Heat-transfer rate from Q̇ = FUAΔTₗₘ.
Thermal engineering · Heat transfer · conduction
Heat exchanger LMTD — duty, area and U
Calculate the logarithmic mean temperature difference from four terminal temperatures, apply correction factor F, then solve Q̇, A or U.
Created v1.0
By Thibaut Grzelak, Mechanical Analysis Engineer
Results
Required or entered heat-transfer area.
Required or entered overall heat-transfer coefficient.
Logarithmic mean temperature difference calculated from the two terminal differences.
LMTD multiplied by the explicit correction factor F.
First terminal temperature difference for the selected flow arrangement.
Second terminal temperature difference for the selected flow arrangement.
Steady operation with known terminal bulk-fluid temperatures. The hot stream cools and the cold stream warms; both terminal differences must remain strictly positive.
01Formulas and symbols
Formulas used
ΔT₁ = Tₕ,in − T𝒸,out ; ΔT₂ = Tₕ,out − T𝒸,inCounterflowΔT₁ = Tₕ,in − T𝒸,in ; ΔT₂ = Tₕ,out − T𝒸,outParallel flow(ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂)LMTD ΔTₗₘF · ΔTₗₘCorrected LMTD FΔTₗₘF · U · A · ΔTₗₘHeat-transfer rate Q̇Q̇ / (F · U · ΔTₗₘ)Heat-transfer area AQ̇ / (F · A · ΔTₗₘ)Overall heat-transfer coefficient U02Assumptions and limits
Scope of validity
- Steady operation with known terminal bulk-fluid temperatures.
- The hot stream cools and the cold stream warms; both terminal differences must remain strictly positive.
- The overall coefficient U is treated as constant over the whole heat-transfer area.
- Heat loss to surroundings, energy storage and property variation are not modeled.
- Correction factor F is an explicit engineering input between 0 and 1; this calculator does not derive F from multipass geometry.
- Phase change, ε-NTU analysis, mass flow rates, heat capacities and pressure drop are outside scope.
03Validation example
Reference numerical case
- Counterflow reference: Tₕ,in = 120 °C, Tₕ,out = 80 °C, T𝒸,in = 20 °C and T𝒸,out = 60 °C.
- ΔT₁ = 60 K and ΔT₂ = 60 K; the analytical limit gives ΔTₗₘ = 60 K.
- With F = 1, U = 500 W/(m²·K) and A = 10 m², Q̇ = 300 kW.
- Inverse modes return A = 10 m² and U = 500 W/(m²·K) for Q̇ = 300 kW.
04References
Technical references
- Incropera et al. — Fundamentals of Heat and Mass Transfer: logarithmic mean temperature difference method.
- Kern — Process Heat Transfer: LMTD heat-exchanger sizing.
- NIST Special Publication 811 — SI units and temperature intervals.