Heat exchangers: LMTD and effectiveness-NTU methods

Why it matters

Heat exchangers move heat between two fluids in radiators, oil coolers, intercoolers, chillers, condensers, battery and drive coolant loops, and process plants. An engineer must either size one for a duty (how much area?) or rate an existing one (what outlet temperatures will it give?). The LMTD method suits the first job, the effectiveness–NTU method the second.

Key ideas

Types. Double-pipe (parallel flow or counterflow), shell-and-tube (with shell and tube passes), and crossflow (compact fin-tube units, car radiators), with each fluid mixed or unmixed. Counterflow is the most effective arrangement: for the same terminal temperatures it has the largest mean temperature difference, and its cold outlet can exceed the hot outlet temperature, which is impossible in parallel flow.

Overall heat-transfer coefficient U. The hot fluid, wall and cold fluid form series resistances. Fouling (scale, deposits, oil films) adds fouling resistances R_f whose values come from your data book or the exchanger standard; they reduce U over time, so designs include a fouling margin.

Energy balance. With no heat loss to the surroundings, the heat given up by the hot fluid equals the heat gained by the cold fluid. The heat-capacity rate C = ṁc_p (W/K) tells how much a stream's temperature changes for a given duty: the fluid with the smaller C (C_min) has the larger temperature change.

LMTD. Because the temperature difference between the fluids varies along the exchanger, the correct mean for Q = UA·ΔT_m is the logarithmic mean of the end differences. Assumptions: steady flow, constant U and c_p, no heat loss to the surroundings, negligible axial conduction and negligible kinetic and potential energy changes. For multipass and crossflow units use ΔT_m = F·LMTD_counterflow, with the correction factor F (≤ 1) from charts in your data book. When one fluid is condensing or boiling at constant temperature, the arrangement no longer matters (F = 1).

Effectiveness–NTU. Effectiveness ε = Q / Q_max, where Q_max = C_min(T_h,in − T_c,in) is the heat an infinitely long counterflow exchanger could transfer. NTU = UA/C_min measures the thermal size of the exchanger. ε depends only on NTU, the capacity ratio C_r = C_min/C_max and the flow arrangement. When only inlet temperatures are known, ε–NTU gives Q directly; LMTD would need iteration.

• For C_r = 0 (one fluid boiling or condensing), every arrangement gives ε = 1 − exp(−NTU).
• Increasing NTU gives diminishing returns: ε saturates, so doubling the area beyond NTU ≈ 3 adds little.

Formulas

Q = ṁ_h c_ph (T_h,in − T_h,out) = ṁ_c c_pc (T_c,out − T_c,in) — Q (W), ṁ (kg/s), c_p (J/kg·K). C = ṁ c_p (W/K), C_r = C_min / C_max 1/(U A) = 1/(h_i A_i) + R_f,i/A_i + ln(r_o/r_i)/(2π k L) + R_f,o/A_o + 1/(h_o A_o) — tube wall; for a thin wall 1/U ≈ 1/h_i + 1/h_o + R_f. Q = U A ΔT_lm ΔT_lm = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂)

• Parallel flow: ΔT₁ = T_h,in − T_c,in, ΔT₂ = T_h,out − T_c,out.
• Counterflow: ΔT₁ = T_h,in − T_c,out, ΔT₂ = T_h,out − T_c,in; if ΔT₁ = ΔT₂, ΔT_lm = ΔT₁. ΔT_m = F · ΔT_lm,cf — multipass or crossflow. ε = Q / Q_max, Q_max = C_min (T_h,in − T_c,in) NTU = U A / C_min ε = [1 − exp(−NTU(1 + C_r))] / (1 + C_r) — parallel flow. ε = [1 − exp(−NTU(1 − C_r))] / [1 − C_r exp(−NTU(1 − C_r))] — counterflow, C_r < 1. ε = NTU / (1 + NTU) — counterflow, C_r = 1. ε = 1 − exp(−NTU) — C_r = 0, any arrangement.

Worked examples

Example 1 (standard, sizing by LMTD). Oil (ṁ = 2 kg/s, c_p = 2100 J/kg·K) is cooled from 110 °C to 70 °C by water (ṁ = 1.5 kg/s, c_p = 4180 J/kg·K) entering at 30 °C. U = 300 W/m²·K. Find the area for counterflow and for parallel flow.

1. Q = ṁ c_p ΔT (oil) = 2 × 2100 × 40 = 168 000 W.
2. Water outlet: 30 + 168 000 / (1.5 × 4180) = 30 + 26.79 = 56.79 °C.
3. Counterflow: ΔT₁ = 110 − 56.79 = 53.21 K; ΔT₂ = 70 − 30 = 40 K; ΔT_lm = 13.21 / ln(1.330) = 46.29 K.
4. A = Q / (U ΔT_lm) = 168 000 / (300 × 46.29) = 12.1 m².
5. Parallel flow: ΔT₁ = 110 − 30 = 80 K; ΔT₂ = 70 − 56.79 = 13.21 K; ΔT_lm = 66.79 / ln(6.056) = 37.08 K; A = 168 000 / (300 × 37.08) = 15.1 m².

A ≈ 12.1 m² (counterflow) versus ≈ 15.1 m² (parallel flow)

Example 2 (GATE level, rating by ε–NTU). A counterflow exchanger has UA = 6000 W/K. Hot water enters at 90 °C with C_h = 4000 W/K; cold water enters at 20 °C with C_c = 8000 W/K. Find Q and both outlet temperatures.

1. C_min = 4000 W/K (hot side), C_r = 0.5; NTU = UA / C_min = 1.5.
2. ε (counterflow) = [1 − exp(−0.75)] / [1 − 0.5 exp(−0.75)] = 0.5276 / 0.7638 = 0.691.
3. Q_max = C_min (T_h,in − T_c,in) = 4000 × 70 = 280 000 W; Q = 0.691 × 280 000 = 193 400 W.
4. T_h,out = 90 − 193 400/4000 = 41.6 °C; T_c,out = 20 + 193 400/8000 = 44.2 °C.
5. Check: the cold outlet (44.2 °C) is above the hot outlet (41.6 °C), possible only in counterflow.

Q ≈ 193 kW, T_h,out ≈ 41.6 °C, T_c,out ≈ 44.2 °C

Common mistakes

• Pairing the wrong end temperatures for counterflow (hot inlet goes with cold outlet).
• Using the arithmetic mean temperature difference; it overestimates Q unless ΔT₁ ≈ ΔT₂.
• Basing Q_max or NTU on C_max instead of C_min.
• Forgetting the F correction for multipass or crossflow units.
• Adding U values instead of resistances when including fouling.
• Mixing kJ and J in ṁc_p, which gives NTU off by a factor of 1000.

For GATE ME

Expect LMTD for parallel and counterflow, the outlet temperature from an energy balance, the area for a given duty, ε–NTU for condensers and evaporators (C_r = 0), counterflow with C_r = 1, and comparisons between arrangements. Practise the special cases quickly; they let many questions be solved in a few lines.

Quick check

1. In counterflow with ΔT₁ = 60 K and ΔT₂ = 20 K, what is ΔT_lm?
2. Which fluid sets Q_max?
3. What is ε for a condenser with NTU = 1?
4. Can the cold outlet temperature exceed the hot outlet temperature in parallel flow?
5. What does fouling do to U and to the required area?

Answers: 1. 40/ln 3 = 36.4 K. 2. The fluid with the smaller ṁc_p (C_min). 3. 1 − e⁻¹ = 0.632. 4. No. 5. It reduces U and increases the required area.