Tank Heating Load Calculation: Start-up, Heat Losses and Steam Demand

 

Heating liquids in tanks is a common requirement across process industries. Tank heat-load calculations should not be limited to the energy needed to raise liquid temperature. Depending on the application, vessel heating, wall losses, exposed-surface losses and cold articles introduced into the process may also be important.

A key distinction is the difference between start-up load and operating load.

Closed and open tanks

Totally enclosed, well-insulated tanks generally have a simpler heat balance. Open-topped tanks can have substantial losses from the exposed liquid surface and may receive cold articles or materials.

Typical applications and temperatures

Application Typical temperature
Boiler feedtank 80–90°C
Hot-water tank 40–85°C
Degreasing tank 90–95°C
Metal treatment / plating 70–85°C
Rust or scale removal 90–95°C
Raw-juice heating 80–85°C
Pharmaceutical wash tank about 70°C
Caustic-oil heating about 140°C
Heavy fuel oil for pumping 30–40°C

Components of total heat demand

  1. Heating the process liquid.
  2. Heating the vessel material.
  3. Heat loss from solid vessel surfaces.
  4. Heat loss from the exposed liquid surface.
  5. Heat absorbed by cold articles or materials.

Heating liquid, vessel and articles

Q̇ = m · cp · ΔT / t

Heat loss from vessel walls

Q̇ = U · A · ΔT

Where a tank bottom lies flat on the ground and is not exposed to ambient air, its corresponding loss may normally be neglected in this type of calculation.

Effect of insulation

  • 25 mm insulation: multiply U by 0.2.
  • 50 mm insulation: multiply U by 0.1.

Effect of air movement

Air velocity Multiplying factor
0 m/s 1.0
0.5 m/s 1.3
1.0 m/s 1.7
2.0 m/s 2.4
3.0 m/s 3.1

Indicative U-values for oil tanks

Position ΔT Unlagged Lagged
Sheltered up to 10°C 6.8 1.7
Sheltered up to 27°C 7.4 1.8
Sheltered up to 38°C 8.0 2.0
Exposed up to 10°C 8.0 2.0
Exposed up to 27°C 8.5 2.1
Exposed up to 38°C 9.1 2.3
Underground any temperature 6.8 —

U-values are in W/(m²·°C).

Heat loss from exposed liquid surfaces

In an open tank, surface heat loss may be the largest heat-loss component. It depends mainly on surface temperature, air velocity and exposed surface area.

Worked example

  • tank dimensions 3 × 3 × 2 m,
  • 2/3 full,
  • 12,000 kg liquid,
  • liquid cp = 3.9 kJ/(kg·°C),
  • 8°C → 60°C in 2 hours,
  • 15 mm mild-steel vessel,
  • 500 kg steel article every 20 minutes.

Start-up load

Component Load
Liquid heating 338 kW
Vessel heating 14 kW
Side losses 7 kW
Surface losses 9 kW
Total 368 kW

Operating load

Component Load
Side losses 14 kW
Surface losses 27 kW
Immersed article 11 kW
Total 52 kW

Why start-up duty is higher

During start-up, both the liquid and vessel must absorb sensible heat. Once operating temperature has been reached, demand mainly consists of environmental losses and heat absorbed by cold material entering the process.

Warm-up time

A longer permitted warm-up period reduces the required mean heat-transfer rate, levels boiler demand and reduces instantaneous demand on the temperature-control system.

Practical calculation sequence

  1. Define tank geometry and fill level.
  2. Determine liquid mass and cp.
  3. Determine vessel mass and cp.
  4. Define initial and operating temperatures.
  5. Define warm-up time.
  6. Calculate liquid and vessel heating duties.
  7. Calculate solid-surface losses.
  8. Calculate exposed-liquid-surface losses.
  9. Add cold-material loads.
  10. Check start-up and operating duties separately.

Once heat demand is known, the heating method may be selected, including internal coils, jackets or a suitable heat exchanger. The actual loads should also be considered when selecting control valves and steam-trapping and condensate equipment.

Conclusion

Total tank duty is the sum of heating energy and the significant real heat losses. Insulation, air movement, exposed liquid surface and cold articles can all materially affect demand.

Start-up and operating loads should be calculated separately because they may differ by several times.

Frequently Asked Questions

What are the main components of tank heat load?

Liquid heating, vessel heating, solid-surface losses, exposed-liquid-surface losses and cold-material heating.

How is liquid heating duty calculated?

Using Q̇ = m·cp·ΔT/t.

How are wall heat losses calculated?

Using Q̇ = U·A·ΔT.

How does insulation affect heat loss?

It reduces the overall U-value and therefore the heat loss.

Why do open tanks lose more heat?

Because heat is also lost directly from the exposed liquid surface.

Why does air movement increase heat loss?

Higher air velocity increases convective heat transfer from exposed surfaces.

Why is start-up duty higher?

Because both the liquid and the vessel must be heated from their initial temperature.

How does warm-up time affect heat input?

A longer warm-up period reduces the required mean heat-transfer rate.

 
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