Heat exchangers: LMTD and effectiveness-NTU methods

Heat exchanger types, overall U with fouling, LMTD for parallel flow and counterflow with correction factors, and the effectiveness–NTU method, with sizing and rating numericals.

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Why it matters

A vehicle carries several heat exchangers: the radiator, the charge-air intercooler, the engine-oil and transmission-oil coolers, the AC condenser and evaporator, the heater core and, in EVs, battery chillers. Two methods size and rate them: the LMTD method when all four terminal temperatures are known, and the effectiveness–NTU method when only the inlets are known. Both appear in almost every GATE paper.

Key ideas

Types. Double-pipe (parallel flow or counterflow), shell-and-tube (multiple passes), compact cross-flow exchangers with fins (radiators, intercoolers, condensers), and plate exchangers (oil coolers, battery chillers). In cross-flow, a stream is "mixed" if it can move sideways between passages and "unmixed" if fins or tubes keep it in channels.

Overall heat transfer coefficient. Heat passes through a chain of resistances: inside convection, inside fouling, wall conduction, outside fouling and outside convection. For a tube, U must be referred to a stated area (UA is the same either way). Fouling (scale, oil film, corrosion, dust on radiator fins) adds a resistance R_f (m²·K/W, from data tables) and lowers U over time.

Energy balance. Q = C_h (T_h,in − T_h,out) = C_c (T_c,out − T_c,in), where C = ṁ c_p is the heat capacity rate (W/K). The stream with the smaller C undergoes the larger temperature change.

LMTD method. The temperature difference between the streams varies along the exchanger, and for constant U and specific heats the correct average is the log-mean of the end differences.

  • Parallel flow: ΔT₁ = T_h,in − T_c,in, ΔT₂ = T_h,out − T_c,out. The cold outlet can never exceed the hot outlet.
  • Counterflow: ΔT₁ = T_h,in − T_c,out, ΔT₂ = T_h,out − T_c,in. The cold outlet can exceed the hot outlet. For the same terminal temperatures, counterflow has the larger LMTD and needs the smallest area.
  • If ΔT₁ = ΔT₂ (counterflow with C_h = C_c), LMTD = ΔT₁.
  • Cross-flow and multipass: Q = U A F ΔT_lm,CF, with the correction factor F (≤ 1) from charts.
  • Condensers and evaporators: one stream is at constant temperature, and the flow arrangement no longer matters.

Effectiveness–NTU method. When the outlet temperatures are unknown (rating an existing exchanger, or a change in flow), LMTD needs iteration. Instead:

  • Q_max = C_min (T_h,in − T_c,in): the most heat that could be transferred, in an infinitely long counterflow exchanger.
  • Effectiveness ε = Q/Q_max.
  • NTU = UA/C_min, a measure of size; capacity ratio C_r = C_min/C_max.
  • ε = f(NTU, C_r, flow arrangement), from formulas or charts. Effectiveness rises with NTU but with diminishing returns.
  • When C_r = 0 (condensing or evaporating stream), ε = 1 − exp(−NTU) for every arrangement.

Design trade-off. Higher effectiveness needs more area (cost, weight, pressure drop). Automotive radiators are compact cross-flow units with both fluids unmixed; their ε is read from charts.

Formulas

Q = U A ΔT_lm — Q: W; U: W/m²·K; A: m²; ΔT_lm: K.

ΔT_lm = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂) — end differences defined as above for parallel flow or counterflow.

1/(UA) = 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 with fouling.

C = ṁ c_p (W/K); C_r = C_min / C_max; NTU = U A / C_min.

Q_max = C_min (T_h,in − T_c,in); ε = Q / Q_max.

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 with C_r = 1.

ε = 1 − exp(−NTU) — any arrangement with C_r = 0.

Worked examples

Example 1 (standard, LMTD sizing). An engine-oil cooler cools 0.5 kg/s of oil (c_p = 2.1 kJ/kg·K) from 120 °C to 80 °C using 0.6 kg/s of water (c_p = 4.18 kJ/kg·K) entering at 30 °C. U = 300 W/m²·K. Find the area for counterflow and for parallel flow.

