There are zero amps in a 6kW 3 phase heater. Amperage is the result of dividing the Watts by the Voltage. A = W/E. Without stating the voltage the heater operates on the amperage can not be calculated.
The equation you are looking for is, Amps = kW x 1000/1.73 x Volts x PF. 6000 watts is the same as 6 kW. The power factor constant to use should be .9.
There are zero amps in 6600 watts. Watts are the product of amps times volts. W = A x V. To find amperage use the following equation, A = W/V, so as you can see a voltage value is needed in the equation to result in an amperage.
I'm assuming the electrical system is single phase, such as a home. If not, you need to hire someone to do the work for you. First, 6000 watts at 120v is 50 amps, so if your load is truly 6000 watts, 30 amps isn't enough. If your 6000 watts is 240v, which would draw 25 amps, then 30 amps is plenty. Ten gauge wire may not be placed an a breaker or fuse larger than 30 amps.
Six KVA is the same as 6000 watts. As you can see, the appliances have to be totaled up to the amount of 6000 watts to see how many can be used. Each device has its own wattage on the manufactures label and it is usually different for different appliances.
You better have a 10,000 watt capable generator.
Efficiency is measured as the ratio of power output to power input. In this case the power input of the generator is 240V * 25A = 6000 VA however the stated losses are 900 W so the power output is 6000 - 900 = 5100W. Then the efficiency would be 5100/6000 = 0.85 or 85% efficient.
The equation that you are looking for is Amps = Watts/Volts. There are 6000 watts in 6kW.
Watts divided by volts equals amps so 6000 divided by 120 = 50
The equation that you are looking for is I = W/E. Amps = Watts/Volts.
There are zero amps in 6600 watts. Watts are the product of amps times volts. W = A x V. To find amperage use the following equation, A = W/V, so as you can see a voltage value is needed in the equation to result in an amperage.
1600 watts is unusually low for electric baseboard heaters! These usually are rated at about 6000 watts.Typical electric baseboard heaters operate on 220 volts thus you heater would pull about 7.3 amperes and 16 gauge wire can easily handle this.A standard 6000 watt baseboard heater pulls about 27 amperes and needs 10 gauge wire.
6000 watts divided by 14.2 volts (12 volt car system operates near 14.2 volts), gives you 422 amps. You would need 4 alternators rated at 110 amps each to make 6000 watts of power at 12 volts. If the system were 24 volts, you would only need 2.
Power is volts times amperes, so 120 V and 50 A would be 6000 watts. However, it also depends on phase angle and power factor, something that is related to reactive loads such as motors and power supplies. As a result, the power may actually be less, so it is more correct to say 6000 volt-amps. The case of 6000 watts being the same as 6000 volt-amps is only true for a purely resistive load such as a toaster.
I'm assuming the electrical system is single phase, such as a home. If not, you need to hire someone to do the work for you. First, 6000 watts at 120v is 50 amps, so if your load is truly 6000 watts, 30 amps isn't enough. If your 6000 watts is 240v, which would draw 25 amps, then 30 amps is plenty. Ten gauge wire may not be placed an a breaker or fuse larger than 30 amps.
Yes, Power (in watts) equals voltage times current (Amps). Amps = Watts/Volts If your generator is 6000 watts that would operate a 120 volt load up to 50 amps (6000/120=50). If you are running a motor or compressor, the initial inrush of current is about 3 times higher so 14 X 3 = 42amps. Should do fine...
You failed to state the input voltage. To calculate the amps, use the formula: Amps * Volts = Watts X * volts = 6000 (If we guess at 12 Volts) X * 12 = 6000 X = 6000/12 X = 500 Which is a pretty big fuses, but to make matters worse, you would need to add about 10% or more for power lost to the inefficiency of the inverter. A 600 Amp fuse is an industrial device; hope you're planning on expensive. <<>> The 6000 watt rating will be the output wattage usually at 120 volts. 6000/120 = 50 amps. If the output is a three wire output at 240 volts the amperage will be 25 amps. For a 50 amp circuit a #6 copper conductor will be needed from a 60 amp breaker. For a 25 amp circuit a #10 copper conductor will be needed 30 amp breaker.
Any two legs of a three phase system are classed as single phase. If a 208 volt device is connected across two legs of a 240 volt system its current draw will be higher. It is not recommended to connect a 208 volt device to a 240 volt supply. Using a constant resistive load of 6000 watts for an example, the resistance of the unit at 208 volts has to be found. R = E (squared)/W = 7.21 ohms. Now using the formula to find amperage at this resistance I = E/R, 208/7.21 = 28.84 amp at 208 volts. If the 208 volt device is connected to a 240 volt supply the following condition happens. I = E/R, 240/7.21 = 33.29 amps. The device in now definitely overloaded and instead of its rated 6000 watts it is outputting W = A x V, 33.29 x 240 = 7989 watts almost 2000 watts more than the unit is rated at.Using a 240 volt rated device on 208 volts is more forgiving and is done quite often. Using the same 6000 watts on 240 volts the resistance is calculated to be R = E (squared)/W = 9.6 ohms. If the 240 volt device is connected to a 208 volt supply the following condition happens. I = E/R, 208/9.6 = 21.67 amps. To check this connection's wattage W = A x V, 21.67 x 208 = 4507 watts or about 1500 watts below its rated output at 240 volts.If the connected load is a constant wattage like a motor, the amperage and voltage will change to maintain the constant wattage. I = W/E. A motor rated at 6000 watts at 240 volts will draw 25 amps. A motor rated at 6000 watts at 208 volts will draw 28.8 amps.The CEC states that if a 208 volt rated motor is connected to a 240 volt supply its full load current has to be increased by 10 % for its overload protection.If the three phase source is delta connected, and the neutral/ground is at the center tap of one of the phase windings, then there is 208 volts available from neutral to the third phase leg. In this case, there would be no mismatch of voltage, and the 208 volt load will operate as designed.This particular configuration is not so common. It is sometimes used in a light industrial setting where a fourth transformer for the 120/240 split phase portion is not used - it provides 240 three phase delta only, 120/240 split single phase, residential style, and 208 single phase.See related links below (High Leg Delta)Note that 240 three phase loads are also not so common. In the case where 480 three phase is required in this configuration, then there is a step up transformer trio provided, ususally by the customer.
The Lasko 6000 Cyclonic Ceramic Heater is for indoor use.