Heat Exchangers: LMTD and NTU

Types of exchanger, the log-mean temperature difference, the effectiveness–NTU method, fouling and a complete oil-cooler example.

On this page

  1. Types of heat exchanger
  2. The energy balance
  3. The log-mean temperature difference
  4. Sizing an oil cooler
  5. The effectiveness–NTU method
  6. Fouling and U
  7. Condensers and evaporators
  8. FAQ

Types of heat exchanger

TypeDescriptionTypical use
Double pipeone pipe inside another, simple counterflowsmall duties, high pressure, fouling fluids
Shell and tubea bundle of tubes inside a shell with baffles; the workhorse of industryprocess, oil, steam condensers, power
Platethin corrugated plates clamped or brazed, very high U in a small spaceheating and cooling water, food, HVAC, district heating
Finned-tube (air-cooled)liquid in tubes, air blown across finsradiators, condensers, coolers
Spiral, scraped surface, regeneratorsspecialist designs for viscous or fouling streams, or periodic heat storageslurries, food, gas turbines, glass furnaces

The streams can run in the same direction (parallel flow), in opposite directions (counterflow), or at right angles (crossflow). A shell-and-tube unit with several passes combines them, and its performance is read from a counterflow calculation with a correction factor F below 1.

The energy balance

With no loss to the surroundings, the heat given up by the hot stream is taken up by the cold stream (for a single phase, with the heat capacity rate C = ṁ cp in W/K):

Q̇ = Ch (Th,in − Th,out) = Cc (Tc,out − Tc,in) = U A ΔTm

The log-mean temperature difference

The temperature difference between the streams varies along the exchanger, so the right mean for U A ΔTm is the log-mean temperature difference, with ΔT₁ and ΔT₂ the differences at the two ends:

ΔTlm = (ΔT₁ − ΔT₂) ÷ ln(ΔT₁ ÷ ΔT₂)

Counterflow and parallel-flow temperature profiles0.00.51.00255075100position along the exchangertemperature (°C)Counterflowoil in 90oil out 54.5water in 20water out 42.70.00.51.00255075100position along the exchangertemperature (°C)Parallel flowoil in 90oil out 57.5water out 40.8water in 20Same exchanger (A = 10 m², U = 350 W/m²K): counterflow transfers 142.0 kW, parallel flow only 130.2 kW.Counterflow is always at least as effective.
Temperature of oil and water along the same exchanger in counterflow and in parallel flow.

Sizing an oil cooler

Worked example: how much area to cool the oil?

  • The job: 2 kg/s of oil (cp = 2 000 J/kg·K, Ch = 4 000 W/K) must be cooled from 90 °C to 60 °C by 1.5 kg/s of water at 20 °C (Cc = 6 270 W/K), with U = 350 W/m²K.
  • Duty: Q̇ = Ch ΔTh = 4 000 × 30 = 120 kW. The water warms to 20 + 120 000 ÷ 6 270 = 39.1 °C.
  • Counterflow: ΔT₁ = 90 − 39.1 = 50.9 K, ΔT₂ = 60 − 20 = 40 K, so ΔTlm = 45.2 K (the arithmetic mean 45.4 K is close). Area A = Q̇ ÷ (U ΔTlm) = 120 000 ÷ (350 × 45.2) = 7.6 m².
  • Parallel flow: the same duty gives ΔTlm = 40.6 K, so the area is 8.4 m²: 11 percent more. In parallel flow the outlet temperatures cannot cross, so the water could never be heated above the oil’s outlet temperature; counterflow can.

The effectiveness–NTU method

When the outlet temperatures are not known (rating an existing exchanger) the LMTD needs iteration. The effectiveness–NTU method avoids that. The effectiveness ε is the heat transferred divided by the most that could be transferred, Cmin (Th,in − Tc,in); the number of transfer units is NTU = U A ÷ Cmin; and the capacity ratio is Cr = Cmin ÷ Cmax. For counterflow:

ε = [1 − e−NTU (1 − Cr)] ÷ [1 − Cr e−NTU (1 − Cr)]    parallel: ε = [1 − e−NTU (1 + Cr)] ÷ (1 + Cr)

Effectiveness against NTU0123450.00.20.40.60.81.0NTU = U A ÷ C_mineffectiveness εcounter C_r=0.0counter C_r=0.5counter C_r=1.0parallel C_r=0.5parallel C_r=1.0worked example: NTU 0.88, C_r 0.64, ε 0.51Solid lines: counterflow. Dashed: parallel flow. For C_r = 0 (boiling or condensing on one side) both give ε = 1− e^(−NTU). Beyond NTU of about 3 more area helps little.
Effectiveness against NTU. Counterflow (solid) outperforms parallel flow (dashed).

