BigBudBalls
Well-Known Member
Its state legal, but not fed legal.
Experienced Electrician! Here to Answer Any and All Growroom Electrical Questions
- Thread starter bricktown73
- Start date
mmaaddmmaann
Active Member
hey big bud balls. Thanks for the heads up on bricktown. I do have my sensor now, a couple of them actually, I'll grab the specs in the morning and post em online for a look. Good hunting (and by hunting, I mean growing) Ladies and Gents.
Bucket head
Well-Known Member
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SupraSPL
Well-Known Member
Hi Bigbud. I have an important question that I rarely see much discussion about. If a 600 watt HPS with magnetic ballast has .5 power factor, is it true that the circuit will need to carry more than 2X as much current to operate it when compared against digital ballast with power factor of 1? If so, are these typical numbers you see on HPS ballasts?
I know this would be true in theory (true power versus apparent power) but I'm wondering how it translates in practice. I have a very cheap magnetic ballast 70w HPS and the power factor did in fact test about .5 on my Kill-A-Watt meter (80 watts true power, 160 watts apparent power).
I know this would be true in theory (true power versus apparent power) but I'm wondering how it translates in practice. I have a very cheap magnetic ballast 70w HPS and the power factor did in fact test about .5 on my Kill-A-Watt meter (80 watts true power, 160 watts apparent power).
I'm no electrician but something doesn't add up here. The way I was taught to calculate consumption is to separate one feed and put an amp clamp on it. Determine the amps going though the line and the line voltage and divide.
So if my 400w magnetic ballast uses say 460w, it would read like this. Watts = Amps x Volts.
460w = 3.83amps X 120volts
I don't understand how the actual power used could be doubled. If there is "X" amps and "X" voltage, math will give you the resulting watts of consumption.
bratva
Member
I have two rooms that are both on a 15 AMP circuit. I'm already running two 600 watt lights in one room and an inline fan so roughly around 12 AMPS are being used on that circuit. Is there a way to upgrade that 15 AMP circuit to a 30 AMP using the existing wiring? From searching around it seems the wiring needs to be changed out to a different gauge wire. I did notice the other room has a ceiling fan that is not connected to this 15 AMP circuit so the best way to go about it is to tap into this ceiling fan wiring/circuit and add two outlets. That is the cheapest and easiest way I can think to get this done without having to have these rooms re-wired or having to run a new circuit two accomadate these rooms.
Let me know what you guys think any and all feedback would be appreciated.
Thanks
Let me know what you guys think any and all feedback would be appreciated.
Thanks
hairyrabbit
Member
Do you have any tips in rigging power boxes or diverting power to cut down my power bill?
IAm5toned
Well-Known Member
the power factor is the amount of power actually used,(true power) vs the power that is used to get the juice to where it needs to go(apparent power).
resisitive loads like heating elements and incadescent lamps have an unbroken physical bath between line voltage, and neutral voltage. because of this, there is no frequency lag, the ac is in harmony so to speak, and the power factor is 1. no adjustment needed, working with resisitive loads is like working with a dc load.
inductive loads, like electronics, motors and ballasts/transformers rely on magnetic inductance (parts that dont actually touch) because of this, with every voltage cycle on an ac circuit, there is an inrush of current. that is the power factor. the current isnt actually used, per se, it just occurs because at the moment of flatline in the voltage cycle (0 volts) there is no electrical resistance to the circuit, so when the voltage cycle starts to climb (or drop) just for the briefest of milliseconds there is no resistance and alot of current flows. THAT, is the power factor
Power factor in linear circuit
Instantaneous and average power calculated from AC voltage and current with a unity power factor (φ=0, cosφ=1)
Instantaneous and average power calculated from AC voltage and current with a zero power factor (φ=90, cosφ=0)
Instantaneous and average power calculated from AC voltage and current with a lagging power factor (φ=45, cosφ=0.71)
In a purely resistive AC circuit, voltage and current waveforms are in step (or in phase), changing polarity at the same instant in each cycle. Where reactive loads are present, such as with capacitors or inductors, energy storage in the loads result in a time difference between the current and voltage waveforms. During each cycle of the AC voltage, extra energy, in addition to any energy consumed in the load, is temporarily stored in the load in electric or magnetic fields, and then returned to the power grid a fraction of a second later in the cycle. The "ebb and flow" of this nonproductive power increases the current in the line. Thus, a circuit with a low power factor will use higher currents to transfer a given quantity of real power than a circuit with a high power factor. A linear load does not change the shape of the waveform of the current, but may change the relative timing (phase) between voltage and current.