  1. Duty: Q = ṁ_h c_p,h ΔT_h = 0.5 × 2100 × 40 = 42.0 kW.
  2. Water outlet: T_c,out = 30 + 42 000/(0.6 × 4180) = 30 + 16.75 = 46.75 °C.
  3. Counterflow: ΔT₁ = 120 − 46.75 = 73.25 K; ΔT₂ = 80 − 30 = 50 K. ΔT_lm = 23.25/ln(1.465) = 60.9 K.
  4. A = Q/(U ΔT_lm) = 42 000/(300 × 60.9) = 2.30 m².
  5. Parallel flow: ΔT₁ = 120 − 30 = 90 K; ΔT₂ = 80 − 46.75 = 33.25 K; ΔT_lm = 56.75/ln(2.707) = 57.0 K; A = 42 000/(300 × 57.0) = 2.46 m², about 7% more.

Example 2 (GATE level, ε–NTU rating). The counterflow cooler of Example 1 (UA = 300 × 2.3 = 690 W/K) now receives 0.8 kg/s of oil at 120 °C, with the same water flow at 30 °C. Find the heat duty and both outlet temperatures.

  1. C_h = 0.8 × 2100 = 1680 W/K; C_c = 0.6 × 4180 = 2508 W/K. C_min = 1680 W/K (oil), C_r = 1680/2508 = 0.670.
  2. NTU = UA/C_min = 690/1680 = 0.411.
  3. exp(−NTU(1 − C_r)) = exp(−0.411 × 0.330) = exp(−0.1356) = 0.873.
  4. ε (counterflow) = (1 − 0.873)/(1 − 0.670 × 0.873) = 0.127/0.415 = 0.305.
  5. Q = ε C_min (T_h,in − T_c,in) = 0.305 × 1680 × 90 = 46.2 kW.
  6. Oil outlet = 120 − 46 190/1680 = 92.5 °C; water outlet = 30 + 46 190/2508 = 48.4 °C.

More oil flow raised the duty only from 42 to 46 kW, and the oil now leaves much hotter; the exchanger is too small for that flow.

Common mistakes

  • Using the parallel-flow end differences for a counterflow exchanger (or vice versa).
  • Using the arithmetic mean temperature difference; it always overestimates the driving force.
  • Taking Q_max = C_max ΔT_max; it must be C_min.
  • Forgetting the correction factor F for cross-flow or multipass exchangers.
  • Assuming flow arrangement matters for a condenser or evaporator; with C_r = 0 it does not.
  • Mixing kJ and J in C = ṁ c_p.

For GATE ME

Expect LMTD calculations for parallel flow and counterflow, area or length of a double-pipe exchanger, finding an unknown outlet temperature from the energy balance, ε–NTU with C_r = 0 (condensers and evaporators) or C_r = 1, and MCQs on why counterflow is better, the meaning of NTU and effectiveness, and fouling. Practise writing both energy balances before any LMTD.

Quick check

  1. End temperature differences are 30 K and 10 K. What is the LMTD?
  2. U = 500 W/m²·K, A = 10 m², LMTD = 20 K. What is Q?
  3. A steam condenser has NTU = 2. What is its effectiveness?
  4. In a counterflow exchanger with C_h = C_c, how do the end temperature differences compare?

Answers: 1. 20/ln 3 = 18.2 K. 2. 100 kW. 3. 1 − e⁻² = 0.865. 4. They are equal, so the LMTD equals either one.

Interview questions

All Thermodynamics and Heat Transfer interview questions

Try answering each one aloud before you open it.

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

    The Log Mean Temperature Difference (LMTD) is a measure used to determine the temperature driving force for heat exchange in flow systems, such as heat exchangers. It is defined as the logarithmic average of the temperature difference between the hot and cold streams at each end of the heat exchanger. LMTD is used in the design and analysis of heat exchangers to calculate the heat transfer rate.

