Effectiveness-NTU method

Heat capacity rates, effectiveness, NTU and the epsilon-NTU relations for counterflow, parallel flow and phase change, applied to design and rating problems.

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Why it matters

The LMTD method is ideal when all four terminal temperatures are known and the area is wanted. In the plant the question is often the reverse: "this exchanger already exists — what outlet temperatures will it give if the flow or inlet temperature changes?" The effectiveness–NTU method answers such rating problems directly, without trial and error, and gives a clear picture of how much more area can actually buy.

Key ideas

Heat capacity rates. C_h = ṁ_h·c_ph and C_c = ṁ_c·c_pc (W/K). The smaller one, C_min, belongs to the fluid that undergoes the larger temperature change. The capacity ratio is C_r = C_min/C_max (0 ≤ C_r ≤ 1). A condensing or boiling stream has effectively infinite C, so C_r = 0.

Maximum possible heat transfer. The largest temperature change any stream could have is the inlet-to-inlet difference T_h,in − T_c,in, and only the C_min fluid could achieve it (in an infinitely long counterflow exchanger). So q_max = C_min·(T_h,in − T_c,in).

Effectiveness ε = q/q_max (0 to 1). Once ε is known, q follows from inlet temperatures alone, and the outlet temperatures from energy balances.

Number of transfer units NTU = U·A/C_min. A dimensionless measure of exchanger "thermal size": how large the conductance UA is relative to the stream's capacity to absorb heat. Larger NTU → higher ε, with diminishing returns.

ε = f(NTU, C_r, flow arrangement). Derived from the same differential balances as the LMTD; the two methods give identical answers when used consistently.

  • C_r = 0 (condensing or boiling on one side): ε = 1 − exp(−NTU) for every arrangement.
  • Counterflow gives the highest ε for given NTU and C_r; it tends to 1 as NTU → ∞ for any C_r ≤ 1.
  • Parallel flow: ε can never exceed 1/(1 + C_r) (0.5 for balanced streams), however large the exchanger.
  • Balanced counterflow (C_r = 1): ε = NTU/(1 + NTU).
  • Shell-and-tube and cross-flow units have their own formulas and charts (take them from your data book).

Design vs rating.

  • Design (sizing): outlet temperatures known → find ε from the duty, then NTU from the inverse relation (or chart), then A = NTU·C_min/U. LMTD is equally convenient here.
  • Rating (performance): A, U and inlet conditions known → NTU → ε → q → outlets. LMTD would need iteration here.

Diminishing returns. Because ε rises steeply at low NTU and then flattens, exchangers are usually designed at NTU of about 1–3; going from ε = 0.9 to 0.95 can nearly double the area.

Formulas

C = ṁ·c_p, C_r = C_min/C_max Heat capacity rate (W/K); ratio (–).

q_max = C_min·(T_h,in − T_c,in), ε = q/q_max Maximum heat rate (W), effectiveness (–).

NTU = U·A/C_min U (W/m²·K), A (m²).

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·(1 + C_r))]/(1 + C_r) (parallel flow) ε = 1 − exp(−NTU) (any arrangement, C_r = 0)

NTU = [1/(C_r − 1)]·ln[(ε − 1)/(ε·C_r − 1)] (counterflow, C_r < 1) NTU = −ln(1 − ε) (C_r = 0) Inverse forms for design.

T_c,out = T_c,in + q/C_c, T_h,out = T_h,in − q/C_h Outlet temperatures from energy balances.

Worked examples

Example 1 (standard, rating). An existing counterflow exchanger (U = 500 W/m²·K, A = 8 m²) heats water (0.5 kg/s, c_p = 4180 J/kg·K) entering at 20 °C using oil (1.2 kg/s, c_p = 2100 J/kg·K) entering at 160 °C. Find the duty and both outlet temperatures.

