LMTD method for double-pipe and shell-and-tube exchangers

Energy balances, derivation and use of the log mean temperature difference, counterflow versus parallel flow, and the F correction factor for multipass shell-and-tube exchangers.

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Why it matters

Sizing an exchanger means answering one question: how much area is needed to transfer a known duty between two streams whose terminal temperatures are specified? The log mean temperature difference (LMTD), with a correction factor for multipass shell-and-tube units, is the standard tool. It also explains why counterflow is preferred and why a "temperature cross" makes a single shell impossible.

Key ideas

Energy balances first. With no heat loss and no phase change, q = ṁ_h·c_ph·(T_h,in − T_h,out) = ṁ_c·c_pc·(T_c,out − T_c,in). These fix the duty and any unknown outlet temperature before the rate equation is used. With phase change, use ṁ·h_fg for the condensing or boiling stream.

Why a log mean. Along the exchanger the local temperature difference ΔT changes. Writing dq = U·dA·ΔT and the two energy balances, and integrating with constant U and c_p, gives q = U·A·ΔT_lm, where ΔT_lm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) and ΔT₁, ΔT₂ are the differences at the two ends of the exchanger. The arithmetic mean always overestimates the true mean (it is within about 1 % when ΔT₁/ΔT₂ < 1.5). When ΔT₁ = ΔT₂, ΔT_lm equals that value.

Counterflow vs parallel flow (double-pipe exchanger).

  • Parallel (co-current): both fluids enter at the same end; ΔT₁ = T_h,in − T_c,in, ΔT₂ = T_h,out − T_c,out. The cold outlet can never exceed the hot outlet. Large ΔT at the inlet, small at the outlet.
  • Counterflow: ΔT₁ = T_h,in − T_c,out, ΔT₂ = T_h,out − T_c,in. For the same terminal temperatures the LMTD is always larger than parallel flow, so less area is needed, and the cold outlet can exceed the hot outlet (temperature cross within one pass is possible).
  • When one stream changes phase at constant temperature (condensing steam, boiling liquid), the arrangement does not matter: both give the same LMTD.

Assumptions. Steady state; no heat loss; constant U along the exchanger; constant specific heats; no axial conduction; each stream at a uniform temperature over any cross-section.

Multipass and cross-flow exchangers. In a 1-shell-pass, 2-tube-pass (1-2) exchanger the tube fluid flows partly co-current and partly counter-current to the shell fluid. The true mean ΔT is F × (LMTD computed as if counterflow). F (≤ 1) is read from charts or formulas in terms of

  • P = (t₂ − t₁)/(T₁ − t₁) — the thermal effectiveness of the tube stream (t tube side, T shell side), and
  • R = (T₁ − T₂)/(t₂ − t₁) — the heat-capacity-rate ratio of tube to shell stream. F is the same whichever fluid is on the shell side. As a design rule, F should be above about 0.75–0.8; below that, the design is sensitive to small changes and more shells in series are used. When one stream is isothermal, F = 1.

Temperature cross. If the cold outlet temperature exceeds the hot outlet temperature, a single 1-2 shell often cannot achieve the duty (F falls steeply or the log becomes undefined); use two or more shells in series or a true counterflow unit.

Formulas

q = ṁ_h·c_ph·(T_h,in − T_h,out) = ṁ_c·c_pc·(T_c,out − T_c,in) Energy balance; ṁ (kg/s), c_p (J/kg·K), q (W).

ΔT_lm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) LMTD (K or °C); ΔT₁, ΔT₂ end temperature differences.

Counterflow: ΔT₁ = T_h,in − T_c,out, ΔT₂ = T_h,out − T_c,in Parallel: ΔT₁ = T_h,in − T_c,in, ΔT₂ = T_h,out − T_c,out

q = U·A·F·ΔT_lm,cf Design equation; F correction factor (–), ΔT_lm,cf the counterflow LMTD.

P = (t₂ − t₁)/(T₁ − t₁), R = (T₁ − T₂)/(t₂ − t₁) Parameters for F (t: tube side, T: shell side).

F = [√(R² + 1)/(R − 1)]·ln[(1 − P)/(1 − P·R)] / ln{[2 − P·(R + 1 − √(R² + 1))]/[2 − P·(R + 1 + √(R² + 1))]} 1-2 (and 1-2n) exchanger, R ≠ 1.