Circuits containing purely resistive heating elements (filament lamps, strip heaters, cooking stoves, etc.) have a power factor of 1.0. Circuits containing inductive or capacitive elements (electric motors, solenoid valves, lamp ballasts, and others ) often have a power factor below 1.0.
[edit] Definition and calculation
AC power flow has the three components: real power (P), measured in watts (W); apparent power (S), measured in volt-amperes (VA); and reactive power (Q), measured in reactive volt-amperes (var).
The power factor is defined as:
. (λ=P/S)
In the case of a perfectly sinusoidal waveform, P, Q and S can be expressed as vectors that form a vector triangle such that:
If φ is the phase angle between the current and voltage, then the power factor is equal to
, and:
Since the units are consistent, the power factor is by definition a dimensionless number between 0 and 1. When power factor is equal to 0, the energy flow is entirely reactive, and stored energy in the load returns to the source on each cycle. When the power factor is 1, all the energy supplied by the source is consumed by the load. Power factors are usually stated as "leading" or "lagging" to show the sign of the phase angle.
If a purely resistive load is connected to a power supply, current and voltage will change polarity in step, the power factor will be unity (1), and the electrical energy flows in a single direction across the network in each cycle. Inductive loads such as transformers and motors (any type of wound coil) consume reactive power with current waveform lagging the voltage. Capacitive loads such as capacitor banks or buried cable generate reactive power with current phase leading the voltage. Both types of loads will absorb energy during part of the AC cycle, which is stored in the device's magnetic or electric field, only to return this energy back to the source during the rest of the cycle.
For example, to get 1 kW of real power, if the power factor is unity, 1 kVA of apparent power needs to be transferred (1 kW ÷ 1 = 1 kVA). At low values of power factor, more apparent power needs to be transferred to get the same real power. To get 1 kW of real power at 0.2 power factor, 5 kVA of apparent power needs to be transferred (1 kW ÷ 0.2 = 5 kVA). This apparent power must be produced and transmitted to the load in the conventional fashion, and is subject to the usual distributed losses in the production and transmission processes.
Do I understand this correctly?
Basically a large current or "wave" is required to carry the "used" energy the device needs to operate, but when all is said and done the extra current is dissipated and the efficiency losses are in relation to the extra resistance required for carrying the extra current?
Basically a large current or "wave" is required to carry the "used" energy the device needs to operate, but when all is said and done the extra current is dissipated and the efficiency losses are in relation to the extra resistance required for carrying the extra current?
IAm5toned
Well-Known Member
no.
if that was the case we would be dealing with a DC circuit, not AC.
you see, AC current (alternating current) never remains at a constant voltage. in north america, on the 60hz system, at the start of the voltage cycle, the voltage is zero. then the voltage starts to rise as time goes forward, untill 177 volts positive is achieved. this is referred to as peak voltage. after peak voltage is achieved in the positive field, the current reverses direction, and the voltage devreases to zero volts, then continues to travel in the opposite direction, or negative field, untill it reaches 177 volts; then the voltage returns to 0, and another cycle begins. that complete cycle is known as 1 hz. it happens 60 times a second, its what the 60hz means you see on most electrical equipment. remember that during one complete cycle, the voltage is at 0 and there is no current flow 3 seperate times. keep that last thought in mind plz...
devices that have inductive loads have a unique property, in thatsome of the parts inside them that conduct electricity do not actual touch, instead the energized parts create a magnetic field that the current actually 'jumps' across. (it doesnt actual jump, its just easy to express it that way) its that magnetic field that causes the inrush of current that the power factor is compensating for, or more properly, it is the collapse and reformation of that magnetic field that uses the excess energy. the reason being is that the parts that utilitize that magnetic field have little or no resistance when they are not energized, think of them as the purest conductor known to man. an induction device operates at 99.997% efficiency, a simple transformer is the most effecient machine that man has ever devised. so in the nanosecond periods that precede and follow the 3 times the voltage cycle is at 0 volts, these inductive devices lose all resistance to electrical current. since resistance is the only thing that actual holds electrical power back, during those nanoseconds where there is little or no resistance because the magnetic field has collapsed, an unbelievable amount of power flows through your device. im talking 100,000 amps or more. however the duration of this current is extremely short, barely even computable, but if you plan on using a peice of equipment for a period greater than 6 hours of use, then it is something that must be accounted for, because the conductors feeding your device will start to heat up if it is not properly compensated for.
