4' T8 - How many per outlet?

[jman103099]

Hi all -

I just moved into a new house and I am ready to get started on the garage. It's a three car with three single bulbs in the ceiling now. I really want to install some type of fluorescent lighting, but my question is how?

I would like the bulbs to remain on the switch they are on now and install the fluorescents on pull chains that plug into standard outlets if I can. So my question is, how many of these 2 bulb 4' T8s can I plug into one outlet? There are two outlets in my ceiling that power the two garage doors. Is this a good idea? I dont know anything about electrical... sorry!

THANKS!

 

[mmhouse]

Well, here's what I did in a similar situation. It may not be the 'right' way but since I'm not much of an electrician either it worked great for me.

I removed the three porcelain fixtures (old incandescent bulbs) and wired in three 4-tube 48" fluorescent fixtures over them. Since these are mounted directly over the existing boxes it was a piece of cake. These fixtures were basically located in the center of each bay at the front of where each vehicle is parked. This alone made a big difference (improvement) in lighting.

Next I mounted four 2-tube fixtures toward the rear of the garage, one near each wall and one between each bay. I wired cords and pull chain switches into each fixture and ran the cords on the ceiling using the plastic, self-adhesive extension cord covers to the garage door opener outlets....so I have two 2-tube fixtures plugged into each opener outlet (only one plug was open in each one so I just used 3-prong splitters in the open plugs-ins).

I also hung another 2-tube shop lite over my bench. This one I wired into the center 4-tube fixture and also put a pull-chain on it. I could have just used a shop lite for this one but wanted all the lights to match.

All the fixtures are the same brand/design with lenses and are surface mounted.

Now when I turn on my original switch the three 4-tube fixtures come on for general lighting. I use the pull chains to turn on the others when I'm working in the garage. It was a very simple install not requiring any fancy electrical skills or tearing into my drywall or attic and I'm very pleased with the result.

Sorry, I can't answer your question directly but I can tell you that two 2-tube lights plugged into each open outlet in my case works fine.

 

[jman103099]

That's pretty much what I want to do... I was just worried about overloading something but it sounds like I will have no worries. Thanks for the reply.

I believe my garage (outlets, openers, bulbs) are all going to one circuit/breaker... does this pose any further problems?

 

[mmhouse]

That's pretty much what I want to do... I was just worried about overloading something but it sounds like I will have no worries. Thanks for the reply.

I believe my garage (outlets, openers, bulbs) are all going to one circuit/breaker... does this pose any further problems?

I'll need to leave that to the experts to answer.

I would suggest you make sure you know how many breakers are involved by turning them off and checking to see what's on each one.

 

[tomstin]

I have a similar pull chain setup as mmhouse. To get to your question, and I am not an electrician but have some basic knowledge. I'm surprised the electricians have not weighed in.

The T8s appear to come in 15 and 32 watt varities. It should be on you bulbs so take a look. Let's assume 32 watt. So thats 64 watts per fixture. What wattage are your current bulbs? Probably 60 or 100 watt.

So if you want to add more fixtures with the pull chain method like mmhouse. At 64 watt per fixture divide by 110 (volts) and you get to .6 amps per fixture. Take a look at the garage door openers and see what they draw (it should be on the label). Add up anything else on that circuit to get the total load. Then, what's the size of the breaker, probably 15 amps (its written on the breaker).

You will probably not overload the circuit with the T8s unless you go nuts. It all depends on what else in on or will be on the circuit when the lights are on. Normally you would not run both garage door openers at the same time, but it's possible. Any other high draw appliances on the circuit? Refrig? Compressor? Heat gun?

Technically, you could overload the switch versus the breaker, but if you installed 6 fixtures, you are still less than 4 amps and I'll bet the switch is rated for at least 10 amps.

 

[mmhouse]

Here's a photo of my setup.....

The three 4-tube fixtures replaced the incandescent bulbs and are on the garage switch. All the 2-tube fixtures are on pull chains and the four at the rear (three visible) are plugged into the open outlets used by the garage door openers.

 

[Scotto]

You'll probably be OK to add more lights, but find what else is on that garage lighting circuit.

 

[jman103099]

I dont plan on going nuts with the lights... currently in the garage there are three 100W bulbs, two garage openers, and three (I wish more) electrical outlets on the circuit. My shop set up is going to be in the basement, so I wont be running any major tools in the garage when the lights are on.

I feel pretty safe putting up three or four 2-bulb T8s out there. Thanks for all of the responses!

 

[ktm450]

to determine actual current draw/fixture you CANNOT assume a resistive load with floresent lamps. (i.e. assume 2 40watt bulbs =80 watts). While the power consumed is 80 watts, the current is NOT 80Watts/120V= .6 A. actual current will be signficantly higher because of power factor and power in the ballast. I use comerical electronic ballast fixtures. For these fixtures you need to mulitply lamps watts by 1.5 to determine total VA draw. So, for example a two tube 32W fixture, you will draw current equivalent to a 100watt bulb,or about 0.9A not 64W. Older magnetic ballasts or cheap electronic ballasts are even worse.

the fixtures you have should have a note near the ballast telling you what the VA requirements are per fixture.

