Fluorescent ballast efficiency 1

[chunia]

Hi,
Does anybody know how to calculate efficiency for fluorescent ballast? I need it for room heat load calculations.

TIA

 

[Senselessticker]

Note that Ballast Efficacy Factor (BEF), not ballast efficiency, can be calculated by:

BEF = BF/input power; where BF = ballast factor

 

[chunia]

Senselessticker,

Ballast Efficacy Factor (BEF) useed to compare two ballasts (AFAIK). Is it any way to get to efficiency from BEF?

 

[Senselessticker]

If you are trying to determine the amount of "heat" creeping out of the ballast and into the "space" that must be cooled.....consider the following:

Its generally the case if you are providing info to an HVAC designer/engineer: to THEM, a Watt is a Watt is a Watt to them regardless of the eff. of the device. I disagree with this approach, but then again, I failed thermo a couple times in college.

That being said: what about the heat being dissapated by the bulbs?

Let's say you have a fluorscent fixture with an electronic ballast that drives two T-8 32W bulbs. You will need to know either the input power to the ballast, or the ballast factor (BF) to determine the amount of total Watts (and don't forget pf).

Example: of two T-8 32W bulbs in fixture with 80% BF and .95 pf ballast:

Total Watts in space = .8 X 2 X 32 X .95 = 48.64 Watts

I don't agree that amount of "heat" is actually released into a room (or space), but in my experience, that is how an HVAC engineer will interpret.

 

[davidbeach]

Input watts into the ballast become heat in the space to the extent the light doesn't go out a window. Some of it will be watts radiated out of the ballast and lamps themselves; the rest of it will be the energy of the light heating up the surfaces lit. Some from an HVAC standpoint they are close enough by considering total power in.

 

[waross]

Hello davidbeach
Have I got this right?
Amps in, times volts in, times power factor?
What goes in, comes out, regardless of efficiency.
An inefficient ballast will draw more current for the same output so the formula does not need to consider efficiency if it considers current in.
respectfully

 

[davidbeach]

Yep.

 

[Senselessticker]

Waross...from heating standpoint....yes...not necessarily from a lighting standpoint when considering bang for buck in lumens per watt.

 

[chunia]

Gentlemen, thank you for answers

I called tech.support at Universal ballasts (

http://www.universalballast.com/productLines/ballast/line_fluor_elec.html)

and they told me to calculate loses on the ballast use formula:

Loses = Watts in – Total lamp wattage X Ballast factor

If we will take B232IUNVHP-B ballast (

http://www.universalballast.com/productLines/ap_sheets/ELFB/B232IUNVHP-B.pdf)

Loses = 59-(2*32)*0.88 = 2.68 Wt (power factor >.99 assume =1)
Efficiency = 95.5%

For ballast B259IUNVHP-A (

http://www.universalballast.com/productLines/ap_sheets/ELFB/B259IUNVHP-A.pdf)

Loses = 113-(2*59)*0.88 = 9.16 Wt (power factor >.99 assume =1)
Efficiency = 91.9%

 

[chunia]

davidbeach, you are right. Light photon emitted by the bulb has energy, and this energy transfers to heat, when object absorbs light.
For heat load calculation amount of heat produced by lamp-ballast combination equal to ballast wattage (IMHO).