Drafting symbol for "High Efficacy Recessed Light Fixture"

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tallgirl

Senior Member
Location
Great White North
Occupation
Controls Systems firmware engineer
Yep. Avoid the hassle, just define it in the schedule. I cannot tell you how many people don't use the standard symbols from NFPA 170 when they lay out a fire alarm project. Well, actually I can; all of them don't.
I could put the schedule in the same CAD layer with the fixtures themselves. I've been avoiding non-standard symbols because I didn't want to have a schedule taking up room on a piece of paper.

Also, yeah, need to lay out the smokes and CO sensors some day.
 

Besoeker3

Senior Member
Location
UK
Occupation
Retired Electrical Engineer
Eight in your BATHROOM?

I'm putting 6 in a BEDROOM. Are the 4 halogens for heat? Serious question.
Yes, eight in the bathroom. It is quite large.
The Halogen ones are GU10.. The LED ones are GU5.3

When I replace the Halogens I will replace them with LEDs
 

gadfly56

Senior Member
Location
New Jersey
Occupation
Professional Engineer, Fire & Life Safety
That would only be true in space, where all the heat loss is radiant, i.e. photons. But some of the heat loss will be conductive or convective, so some of the energy loss will be as heat. That is, increased average kinetic energy of the molecules of a substance.

Cheers, Wayne
Yes, infrared photons. That's what we generally sense as "heat". That's why radiant heaters work.
 

gadfly56

Senior Member
Location
New Jersey
Occupation
Professional Engineer, Fire & Life Safety
Correct, but not all heat transfer is radiant, so not all heat-based energy loss will be in the form of photons.

Cheers, Wayne
In the end, all heat transfer is radiant. There's no convection or conduction in inter-galactic space.
 

kwired

Electron manager
Location
NE Nebraska
In the end, all heat transfer is radiant. There's no convection or conduction in inter-galactic space.
We can maybe say that if convection or conduction ability is there there can be more heat lost in a particular application that isn't about producing heat? However depending on the application that may or may not be desirable when it comes to protecting a particular component from overheating as well and you need to factor convection, conduction, and/or radiant effects into your cooling scheme.
 

gadfly56

Senior Member
Location
New Jersey
Occupation
Professional Engineer, Fire & Life Safety
We can maybe say that if convection or conduction ability is there there can be more heat lost in a particular application that isn't about producing heat? However depending on the application that may or may not be desirable when it comes to protecting a particular component from overheating as well and you need to factor convection, conduction, and/or radiant effects into your cooling scheme.
As usual, we find ourselves down a rabbit hole. Yes, locally, convection, conduction, and radiation will all play a part depending on conditions, and only at the end of a variable chain is it all reduced to thermalized photons. The following is from our good friends at Wiki and seems to cover all cases:

Luminous efficacy is a measure of how well a light source produces visible light. It is the ratio of luminous flux to power, measured in lumens per watt in the International System of Units (SI). Depending on context, the power can be either the radiant flux of the source's output, or it can be the total power (electric power, chemical energy, or others) consumed by the source. Which sense of the term is intended must usually be inferred from the context, and is sometimes unclear. The former sense is sometimes called luminous efficacy of radiation, and the latter luminous efficacy of a light source or overall luminous efficacy.
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
In the end, all heat transfer is radiant. There's no convection or conduction in inter-galactic space.
Yes, but we are not operating the bulb in outer space. So if you use any reasonable method of measuring the net photon flux produced by the bulb in an air-filled room, you are going to get a rate of energy transfer via photons that is less than the input electrical power.

Cheers, Wayne
 
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