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What else can I help you with?
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.
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).
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.
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.
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 would a machine that had 100 percent mechanical efficiency be called?
A machine with 100 percent mechanical efficiency would be called an ideal machine, as it would have no energy losses due to friction, heat, or other inefficiencies.
What is the efficiency of a generator that produces 80 kilowatts of electric energy and has electric and mechanical losses of 20 kilowatts?
The efficiency of the generator is calculated as the output power divided by the input power. In this case, the input power is the sum of the output power and losses, which is 100 kilowatts. So, the efficiency would be 80 kW / 100 kW = 0.8 or 80%.
Why is is that is impossible to have a 100 percent efficiency?
It is impossible to achieve 100 percent efficiency due to various factors such as friction, heat loss, and energy conversion losses. These factors result in some amount of energy being wasted in any process, making perfect efficiency unattainable.
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 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..
If efficiency of converting chemical to thermal energy was 90 percent what would the overall efficiency be?
In saying what the overall efficiency would be, I suppose you mean for other processes, creating the chemical energy for example, and using the thermal energy. This is impossible to answer, not knowing what these processes are.
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.
How can the efficiency of electrical generators be optimized for maximum performance?
The efficiency of electrical generators can be optimized for maximum performance by ensuring proper maintenance, using high-quality materials, and implementing advanced technologies such as variable speed drives and power factor correction. Regular maintenance, such as cleaning and lubricating components, can help reduce energy losses and improve overall efficiency. Using high-quality materials can also help minimize energy losses and increase the lifespan of the generator. Additionally, implementing advanced technologies like variable speed drives can help adjust the generator's speed based on the required load, while power factor correction can help improve the power quality and efficiency of the generator. By incorporating these strategies, the efficiency of electrical generators can be optimized for maximum performance.
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
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.
