losses in 3phase induction motor?
370576
Here are some that I can think of:\n\nwindage losses (air resistance to spinning rotor)\nreactive losses (coils/inductive)\nfriction losses (bearings)\nhysteresis\nleakage current (insulaton)\ncooling fan\n\nI would check the references below for more insight.  The europa site calls the cooling fan a windage loss--I disagree.
458600
You say 3 phase, so the simple amps times volts doesn't work.\n\nThe complete formula is KW= 1.73 * amps * volts * power factor /1000\n\nThe 1.73 is the square root of 3.  The amps is your line current (same as phase current if it is wye, not the phase current of a delta) and the volts is the phase to phase voltage (not phase to neutral of a wye).  Power factor is 1 for simple resistive loads, and is less than 1 for inductive loads such as motors and fluorescent light ballasts.\n\nIf you take the power factor out of the formula, you get apparent power measured in KVA.
711069
Power in = Power out + Power losses\n\nPower into the fan is electrical power V*I (voltage times current). Power out in mechanical terms is T*w (N*m torque times rotational speed in rads/sec). Your losses are those of friction/air, reactive losses in the coils, and I^2*R electrical losses.\nIF you were to add another load to the shaft to generate electricity, it must be added to the right side of the upper equation, and require more initial input power. If more input power is not available, the fan would just be slower.\n\nIt has been suggested elsewhere to do this and feel the generated electricity back into the first motor. Even if these devices were 100% effecient with no losses, it wouldn't even be worth it. Thats like adding the same value to both sides of the upper equation.
756481
Normally this is a "Restive" not voltage question/answer, and the proper neutral to ground = 2 Megg ohms (min.) this being incorporated in a residential application.\nHigher voltage in let us say a 3-phase utilization this # is on the low side. (assuming a.c. source. d.c. is a completely different animal!
732893
we know that for power, P=IV. we also know that whatever power goes into the transformer comes out of the transformer(conservation of energy). this means that V is inversely proportional to I. or I is inversely proportional to V.\n\nif you already know how much V was stepped up, then just step down the I by the same ratio. for eg, if 110V 10A was transformed to 220V, then the step up ratio is 2. therefore current is stepped down by 2. so current will be 5. try equating 110x10 with 220x5, it is constant.\n\nall of this assumes that the transformer is 100% efficient. if there are losses due to heat, then you multiply the exit current by the efficiency factor. eg, if the trans is said to be 80% effiecient, then the actual exit current would not be 5A but 4A. voltage is not affected by effiency issues.\n\nby the way transformers only work with AC, the guy who mentioned direct currents is thinking of current rectifer not transformer.
910581
Go to your local hardware store and purchase a one of those small plug in circuit testers with the three LEDs. Plug it in and in less than a second, you will know exactly what is wrong (reverse polarity, open neutral, open ground, etc....) I think they cost around $3.50 - and they are very handy.\n\nBe sure to cut the power before attempting any repairs.\n\nI suspect you'll find that the three outlets are on the same line (circuit) and the originating plug has lost the neutral wire - just a guess based on 35 years experience.
1063180
first...identify the power source (AC or DC)\nthen...isolate the power source (lockout)\nnext...verify the circuit is not live (test)\n\nfinally...if it is AC three phase reverse any two of the three motor leads, if it is DC just reverse the two leads (if shouldn't be DC)\n\nif it is a european piece of equipment the wires should be black-black-brown, swap the blacks\n\notherwise, it should be brown-orange-yellow or black-red-white\n\nDO NOT ATTEMPT IF YOU ARE NOT QUALIFIED\n\ngood luck
370881
try this website http://www.howstuffworks.com/search.php?search=electrical%20motors%20and%20generators&ref=hpaw&page_ref=http://www.howstuffworks.com/
890746
Loose wires or bad grounding. Motors,such as blowers for the furnace, that need replaced.
807450
For a wye-connected machine:\nThe phase voltage (measured between a-b, b-c, or a-c) is (SQRT(3))* the line voltage (a-n, b-n, or c-n). In this case, connect your invertor across a-n, b-n, c-n (this may mean breaking the "common" neutral connection).\n\nFor a delta-connected machine:\nThe phase voltage equals the line voltage. In this case, connect your invertor across a-b, b-c, and a-c.\n\nWhich polarity you choose for the three connections will determine rotation - make sure your resultant rotation matches that on the machine nameplate, or you'll likely burn it up due to insufficient cooling...
828167
You might start by contacting the power company.  If you have concerns about eletromagnetic interference, they should be willing to help you.
572836
Three phase power is delivered perhaps to a machine shop with large electric motors.  The motors and transformers (if needed) are often connected in "Y" or "Delta" configurations using all three phases.  The machine shop may also use single phase (for light bulbs?) and two phase (for heaters?).  A single phase to ground provides 110 volts.  Two phase connected together provide 220 volts.  When all three phases are connected 440 volts is provided.  You can often see all three phases up on most power poles and only two phases are connected to most homes (for 110 and 220 volt service).\n\nConsider the analogy between the flow of electric current and the flow of water.  Voltage is like water pressure and current (amps) is like flow rate (gpm).  The three phases represent three different "pipes" between the generator plant and the customer.  Each phase is like a piston pump at the generator that shoves a pulse of water (amps) to the customer then sucks it back (alternating current).  The three phases are connected to the same shaft (like pistons in a three cylinder car) but are evenly spaced (360 degrees/3 = 120 degrees out of phase).  The electrons shoved into the pipe at the generator need not reach the customer before being sucked back.  A like quantity of electrons are pushed into the ground at the customer's installation (communicated at the speed of light!) then pulled back out of the "ground" for reuse.\n\nObviously the power that can be transmitted to the customer depends on whether he/she is using single, two or three phases where P = EI (Power = Voltage x Current).  Hope that helps a little.
398366
How many power leads does the motor have? It will have four or five if it's a three phase motor (Three phases, common, and probably a ground)\n\nWhat color are the leads?\nThere is a key to the colors in the referenced link under sources . . .
482056
I am not sure I understand your question. Are you asking, what are the effects of core losses in relationship with frequency and supply voltage?\n\nThe total power loss in a transformer is a combination of three types of losses. One is due to the DC resistance in the primary and secondary windings. This is called Copper loss. The two other losses are due to Eddy Currents and Hysteresis in the core of the transformer. Copper loss, Eddy Current and Hysteresis loss result in unwanted conversion of electrical energy into heat energy.
995705
Yes and no.  A 3 phase load is balanced, like a 3 phase motor, and doesn't even connect to the neutral, so there is no neutral current for that case.  Also, single phase loads that are connected from one phase to another will have no neutral current.  The neutral is the return current for single phase loads that are connected between a phase and neutral.  The NET neutral current will be the unbalanced current (if all the phases had the same current the neutral would be 0), and will be no more than the largest phase (ignoring harmonics).
315859
you got Ohm law : Ur=R*i(where i stands for current) and what's more, for the coil we have Ul=-L*di/dt . And for the condenser we have C*Uc=q and i = dq/dt.Now you have a differential equation with only one variable i.\n note: if the circuit is parallel then U should be a constant , in this case you shuold use U instead of i.Goodluck
1041136
Read in Detail at this site\nhttp://books.google.com/books?vid=ISBN0768917425&id=Ca58uytAJoAC&printsec=toc&dq=mechanical+engineering%3F+CIVIL+%26+ELECTRICAL%3F