Heat Exchanger.

The machine and its temperature profile on one shared axis, so a point on the tube sits directly above the same point on the curve. Drag the cursor along the length and read the driving difference where it actually is.

Mode

Units

Arrangement

Shell-and-tube exchanger — counter flow, Water to Water

Layers

Heat duty

202.6 kW

LMTD

28.14 K

Effectiveness

48.4%

NTU

0.86

Hot out

55.8 °C

Cold out

49.4 °C

Cmin = 8.38 kW/K · Cr = 0.80 · Qmax = 418.9 kW · Q = ε · Cmin · ΔTmax = 48.4% × 8.38 × 50.0 = 202.6 kW

Drag anywhere on the chart to move the position cursor along the exchanger. With the chart focused, the left and right arrow keys step it, Shift moves ten steps at a time, and Home and End jump to the two ends.

Start from a real machine

Six machines, each one you have met. The note explains why its numbers are what they are.

The two streams

Hot fluid

Cold fluid

Hot inlet temperature °C

Cold inlet temperature °C

Hot mass flow kg/s

Cold mass flow kg/s

The exchanger

Clean overall coefficient U W/m²·K

Surface area m²

Fouling, hot side m²·K/W

Fouling, cold side m²·K/W

Size it instead

Turn the problem round: name the duty you need and solve for the surface area — or for the coefficient — that delivers it at these conditions.

Required duty kW

Solve for

The two streams, in numbers

Stream	Fluid	In (°C)	Out (°C)	ṁ (kg/s)	cp (kJ/kg·K)	C (kW/K)	Pr
Hot	Water (liquid)	80.0	55.8	2.00	4.189	8.378	2.7
Cold	Water (liquid)	30.0	49.4	2.50	4.178	10.446	4.4

ΔT₁

30.6 K

ΔT₂

25.8 K

Fouled U

900.0 W/m²·K

Correction F

1.000

All three arrangements at these conditions

Same fluids, same flows, same surface — only the direction changes. Recomputed every time you move anything.

Arrangement	Duty (kW)	Effectiveness	LMTD (K)	Hot out (°C)	Cold out (°C)
Counter flow	202.6	48.4%	28.14	55.8	49.4
Parallel flow	183.1	43.7%	25.43	58.2	47.5
Cross flow	192.3	45.9%	29.26	57.0	48.4

The equations, with your numbers in them

Both methods, evaluated at the exchanger on screen right now. The last two lines are the same duty computed two different ways — they agree because they are the same energy balance.

Capacity rates — what each stream can carry: C = ṁ·c_p—
Capacity ratio: C_r = C_min / C_max—
Fouled coefficient — resistances in series: 1/U = 1/U_clean + R_f,h + R_f,c—
Log mean difference — ends paired for counter flow: ΔT_lm = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂)—
Number of transfer units — a dimensionless size: NTU = U·A / C_min—
Effectiveness — counter flow: ε = [1 − e^{−NTU(1−Cr)}] / [1 − C_re^{−NTU(1−Cr)}]—
Duty, the LMTD way: Q = U·A·F·ΔT_lm—
Duty, the ε-NTU way: Q = ε·C_min·(T_h,in − T_c,in)—

—

The link carries both fluids, both flows, the arrangement and the whole exchanger, in plain readable parameters.

How to calculate LMTD and NTU for a heat exchanger

Set the two inlet temperatures and the two mass flows. Those four numbers fix the capacity rates C = ṁ·cp, and the smaller of the two — Cmin — is the stream that limits everything the exchanger can do.
Set the surface area and the overall coefficient U. Their product over Cmin is the NTU, a dimensionless measure of how big the exchanger is relative to the job you have given it.
Read the profile. The two curves are the stream temperatures along the length, and the gap between them is the local driving difference — drag the cursor to read it anywhere, and watch it collapse toward the closed end.
Take the duty either way. Q = U·A·F·ΔTlm needs all four terminal temperatures; Q = ε·Cmin·(Th,in − Tc,in) needs only the two inlets. The equations panel computes both and they agree, because they are the same energy balance.
Switch the arrangement to see what direction alone is worth, and add fouling to see how much of it you lose in service.

