🔀 Heat Exchanger — ε-NTU Method
Determine fluid outlet temperatures, heat transfer rates, and overall effectiveness for four flow configurations using the Effectiveness-NTU analysis.
📝 Configuration
ε-NTU Method:
Q = ε × Q_max
Q_max = C_min × (T_hi - T_ci)
NTU = U × A / C_min
C_r = C_min / C_max
Parallel Flow:
ε = [1 - exp(-NTU(1+Cr))] / (1+Cr)
Counter Flow:
ε = [1 - exp(-NTU(1-Cr))] / [1 - Cr exp(-NTU(1-Cr))]
Q = ε × Q_max
Q_max = C_min × (T_hi - T_ci)
NTU = U × A / C_min
C_r = C_min / C_max
Parallel Flow:
ε = [1 - exp(-NTU(1+Cr))] / (1+Cr)
Counter Flow:
ε = [1 - exp(-NTU(1-Cr))] / [1 - Cr exp(-NTU(1-Cr))]
📊 Results & Visualization
Configure the inputs and click "Rate Heat Exchanger" to see results.
ℹ️ About the Effectiveness-NTU (ε-NTU) Method
The ε-NTU method is a powerful thermal rating approach when fluid inlet temperatures and mass flow rates are specified, but outlet temperatures are unknown.
Supported Configurations:
• Parallel Flow: Both fluid streams flow in the same direction.
• Counter Flow: Fluid streams flow in opposite directions. The most thermally effective layout.
• Shell-and-Tube (1-2 Pass): One shell pass with two tube passes inside. Highly common in industrial processes.
• Cross-Flow (Unmixed): Fluids flow perpendicular to each other with no mixing along the flow paths.
The ε-NTU method is a powerful thermal rating approach when fluid inlet temperatures and mass flow rates are specified, but outlet temperatures are unknown.
Supported Configurations:
• Parallel Flow: Both fluid streams flow in the same direction.
• Counter Flow: Fluid streams flow in opposite directions. The most thermally effective layout.
• Shell-and-Tube (1-2 Pass): One shell pass with two tube passes inside. Highly common in industrial processes.
• Cross-Flow (Unmixed): Fluids flow perpendicular to each other with no mixing along the flow paths.
📘 Calculation Methodology
Mathematical Model & Theory
The Effectiveness-NTU ($\varepsilon$-NTU) method is used to analyze heat exchangers when outlet temperatures are unknown:
$$Q_{max} = C_{min} (T_{hi} - T_{ci})$$
$$Q = \varepsilon Q_{max}, \quad NTU = \frac{U A}{C_{min}}$$
$$\varepsilon = f(NTU, C_r) \quad \text{where } C_r = C_{min}/C_{max}$$
Assumptions & Boundary Conditions:
- Steady-state operation.
- Negligible heat transfer between the heat exchanger outer shell and surroundings (adiabatic boundaries).
- Constant fluid physical properties (such as specific heat $C_p$) throughout the exchanger.
- Constant overall heat transfer coefficient ($U$).
- Axial heat conduction along the tubes and shell wall is negligible.
- Uniform flow distribution at cross-sections.
Academic References:
- Incropera, F. P., Dewitt, D. P., Bergman, T. L., & Lavine, A. S. (2011). Fundamentals of Heat and Mass Transfer (7th ed.). John Wiley & Sons.
- Kays, W. M., & London, A. L. (1984). Compact Heat Exchangers (3rd ed.). McGraw-Hill.
Worked Engineering Example
Problem Statement:
A counter-flow heat exchanger ($U = 250$ W/m²·K, $A = 15$ m²) uses hot gas ($C_{p} = 1000$ J/kg·K, 1.0 kg/s) entering at 150°C to heat cold water ($C_p = 4180$ J/kg·K, 0.5 kg/s) entering at 15°C. Calculate the actual heat transfer rate.
Step-by-step Solution:
1. Calculate heat capacity rates:
$$C_h = 1.0 \times 1000 = 1000 \text{ W/K} \quad (C_{min})$$ $$C_c = 0.5 \times 4180 = 2090 \text{ W/K} \quad (C_{max})$$ $$C_r = C_{min}/C_{max} = 0.478$$ 2. Calculate NTU:
$$NTU = \frac{U A}{C_{min}} = \frac{250 \times 15}{1000} = 3.75$$ 3. Calculate counter-flow effectiveness $\varepsilon$:
$$\varepsilon = \frac{1 - e^{-NTU(1 - C_r)}}{1 - C_r e^{-NTU(1 - C_r)}} = \frac{1 - e^{-3.75(1 - 0.478)}}{1 - 0.478 e^{-3.75(1 - 0.478)}} = 0.906$$ 4. Calculate maximum and actual heat transfer:
$$Q_{max} = 1000 \times (150 - 15) = 135,000 \text{ W}$$ $$Q = 0.906 \times 135,000 = 122,310 \text{ W}$$
Final Result:
The heat transfer rate is 122.3 kW.
A counter-flow heat exchanger ($U = 250$ W/m²·K, $A = 15$ m²) uses hot gas ($C_{p} = 1000$ J/kg·K, 1.0 kg/s) entering at 150°C to heat cold water ($C_p = 4180$ J/kg·K, 0.5 kg/s) entering at 15°C. Calculate the actual heat transfer rate.
Step-by-step Solution:
1. Calculate heat capacity rates:
$$C_h = 1.0 \times 1000 = 1000 \text{ W/K} \quad (C_{min})$$ $$C_c = 0.5 \times 4180 = 2090 \text{ W/K} \quad (C_{max})$$ $$C_r = C_{min}/C_{max} = 0.478$$ 2. Calculate NTU:
$$NTU = \frac{U A}{C_{min}} = \frac{250 \times 15}{1000} = 3.75$$ 3. Calculate counter-flow effectiveness $\varepsilon$:
$$\varepsilon = \frac{1 - e^{-NTU(1 - C_r)}}{1 - C_r e^{-NTU(1 - C_r)}} = \frac{1 - e^{-3.75(1 - 0.478)}}{1 - 0.478 e^{-3.75(1 - 0.478)}} = 0.906$$ 4. Calculate maximum and actual heat transfer:
$$Q_{max} = 1000 \times (150 - 15) = 135,000 \text{ W}$$ $$Q = 0.906 \times 135,000 = 122,310 \text{ W}$$
Final Result:
The heat transfer rate is 122.3 kW.