Difference between revisions of "Generator Efficiency"
From DIDEAS Wiki
m (→Power Efficiency) |
m (→Electrical Efficiency) |
||
| Line 5: | Line 5: | ||
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: | ||
| − | + | <pre> | |
| − | + | 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% | ||
| + | </pre> | ||
=Power Efficiency= | =Power Efficiency= | ||
Revision as of 08:12, 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 generator resistance, a fraction of the maximum output power is obtained. If the load equals the generator resistance, then the maximum output power is obtained.
Let the power efficiency be: Pe = Pout / Pmax some math... Pe = 4k/(1+k)^2 64% -> k = 4 (80% electrical efficiency) 36% -> k = 9 (90% electrical efficiency)
