Lighting Question

[Gnasher]

When calculating the cable size don't forget to multiply the load on each circuit/phase by 1.8 to take into account the starting current of discharge lighting. This is the current you need to cater for for your cable and protection device. On top of this you need to take into account correction factors such as installation method, grouping factors, thermal insulation, and any other relevant correction factors.

The calculation would be the highest of the 3 phases. Number of lights on each phase

Number of lights 10 multiplied by wattage 400 multiplied by 1.8 divided by 230 this will give you your load.
You will then need to take into account the correction factors that will ultimately determine your cable size.

Next step is confirm volt drop and calculated Zs value for selected protection device.

Would be interesting to see a photograph of termination method and any local fusing as you may have a large value fuse at the origin of the circuit.

 

[HandySparks]

When calculating the cable size don't forget to multiply the load on each circuit/phase by 1.8 to take into account the starting current of discharge lighting. This is the current you need to cater for for your cable and protection device.

If the 1.8 multiplier was to allow for power factor, then yes, it needs to be allowed for when calculating cable size. If it's for short term starting current, then surely you just need to check that the volt drop during starting won't be so great that the lights fail to start.

 

[Gnasher]

The 1.8 is not for power factor, it is used to calculate the maximum current requirements for the circuit being designed regardless of it is only under those conditions for a short time. This calculated current will be used in calculating the volt drop when starting. Amongst others.
The problem is that on normal domestic installs standard circuits tend to be used. 2.5 with a 32A MCB for a ring. 1.5 with a 6A MCB for lighting. On the whole this is fine, but during install the installer should be aware of the correction factors and reasons for them. For instance if the cable passes through thermal insulation all around the cable for more than 500mm the current carrying capacity is reduced by a huge amount.
The lighting circuit in question is not a standard domestic circuit.
So the correct way is a full design. That includes using all the correction factors stated in BS7671. The outcomes of thes calculations can be supprising.
Don't forget that on an EIC there is a requirement to sign for design not best guess.

Sorry this is a not giving a full design example but there are so many variants that it is not an easy thing to do in a few words, but correct design is often overlooked or not fully understood. I can understand this as it is not always easy and is a subject in itself. One course worth doing is the City and Guilds design and verification this is a 10 evening course with a four hour written exam. Not easy but well worth it. Not my intention to insult or upset anyone on here just trying to help and share my experience.

One book worth getting hold of is the IEE Electrical installation guide if they still do it it is the same format as the onsite guide.

 

[Darkwood]

Q - How many 400w metal halide lamps can I have on 1 6amp(b) breaker, the lamps will be switched on together with one switch, do you calc's with and without the correction factor and you will start to understand why its neccessary to apply to inductive loads. There are also added issues with my example question that colleges just don't teach anymore that can lead you to make a very expensive design flaw if you can guess it.

 

[Pete999]

We have a 3 ph DB. I am a little confused with the cable sizing that has been carried out and would like to verify it for my self.

A 4core 16mm (3 ph and N) goes from this DB to a jb in the field, through an underground cable route. At this JB the cable splits. One cable goes off and feeds 14 400W HQI lights and another cable feeds 12 400 W HQI Lights and three street lights.

From a sizing point of view is this treated as one three phase circuit with 29 lights on it?

Thanks

This JB in the field, is it just a joint or is there some kind of feeder pillar, whereby the cables going to the lighting circuits are fused down?

 

[Leesparkykent]

The power factor is usually less than unity so current is greater, So long as the power factor of a discharge lighting circuit is not less than 0.85, the current demand for the circuit can be calculated from (Watts x 1.8)/V. How ever the 1.8 value is a general rule of thumb and should only really be used in the absence of the manufacturers instructions .To calculate if a circuit breaker could withstand the inrush on start up you would need to use crest factor and the instantaneous tripping current.