10, maybe 12 of these fixtures would be the most you would want to use on a 15A circuit, since you normally don't want the continous current load on a circuit to be more than about 80% of the circuit rating to comply with NEC and most local code requirements.

 

[Marcusr13]

A ballast converts AC to DC. Watts used is watts used based off from voltage. All ballasts have an amperage rating, but with T8's they are usually milliamps, which are stated on the ballast itself. Out of curiosity, how do you, KTM450 figure that a ballast would be drawing so much amperage?

Lumens produced as far as an equivalence to incandescent is not relevant since fluorescent is 1/4 of the energy usage? More energy efficient doesn't matter about the potential energy that a ballast can consume it's all about power output. If a bulb is 40 watts, that's the power it consumes. IE, 40watts=.33 amps. The ballasts are always rated over what the lamp can handle, at least what I have seen. I only see that voltage drop should be the only concern. Correct me if I am wrong. I just don't see the point you are trying to make. Is the power used differ in the equation because of the change to DC? Please let me know if anyone knows the answer, I really want to learn this stuff. According to AC rules, fluorescent fixtures pull almost nothing.

 

[Fast Orange]

Marcus-
You sir,have applied but a mere tidbit of correct information,mixed in a bit of misinformed heresay,tossed in some non-applicable theory and gotten the whole mess totally wrong.
First-a ballast does not covert AC to DC. A ballast,whether for flourescent or HID application,changes the incoming voltage to the voltage needed by the lamps to light,and by use of capacitors,creates the voltage spikes to initially cause the lamp to ignite.
Next-ballasts do have an amperage rating.This rating may be stated in milliamps,but it is still amperage.As an example,600mA=.6A This rating will also always be more than the total power consumed by the lamps.The difference is the amount of power lost by the ballast,usually in the form of heat.The idea is- power in =power out +ballast loss.
According to the NEC,lighting circuits have to be derated because they are a continuos load(IE-in use for more than 3 hrs at a time).The derating factor is .8,so a
15A circuit can only be loaded to 12A,and a 20A circuit to 16A.Using KTM450 's approximation of .9A per fixture and a fudge factor,I would put 12 2 lamp fixtures on a 15A circuit and 16 on a 20A circuit,providing that the circuits are dedicated lighting (only)circuits.
Marcus-there is much that can be learned here "about this stuff",but please post in the form of a question,instead of a challenge to someone who has a better handle on the subject than you do-you'll get better info that way.

 

[ktm450]

A ballast converts AC to DC. Watts used is watts used based off from voltage. All ballasts have an amperage rating, but with T8's they are usually milliamps, which are stated on the ballast itself. Out of curiosity, how do you, KTM450 figure that a ballast would be drawing so much amperage?

Lumens produced as far as an equivalence to incandescent is not relevant since fluorescent is 1/4 of the energy usage? More energy efficient doesn't matter about the potential energy that a ballast can consume it's all about power output. If a bulb is 40 watts, that's the power it consumes. IE, 40watts=.33 amps. The ballasts are always rated over what the lamp can handle, at least what I have seen. I only see that voltage drop should be the only concern. Correct me if I am wrong. I just don't see the point you are trying to make. Is the power used differ in the equation because of the change to DC? Please let me know if anyone knows the answer, I really want to learn this stuff. According to AC rules, fluorescent fixtures pull almost nothing.

the amp draw I quoted was taken directly from the nameplate in the fixture.

It states, as I mentioned, to calcuate VA draw (volts * amps) to multiply the lamps watts * 1.5. So, for example in the case of a two tube 32W T8 fixture which does indeed disipate very close to 64 watts, (bulbs + some loss in the ballast) the current is not 64W/120V = 0.5A, but ( 64*1.5)/120V = 0.8A. so while the fixture consumes 64 Watts, it draws the same line current as a 96 watt resistive load. The power meter on the side of your house will charge you for the 64 watts you are using, but if you just measured the current going into the house you would calculate you were using 96 watts of power.

The thing to know is that current draw from a non resisitve load (i.e. motors, florescent lights, most electronic equipment, TV's etc) is HIGHER than the current draw you would calculate using from the power draw. In your house plain lightbulbs and resistive heaters, toasters, ovens are two examples of almost pure resistive loads. Without going into lots of detail, current and voltage in these applications are NOT in phase due to the reactive load, either capacitive or inductive or a combination. This results in higher current than you would expect. It also means for these kinds of loads you can't calcuate power by multiplying volt * amps, nor can you calculate current draw in the line by dividing watts by line volts. The first overstates the power, the second understates the current draw.

As an electrical engineering student decades ago, we spent almost 1 full semester on this and related topics, it's not a trivial topic to understand, don't feel bad. It is however, a big issue for the power companies since they don't get paid to supply the extra current, but transformers, generator and lines need to be sized to provide it. Big industrial users that have these kinds of loads are very concerned as well, since without correction for this effect (know as power factor and power factor correction) the power company will charge them as if they were actually consuming the 96 watts, resuting in a big increase in their power bill.

hope this helps.

 

[Shocker]

Thanks a bunch. Helps me out a lot!

 

[nissan_crawler]

My dual bulb, 4' t8 32w fixtures pull .8a, IIRC.