What a heat exchanger is, and why one number is not enough

A heat exchanger moves thermal energy between two fluids without letting them mix. It is one of the most common pieces of equipment in engineering — a car radiator, a domestic boiler, the condenser behind a fridge, the intercooler on a turbocharged engine, the vast condensers under a power station turbine hall are all the same device with different plumbing. Two streams, one wall, and a temperature difference doing the work.

The complication that makes the subject interesting is that the temperature difference is not one number. Both fluids change temperature as they travel, so the driving force varies continuously from one end to the other. At the inlet end of a counter-flow unit the hot stream is at its hottest and the cold stream is on its way out at its warmest; at the far end both are cooler. The heat crossing the wall at any point depends on the gap at that point, and the gap is different everywhere.

That is why this page draws the apparatus and the temperature profile on the same horizontal axis. The vertical distance between the two curves is the local driving difference, and you can watch it collapse toward the closed end. Once you have seen that, the log mean temperature difference stops being a formula to memorise and becomes the obvious answer to a question you can see being asked.

LMTD and ε-NTU — two ways to write one energy balance

There are two standard methods, and students often learn them as rivals. They are not. They are the same conservation statement rearranged for two different questions, and for a given exchanger they return the same duty — which is why the equations panel above computes it both ways and prints both answers.

Use LMTD when you know all four terminal temperatures. Take the difference at each end, combine them logarithmically, and Q = U·A·F·ΔTlm gives the duty — or, rearranged, the area a required duty needs. This is the sizing calculation, and it is a direct computation with no iteration.
Use ε-NTU when you do not. Rating an exchanger that already exists, you know the geometry and the two inlets but not the outlets. Define NTU = U·A/Cmin, read effectiveness from the relation for your arrangement, and the duty follows from Q = ε·Cmin·(Th,in − Tc,in). Trying to use LMTD here forces you to guess an outlet and iterate.
The log mean is not the arithmetic mean, and the difference matters. The driving force decays exponentially along the length, so the correct average sits below the midpoint of the two ends — always. Using the arithmetic mean overstates the driving force and therefore undersizes the exchanger, which is the unsafe direction to be wrong in.
Both methods have a singular point, and it is the same one. In balanced counter flow the two end differences are equal, the log-mean expression becomes 0/0, and the general effectiveness relation does too. Both have clean limits — the common gap, and NTU/(1+NTU) — and this tool takes them explicitly rather than returning a blank.

Counter, parallel and cross flow — why the direction is worth so much

Reverse one stream and nothing about the hardware changes: same tubes, same area, same coefficient, same fluids at the same flows. Yet the duty changes, sometimes by a fifth. The reason is entirely about how the driving force is spent along the length.

Parallel flow starts with the largest possible difference — the full inlet-to-inlet gap — and then throws it away, because both streams move toward each other and the gap collapses. Counter flow starts smaller but stays more even, and the integral of a moderate difference over the whole length beats a large difference that vanishes after a third of it. That is also why counter flow can do something parallel flow cannot at any price: deliver a cold outlet hotter than the hot outlet. Parallel flow is capped at an effectiveness of 1/(1+Cr) and stops improving no matter how much surface you buy, which rules it out of any application needing a close approach temperature.

Cross flow, where the streams meet at right angles, sits between the two. It is the arrangement of a finned radiator or an air-cooled condenser, chosen because ducting air along a tube bundle is impractical, not because it is thermally preferable. Its effectiveness has no closed-form solution — the exact answer is an infinite series — so the relation used here, and everywhere else, is a correlation good to about one percent.

What decides U, and why fouling is a design problem rather than a maintenance one

The overall coefficient bundles every thermal resistance between the two streams into one number: 1/U = 1/hᵢ + Rfᵢ + t/k + Rfₒ + 1/hₒ. Because those resistances are in series, the largest one dominates completely. A water-to-water exchanger can reach well past 1,000 W/m²·K; a gas-to-gas unit struggles to pass 40. If either side is a gas, that side is your U, and effort spent improving the liquid side is effort wasted.

