KEVOS® MAKEOS · Hydraulics
Tank Heating or Cooling Coil Plan (Area, Tube Length, Turns, Pressure Drop)
How many turns of tube does my tank heating or cooling coil need for the duty, how long is the tube, and what pressure drop will the water see through it?
Plan for a helical tube coil in a tank: heat-transfer area needed for the duty from the overall coefficient and mean temperature difference, the tube length and turns that give it, the developed length of the coil you choose with its straight ends (as the helix length tool), the area it provides, and the velocity, friction factor and pressure drop of the fluid through the tube (as the pipe friction tool), checked against the duty and your pressure-drop limit.
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How this tool works
Inputs, units and where to find them
| Input | Unit | Typical source |
|---|---|---|
| Heat duty | kW | Heat to transfer. |
| Overall heat-transfer coefficient | W/(m²·K) | Overall coefficient for the coil, tube and fluids (from your heat-transfer data). |
| Mean temperature difference | K | Log-mean (or mean) temperature difference between the fluid in the tube and the tank. |
| Coil | ||
| Mean coil diameter | mm | Mean diameter of the coil. |
| Coil pitch | mm | Axial pitch of the coil. |
| Turns | turns | Number of turns you plan. |
| Straight ends (total) | mm | Total length of the straight inlet and outlet legs. |
| Tube outside diameter | mm | Outside diameter of the tube. |
| Tube wall | mm | Wall thickness of the tube. |
| Fluid in the tube | ||
| Flow | L/min | Flow of the heating or cooling fluid through the tube. |
| Kinematic viscosity | cSt | Kinematic viscosity of that fluid. |
| Density | kg/m³ | Density of that fluid. |
| Tube roughness | mm | Roughness of the tube bore. |
| Sum of fitting loss coefficients | Loss coefficients for the entry, exit and fittings. | |
| Largest pressure drop | bar | Largest pressure drop you accept across the coil. |
Method and formulas
- Area needed = Q ÷ (U ΔT_m); tube length needed = area ÷ (π × tube OD).
- Developed length = turns × √[(π D_c)² + p²] + ends (as the helix length tool); area provided = π × OD × length.
- Pressure drop = (f L ÷ d + ΣK) ρ v² ÷ 2 with the Colebrook friction factor for the tube bore (as the pipe friction tool).
- Example: 20 kW needs 1 m², about 12.961 turns; 14 turns give 1.077 m² of 13.713 m tube with 0.085323 bar drop.
Limitations
- Straight-tube friction factor; coil curvature raises the pressure drop somewhat.
- Overall coefficient must allow for fouling and the tank side agitation.
- Example values are illustrations. Replace them with values for your actual task.
Typical uses
- Size a heating or cooling coil for a tank or bath
- Check an existing coil for a new duty
- Find the pressure drop through a coil
Who it is for
- Plant engineers
- Process engineers
- Maintenance engineers
What you get
- Area and tube length needed, turns needed
- Developed length and area provided
- Velocity and pressure drop
Interpreting the result
- If the drop is too high, split the coil into parallel circuits rather than shortening it.
References
- Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 2060–2062, 3024–3027, 371–378. Topic reference; no source tables or recommended settings are reproduced.
WORKOS tools are screening and decision-support utilities. Verify important financial, engineering, safety, legal or operational decisions using appropriate professional standards and your organisation’s approved data.
Open the Tank Heating or Cooling Coil Plan (Area, Tube Length, Turns, Pressure Drop) calculator