thats power factor in a nutshell, cliffnotes version. it usually takes a good hour long lecture with notes for people to grasp i properly, so i hope i didnt confuse you to much with the laymans version...
if that was the case we would be dealing with a DC circuit, not AC.
you see, AC current (alternating current) never remains at a constant voltage. in north america, on the 60hz system, at the start of the voltage cycle, the voltage is zero. then the voltage starts to rise as time goes forward, untill 177 volts positive is achieved. this is referred to as peak voltage. after peak voltage is achieved in the positive field, the current reverses direction, and the voltage devreases to zero volts, then continues to travel in the opposite direction, or negative field, untill it reaches 177 volts; then the voltage returns to 0, and another cycle begins. that complete cycle is known as 1 hz. it happens 60 times a second, its what the 60hz means you see on most electrical equipment. remember that during one complete cycle, the voltage is at 0 and there is no current flow 3 seperate times. keep that last thought in mind plz...
devices that have inductive loads have a unique property, in thatsome of the parts inside them that conduct electricity do not actual touch, instead the energized parts create a magnetic field that the current actually 'jumps' across. (it doesnt actual jump, its just easy to express it that way) its that magnetic field that causes the inrush of current that the power factor is compensating for, or more properly, it is the collapse and reformation of that magnetic field that uses the excess energy. the reason being is that the parts that utilitize that magnetic field have little or no resistance when they are not energized, think of them as the purest conductor known to man. an induction device operates at 99.997% efficiency, a simple transformer is the most effecient machine that man has ever devised. so in the nanosecond periods that precede and follow the 3 times the voltage cycle is at 0 volts, these inductive devices lose all resistance to electrical current. since resistance is the only thing that actual holds electrical power back, during those nanoseconds where there is little or no resistance because the magnetic field has collapsed, an unbelievable amount of power flows through your device. im talking 100,000 amps or more. however the duration of this current is extremely short, barely even computable, but if you plan on using a peice of equipment for a period greater than 6 hours of use, then it is something that must be accounted for, because the conductors feeding your device will start to heat up if it is not properly compensated for.
thats power factor in a nutshell, cliffnotes version. it usually takes a good hour long lecture with notes for people to grasp i properly, so i hope i didnt confuse you to much with the laymans version...
SupraSPL
Well-Known Member
"This apparent power must be produced and transmitted to the load in the conventional fashion, and is subject to the usual distributed losses in the production and transmission processes."
So it is true then , my magnetically ballasted 70 watt HPS requires 160 (VA) of apparent power and I would select the wire gauge based on that current (1.35A). My electric bill will be based on 80 watts. Same thing applies to compact fluorescent bulbs or magnetically ballasted fluoro tubes, most of which have a power factor of .5.
It is worth mentioning, according to the Kill-A-Watt, my fluoro backlit LCD TV has a power factor of .99. Because the backlight is dimmable, it is electronically ballasted.
So it is true then , my magnetically ballasted 70 watt HPS requires 160 (VA) of apparent power and I would select the wire gauge based on that current (1.35A). My electric bill will be based on 80 watts. Same thing applies to compact fluorescent bulbs or magnetically ballasted fluoro tubes, most of which have a power factor of .5.
It is worth mentioning, according to the Kill-A-Watt, my fluoro backlit LCD TV has a power factor of .99. Because the backlight is dimmable, it is electronically ballasted.
IAm5toned
Well-Known Member
you would also have to rate the ampacity for continuous duty, which is 120% of the true power. so if you had a load of 10 amps, you would size your conductor for a 12 amp load, and that would be compensating for a continuous duty cycle."This apparent power must be produced and transmitted to the load in the conventional fashion, and is subject to the usual distributed losses in the production and transmission processes."
So it is true then , my magnetically ballasted 70 watt HPS requires 160 (VA) of apparent power and I would select the wire gauge based on that current (1.35A). My electric bill will be based on 80 watts. Same thing applies to compact fluorescent bulbs or magnetically ballasted fluoro tubes, most of which have a power factor of .5.
