No, you should not load a panel beyond its rated capacity. In this case, a 30A load per phase in a 100A panel would exceed its capacity and could cause overheating or other safety hazards. It's important to properly size the panel based on the expected load requirements.
To calculate the current per phase, use the formula: Current (I) = Power (P) / (Square Root(3) x Voltage (V)). For an 18kW motor at 415V, the current per phase will be: 18,000W / (1.732 x 415V) ≈ 24.5A per phase.
The breaker will trip at the amperage notated on the breaker. If it's 100A...it will trip at or around 100A. It does not matter if that breaker is physically tied to another 100A breaker. To understand this, imagine that you remove the mechanical tie from the two-pole breaker. Now you just have two 100A breakers. In actuality, you always had two 100A breakers. The mechanical tie does not change that. If you then powered two, separate 120 volt devices from the two breakers, each breaker would allow 100 amperes to pass to each of the devices before tripping. So why are they tied together? That is done when the two-pole breaker is to be used to power a 240 volt circuit. In AC current, electricity flows in both directions. In a 120 volt circuit, it flows "out" toward the device via the hot (generally the black wire) and "back" via the neutral (generally the white wire). Then the cycle reverses. It does this 60 times per second (60Hz). The amperage in the hot and neutral wires are the same (in the perfect world). Only the hot wire is connected to the breaker. In a 240 volt circuit, there is no neutral wire. You are using two "legs" of 120 volts each that are 180 degrees out of phase with each other. In other words, as leg 1 is flowing "out", leg 2 is flowing "back". Because they are out of phase, the potential difference is twice the voltage of each line or 240 volts. The current flows out and back at the same 60 Hz but this time via the two hot wires (generally black and red). Each of these hot wires are connected to the two terminals of the two-pole breaker. Due to mechanical tolerances, one breaker will most likely trip before the other. Therefore, if the rated current, (100 amps), is exceeded on either breaker, that breaker will trip and the other breaker will trip via the mechanical tie. This ensures that all power to the outlet is disconnected. If you removed the tie and only one breaker tripped, there would still be 120 volts connected to the outlet. In summary, each leg of a single, double (2 phase) or triple (3 phase) breaker is capable of allowing the amount of current denoted on the breaker. The connected circuit, regardless of voltage is protected from exceeding that amperage.
Still 30 amps, but at 240 V you'll have twice the watts that you would on a 120 V, 30 amp circuit, and after all, watts are what actually does the work.
It is not in parallel. You put a breaker in existing panel and use that to feed the subpanel. The Amperage of this subfeed breaker should match the rating of the new panel. For example a 100A breaker might be typical. Remember that ground and neutral are only "bonded" at the main panel. Usually a subpanel has a means to separate the neutral and ground in a subpanel. Be careful since everything about doing this is dangerous.Another AnswerYes, you can have two breaker panels in parallel. If you had a 100 amp panel on a 200 amp service (or increase the size of the service, check with your power company on the size of your service, you could add a second 100 amp panel in parallel with the first. You can have up to 6 disconnects per service, but they must be located adjacent to each other or in the same enclosure. So either install the second panel beside the first or a 100 amp disconnect beside the first panel and feed out of the disconnect to the new panel located where you need it.
It depends on the internal circuit of the machine. If it is star operated its full load current will be same as rated current. If it is Delta operated its full load per phase current will be as given below : Phase current = Line Current / 1.732
It is very seldom that a main breaker trips. If it does it again make a mental note as to what was running at the time. As you can appreciate the total amperage available to the loads exceeds the main breaker total. 1 - 40A, 1 - 30A, 1 - 20A and a few 15A. Circuitry design when wiring a house goes on the assumption that not everything will be loaded at the same time. An average 100A panel usually loads up at about 50 - 55A during a normal day. With more appliances and electronics these days most new installation are in the range of 150 to 200A at 42 circuits per panel. Your 75A suggests that you may have more than the normal loading of the panel. Maybe electric heat in a older home, welding plug in the garage. Usually there would be a difference in amperage on the two hot legs due to the 120V loads. Amperage being the same suggests that it is a 240 volt load that is the culprit. I have seen an installation where the main breaker load side has been loose and corrosion has set in from a loose connection to the buss bars. This situation had to be rectified by a quailfied electrician because the incoming lines to the house had to be disconnected from the power supplier. 100 amp main, and yes single phase What is the size of the main breaker and is it a single phase panel?
The kV.A (not 'kva') rating is the total apparent power of the machine. So a 75 kV.A machine is 25 kV.A per phase.
Regulation =(Obtained voltage per phase - rated voltage per phase)/rated voltage per phase *100
400V
Unit of Phase constant is Radians per metre .
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Total slots per phase =12/3=4 slots to be occupied. Slots per pole per phase = 12/(3*4)=1 hence for A phase, the slots to be occupied are 1,4,7,10;for B phase, the slots to be occupied are 5,8,11,2;for C phase, the slots to be occupied are 9,12,3,6 and winding to we wave type.
To calculate the current per phase, use the formula: Current (I) = Power (P) / (Square Root(3) x Voltage (V)). For an 18kW motor at 415V, the current per phase will be: 18,000W / (1.732 x 415V) ≈ 24.5A per phase.
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You will need to determine the power per phase, and add them up to give the total power of the three-phase load. To do this, you will need to multiply the phase-voltage by the phase current by the power factor -for each phase.
I(per Phase)=4000/(230*3) = 5.8A
As per www.totalrecallinfo.com there are 5 recalls on this itme, but not on the instrument panel.