Engineering guide
This module calculates overall heat transfer and the overall heat transfer coefficient to the VDI Heat Atlas, chapter C2 (12th edition, 2019). The overall heat transfer coefficient k combines the series resistances of the heat transport: convective heat transfer on both sides of the separating wall, conduction through the wall — including multi-layer walls, for instance with a lining or an insulation layer — and, where applicable, fouling resistances.
Through its selectable calculation options, the module covers the geometries relevant in practice: the plane single- and multi-layer wall, the tube with the logarithmic conduction resistance of the cylindrical wall and the question of the reference area (inside or outside), and insulated components. This determines the heat flow through the wall or the k value referred to a chosen reference area, which then feeds into the heat exchanger design per chapter C1.
In practice the calculation is needed throughout equipment and plant engineering: for designing heat exchangers and vessel heating systems, for evaluating heat losses of insulated piping and equipment, or for analysing which individual resistance limits a heat duty and where an improvement should start.
Scope
- Heat Exchangers: Heat Transfer and Heat Transfer Coefficients
Typical workflow
Choose geometry and calculation option
First it is established whether the case is a plane wall, a tube or a multi-layer construction (e.g. wall with insulation). For tubes, the reference area to which the k value is referred must also be chosen — usually the outside surface.
Determine the heat transfer coefficients on both sides
The convective heat transfer coefficients inside and outside come from the relevant Nusselt correlations of the VDI Heat Atlas (forced or free convection, condensation, evaporation) or from empirical values for the application at hand.
Sum the conduction resistances of the wall layers
For each layer, the conduction resistance is formed from layer thickness and thermal conductivity; for a tube, the plane-wall term s/λ is replaced by the logarithmic cylindrical wall term with the radius ratios. Fouling resistances on both sides are applied as additional series resistances.
Calculate the k value and the heat flow
The reciprocal of the resistance sum yields the overall heat transfer coefficient referred to the chosen area. With the driving temperature difference and the area, the heat flow follows; if required, the wall temperatures at the layer interfaces can be back-calculated from it.
Frequently asked questions
Why does the smallest heat transfer coefficient often dominate the k value?
The resistances are in series, and the largest individual resistance governs the sum. A gas-side heat transfer coefficient of, say, 30 W/(m²·K) contributes a resistance orders of magnitude above that of a water-cooled side with 4,000 W/(m²·K) — the k value can then hardly exceed the smallest transfer coefficient. Improvements therefore pay off only on the limiting side, for example through fins or a higher flow velocity.
What role does the reference area play for a tube?
For a tube, the inside and outside surfaces differ in size; the k value is therefore unambiguous only together with its reference area. If k is referred to the outside surface, the inner film resistance must be converted with the area ratio da/di and the cylindrical wall with the logarithmic term. When comparing k values from different sources, always check which area they are referred to.
How are fouling resistances taken into account?
Fouling resistances are applied as additional series resistances on the respective side; typical guide values lie between about 0.0001 and 0.0005 (m²·K)/W depending on the medium (e.g. TEMA values). They can reduce the k value of clean equipment considerably and are the reason why the design duty and the clean-condition duty of a heat exchanger differ.
When is the plane-wall approximation sufficient for a tube?
For thin-walled tubes with da/di close to 1 (roughly below 1.1), the logarithmic cylindrical term deviates only slightly from the plane s/λ, and the approximation is usually adequate. For thick-walled tubes, multi-layer constructions or insulation with a large radius ratio, however, the cylindrical geometry must be used, otherwise the conduction resistance is significantly misjudged.
Worked example
For a water-cooled heat exchanger, determine the overall heat transfer coefficient of a plane steel wall in the clean condition — a worked example of an overall heat transfer coefficient calculation per the VDI Heat Atlas. Cooling water with a high heat transfer coefficient flows on the inside, an organic medium on the outside.
Given
| Heat transfer coefficient inside (water) αi | 4,000 W/(m²·K) |
|---|---|
| Heat transfer coefficient outside αa | 800 W/(m²·K) |
| Wall thickness s (steel) | 2 mm |
| Thermal conductivity of steel λ | 50 W/(m·K) |
Solution
Sum the series resistances
For the plane wall: 1/k = 1/αi + s/λ + 1/αa
1/k = 1/4,000 + 0.002/50 + 1/800 = 0.00025 + 0.00004 + 0.00125 = 0.00154 (m²·K)/W
Overall heat transfer coefficient
k = 1 / 0.00154 (m²·K)/W ≈ 649 W/(m²·K)
At 0.00125 (m²·K)/W, the outside film resistance accounts for about 81 % of the total resistance — so any improvement of the k value must start on the outside; the conduction resistance of the thin steel wall is practically negligible. Fouling resistances would enter as additional summands and reduce k further.
Result
| Sum of resistances 1/k | 0.00154 (m²·K)/W |
|---|---|
| Overall heat transfer coefficient k | 649 W/(m²·K) |
Technical content source: LV-Solved-Problems-Content 628d313247e7