Worked example: rating an existing cooler

  • The oil cooler has A = 10 m² and U = 350 W/m²K (U A = 3 500 W/K). Oil in at 90 °C (Ch = 4 000 W/K is the minimum), water in at 20 °C. What does it do?
  • NTU = U A ÷ Cmin = 3 500 ÷ 4 000 = 0.875 and Cr = 0.638. Counterflow effectiveness ε = 0.507.
  • Q̇ = ε Cmin (Th,in − Tc,in) = 0.507 × 4 000 × 70 = 142.0 kW; the oil leaves at 54.5 °C and the water at 42.7 °C.
  • Run in parallel flow the same exchanger has ε = 0.465 and transfers only 130.2 kW: connecting the pipes the wrong way costs 8 percent of the duty.

Fouling and U

The overall coefficient combines all the resistances between the streams: 1 ÷ U = 1 ÷ hi + Rf,i + Rwall + Rf,o + 1 ÷ ho. Over time scale, rust, sludge, biological growth and coke deposit on the surfaces and add fouling resistance Rf. Typical values (m²·K/W): clean treated water 0.0001, river or cooling-tower water 0.0002 to 0.0005, fuel oil 0.0009, heavy fouling streams 0.001 and more.

In the oil cooler, adding a total fouling resistance of 0.0004 m²K/W turns U = 350 into 307 W/m²K. The same duty then needs 8.6 m², 14 percent more than the clean 7.6 m². Designers add that margin and the plant cleans the exchanger when the temperatures drift. A rising approach temperature or pressure drop is the signal.

Condensers and evaporators

When one stream changes phase at constant temperature (steam condensing, a refrigerant evaporating) its C is effectively infinite, Cr = 0, and the stream stays at the saturation temperature. The LMTD formula still works with that fixed temperature.

Worked example: a steam condenser

  • Cooling water, 5 kg/s, enters at 15 °C and leaves at 35 °C. Duty Q̇ = 5 × 4 180 × 20 = 418 kW, condensing 667 kg/h of steam (latent heat 2 257 kJ/kg).
  • Steam at 100 °C: ΔT₁ = 85 K, ΔT₂ = 65 K, ΔTlm = 74.6 K. With U = 2 000 W/m²K, A = 418 000 ÷ (2 000 × 74.6) = 2.80 m². NTU = 0.27 and ε = 1 − e−NTU = 0.24.

Related

FAQ

Which is better, LMTD or NTU?

They are the same physics. Use LMTD when the inlet and outlet temperatures are known (sizing), and NTU when you know the exchanger and the inlets and want the outlets (rating).

Why does counterflow outperform parallel flow?

Because the coldest cold fluid meets the coldest hot fluid, keeping a temperature difference along the whole length. In parallel flow the difference is large at the entry and tiny at the exit, so much of the area does little.

What is the correction factor F?

Multi-pass shell-and-tube and crossflow exchangers are not pure counterflow, so the effective temperature difference is F × LMTD(counterflow), with F from charts (about 0.8 to 1). Below about 0.75 use more shells in series.

Can the cold stream leave hotter than the hot stream?

In counterflow, yes: this “temperature cross” happens when the cold stream has a lower heat capacity rate than the hot one, and it is one of the advantages of counterflow. In parallel flow the outlets can approach each other but never cross.

References

  • Kakaç, S., Liu, H. and Pramuanjaroenkij, A., Heat Exchangers: Selection, Rating, and Thermal Design, CRC Press.
  • Incropera, F. P., et al., Fundamentals of Heat and Mass Transfer, Wiley; Kays, W. M. and London, A. L., Compact Heat Exchangers.
  • TEMA Standards (Tubular Exchanger Manufacturers Association) for fouling factors and design; ASME Section VIII.
  • Hewitt, G. F. (ed.), Heat Exchanger Design Handbook (HEDH).

Last reviewed: October 2026. Coefficients and fouling factors are typical.