# Pavements

##### Dimensioning heating cables for a pavement covered with ground bricksl When calculating the cable resistance and length this is the best way:

1. The required power is decided.
2. Distribution into an appropriate number of cable runs.
For each run you then calculate:
3. The total resistance (Rt).
4. The shortest cable length possible (l).
5. The cable resistance (W/m) is chosen.
6. The real length of the cable (l).
7. Check calculations.
8. The distance between the cable runs (c/c).
9. The solution.

Example: A pavement outside a shop, 3 x 12m, is to be applied with heating cables to be kept dry and non-slippery all the year around. The cables are to be applied in a bed of sand and the covering of the pavement is of ground bricks.

1. Required power: surface power x surface
In this example the surface power is decided to 300 W/m², the surface is 36 m²
The required power will thus be 300 x 36 = 10.8 kW.
2. Distribution on number of cable runs:
10.8 kW distributed into 2 three-phase groups gives: 10 800 / 6 = 1800 W. From the installation point of view this is a suitable distribution as it gives a possibility to design the application for 230V and to fuse with 10A.
3. The total resistance per meter: Rt out of Ohms law for power R = U² / P
Rt = 230² / 1800 Rt= 29 W
4. The shortest possible length: The best cable in this example is, because of the expense, the TCPR-cable. In the data sheets you will find that the max. power per meter at an application in a bed of sand is 20 W/m for the TCPR.
This means shortest cable length of 1800 / 20 = 90m
5. The cable resistance per meter (R/m): R/m = Rt / l
29 / 90 = 0.3
The theoretical value will then be 90m heating cable 0.3 Ohm/m.
In the data sheets you will find that there is no cable with this resistance value. 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.
You then choose 0.25 Ohm/m
6. The real length of the cable: l = Rt / R/m
29 / 0.25 = 116
7. Check calculations: the total power Pt = U² / Rt
230² / (116 x 0.25 ) = 1824 W
Power/m: P/m = Pt / l.
1824 / 116 = 15.7 W/m
This power per meter is somewhat low and a new calculation is made with a shorter cable.
230² / (100 x 0.25) = 2116 W.
2116 / 100 = 21.16 W/m.
This is a better value but it certainly gives a higher surface power but a lower price, as the cables are shorter.
8. c/c-distance: c/c = surface x 100 / l
The answer is obtained in cm if you put the surface into m² and l into meter
6 x 100 / 100 = 6 cm
9. The solution: 6 cable runs each 100 m TCPR 0.25 Ohm/m are applied with a c/c-distance of c/c 6 cm.
The total power: 6 x 1824 = 10.94 kW
Surface power: 10.94 / 36 = 303 W/m²