Is power factor relative to heat output?

[ScrubNinja]

Would a 23w cfl with a pf of .4 produce roughly twice as much heat in cab as a 23w cfl with .8 pf?

From the wiki page:

The power factor of an AC electric power system is defined as the ratio of the real power flowing to the load to the apparent power, and is a number between 0 and 1 (frequently expressed as a percentage, e.g. 0.5 pf = 50% pf). Real power is the capacity of the circuit for performing work in a particular time. Apparent power is the product of the current and voltage of the circuit. Due to energy stored in the load and returned to the source, or due to a non-linear load that distorts the wave shape of the current drawn from the source, the apparent power can be greater than the real power.

In an electric power system, a load with low power factor draws more current than a load with a high power factor for the same amount of useful power transferred. The higher currents increase the energy lost in the distribution system, and require larger wires and other equipment. Because of the costs of larger equipment and wasted energy, electrical utilities will usually charge a higher cost to industrial or commercial customers where there is a low power factor.

That extra energy it's drawing must get expelled as heat inside the cab, correct? Thank you.

 

[JustanotherDave]

The only debates I've ever seen about power factor involve a minimum of 6Kw....

 

[ScrubNinja]

Thanks Dave. It is a common electrical thing that effects any appliance. My CFLs have a pf of .6 stamped on the side. My PLLs are rated >.9

Just trying to establish if my thinking is correct instead of firing off half assed knowledge. Would be great if an electrician could verify - this information is very important to CFL growers if what I'm saying is correct. Thank you for reading.

 

[madpenguin]

Your thinking is correct but hardly worth the effort involved to think about it in the first place. This is of course my opinion and not grounded in fact or studies. I don't think your going to get that much more heat with differing power factors. Nor are you likely to save any significant amount of money with differing power factors. This is all on par with watt loss. Not worth the effort in thinking about unless you have a massive grow. Atleast on the savings side anyway.

Try to find some CFL's of the same wattage but different PF's... Put an amp meter around each hot conductor feeding them in unison with a digital thermometer. Probably about the only way you'll get any real answers unless you google it and try to find some real case studies involving PF and heat.

 

[qbert]

short of googling equations for power dissipatation vs heat, I can say yes more heat produced. But remember, if you have 2 devices one with a .4 pf and one with .8 - the .4 isn't drawing twice as much current, its only wasting twice as much current. So while more heat will be produced by the less efficient device, it is unlikely to be anywhere near twice as much in most devices.

 

[Philosophelon]

First of all, I'm not an electrician, and it's been awhile since I've taken a physics class .

Some of the inefficiency is just caused by inductive loads (ballasts, fans, etc.), and the extra electricity for which you are billed is not actually used, it only appears as if it were used. Since some of the energy is stored in the load, it distorts the waveforms returning from the load so that they are out of step with the source energy waves, and it appears (to the elec. co.) that you are using more power than you are. This is not a big deal in most grows, but for a very large grow one can install a power conditioner that adjusts waveform for electricity returning to the source. You won't lower temps in your gardens by using a power conditioner, but if you are running a shitload of ballasts, it makes a difference in your electric bill, because it corrects the waveform returning to the source, and raises your power factor back to close to one. Here are a couple of graphs I stole from Wikipedia.


PF=1


PF=0.7


PF=0

As you can see, the waveform for for a power factor of zero is way different than the waveform for a power factor of one. The electric co. interprets this difference as increased electrical consumption (apparent power), even though the actual power consumed (real power) is the same. Electric companies bill based on apparent power, not real power.

 

[magiccannabus]

But remember, if you have 2 devices one with a .4 pf and one with .8 - the .4 isn't drawing twice as much current, its only wasting twice as much current.

Yes this is the key! If the one CFL with the .8 is *wasting* 2W, then the .4 would be wasting 4W. You just double the wasted power, which is why you also need to know the ballast factor and the efficacy of the lamp. Ballast factor data is easy to get, but lamp stats can sometimes be hard to find.

 

[ScrubNinja]

Thank you very much guys, really well explained.

So do I understand magiccannabus right that the pf stamped on my integrated ballast CFL is only for the ballast, and the coiled bulb part has it's own pf? Or were you referring to seperately ballasted shop light style fluoros?

 

[magiccannabus]

Any lamp driven by a ballast has losses at both the ballast, and the bulb. I guess though it does depend on how you define losses. Heat is generated instead of usable light. In some applications, like the Easy Bake oven(100W incandescent), bulb losses are actually not losses at all since you want all that heat. I think it's safe to say for our purposes that any power that makes heat = loss. Finding honest numbers on this for the lamp itself can be hard though. Any stage of a circuit inevitably has some losses. Even super-conductors have losses due to interfacing with power sources.