $\dot m c_p$ of tube side over that of shell side. $R$ may be above or below 1.

F factor, 1 shell pass / 2+ tube passes

F = \dfrac{\sqrt{R^2+1}\,\ln\!\dfrac{1-P}{1-PR}}{(R-1)\,\ln\!\dfrac{2-P\left(R+1-\sqrt{R^2+1}\right)}{2-P\left(R+1+\sqrt{R^2+1}\right)}}

Closed form for the most common shell-and-tube geometry (valid $R\ne1$ ). The exam usually supplies $F$ from a chart; know how $P$ and $R$ feed it.

Equal-end LMTD limit

\Delta T_{lm} = \Delta T_1 = \Delta T_2 \quad(\text{when }\Delta T_1=\Delta T_2)

Avoids the 0/0 logarithm; occurs in balanced counterflow ( $C_H = C_C$ ), where $\Delta T$ is constant along the length — not in a condenser, whose ends differ by the other stream's temperature change.

Required area

A = \dfrac{Q}{U\,\Delta T_{lm}\,F}

Sizing form. $A$ is on the same reference surface as $U$ (inside or outside).

Worked examples

Counterflow vs parallel LMTD

Problem. Hot water enters a double-pipe exchanger at $200^\circ\text{F}$ and leaves at $150^\circ\text{F}$ ; cold water enters at $80^\circ\text{F}$ and leaves at $120^\circ\text{F}$ . Find the LMTD for (a) counterflow and (b) parallel flow.

Solution. (a) Counterflow:

\Delta T_1 = T_{Hi}-T_{Co} = 200-120 = 80^\circ\text{F}

\Delta T_2 = T_{Ho}-T_{Ci} = 150-80 = 70^\circ\text{F}

\Delta T_{lm} = \frac{80-70}{\ln(80/70)} = \frac{10}{0.13353} = 74.9^\circ\text{F}.

(b) Parallel:

\Delta T_1 = 200-80 = 120^\circ\text{F}

\Delta T_2 = 150-120 = 30^\circ\text{F}

\Delta T_{lm} = \frac{120-30}{\ln(120/30)} = \frac{90}{1.3863} = 64.9^\circ\text{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^\circ\text{F}

— the log mean is always the smaller, more conservative driving force.

Sizing a 1-2 shell-and-tube heater with F

Problem. A 1-shell-pass, 2-tube-pass exchanger uses $20{,}000\ \text{lbm/h}$ of hot water (shell side) entering at $250^\circ\text{F}$ and leaving at $175^\circ\text{F}$ to heat a tube-side stream from $100^\circ\text{F}$ to $160^\circ\text{F}$ . With $U = 120\ \text{Btu/h·ft}^2\text{·}^\circ\text{F}$ , find the required surface area.

Solution. Duty from the shell (hot) stream,

c_p = 1.0

Q = 20{,}000(1.0)(250-175) = 1.50\times10^6\ \text{Btu/h}

. Counterflow LMTD:

\Delta T_1 = 250-160 = 90

\Delta T_2 = 175-100 = 75

\Delta T_{lm} = \frac{90-75}{\ln(90/75)} = \frac{15}{0.18232} = 82.3^\circ\text{F}.

Correction factor:

P = \dfrac{160-100}{250-100} = 0.400

R = \dfrac{250-175}{160-100} = 1.25

. From the 1-2 chart (or the closed form)

F = 0.877

. Area:

A = \frac{Q}{U\,\Delta T_{lm}\,F} = \frac{1.50\times10^6}{120(82.3)(0.877)} = 173\ \text{ft}^2.

**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

152\ \text{ft}^2

, a 12% undersize.

Common pitfalls

•Using the arithmetic mean $\tfrac12(\Delta T_1+\Delta T_2)$ 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 $\Delta T_{lm}$ to absolute temperature. It is a temperature difference: $74.9^\circ\text{F} = 74.9^\circ\text{R}$ . 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.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

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.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

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.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

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.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders