Difference between revisions of "Generator Efficiency"
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| − | = | + | =Electrical Efficiency= |
Efficiency as a fraction input mechanical power that is available at the output terminals: | Efficiency as a fraction input mechanical power that is available at the output terminals: | ||
E = Pout / (Pout + Ploss) | E = Pout / (Pout + Ploss) | ||
| − | + | . | |
| − | Pout = I^2 * Rl | + | Pout = I^2 * Rl and Ploss = I^2 * Rm thus |
| − | Ploss = I^2 * Rm | + | . |
| − | |||
| − | |||
| − | |||
E = I^2 * (Rl / (Rl+Rm)) | E = I^2 * (Rl / (Rl+Rm)) | ||
| − | + | . | |
let Rl = k * Rm then | let Rl = k * Rm then | ||
| − | + | . | |
E = k/(1+k) | E = k/(1+k) | ||
| − | + | . | |
for a given desired efficiency keep | for a given desired efficiency keep | ||
| − | + | . | |
Rl >= Rm * E/(1-E) | Rl >= Rm * E/(1-E) | ||
| − | + | . | |
| + | some results: | ||
k = 4 -> 80% | k = 4 -> 80% | ||
k = 9 -> 90% | k = 9 -> 90% | ||
| + | |||
| + | =Power Efficiency= | ||
| + | |||
| + | For a load that is a multiple of the genetator resistance, a fraction of the maximum output power is obtained. If the load equals the genetator resistance, then the maxium output power is obtained. | ||
| + | |||
| + | Let the power efficiency be: | ||
| + | . | ||
| + | Pe = Pout / Pmax | ||
| + | . | ||
| + | some math... | ||
| + | . | ||
| + | Pe = 4k/(1+k)^2 | ||
| + | . | ||
| + | Pe of 64% -> k = 4 (80% electrical efficiency) | ||
| + | 36% -> k = 9 (90% lectrical efficiency) | ||
Revision as of 08:08, 20 April 2007
Electrical Efficiency
Efficiency as a fraction input mechanical power that is available at the output terminals:
E = Pout / (Pout + Ploss) . Pout = I^2 * Rl and Ploss = I^2 * Rm thus . E = I^2 * (Rl / (Rl+Rm)) . let Rl = k * Rm then . E = k/(1+k) . for a given desired efficiency keep . Rl >= Rm * E/(1-E) . some results: k = 4 -> 80% k = 9 -> 90%
Power Efficiency
For a load that is a multiple of the genetator resistance, a fraction of the maximum output power is obtained. If the load equals the genetator resistance, then the maxium output power is obtained.
Let the power efficiency be: . Pe = Pout / Pmax . some math... . Pe = 4k/(1+k)^2 . Pe of 64% -> k = 4 (80% electrical efficiency) 36% -> k = 9 (90% lectrical efficiency)