Fouling is the resistance that was not there on the day of commissioning. Scale, biological growth, corrosion products and hydrocarbon deposits build a thin insulating layer on the wall, and because it sits in series with everything else it subtracts directly from U. The numbers are unforgiving: a clean 350 W/m²·K unit running untreated river water on one side and heavy oil on the other loses about 40% of its coefficient. This is why fouling is a design decision rather than a maintenance afterthought — the exchanger is deliberately oversized at specification time so that it still meets duty when dirty, and cleaned before performance falls below that fouled design point.

Move the two fouling sliders above and watch the duty fall. The gap between the clean and fouled cases is exactly the margin a designer has to buy up front.

Who this is for, and what it is not

It is built for people who need the exchanger in front of them rather than in a textbook: mechanical, chemical and building-services students working through a heat-transfer course, HVAC and process engineers checking a duty or an area, and anyone who wants to understand why a radiator is finned on one side only. The Practice and Quiz modes exist because reading a temperature profile is a skill that comes from doing it — and the instrument deliberately stays on screen in both, because a question about an exchanger should not be asked with the exchanger hidden.

It is a teaching and checking instrument. The model is one-dimensional with a constant overall coefficient, which is the standard engineering treatment and is what every textbook method above assumes; a real unit has an entry region, a variable local coefficient and, in a multi-pass shell, temperature cross-over between passes. Fluid properties are interpolated from published tables and evaluated at each stream's bulk mean temperature, so they are better than a single fixed specific heat but they are not an equation of state. Cross flow uses the standard both-unmixed correlation, accurate to roughly one percent. Cite Çengel or Incropera in a report; use this to understand the answer, and to catch the times a number does not look right.

Frequently Asked Questions

What is the LMTD method for heat exchangers?

The Log Mean Temperature Difference method computes the duty directly from Q = U·A·F·ΔT_lm, where ΔT_lm = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂). For counter flow the end differences are ΔT₁ = T_h,in − T_c,out and ΔT₂ = T_h,out − T_c,in; for parallel flow they are ΔT₁ = T_h,in − T_c,in and ΔT₂ = T_h,out − T_c,out. Getting that pairing the wrong way round is the most common slip in the whole subject, which is why this tool labels the two ends for whichever arrangement you have selected.

Why is the log mean used instead of the ordinary average?

Because the driving temperature difference decays exponentially along the exchanger, not linearly. More of the surface therefore operates nearer the small end than a midpoint average suggests, and the correct mean sits below it. The arithmetic mean is always the higher of the two — on a typical counter-flow cooler with end differences of 55 K and 30 K it reads 42.5 K against a true log mean of 41.25 K, about 3% high. Because it overstates the driving force, using it undersizes the exchanger, which is the unsafe direction to be wrong in. The two means agree in exactly one case: when the end differences are equal.

What is the NTU-effectiveness method, and when should I use it instead?

NTU rates an exchanger without knowing its outlet temperatures. Define NTU = U·A/C_min and C_r = C_min/C_max, read ε from the relation for your arrangement, then take Q = ε·C_min(T_h,in − T_c,in).

Use LMTD when all four terminal temperatures are known and you want the area or the duty — the sizing problem. Use NTU when the outlets are unknown, which is the usual case when rating existing equipment, because it avoids iterating. They are not rival models: both are the same energy balance written differently, and for a given exchanger they return the same Q.

Why can effectiveness never exceed 1?

Effectiveness is actual heat transfer over the thermodynamic maximum, ε = Q/Q_max, where Q_max = C_min(T_h,in − T_c,in). Q_max is what you would transfer if the stream with the smaller capacity rate reached the other stream's inlet temperature in an exchanger of infinite area. Beating it would require heat to flow from cold to hot somewhere along the length, which the second law forbids. So ε is bounded by 1, and ε = 1 is only approached as NTU → ∞.

What is the difference between counter-flow and parallel-flow heat exchangers?

