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What does dB equals to?
dB (decibel) is a logarithmic measure of the ratio of two power values, for example, two signal strengths. This is often used for power gain or power loss. For example, a loss of 10 dB means that the signal degrades by a factor of 10, a loss of 20 dB means that the signal degrades by a factor of 100, and a loss of 30 dB means that the signal degrades by a factor of 1000.
What Equation is used to determine power loss?
Power (Joules) = the square root of the voltage squared divided by the resistance
Why the corrosion rate expressed in penetration rate is more useful than that expressed in the rate of metal loss per unit area?
In most cases you'd be concerned with loss of strength of an item due to reduction in thickness or in cross section area rather than with how much metal you've lost (weightwise) from the whole of the item surface. It is easy to convert from one to the other though. The imperial unit of mpy (mils per year) equates to a penetration rate of about ten atom spacings per hour :)
How do you find the loss factor in hysteretic damping?
Loss factor is best obtained by dynamically loading (extensional, torsional etc.) a specimen of the material and plotting the hysteresis curve in stress-vs strain plane. If the total area under the hysteresis loop is D, the loss factor is computed from the following formula Loss factor=D/(2*pi*max stress* max strain) For lightly damped materials, loss factor is just twice the daming factor 'zeta' which obtained either by log-decrement method or half-power bandwidth method. Loss factor is best obtained by dynamically loading (extensional, torsional etc.) a specimen of the material and plotting the hysteresis curve in stress-vs strain plane. If the total area under the hysteresis loop is D, the loss factor is computed from the following formula Loss factor=D/(2*pi*max stress* max strain) For lightly damped materials, loss factor is just twice the daming factor 'zeta' which obtained either by log-decrement method or half-power bandwidth method.
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'.