Worked examples

Example 1 (standard). Oil (ṁ = 2 kg/s, c_p = 2.2 kJ/kg·K) is cooled from 150 °C to 90 °C by water (ṁ = 1.5 kg/s, c_p = 4.18 kJ/kg·K) entering at 25 °C in a double-pipe exchanger with U = 350 W/m²·K. Find the water outlet temperature and the area for counterflow and for parallel flow.

  1. Duty: q = 2 × 2200 × (150 − 90) = 264 000 W.
  2. Water outlet: T_c,out = 25 + 264 000/(1.5 × 4180) = 25 + 42.1 = 67.1 °C.
  3. Counterflow: ΔT₁ = 150 − 67.1 = 82.9 K; ΔT₂ = 90 − 25 = 65.0 K; ΔT_lm = 17.9/ln(1.2753) = 73.6 K.
  4. A_cf = q/(U·ΔT_lm) = 264 000/(350 × 73.6) = 10.25 m².
  5. Parallel: ΔT₁ = 150 − 25 = 125 K; ΔT₂ = 90 − 67.1 = 22.9 K; ΔT_lm = 102.1/ln(5.460) = 60.2 K.
  6. A_pf = 264 000/(350 × 60.2) = 12.54 m².

Answer: T_c,out ≈ 67.1 °C; A ≈ 10.3 m² counterflow vs 12.5 m² parallel flow (22 % more).

Example 2 (GATE level). A 1-2 shell-and-tube exchanger must transfer 500 kW. The shell-side fluid cools from 200 °C to 120 °C; the tube-side fluid heats from 40 °C to 100 °C. U = 400 W/m²·K. Find P, R, F and the area; compare with a true counterflow unit.

  1. Counterflow LMTD: ΔT₁ = 200 − 100 = 100 K, ΔT₂ = 120 − 40 = 80 K → ΔT_lm = 20/ln(1.25) = 89.6 K.
  2. P = (t₂ − t₁)/(T₁ − t₁) = 60/160 = 0.375; R = (T₁ − T₂)/(t₂ − t₁) = 80/60 = 1.333.
  3. √(R² + 1) = 1.6667. Numerator: (1.6667/0.3333) × ln(0.625/0.5) = 5.0 × 0.2231 = 1.1157.
  4. Denominator: ln{[2 − 0.375 × (2.3333 − 1.6667)]/[2 − 0.375 × (2.3333 + 1.6667)]} = ln(1.75/0.5) = 1.2528.
  5. F = 1.1157/1.2528 = 0.891 (above 0.8, acceptable).
  6. A = q/(U·F·ΔT_lm) = 500 000/(400 × 0.891 × 89.6) = 15.66 m². Pure counterflow would need 500 000/(400 × 89.6) = 13.95 m².

Answer: P = 0.375, R = 1.33, F ≈ 0.89, A ≈ 15.7 m² (about 12 % more than true counterflow).

Common mistakes

  • Pairing the wrong terminal temperatures: in counterflow, hot-in goes with cold-out.
  • Applying F to a parallel-flow LMTD; F always multiplies the counterflow LMTD.
  • Using the arithmetic mean when the end differences are very unequal.
  • Forgetting that with one isothermal stream (condensing steam) F = 1 and flow direction does not matter.
  • Swapping P and R definitions or mixing shell and tube temperatures in them.
  • Using LMTD when outlet temperatures are unknown (rating problems) — the ε-NTU method avoids iteration.

For GATE CH

Expect energy-balance-plus-LMTD area calculations, counter vs parallel comparisons, outlet temperature from a given area (sometimes requiring iteration or ε-NTU), F-factor use with given P and R, and conceptual questions on temperature cross and isothermal streams. Practise sketching the temperature profiles for both arrangements and labelling ΔT₁ and ΔT₂ before writing any formula.

Quick check

  1. End differences 40 K and 20 K. LMTD?
  2. End differences both 30 K. LMTD?
  3. Steam condenses at 120 °C heating water from 30 °C to 80 °C. LMTD?
  4. Which arrangement allows the cold outlet to exceed the hot outlet?
  5. F for a 1-2 exchanger when one fluid condenses?

Answers: 1. 28.9 K. 2. 30 K. 3. 50/ln(90/40) = 61.7 K. 4. Counterflow. 5. 1.