  1. C_c = 0.5 × 4180 = 2090 W/K; C_h = 1.2 × 2100 = 2520 W/K → C_min = 2090 W/K (water), C_r = 0.829.
  2. NTU = U·A/C_min = 500 × 8/2090 = 1.914.
  3. Counterflow: NTU(1 − C_r) = 0.3266; exp(−0.3266) = 0.7214. ε = (1 − 0.7214)/(1 − 0.829 × 0.7214) = 0.2786/0.4020 = 0.6936.
  4. q_max = 2090 × (160 − 20) = 292 600 W; q = 0.6936 × 292 600 = 202 900 W.
  5. T_c,out = 20 + 202 900/2090 = 117.1 °C; T_h,out = 160 − 202 900/2520 = 79.5 °C.
  6. Check with LMTD: ΔT₁ = 160 − 117.1 = 42.9 K, ΔT₂ = 79.5 − 20 = 59.5 K, ΔT_lm = 50.7 K; U·A·ΔT_lm = 4000 × 50.7 ≈ 202.9 kW. ✓

Answer: q ≈ 203 kW; water leaves at ≈ 117 °C, oil at ≈ 79.5 °C. (The water outlet exceeds the oil outlet — possible only in counterflow.)

Example 2 (GATE level). Steam condenses at 100 °C on the shell side of a condenser; cooling water (1 kg/s, c_p = 4180 J/kg·K) enters the tubes at 25 °C. U = 2000 W/m²·K, A = 3 m². (a) Find the water outlet temperature and the duty. (b) If the area is doubled, by what factor does the duty rise?

  1. Condensing steam: C_r = 0, C_min = C_water = 4180 W/K.
  2. NTU = 2000 × 3/4180 = 1.435; ε = 1 − exp(−1.435) = 0.762.
  3. T_c,out = 25 + 0.762 × (100 − 25) = 82.1 °C; q = 0.762 × 4180 × 75 = 238.9 kW.
  4. Doubled area: NTU = 2.871; ε = 1 − exp(−2.871) = 0.943; T_c,out = 95.8 °C.
  5. Ratio of duties = 0.943/0.762 = 1.24.

Answer: (a) 82.1 °C and ≈ 239 kW; (b) doubling the area raises the duty by only about 24 %.

Common mistakes

  • Using C_max (or the hot stream by default) in NTU or q_max; both use C_min.
  • Defining q_max with an outlet temperature; it uses the two inlet temperatures.
  • Using the counterflow formula for C_r = 1 (it becomes 0/0); use NTU/(1 + NTU).
  • Forgetting that with phase change C_r = 0 and the arrangement does not matter.
  • Expecting a parallel-flow exchanger to reach ε above 1/(1 + C_r).
  • Mixing units: C in W/K with c_p in kJ/kg·K.

For GATE CH

Expect rating problems (outlet temperature or duty of a given exchanger), condenser/evaporator problems with C_r = 0, balanced counterflow, ratio questions (what happens when area or flow changes), and identification of C_min and q_max. Practise computing ε for counterflow and parallel flow quickly, and cross-checking one answer with the LMTD.

Quick check

  1. C_h = 3000 W/K, C_c = 1500 W/K. Which is C_min and what is C_r?
  2. NTU = 1 with condensing steam. ε?
  3. Balanced counterflow with NTU = 3. ε?
  4. Maximum possible ε for balanced parallel flow?
  5. U = 250 W/m²·K, A = 12 m², C_min = 1000 W/K. NTU?

Answers: 1. C_min = 1500 W/K (cold), C_r = 0.5. 2. 0.632. 3. 0.75. 4. 0.5. 5. 3.

Interview questions

All Heat Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is the Effectiveness-NTU method in heat exchanger analysis?Concept

    The Effectiveness-NTU method is a technique used to evaluate the performance of heat exchangers. It relates the actual heat transfer to the maximum possible heat transfer. The method uses two key parameters: effectiveness (ε), which is the ratio of actual heat transfer to the maximum possible heat transfer, and the Number of Transfer Units (NTU), which is a dimensionless parameter that represents the size of the heat exchanger relative to the heat capacity rates of the fluids.

  2. 2.Explain the significance of the Number of Transfer Units (NTU) in the Effectiveness-NTU method.Concept

    The Number of Transfer Units (NTU) is a dimensionless parameter that indicates the effectiveness of a heat exchanger. It is defined as the ratio of the heat exchanger's thermal conductance to the minimum heat capacity rate of the fluids. A higher NTU value generally indicates a more effective heat exchanger, as it suggests a larger surface area for heat transfer relative to the fluid flow rates.

