Engineering task and calculation objective
The TANK module calculates the heat loss of storage tanks – insulated or non-insulated, with a round or rectangular ground plan. Anyone who wants to calculate the heat loss of a storage tank needs the result for two central questions of plant design: how large must the heating system (steam, thermal oil, hot water via half-pipe coils or tube spirals) be sized to maintain the storage temperature? And how fast does the tank contents cool down if the heating fails?
The module balances the heat flows through the tank wall, roof and bottom, taking into account the respective heat transfer conditions: free convection of the stored product (e.g. heavy fuel oil, water, thermal oil) on the inside, convection to the ambient air with wind influence on the outside, and heat conduction through wall and insulation. Besides the steady-state heat loss, transient processes can also be examined, i.e. the temperature change of the tank contents over time during cool-down.
Typical applications are heavy fuel oil and bitumen tanks whose contents must be kept pumpable, heated process and feed vessels, and the optimization of the insulation thickness: the calculated heat loss directly yields the operating cost of the heating and thus the basis for deciding on the economics of additional insulation.
Calculation workflow
- Enter geometry and construction: Ground plan (round or rectangular), dimensions, filling level and the wall construction of shell, roof and bottom are entered – including insulation layers with thickness and thermal conductivity.
- Define the operating conditions: Storage temperature of the medium, ambient temperature (design winter conditions where applicable) and wind velocity are specified; for the bottom, the soil or foundation temperature enters the calculation.
- Determine the heat transfer coefficients: On the inside, the heat transfer by free convection of the stored product to the wall is calculated; on the outside, the superposition of convection (with wind influence) and radiation; together with the heat conduction through wall and insulation, this yields the overall heat transfer coefficient of each partial surface.
- Balance the heat loss: The losses through the wetted shell, gas space, roof and bottom are calculated individually and summed to the total heat loss. From this follows directly the heating power required to maintain the temperature.
- Assess the transient cool-down: Optionally, the temperature-time curve during cool-down is determined from the heat loss and the heat capacity of tank contents and tank body – for example to state the permissible standstill time until a minimum temperature is reached (e.g. the pumpability limit of heavy fuel oil).
Input quantities
| Quantity | Symbol | Unit |
|---|---|---|
| Product temperature | ϑP | °C |
| Air temperature | ϑL | °C |
| Ground temperature | ϑB | °C |
| Tank inside diameter | DT | m |
| Tank height | HT | m |
| Filling level | HF | m |
| Heat transfer coefficient inside | αi,B | W/(m²·K) |
| Thermal resistance of the wall | βW,B | m²·K/W |
| Thermal resistance of the insulation βInsu,B | Boden | m²·K/W |
| Area | AB | m² |
| Temperature (wall-insulation) | ϑg,B | °C |
| Wall temperature inside, bottom | ϑi,B | °C |
| Temperature outside, bottom | ϑa,B | °C |
| wet part | Qa,B | W |
| Thermal resistance of the wall | βW,M | m²·K/W |
| Thermal resistance of the insulation βInsu,M | Mantel | m²·K/W |
| Heat transfer coefficient outside | αa,M | W/(m²·K) |
| Heat transfer coefficient inside | αi,M,b | W/(m²·K) |
| Contact surface | AM,b | m² |
| Temperature (wall-insulation) | ϑg,M,b | °C |
| Wall temperature inside, shell | ϑi,M,b | °C |
| Temperature insulation outside, shell | ϑa,M,b | °C |
| Heating performance | Qh,M,b | W |
| Heat flow from inside | Qi,M,b | W |
Calculation options
Sketch
Round sketch · Rectangular sketch
Worked example
For the cylindrical shell of a vertical storage tank (diameter 4 m, wetted height 6 m) with 100 mm of mineral wool insulation, the steady-state heat loss is to be estimated in a worked example. The stored product is at 80 °C, the ambient air at 0 °C. The curvature of the shell is neglected for the thin insulation layer (flat wall); roof and bottom are not considered here.
Given values
| Tank diameter D | 4 m |
| Wetted shell height H | 6 m |
| Insulation thickness s | 100 mm |
| Thermal conductivity of insulation λ | 0.045 W/(m·K) |
| Internal heat transfer coefficient αi | 200 W/(m²·K) |
| External heat transfer coefficient αa | 15 W/(m²·K) |
| Temperature difference ϑi − ϑa | 80 K |
Solution
Overall heat transfer coefficient
The resistances of the steel wall and the paint coat are negligibly small. For the flat wall:
1/U = 1/αi + s/λ + 1/αa = 1/200 + 0.100/0.045 + 1/15 = 0.005 + 2.222 + 0.067 = 2.294 m²·K/W
U ≈ 0.436 W/(m²·K) – the insulation resistance clearly dominates.
Shell area
A = π · D · H = π · 4 · 6 ≈ 75.4 m²
Heat loss
Q̇ = U · A · (ϑi − ϑa) = 0.436 · 75.4 · 80 ≈ 2,630 W
The shell thus loses around 2.6 kW; for sizing the heating system, the losses through roof, bottom and the unwetted wall region are added.
Result
| Overall heat transfer coefficient U | ≈ 0.44 W/(m²·K) |
| Shell area A | ≈ 75.4 m² |
| Heat loss of shell Q̇ | ≈ 2.6 kW |
All values are illustrative. The applicable standard and project-specific boundary conditions remain authoritative.
Frequently asked questions
Why is the wind velocity so important for the heat loss?
For non-insulated or poorly insulated tanks, the external heat transfer resistance is of the same order of magnitude as the rest of the chain; the external heat transfer coefficient rises strongly with wind velocity and can multiply the loss of an uninsulated tank. For well-insulated tanks, on the other hand, the insulation resistance dominates and the wind influence on the total loss becomes small.
Does the gas space above the liquid level have to be considered separately?
Yes. Above the filling level, the internal heat transfer (gas instead of liquid) is considerably poorer, the wall temperature is lower and the area-specific loss is smaller than in the wetted region. For partly filled tanks, a calculation assuming a fully wetted wall therefore overestimates the losses; conversely, condensation on cold walls in the gas space can be a problem in its own right.
How is the heat loss through the tank bottom accounted for?
The bottom releases heat to the foundation and the soil; governing factors are the foundation construction, any bottom insulation and the soil temperature, which is considerably more sluggish than the air temperature. For large flat-bottom tanks, the bottom loss per unit area is usually small compared with the shell, but can be relevant in absolute terms because of the large area.
For which design case should the heating power be sized?
The usual approach is the most unfavorable steady-state case: lowest applicable ambient temperature, design wind, target temperature of the stored product. In addition, it should be checked whether the heating can also reheat the contents after a standstill within an acceptable time – this often requires more power than merely covering the losses.