0
DC generator has an efficiency of 94 percent has an output of 240 volts at 100 amperes what would be the pover losses?
Wiki User
∙ 12y ago
Assuming that the output you've listed is after the power factor has been taken into account:
Then 240x100 =24000w.
The power factor is 0.94, therefore 24000 divided by 0.94 = 25531.915 VA.
Power loss 25531.915 - 24000 = 1531.9149 VA.
Wiki User
∙ 12y ago
Add your answer:
Earn +20 pts
What are the constant losses in a DC machine?
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
Can you use a dc generator as a dc motor?
Yes you can turn a motor into a generator, if it is a permanent magnet motor.
Why are distribution transformers frequently designed to develop maximum efficiency at loads that are somewhat lower than rated value?
It is always desirable to run any equipment or device at maximum efficiency for that matter, not only the power transformer. Power transformer maximum efficiency occurs when copper loss is equal to iron loss. (or no load loss equals to load loss). This does not necessariliy mean that maximum efficiency occurs at maximum or full load. Generally the maximum efficiency occurs at relatively less than full load of the transformer.
Can a motor drive a generator which can produce power more than the power used by the motor to drive the generator?
Claims about perpetual motion machines are always false because they ignore the inevitable losses of energy which must be overcome to make any machine operate so that it can do its intended work. Such losses mean that any machine, whether electrical or mechanical, must be supplied with more input power than it can convert into output power. In the case of electrical machines there are many reasons for energy losses: Bearing friction losses; "windage" losses caused by air having to be moved out of the way by spinning parts such as rotating armatures and cooling fans; magnetic hysteresis losses absorbed by eddy currents flowing in rotor armature and stator laminations; commutation losses in dc machines and slip ring losses in ac machines; excitation losses caused by the need to supply energy to the coils in the stator of a machine in order to magnetize it; general electrical resistance losses caused by the need to use energy to force electric currents to flow through the windings of any electrical machine. Whilst engineers have found ways to reduce each one of those types of loss to a minimum, it is still a fact of life that no electrical machine can be 100% efficient. In the case of a motor, it will always require more energy to be supplied to it than it can convert to mechanical power. In the case of a generator, it will always require more mechanical power to be supplied to it than it can generate as electricity. Some electrical machines have a power-conversion efficiency of up to 80% but most have efficiencies much less than that. If a motor and a generator were linked together to form a motor-generator set (the motor driving the generator driving the same motor) then, if both machines were as good as 80% efficient, that would mean the combined efficiency of the system of would only be .8 x .8 = .64 i.e. 64%. In other words, to keep the motor-generator set running, additional power equal to at least 36% of the motor-generator set system's own power would need to be supplied from an external source. Those facts make the set-up described in the answer to this question shown below pure nonsense. Period.
How do you change megawatts to kilovolts or What is the relation to MW produced to KV transmitted?
There is no direct relationship. It is a design, an engineering, decision, what voltage to use. A high voltage means less amperes are required; and alectrical losses are proportional to the square of the current (amperes).
A generator produces 240volts at 25 amperes. If the mechanical and electrical power losses total 900 watts what is the efficiency of the 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.
Why the efficiency of generator is more than the motor in case of swinburne's test?
the generator windings are made of more thicker windings . Hence lesser resistance and lower Cu losses
What is the condition for maximum efficiency of a DC machine?
The condition for maximum efficiency of a d.c. machine is that VARIABLE LOSSES must be equal to CONSTANT LOSSES i.e., variable losses = constant losses..
What are the constant losses in a DC machine?
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
What are the advantages of Swinburne's test?
1. swinburne's test is economical since power required to test a large machine is very small (i.e.,)no load input power. 2.The efficiency of the machine can be predicted at any load, since constant losses are known. 3.This method enables us to determine the losses and efficiency without actually loading the machine
How many watts the 10kw generator will consume under no load condition?
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.
A generator produces 240volts at 25 amperes If the mechanical and electrical power losses total 900 watts what is the efficiency of the generator?
If the generator produces 240v at 25 amps, this amounts to 6000 watts, so if the losses are 900 watts, the total power generated must be 6900 watts. The efficiency is then 6000/6900 = 0.87, or 87 percent. You have not put in heat losses however, or are they included in 'mechanical? ****The previous answer above is incorrect. Read and understand the question first. The general rule of thumb when calculating power equations is to factor in power loss as an inclusive amount of the total power being generated. unless stated otherwise Therefore 240v at 25amps gives 6000watts power loss is 900watts The question says that the power the generator produces is 240v at 25amps which is not the useful power it delivers ( this is a very common mistake in learning and real life applications) and that is why we do not add power loss to the total power generated, rather we subtract power loss from the total power generated. Efficiency of the generator is actually the power the generator is delivering for useful work compared to the total power generated. In this case 6000 watts (Total power produced) - 900watts ( power loss) = 5100watts Therefore the geneator's efficiency is 5100/6000 x 100% = 85% For the last part of the previous answer, For a generator, mechanical movement is counter reacted by friction which produces heat and noise while resistance to electrical current flow and electomagnetic induction generate heat. Of course this is just a general description, it gets quite complicated in advanced level analysis where back e.m.f, magnetic resonance, physical construction (generator type) and purity of construction materials just to mension a few are involved. I'm only a simple physicist, but I feel the word 'produced' is ambiguous, and I took it to mean the same as 'delivered'. If I went out and bought a small generator, and the sales people said it was a 2 Kw generator, I would expect it to deliver 2 Kw, and not 85 percent of that figure. So I think you could take this either way. To use the word 'produced' is not ideal, better to use 'generated power' and 'output power'.
Why cant a machine be one hundred percent affective?
The reason why amachine cant be one hundred percent effective is as a result of losses.For Electrical machine such as alternator we have Mechanical losses(such as windage and frictional losses)and Electrical losses(such as copper loss in the rotor winding and stator winding).In Mechanical machine the main losses are windage and frictional losses.NOTE;ALL THIS LOSSES RESULT IN THE PRODUCTION OF HEAT,THUS REDUCING THE EFFICIENCY OF THE MACHINE.
What is the percentage load at which maximum efficiency occurs for the single phase transformer?
That is the maximum efficiency occurs when the copper losses are equal to the core losses of the transformer.
Why is the efficiency of a machine less than 1?
An efficiency factor of 1 (or 100% efficiency) is not possible due to losses that cannot be reduced to zero. These losses take the form of friction, heat loss etc.
What will happen if iron losses and copper losses are equal for any transformer?
The transformer will have the maximum efficiency.
Why the efficiency of induction motor is less than that of transformer?
Because, the losses of IM is more due to the contribution of friction losses at shaft bearings and wind age losses in rotor air gap, this reduces the efficiency of the equipment, Since transforms is a static equipment zero mechanical loss so efficiency of the equipments in high .
