TERCER_PROBLEMARIO_DE_BALANCE_DE_MATERIA.pdf - PROBLEMA 6 Se desean obtener 50 kg de agua a 38\u00b0C Para ello se tiene un recipiente con agua a 90\u00b0C y

# TERCER_PROBLEMARIO_DE_BALANCE_DE_MATERIA.pdf - PROBLEMA 6...

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H 2 O (v) H 2 O (v) P = 0.6 bar P = 0.2 bar H 2 O (V) Saturada P = 4 bar H 2 O (l) saturado P = 4 bar Salmuera 5.5% sal Agua de mar 5000 kg/h, 3.5% sal, T 1 =300 K 1 2 3 4 5 6 7 8 9 10 Como únicamente condensa el vapor de calentamiento T 3 = 416.75 K A la presión de 0.6 bares del primer efecto, T 4 = 359.15 K, T 5 = 359.5 K Lo mismo sucede para el efecto II que se encuentra a 0.2 bares T 6 = 333.25 K, T 8 = 333.25 K A la salida del efecto II T 7 = 359.15 K Balance de sólidos en efecto I 5 000 0.035 = M 4 0.055 M 4 = 3 181.82 kg/h Balance total de materia en el mismo efecto 5 000 = 3 181.82 + M 5 M 5 = 1 818.18 kg/h Para obtener el flujo de vapor de calentamiento alimentado, es conveniente efectuar un balance de energía en el efecto I E 1 + E 2 = E 3 + E 4 + E 5 M 1 H 1 + M 2 H 2 = M 3 H 3 + M 4 H 4 + M 5 H 5 En donde M 2 = M 3 H 1 = cp (T 1 – T 0 ) = 112.34 kJ/kg H 2 = H L + λ 4 bares = 2 737.6 kJ/kg H 3 = cp (T 3 – T 0 ) = 604.7 kJ/kg H 4 = cp (T 4 – T 0 ) = 359.9 kJ/kg H 5 = H L + λ 0.6 bares = 2 653.6 kJ/kg Sustituyendo en el balance de energía
5 000 112.34 + M 2 2 737.6 = M 2 604.7 + 3 181.82 359.9 + 1 818.18 2 653.6 M 2 = 2 535.59 kg/h

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• Spring '18
• Delia del Carmen Gamboa

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