In parallel flow both streams enter the same end, so the driving difference starts at its largest — the full inlet-to-inlet gap — and then collapses as the outlets converge. In counter flow they enter at opposite ends, the difference stays more even along the length, and the cold outlet can leave hotter than the hot outlet. For the same U, A and capacity rates, counter flow always gives the higher effectiveness, with no exceptions at any NTU or capacity ratio. Switch arrangement in the simulator and watch the curves change from converging to running roughly parallel.

Why does parallel flow have a maximum effectiveness below 1?

Because both outlets converge toward the same intermediate temperature. Once they meet there is no driving difference left anywhere in the exchanger, and adding surface does nothing at all. The ceiling works out at ε_max = 1/(1 + C_r) — so at C_r = 1 a parallel-flow exchanger can never pass 50%, however large you build it. Any application needing a close approach temperature, such as refrigeration or cryogenics, is therefore ruled out of parallel flow on principle rather than on cost.

Why do the temperature profiles curve instead of running straight?

The local driving difference decays exponentially along the exchanger, so each stream follows an exponential approach rather than a straight line. Straight lines are correct in exactly one case — balanced counter flow, where the two capacity rates are equal and the gap between the curves is the same at both ends. In every other case the curvature is real, and it is precisely why the log mean rather than the arithmetic mean is the right average. Drag the mass-flow sliders to unbalance the exchanger and the bend becomes unmistakable.

What happens when one fluid is boiling or condensing?

A phase-changing stream absorbs or releases latent heat at constant temperature, so its temperature does not change along the exchanger and its capacity rate is effectively infinite. That drives the capacity ratio to zero, and at C_r = 0 every effectiveness relation collapses onto the same curve, ε = 1 − e^−NTU. One consequence surprises people: with one temperature fixed, the flow arrangement stops mattering entirely — counter, parallel and cross flow all give identical answers. Pick the Steam condenser or Chiller evaporator preset and try switching arrangement to see it.

What does the correction factor F do?

A cross-flow exchanger, or a shell-and-tube unit with more than one tube pass, is neither pure counter flow nor pure parallel flow, so its true mean driving force falls short of the counter-flow log mean. F is the ratio that closes the gap, giving Q = U·A·F·ΔT_lm. It is exactly 1 for pure counter flow, for pure parallel flow, and for any exchanger with one isothermal stream. Textbooks have you read F off a chart of curves; this tool computes it from the solution itself, so it cannot disagree with the duty shown beside it. As a design rule, keep F above about 0.75 — below that the curve is so steep that a small error in an inlet temperature swings the required area wildly.

What is a typical overall heat transfer coefficient?

U bundles every resistance in series: 1/U = 1/h_i + R_fi + t/k + R_fo + 1/h_o. Clean-surface ranges for preliminary sizing, in W/m²·K:

Water to water: 850–1700
Steam to water, condensing: 1000–6000
Water to oil: 110–350
Gas to water, air cooler: 25–60
Gas to gas: 10–40

The largest resistance dominates completely, which is why gas-side coefficients set the total and why air-cooled units need so much more surface.

How much capacity does fouling cost?

Deposits add a thermal resistance R_f in series with everything else, so they subtract directly from U. Typical TEMA fouling factors run from 0.0001 m²·K/W for distilled water to 0.0010 for river water and 0.0009 for heavy oil above 80 °C. A clean 350 W/m²·K oil-and-river-water cooler carries 1/U = 0.00286; add about 0.0019 of fouling and U falls to roughly 210 W/m²·K — a 40% loss. This is why fouling is a design decision rather than a maintenance afterthought: the area is deliberately oversized at specification time so the unit still meets duty when dirty.

Why does the tool sometimes report the LMTD as indeterminate?

Because at very high NTU the effectiveness approaches 1, the limiting stream reaches the other stream's inlet temperature, and one of the two end differences closes to zero. Q = U·A·ΔT_lm then becomes 0 × ∞, and the ratio computed from what is left is numerical noise rather than a temperature. Rather than print a confident number derived from cancelled digits, the tool says so. The ε-NTU result beside it is completely unaffected — which is a neat demonstration of why the second method exists at all.