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TRADICIÓN E INNOVACIÓN S.A. de C.V. Calculando Mp Balance de agua global ⎛ ( Mf )( Xf ) ⎞ Mp = ⎜ ⎟ ⎝ Xf ⎠ Mp=246 Kg/h MfYf , H 2O + Ma, eYa, eH 2O = MpYpH 2O + Ma, sYa, sH 2O − − − − − − − Fracción de agua Ma , e = Ma, s − − − − − − − − − − − − − −( 4) Yf , H 2O + Ya, eH 2O = YpH 2O + Ya, sH 2O − − − − − − − − − Sustituyendo valores en el balance de agua: Sustituyendo valores en la ecuación 3 Ya,sH2O= 0.66 Ma,e = 703Kg/h Masa de aire Masa de agua evaporada del producto en el secador: Mf = Mp − M M M H 2Oev H 2Oev H 2O ev = Mf − Mp = Mf − Mp − − − − − − − − − − − − − − − − − ( 6) MH2Oev= 422Kg/h 303 − ( 5) − ( 3)
TRADICIÓN E INNOVACIÓN S.A. de C.V. BALANCE EN EL TANQUE DE REFRIGERACIÓN Q6 Mf=1023 Kg/h δ= 1023 Kg/m 3 T= 60 ºC ºBrix= 14 Q7 MH2O ENFRIAMIENTO= 2675 Kg/h T= 7ºC Mf= gasto másico de producto alimentado (kg/h) Mp= gasto másico de producto concentrado (kg/h) Mv= gasto másico de agua evaporada (kg/h) MH2Oe= gasto másico de agua de enfriamiento (kg/h) Q Calor (Q = kcal/h) λ Calor latente (kcal/kg) ρ densidad (kg/m^3) °BRIX % de Sólidos Solubles Ts Temperatura de vapor Tf Temperatura del producto Cp Calor sensible (kcal/kg°C) 304 Q8 MH2O ENFRIAMIENTO= 2675 Kg/h T= 25ºC Q9 Mp= 1023 Kg/h ºBrix = 14 T = 7ºC Calor requerido 48,146Kcal/h
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EQUIPO TRADICIÓN E INNOVACIÓN S.A
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CAPÍTULO UNO

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