ThermoFluidCalc
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Engineering Calculation Report
Date: 2026-05-30 00:15
🔀 Heat Exchanger — LMTD Method
Design and analyze heat exchangers using the Log Mean Temperature Difference method for parallel and counter flow configurations.
📝 Configuration
LMTD Method:
Q = U × A × LMTD
LMTD = (ΔT₁ - ΔT₂) / ln(ΔT₁/ΔT₂)
Parallel Flow:
ΔT₁ = T_hi - T_ci, ΔT₂ = T_ho - T_co
Counter Flow:
ΔT₁ = T_hi - T_co, ΔT₂ = T_ho - T_ci
ε = Q / Q_max = Q / [C_min(T_hi - T_ci)]
Q = U × A × LMTD
LMTD = (ΔT₁ - ΔT₂) / ln(ΔT₁/ΔT₂)
Parallel Flow:
ΔT₁ = T_hi - T_ci, ΔT₂ = T_ho - T_co
Counter Flow:
ΔT₁ = T_hi - T_co, ΔT₂ = T_ho - T_ci
ε = Q / Q_max = Q / [C_min(T_hi - T_ci)]
📊 Results & Visualization
Configure the inputs and click "Analyze Heat Exchanger" to see results.
ℹ️ About Heat Exchangers
Heat exchangers transfer thermal energy between two fluid streams. The LMTD method is commonly used for design problems where temperatures are known.
Configurations:
• Parallel Flow: Both fluids enter from the same end
• Counter Flow: Fluids enter from opposite ends — more effective, higher LMTD
Applications:
• Power plants (condensers, feedwater heaters)
• HVAC systems
• Chemical processing
• Automotive radiators
Heat exchangers transfer thermal energy between two fluid streams. The LMTD method is commonly used for design problems where temperatures are known.
Configurations:
• Parallel Flow: Both fluids enter from the same end
• Counter Flow: Fluids enter from opposite ends — more effective, higher LMTD
Applications:
• Power plants (condensers, feedwater heaters)
• HVAC systems
• Chemical processing
• Automotive radiators
📘 Calculation Methodology
Mathematical Model & Theory
The Log Mean Temperature Difference (LMTD) method calculates heat transfer in heat exchangers based on inlet and outlet temperatures:
$$Q = U A F \Delta T_{lm}$$
$$\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}$$
$$\text{Counter-flow: } \Delta T_1 = T_{hi} - T_{co}, \quad \Delta T_2 = T_{ho} - T_{ci}$$
$$\text{Parallel-flow: } \Delta T_1 = T_{hi} - T_{ci}, \quad \Delta T_2 = T_{ho} - T_{co}$$
Where $F$ is the multi-pass correction factor ($F=1.0$ for single-pass concentric exchangers).
Assumptions & Boundary Conditions:
- Steady-state flow and thermal conditions.
- Negligible heat losses to the ambient surroundings (fully insulated shell).
- Constant specific heats and other physical properties for both fluids.
- Constant overall heat transfer coefficient ($U$) throughout the entire surface area.
- One-dimensional flow with temperature varying only in the axial direction.
- Negligible potential and kinetic energy changes.
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.
- Çengel, Y. A., & Ghajar, A. J. (2015). Heat and Mass Transfer: Fundamentals and Applications (5th ed.). McGraw-Hill Education.
Worked Engineering Example
Problem Statement:
A counter-flow concentric tube heat exchanger ($U = 350$ W/m²·K) cools hot oil ($C_{p,h} = 2200$ J/kg·K, 2 kg/s) from 120°C to 70°C using water ($C_{p,c} = 4180$ J/kg·K, 1.5 kg/s) entering at 20°C. Calculate the required heat transfer area.
Step-by-step Solution:
1. Calculate heat duty $Q$ from hot fluid energy balance:
$$Q = \dot{m}_h C_{p,h} (T_{hi} - T_{ho}) = 2 \times 2200 \times (120 - 70) = 220,000 \text{ W}$$ 2. Calculate water outlet temperature $T_{co}$:
$$Q = \dot{m}_c C_{p,c} (T_{co} - T_{ci}) \Rightarrow 220,000 = 1.5 \times 4180 \times (T_{co} - 20) \Rightarrow T_{co} = 55.09 \text{°C}$$ 3. Calculate LMTD for counter-flow:
$$\Delta T_1 = T_{hi} - T_{co} = 120 - 55.09 = 64.91 \text{ K}$$ $$\Delta T_2 = T_{ho} - T_{ci} = 70 - 20 = 50.00 \text{ K}$$ $$\Delta T_{lm} = \frac{64.91 - 50.00}{\ln(64.91 / 50.00)} = 57.17 \text{ K}$$ 4. Calculate required area $A$ ($F=1.0$):
$$A = \frac{Q}{U \Delta T_{lm}} = \frac{220,000}{350 \times 57.17} = 11.0 \text{ m}^2$$
Final Result:
The required surface area is 11.0 m².
A counter-flow concentric tube heat exchanger ($U = 350$ W/m²·K) cools hot oil ($C_{p,h} = 2200$ J/kg·K, 2 kg/s) from 120°C to 70°C using water ($C_{p,c} = 4180$ J/kg·K, 1.5 kg/s) entering at 20°C. Calculate the required heat transfer area.
Step-by-step Solution:
1. Calculate heat duty $Q$ from hot fluid energy balance:
$$Q = \dot{m}_h C_{p,h} (T_{hi} - T_{ho}) = 2 \times 2200 \times (120 - 70) = 220,000 \text{ W}$$ 2. Calculate water outlet temperature $T_{co}$:
$$Q = \dot{m}_c C_{p,c} (T_{co} - T_{ci}) \Rightarrow 220,000 = 1.5 \times 4180 \times (T_{co} - 20) \Rightarrow T_{co} = 55.09 \text{°C}$$ 3. Calculate LMTD for counter-flow:
$$\Delta T_1 = T_{hi} - T_{co} = 120 - 55.09 = 64.91 \text{ K}$$ $$\Delta T_2 = T_{ho} - T_{ci} = 70 - 20 = 50.00 \text{ K}$$ $$\Delta T_{lm} = \frac{64.91 - 50.00}{\ln(64.91 / 50.00)} = 57.17 \text{ K}$$ 4. Calculate required area $A$ ($F=1.0$):
$$A = \frac{Q}{U \Delta T_{lm}} = \frac{220,000}{350 \times 57.17} = 11.0 \text{ m}^2$$
Final Result:
The required surface area is 11.0 m².