Turbomaquinas Hidraulicas Claudio Mataix | Solucionario De

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El caudal en m³/s es: Q = 1200/3600 = 0.3333 m³/s. Las velocidades se calculan a partir del caudal y las secciones: vs = 4Q/(π·Ds²) = (4·0.3333)/(π·0.375²) = 3.0180 m/s. vE = 4Q/(π·DE²) = (4·0.3333)/(π·0.400²) = 2.6526 m/s. Las alturas dinámicas son: vs²/2g = 0.4643 m; vE²/2g = 0.3586 m. Finalmente, la altura útil H se obtiene como: H = (Ps – PE)/(ρg) + (Zs – ZE) + (vs² – vE²)/2g = (12 – 2) + (0.5 + 0.08) + (0.4643 – 0.3586) = 14.686 m . solucionario de turbomaquinas hidraulicas claudio mataix