Interview questions

All Heat Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is the Log Mean Temperature Difference (LMTD) method in heat exchangers?Concept

    The Log Mean Temperature Difference (LMTD) method is a mathematical approach used to determine the temperature driving force for heat exchange in flow systems, particularly in heat exchangers. It is used to calculate the average temperature difference between the hot and cold fluids across the heat exchanger. The LMTD is especially useful for designing and analyzing heat exchangers where the temperature difference varies along the length of the exchanger.

  2. 2.Explain the significance of LMTD in the design of double-pipe heat exchangers.Concept

    In double-pipe heat exchangers, the LMTD is crucial for determining the heat transfer rate. It provides a more accurate representation of the temperature difference between the fluids compared to using a simple arithmetic mean. This is because the temperature difference between the fluids changes along the length of the exchanger. By using LMTD, engineers can design more efficient heat exchangers by ensuring that the heat transfer surface area is optimized for the actual temperature profile.

  3. 3.How is the LMTD calculated for a counterflow heat exchanger?Concept

    For a counterflow heat exchanger, the LMTD is calculated using the formula: LMTD = (ΔT1 - ΔT2) / ln(ΔT1/ΔT2), where ΔT1 is the temperature difference between the hot and cold fluids at one end, and ΔT2 is the temperature difference at the other end. This formula accounts for the logarithmic nature of the temperature change along the heat exchanger.

  4. 4.Why is the LMTD method preferred over the arithmetic mean temperature difference in heat exchanger analysis?Application

    The LMTD method is preferred because it provides a more accurate representation of the temperature driving force in heat exchangers where the temperature difference between the fluids is not constant. The arithmetic mean temperature difference assumes a linear temperature profile, which is often not the case in real-world applications. The LMTD accounts for the exponential nature of heat transfer, leading to more precise calculations and efficient designs.

  5. 5.What happens if the flow arrangement in a heat exchanger is changed from counterflow to parallel flow?Application

    If the flow arrangement is changed from counterflow to parallel flow, the LMTD will generally decrease. In parallel flow, both fluids enter the heat exchanger at the same end and flow in the same direction, leading to a smaller temperature difference between the fluids along the length of the exchanger. This results in a lower heat transfer rate compared to counterflow, where the temperature difference is maintained over a larger portion of the exchanger.

  6. 6.Describe how the LMTD method is applied in shell-and-tube heat exchangers.Concept

    In shell-and-tube heat exchangers, the LMTD method is used to calculate the average temperature difference between the shell-side and tube-side fluids. Due to the complex flow patterns, a correction factor (F) is often applied to the LMTD to account for deviations from ideal counterflow conditions. The corrected LMTD is then used to determine the required heat transfer area and other design parameters.

  7. 7.Why is a correction factor needed when using the LMTD method for shell-and-tube heat exchangers?Application

    A correction factor is needed because shell-and-tube heat exchangers often do not operate under ideal counterflow conditions. The presence of baffles, multiple shell passes, and other design features can create complex flow patterns that deviate from simple counterflow or parallel flow. The correction factor adjusts the LMTD to account for these deviations, ensuring accurate heat transfer calculations.

  8. 8.Calculate the LMTD for a counterflow heat exchanger where the inlet temperatures are 150°C for the hot fluid and 50°C for the cold fluid, and the outlet temperatures are 100°C for the hot fluid and 80°C for the cold fluid.Numerical

    First, calculate the temperature differences: ΔT1 = 150°C - 80°C = 70°C and ΔT2 = 100°C - 50°C = 50°C. Then, use the LMTD formula: LMTD = (ΔT1 - ΔT2) / ln(ΔT1/ΔT2) = (70 - 50) / ln(70/50) = 20 / ln(1.4) ≈ 20 / 0.3365 ≈ 59.45°C.

  9. 9.What are the limitations of the LMTD method in heat exchanger analysis?Application

    The LMTD method assumes steady-state conditions and constant specific heat capacities, which may not be valid in all situations. It also requires knowledge of the inlet and outlet temperatures, which may not always be available. Additionally, the method can become complex when dealing with multi-pass or cross-flow heat exchangers, requiring correction factors to account for deviations from ideal flow patterns.

  10. 10.A shell-and-tube heat exchanger has a correction factor of 0.85. If the calculated LMTD is 60°C, what is the corrected LMTD?Numerical

    The corrected LMTD is calculated by multiplying the LMTD by the correction factor. Therefore, the corrected LMTD = 60°C × 0.85 = 51°C.

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