Select ΔT1, ΔT2, U, area or Q to solve using the LMTD method. Enter the other four quantities with their units. Use positive terminal temperature differences, with ΔT1 ≥ ΔT2, and the same area basis for U and A.
- All Physics Calculators
- Heat Exchanger Effectiveness Calculator
- LMTD (Logarithmic Mean Temperature Difference) Calculator
- Thermal Efficiency Calculator
Coaxial Heat Exchanger Formula
The calculator uses the log mean temperature difference method for steady parallel or counterflow operation with constant overall U. Other flow arrangements can require a correction factor. It is not a complete exchanger design. The main heat transfer equation is:
The log mean temperature difference is calculated as:
If the two terminal temperature differences are equal, the calculator uses the limiting case:
Rearranged forms used to solve for missing values are:
- Q = heat transfer rate, in W, kW, or MW
- U = overall heat transfer coefficient, in W/m²K or BTU/hr·ft²·°F
- A = heat transfer area, in m² or ft²
- LMTD = log mean temperature difference
- DeltaT1 = higher terminal temperature difference
- DeltaT2 = lower terminal temperature difference
- ln = natural logarithm
To calculate Q, the calculator first finds LMTD from DeltaT1 and DeltaT2, then multiplies U, A, and LMTD.
To calculate U, it divides Q by the product of A and LMTD.
To calculate A, it divides Q by the product of U and LMTD.
To calculate DeltaT1 or DeltaT2, it first calculates the required LMTD from Q / (U*A), then solves the LMTD equation numerically. The calculator assumes DeltaT1 is greater than or equal to DeltaT2.
Common Unit Conversions for Coaxial Heat Exchanger Calculations
The calculator converts all inputs to base SI units before applying the formula.
| Quantity | Input unit | Base unit | Conversion |
|---|---|---|---|
| Temperature difference | °F | °C or K difference | ΔT(°C) = ΔT(°F) × 5/9 |
| Overall heat transfer coefficient | BTU/hr·ft²·°F | W/m²K | 1 BtuIT/(hr·ft²·°F) ≈ 5.678263 W/(m²·K) |
| Area | ft² | m² | 1 ft² = 0.09290304 m² |
| Heat transfer rate | kW | W | 1 kW = 1000 W |
| Heat transfer rate | MW | W | 1 MW = 1,000,000 W |
Overall Heat Transfer Coefficient Considerations
Use measured or appropriately calculated U for the actual exchanger. A service label alone does not supply a validated range. U depends on fluid properties, flow regime, wall material, fouling, phase change and area basis.
| Service type | U input basis | Notes |
|---|---|---|
| Liquid to liquid, water-like fluids | Use exchanger-specific data | Common range for clean, turbulent flow |
| Oil to water | Use exchanger-specific data | Oil-side resistance often controls the result |
| Gas to liquid | Use exchanger-specific data | Gas-side heat transfer is usually lower |
| Condensing vapor to liquid | Use exchanger-specific data | Can be high when condensation is efficient |
Example Calculations
Example 1: Calculate heat transfer rate
Suppose a coaxial heat exchanger has:
- DeltaT1 = 60 °C
- DeltaT2 = 30 °C
- U = 750 W/m²K
- A = 4 m²
First calculate LMTD:
Then calculate Q:
The heat transfer rate is about 129.84 kW.
Example 2: Calculate required heat transfer area
Suppose you need:
- Q = 100 kW
- U = 500 W/m²K
- DeltaT1 = 50 °C
- DeltaT2 = 25 °C
Convert Q to watts:
Calculate LMTD:
Calculate area:
The required heat transfer area is about 5.55 m².
FAQs
What are DeltaT1 and DeltaT2 in a coaxial heat exchanger?
DeltaT1 and DeltaT2 are hot–cold differences at the two ends, not absolute fluid temperatures. Counterflow pairs hot inlet with cold outlet and hot outlet with cold inlet; parallel flow pairs the inlets and the outlets. Enter the larger difference as DeltaT1 and the smaller as DeltaT2. Both must be positive. U and A must refer to the same heat transfer surface.
Why does the calculator use LMTD instead of a simple average temperature difference?
The temperature difference between the two fluids changes along the exchanger length. LMTD gives the correct effective temperature difference for steady-state heat exchanger calculations when the terminal temperature differences are known. A simple arithmetic average can give a less accurate result, especially when DeltaT1 and DeltaT2 are far apart.
What does it mean if there is no solution for DeltaT1 or DeltaT2?
When solving for a missing terminal temperature difference, the required LMTD must be physically consistent with the known terminal difference. With DeltaT1 greater than or equal to DeltaT2, LMTD cannot be less than DeltaT2 or greater than DeltaT1. If the calculator reports no solution, check Q, U, A, and the terminal temperature difference values.
