I also have the same question. After some research, I understand that you cant just jump in and calculate this. You have to have alot of information about that spesific generator. I have found out that you can use some formulas from losses in transfomers.
Click the link. There you can figure the size generator you need.
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.
How to calculate iron losses in dc machine
Because if we use a generator of 33kv or higher then the radii must be of 2-3 meters that is not good. And also the area of the generator will be very high and losses and heat paameters will also be high. The main point is that we use 11kv generator at power plants and we have to step up the power to 220KV or 500KV etc. So we use 11kv or 22kv instead of 33 kv or higher.. By doing so we save the cost of area required, radii of rotor, and losses produced by higher power generators.
if a generator is overexcited it is producing a large amount of reactive power which requires an increase in the alternator emf. To produce this emf, the rotor currents must increase resulting in extreme ohmic rotor losses.
Click the link. There you can figure the size generator you need.
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.
No. In an amplifier, Power Out > Power in. In a transformer Power Out ~= Power In (minus internal losses). An AC generator is more like an amplifier than a transformer.
Take the power output of the generator and divide it by the voltage output. I = W/E.
Assuming the generator converts 90% of the mechanical power into electrical power, it has an efficiency of 90%, which means it consumes 11 kW of mechanical power under a full electrical load of 10 kW. Under no load the frictional losses will still apply, but the resistive losses in the windings will not be present. Therefore the no-load losses can be estimated as 500 watts in this conditon.
The field excitation could have been lost. Check the output from the voltage regulator.
Constant losses Those losses in a d.c. generator which remain constant at all loads are known as constant losses. The constant losses in a d.c. generator are: (a) iron losses (b) mechanical losses (c) shunt field losses
How to calculate iron losses in dc machine
Because if we use a generator of 33kv or higher then the radii must be of 2-3 meters that is not good. And also the area of the generator will be very high and losses and heat paameters will also be high. The main point is that we use 11kv generator at power plants and we have to step up the power to 220KV or 500KV etc. So we use 11kv or 22kv instead of 33 kv or higher.. By doing so we save the cost of area required, radii of rotor, and losses produced by higher power generators.
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copper losses are power losses due to flow of current in the wires or resistances,if the resistance is R, current is I then copper losses are I2R. for a 3-phase system; copper losses are same but for a single line, total losses are 3I2R.
if a generator is overexcited it is producing a large amount of reactive power which requires an increase in the alternator emf. To produce this emf, the rotor currents must increase resulting in extreme ohmic rotor losses.