So i was thinking of doing one my my home brew operations again.... I want to build a compact linear HPS system with an auxillary lamp and a shutter system. I cant see this coming to fruition any time soon due to the fabrication skills that I still need to develop more.... but it has been on the books for a minute.


What I am curious about specifically is the effects of having a 1000 or 1500 watt halogen lamp in close proximity to an HPS lamp/arctube while it is warming up. Would it heat up faster noticably? Would it cause life shortening due to some form of stress I am not thinking about? The application does not call for much life anyway. 500-2000 hours is acceptable.. but i am still curious. The auxillary lamps would only function up until complete warmup and then shut down. The hps lamps in question would just be some generic tubular grow light lamps from alibaba. I would not abuse name brand lamps like this.

That halogen in close proximity would almost certainly help the HPS lamp warm up faster to an extent, but I don't think that is a bad thing. In theory this would actually prolong lamp life, because the less time it spends running cold, the less the electrode material sputters away. But I kinda doubt that it would be to the extent of being noticeable.

Interesting sounding project, I would love to hear more about it

If you shut down halogen right at the point when HPS reaches it working temperature, I bet there is no risk of arc tube overheating, run-up time likely wil be somewhat shorter. But you must respect outer bulb and base temperature maximums. 1500W halogen emits a *lot* of heat, I heard back in 1980's there was even an idea of making a crematorium using a battery of IR halogens!

 

Both lamps emit their IR as somwhere between a "line source" and "point source". Which one of them closer describes your case depends on how close the lamps are to each other. If they are close, let's assume it is line source

Line source : Irradiation is reciprocal of the distance from the source

Point source : Irradiation is reciprocal of the distance from the source squared

Assuming an 1.5KW linear Halogen lamp within 1" from the outer glass of a 400W SON, which itself is a ~T17 tube containing a 8mm thick arctube. The arctube is about 1/2 of the length of the Halogen. (I am thinking here in terms of lamps common here, your exact lamps and sizes may vary)

So we have 2 possible paths of heat transfer by IR :

 - From halogen lamp to arctube (for wavelengths for which the SON glass is transparent)

 - From halogen lamp to SON outer glass, then separately from it to arctube (for wavelengths for which the SON glass is opaque)

We can assume (and it is close enough to reality) that the entire 1.5kW output of the Halogen is in the IR. Though i dont know how it then divides between IR for which the SON glass is transparent vs opaque



Transparent glass :

Arctube's part of the projection from the halogen :

8/(2∙50.8∙3.14)/2 = 0.013
Where :
8 - Arctube diameter
50.8 - Radius at which arctube is from the halogen
/2 - The arctube is 1/2 the length of the halogen

0.013∙1500 = 19.5



Opaque glass :

SON outer's part of the projection from the halogen :
60/360/2 = 0.083
Where :
60 - For T17 SON and a Halogen at 1" from it, the SON obstructs 60deg of the perimeter of the halogen
/2 - The SON arctube is 1/2 the length of the halogen. (The outer is more, but then not all the radiation inside will get to the arctube, so this can be still factored in here)

Arctube's part of the projection from the inside of the SON outer glass :
8/(2∙21.4)/2 = 0.093
Where :
8 - Arctube diameter
21.4 - Radius at which arctube is from a linear strip of the glass
No pi this time as the strip of glass is flat emitter, so the intensity seen by the arctube is according to cosine law
/2 - Half of the radiation is radiated to outside the SON

0.083∙0.093∙1500 = 11.6



So, the overall power gained by the arctube from the halogen by IR will be : X*19.5 + (1-X)*11.6 , with X between 0 and 1 depending on which part of the IR spectrum the SON glass is transparent to. Worst case is X=1

So the SON arctube gains extra 20W from the irradiation

The additional delta T caused by heat transfer through other means (convection heating the SON outer, and then by IR from it to the arctube) is probably insignificant

I'd say the effect is about same as adding +20W electrically to the arctube