DIMENSIONING HEATING CABLES FOR A FOOT-BRIDGE
When calculating the cable resistance and length this is the best way:
- The required power is decided.
- Distribution into an appropriate number of cable runs.
- The best cable and voltage feed are decided.
For each run you then calculate:
- The total resistance (Rt).
- The shortest cable length possible (l).
- The cable resistance (W/m) is chosen.
- The real length of the cable (l).
- Check calculations.
- The distance between the cable runs (c/c).
- The solution.
Example: It is difficult to keep a foot-bridge made out of concrete free from snow and ice. It has thus been decided to install heating cables and then apply asphalt on the surface.
- Required power: The bridge is situated in southern parts of Sweden and the required power is actually 200-250 W/m² but as it is a suspended bridge and subject to cold weather even from below the required power must be adjusted upwards with approx. 50%, which gives approx. 350 W/m².
The surface power will then be approx. 350 W/m²
The total power: The bridge is 9 x 3m², the total required power will then be 27 x 350 = 9.450 W.
- Distribution on cable runs: Seen from the installation point of view as well as from the power it is best to distribute the power on three runs each 3.150 W. Each run will then heat a surface of 9m²
- Voltage feed: If you calculate the operating power of the cable runs at a feed voltage of 400V it will be possible to supply also with 230V via and Y/D-change-over switch and thus a power of 1/3. This can be of advantage if you for example want a basic heating, which is always below an adjusted value. This means a much faster heating up when there is a change in the weather and at precipitations.
- The total resistance of the cable (Rt): is obtained out of Ohms law for power P = U² / R gives R = U² / P
Example: 400² / 3.150 = 50.8 W
- The shortest possible length of the cable (l).
Example: 3150W/30W/m = 105 m
- Required resistance per meter cable (R/m):
R/m=Rt / l
Example: 50.8 W / 105 m = 0,48 Ohm/m
In the TCPR cable data sheet you will find that there is no cable with this resistance. You then have to choose the closest lower value, as a higher value would mean too short a cable and too high a power per meter cable. You then choose the 0.36 Ohm.
- The real length of the cable: l = Rt / R/m
Example: 50.8 W / 0,36 = 141 meter
- Check calculations: total power Pt = U² / Rt
Example: 400² / (0.36 x141) = 3.152 W – power/m cable P/m = Pt / 1
Example: 3.152 / 141 = 22.35 W/m
- c/c-distance; c/c = surface x 100 / 1
The answer is obtained in cm if you put the surface into m² and l into meter.
Example: 9 x 100 / 141 = 6,5 cm
- The solution: 3 lengths of heating cable each 128m TCPR 0.36 Ohm/m fare applied with a c/c-distance of 7cm.