  3. 3.How is the effectiveness (ε) of a heat exchanger defined in the Effectiveness-NTU method?Concept

    In the Effectiveness-NTU method, the effectiveness (ε) of a heat exchanger is defined as the ratio of the actual heat transfer to the maximum possible heat transfer. Mathematically, it is expressed as ε = Q_actual / Q_max, where Q_actual is the heat transfer rate achieved by the heat exchanger, and Q_max is the maximum possible heat transfer rate if the heat exchanger were perfect.

  4. 4.Why is the Effectiveness-NTU method preferred over the LMTD method in certain situations?Application

    The Effectiveness-NTU method is often preferred over the Log Mean Temperature Difference (LMTD) method when the inlet and outlet temperatures of the fluids are not known. This is because the Effectiveness-NTU method requires only the inlet temperatures and the heat capacity rates of the fluids, making it more convenient for preliminary design and analysis. Additionally, it is useful for comparing different heat exchanger designs without detailed temperature data.

  5. 5.What happens to the effectiveness of a heat exchanger as the NTU approaches infinity?Application

    As the NTU approaches infinity, the effectiveness of a heat exchanger approaches 1. This means that the heat exchanger is transferring the maximum possible amount of heat between the fluids, effectively acting as a perfect heat exchanger. In practical terms, this would require an infinitely large heat exchanger, which is not feasible, but it serves as a theoretical limit.

  6. 6.How does the flow arrangement (counterflow vs. parallel flow) affect the effectiveness of a heat exchanger?Application

    The flow arrangement significantly affects the effectiveness of a heat exchanger. In a counterflow arrangement, the fluids flow in opposite directions, which generally results in higher effectiveness compared to a parallel flow arrangement, where the fluids flow in the same direction. This is because counterflow allows for a more uniform temperature gradient along the length of the heat exchanger, facilitating better heat transfer.

  7. 7.Calculate the effectiveness of a counterflow heat exchanger with NTU = 2 and a heat capacity rate ratio C_min/C_max = 0.5.Numerical

    For counterflow, ε = [1 − exp(−NTU(1 − C_r))]/[1 − C_r·exp(−NTU(1 − C_r))]. Here NTU(1 − C_r) = 1, exp(−1) = 0.3679, so ε = (1 − 0.3679)/(1 − 0.5 × 0.3679) = 0.6321/0.8161 = 0.775. The exchanger transfers about 77 % of the thermodynamic maximum C_min(T_h,in − T_c,in).

  8. 8.What is the impact of increasing the heat capacity rate ratio (C_min/C_max) on the effectiveness of a heat exchanger?Application

    At a fixed NTU, effectiveness falls as C_r rises from 0 to 1. With C_r = 0 (one stream condensing or boiling) the other stream sees a constant driving temperature and ε = 1 − exp(−NTU) is the highest possible. As C_r rises, the C_max stream also changes temperature appreciably, so the local temperature difference shrinks along the exchanger; for counterflow at NTU = 2, ε drops from 0.865 (C_r = 0) to 0.667 (C_r = 1). The penalty is much larger in parallel flow, where ε cannot exceed 1/(1 + C_r).

  9. 9.Explain how the Effectiveness-NTU method can be used to design a heat exchanger.Application

    The Effectiveness-NTU method can be used to design a heat exchanger by first determining the required effectiveness based on the desired heat transfer rate and the available temperature differences. Once the effectiveness is known, the NTU can be calculated using the relationship between effectiveness, NTU, and the heat capacity rate ratio. With the NTU value, the necessary surface area and configuration of the heat exchanger can be determined to achieve the desired performance.

  10. 10.A parallel flow heat exchanger has an NTU of 1.5 and a heat capacity rate ratio of 0.7. Calculate its effectiveness.Numerical

    For a parallel flow heat exchanger, the effectiveness (ε) can be calculated using the formula: ε = (1 - exp(-NTU * (1 + C_min/C_max))) / (1 + C_min/C_max). Substituting NTU = 1.5 and C_min/C_max = 0.7, we get: ε = (1 - exp(-1.5 * (1 + 0.7))) / (1 + 0.7) = 0.541. Therefore, the effectiveness is approximately 0.541.

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