It is worth mentioning, according to the Kill-A-Watt, my fluoro backlit LCD TV has a power factor of .99. Because the backlight is dimmable, it is electronically ballasted.
however going this far is usually not nessecary when dealing with small loads you would find connected to branch circuits; (less than 20 amps) unless your trying to max out a circuit's available ampacity, and that is something done primarily in a remodel type situation where replacing the wire is not an option.
as a rule of thumb, #12 is good for up to 18 amps continuous duty, and #14 is good for up to 12 amps of continuous duty usage. i would not reccomend to anyone the use of wire smaller than #14, unless it is a specialized limited use application.
hannsp4100
Member
Hi, I just bought a full package of the BC northern lights grow box and im really bad with the whole electric part of the usage.
altogether the system will be using 800 KWh. and usually the system has to run around 14 hrs.
Now A - how much will by electricity be if its .14 cents
B - just for my knowledge so that I dont have to bother RIU ppl again, 800 KWh means KW per hour? or per month? or how is it calculated?
and would this generator provide enough electricity to run the 800 KWh system:
http://cgi.ebay.com/Triple-Fuel-Honda-8750-W-Generator-ELECTRIC-START_W0QQitemZ390140254217QQcmdZViewItemQQptZBI_Generators?hash=item5ad62bd809#ht_4057wt_1165
or how about 2 of these 300 KWh solar panels? will the 2 provide almost 60% of the electricity?
http://cgi.ebay.com/300kWh-Month-Large-Off-Grid-Solar-Panel-Kit-w-Batteries_W0QQitemZ250397167443QQcmdZViewItemQQptZLH_DefaultDomain_0?hash=item3a4cd58f53#ht_4954wt_939
thanks!
altogether the system will be using 800 KWh. and usually the system has to run around 14 hrs.
Now A - how much will by electricity be if its .14 cents
B - just for my knowledge so that I dont have to bother RIU ppl again, 800 KWh means KW per hour? or per month? or how is it calculated?
and would this generator provide enough electricity to run the 800 KWh system:
http://cgi.ebay.com/Triple-Fuel-Honda-8750-W-Generator-ELECTRIC-START_W0QQitemZ390140254217QQcmdZViewItemQQptZBI_Generators?hash=item5ad62bd809#ht_4057wt_1165
or how about 2 of these 300 KWh solar panels? will the 2 provide almost 60% of the electricity?
http://cgi.ebay.com/300kWh-Month-Large-Off-Grid-Solar-Panel-Kit-w-Batteries_W0QQitemZ250397167443QQcmdZViewItemQQptZLH_DefaultDomain_0?hash=item3a4cd58f53#ht_4954wt_939
thanks!
IAm5toned
Well-Known Member
look at the fan closely. somewhere on it it will tell you the amperage of the fan motors. add them up, if the total fan motor amperage is greater than 450 ma, then the wall adapter is too small
SupraSPL
Well-Known Member
@HannsP. 800 kilo watt hours could be a monthly usage figure who knows
An example of how to calculate cost, running 800 watts true power for 14 hours a day would be :
800 watts * .58day = 464 average watts per hour
464 watts * 24 hours = 11100 (11.1 KWh)
11.1 * $0.14 = $1.50 per day
The tricky thing is to try and find out how many watts you are actually drawing. I bought a Kill-A-Watt meter to help sort that out $23 shipped.
You can estimate if you list every piece of equipment, lights fans etc and read their AC input power. If you have AC or heating power usage then a Kill-A-Watt would be helpful because it can keep track of your accumulated energy consumption and then you can average it.
An example of how to calculate cost, running 800 watts true power for 14 hours a day would be :
800 watts * .58day = 464 average watts per hour
464 watts * 24 hours = 11100 (11.1 KWh)
11.1 * $0.14 = $1.50 per day
The tricky thing is to try and find out how many watts you are actually drawing. I bought a Kill-A-Watt meter to help sort that out $23 shipped.
You can estimate if you list every piece of equipment, lights fans etc and read their AC input power. If you have AC or heating power usage then a Kill-A-Watt would be helpful because it can keep track of your accumulated energy consumption and then you can average it.
handyman23
Member
i've got a question for u m8 
Regarding the wattage use of bulbs, i'm gonna use 5x 250W hps lamps. Well the question is, what do you think will be the maximum use of wattage i can use on my growroom without my electrition bill getting too high, and the electro company will be wondering about me growin, like reporting me t2 the cops... any suggestions?
Regarding the wattage use of bulbs, i'm gonna use 5x 250W hps lamps. Well the question is, what do you think will be the maximum use of wattage i can use on my growroom without my electrition bill getting too high, and the electro company will be wondering about me growin, like reporting me t2 the cops... any suggestions?