Green-houses

Soil-heating in greenhouses

If heating cables are applied in greenhouses or hotbeds, the season of cultivation can be extended. With an earlier start of growing in the spring and a longer autumn many so-called impossible plants can be cultivated and this despite the severe climate.

Earth heating can be arranged in an easy way and without too much encroachment by applying heating cables below the bed of earth. The heating cables are then controlled to obtain the required earth temperature (6-14°C).

Miscellaneous

Power required: 60-140 W/m²

Cable types to be used: TCPR

Dimensioning: The power should not exceed 15 W/m cable to prevent a drying up of the roots.

Control: The heating cables are controlled by a thermostat and the sensor is placed in the bed of plants (temperature zone 0-40°C).

Dimensioning

Dimensioning heating cables for soil heating in greenhouses

When calculating the cable resistance and length this is the best way:

  1. The required power is decided (Pt).
  2. Distribution on a suitable number of cable lengths.
  3. The best cable and voltage feed are decided.
    For each length the following is then decided:
  4. The total resistance (Rt).
  5. The shortest cable length possible (l).
  6. The cable resistance (Ohm/m) is chosen.
  7. The real length of the cable (l).
  8. Check calculations.
  9. The distance between the cable runs (c/c).
  10. The solution.

Example: In a hothouse there are two cultivating areas 6 x 2m that are to be provided with heating cables for earth heating.

  1. Required power: The surface x required surface power
    Example: 12 x 100 = 1.200 W.
  2. Number of cable lengths: The relatively low total power of 1.200W you can without problem put on one cable length.
  3. Type of cable: TCP or TCPR are two types of cables that you can choose between. Because of the expense the TCP is the cable that is most often chosen.
    Voltage feed: There is no reason to choose a higher voltage than 230V.
  4. The total resistance of the cable length: (Rt) is obtained out of Ohms law for power
    P = U² / R gives R = U² / P
    Example: 230² / 1.200 = 44.08 Ohm.
  5. The shortest possible length of the cable length (l):
    l = Pt / P/m
    Example: 1.200 W / 15 W/m = 80 m.
  6. Required resistance per meter cable (R/m):
    R/m = Rt / l
    Example: 44.08 Ohm/ 80 m = 0.55 Ohm/m
    In the cable data sheet you will find that the closest lower value is 0.45 Ohm/m. You thus choose a cable with this resistance and then calculate what length is needed.
  7. The real length of the cable length (l):
    l = Rt / R/m
    Example: 44.08 / 0.45 = 98 meter.
  8. Check calculations: 98 m TCP/R 0.45; 12 m²
    P = 230² / (98 x 0.45) = 1.200 W
    P/m = 1.200 / 98 = 13.33 W/m
    P/m² = 1.200 / 12 = 100 W/m²
  9. c/c-distance (c/c): c/c = surface x 100 / l
    You will obtain the answer in cm if you put the surface in m² and l in meter.
    Example: 12 x 100 / 98 = 13 cm.
  10. The solution: A heating cable length of 98 m TCP/R 0.45 Ohm/m is applied with a c/c-distance of 13 cm.