Vincent Del Toro !!install!! - Solucionario Maquinas Eletricas
$$s = \fracN_s - N_rN_s = \frac1800 - 17101800 = \frac901800 = 0.05 \text (5%)$$
El entrehierro domina la reluctancia del circuito (a pesar de ser muy pequeño, su reluctancia es el doble que la del hierro). Solucionario Maquinas Eletricas Vincent Del Toro
| Feature | Official Instructor’s Manual | Unofficial / Community-Made | | :--- | :--- | :--- | | | Very high (verified by author) | Variable (some errors common) | | Completeness | Typically includes all problems | Often only odd-numbered or selected problems | | Step Detail | Full derivations | Sometimes just final answers | | Availability | Restricted (instructor-only) | Widely shared on file-sharing sites | | Legal Status | Copyrighted | Often pirated | $$s = \fracN_s - N_rN_s = \frac1800 -
If you are unable to find a specific solution in Del Toro's manual, many students use the following widely available solution manuals for similar curriculum: Stephen Chapman Electric Machinery Fundamentals (a) Calcule el deslizamiento
Un motor de inducción trifásico, 4 polos, $60 \text Hz$ funciona a una velocidad de rotor $N_r = 1710 \text rpm$. La potencia de entrada es $20 \text kW$ y las pérdidas totales (estator + rotóricas + mecánicas) son $2 \text kW$. (a) Calcule el deslizamiento. (b) Calcule la potencia mecánica de salida y el par en el eje.