  2. 2.Explain the effectiveness-NTU method used in heat exchangers.Concept

    The effectiveness-NTU method is a technique used to analyze heat exchangers. It relates the actual heat transfer to the maximum possible heat transfer. The effectiveness (ε) is defined as the ratio of the actual heat transfer to the maximum possible heat transfer. The Number of Transfer Units (NTU) is a dimensionless parameter that represents the size of the heat exchanger. This method is particularly useful when the inlet and outlet temperatures are not known.

  3. 3.Why is the LMTD method preferred over the effectiveness-NTU method in certain situations?Application

    The LMTD method is preferred when the inlet and outlet temperatures of both fluids are known, as it provides a straightforward calculation of the heat transfer rate. It is particularly useful in the design phase of heat exchangers where these temperatures are specified. In contrast, the effectiveness-NTU method is more suitable when the temperatures are not known, such as in performance analysis.

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

    In a counterflow heat exchanger, the hot and cold fluids move in opposite directions, which allows for a higher temperature difference across the heat exchanger and thus more efficient heat transfer. If the flow arrangement is changed to parallel flow, where both fluids move in the same direction, the temperature difference between the fluids decreases along the length of the heat exchanger, reducing the overall heat transfer efficiency.

  5. 5.How does fouling affect the performance of a heat exchanger?Application

    Fouling refers to the accumulation of unwanted materials on the heat transfer surfaces, which acts as an additional thermal resistance. This reduces the overall heat transfer coefficient, leading to decreased efficiency and performance of the heat exchanger. Fouling can also increase pressure drop and operational costs due to the need for more frequent cleaning and maintenance.

  6. 6.What is the significance of the NTU in the effectiveness-NTU method?Concept

    The Number of Transfer Units (NTU) is a dimensionless parameter that indicates the size of the heat exchanger relative to the heat capacity rate of the fluids. A higher NTU value generally means a more effective heat exchanger, as it implies a larger surface area for heat transfer. NTU is crucial in determining the effectiveness of a heat exchanger when using the effectiveness-NTU method.

  7. 7.Calculate the LMTD for a counterflow heat exchanger with inlet temperatures of 150°C for the hot fluid and 30°C for the cold fluid, and outlet temperatures of 80°C for the hot fluid and 60°C for the cold fluid.Numerical

    For counterflow the end differences pair the hot inlet with the cold outlet and the hot outlet with the cold inlet: ΔT₁ = 150 − 60 = 90 K and ΔT₂ = 80 − 30 = 50 K. Then LMTD = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = 40/ln(1.8) = 40/0.5878 ≈ 68.1 K. As a check, the LMTD must lie between the two end differences, and it is a little below their arithmetic mean of 70 K.

  8. 8.What are the advantages of using a shell-and-tube heat exchanger?Application

    Shell-and-tube heat exchangers are versatile and can handle a wide range of temperatures and pressures. They are robust and can be designed to accommodate high-pressure applications. Additionally, they offer a large surface area for heat transfer and can be easily cleaned and maintained. These features make them suitable for various industrial applications.

  9. 9.Explain how the heat capacity rate affects the performance of a heat exchanger.Concept

    The heat capacity rate is the product of the mass flow rate and the specific heat capacity of a fluid. It determines how much heat a fluid can carry. In a heat exchanger, the fluid with the lower heat capacity rate will experience a larger temperature change. The performance of the heat exchanger is influenced by the balance between the heat capacity rates of the hot and cold fluids, affecting the overall heat transfer efficiency.

  10. 10.A heat exchanger has an effectiveness of 0.75. If the maximum possible heat transfer is 2000 W, what is the actual heat transfer?Numerical

    The actual heat transfer can be calculated using the formula: Actual Heat Transfer = Effectiveness × Maximum Possible Heat Transfer. Given that the effectiveness is 0.75 and the maximum possible heat transfer is 2000 W, the actual heat transfer is 0.75 × 2000 W = 1500 W.

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