# Calculos de evaporador efecto simple

Páginas: 3 (747 palabras) Publicado: 21 de marzo de 2011
Tabla de Datos Experimentales
Tanques Diámetro (cm)
Altura (cm) en Tiempo de ( 5min ) Altura (cm) en Tiempo de ( 5min ) Altura (cm) en Tiempo de ( 5min )
1ra Corrida 2da Corrida 3ra Corrida
Talimentación
59.6 4.9 4.8 4.9
T vapor de caldera
40.2 3.6 3.2 3.9
T producto concentrado 34.6 10.7 10.6 10.6

Datos de Tanque Agua de EnfriamientoTiempo Altura Diámetro
20 seg 6.5 cm 56 cm

Tabla de Temperaturas
T1 T2 T3 T4 T5
112 ºC 74 ºC 74 ºC 34 ºC 32 ºC

Datos de Operación
Rotámetro 20%
P calandria 0.7 Kg/cm2
T calandria 60 ºCP vacio 31 cm Hg

CALCULOS DEL EVAPORADOR DE SIMPLE EFECTO

M_A1=(π*D^2*∆h*ρ)/4ө=(π(0.596m)^2 (0.049m)(1000 kg/〖cm〗^3 ))/(4(0.083h))=164.7 kg/h
M_A2=(π*D^2*∆h*ρ)/4ө=(π(0.596m)^2 (0.048m)(1000kg/〖cm〗^3 ))/(4(0.083h))=161.34 kg/h
M_A3=(π*D^2*∆h*ρ)/4ө=(π(0.596m)^2 (0.049m)(1000 kg/〖cm〗^3 ))/(4(0.083h))=164.7 kg/h

M_P=(π*D^2*∆h*ρ)/4ө=(π(0.346m)^2 (0.106m)(1000 kg/〖cm〗^3))/(4(0.083h))=120.079 kg/h

E=(π*D^2*∆h*ρ)/4ө=(π(0.346m)^2 (0.037m)(1000 kg/〖cm〗^3 ))/(4(0.083h))=41.91 kg/h

M_V=(π*D^2*∆h*ρ)/4ө=(π(0.102m)^2 (0.032m)(1000 kg/〖cm〗^3 ))/(4(0.083h))=48.93 kg/h

MA=MP+E161.34kg/h = (120.08+41.91)kg/h
161.34kg/h =161.99kg/h

W=(π*D^2*∆h*ρ)/4ө=(π(0.560m)^2 (0.065m)(1000 kg/〖cm〗^3 ))/(4(5.55X〖10〗^(-3) h))=192.88 kg/h

CALOR TRANSFERIDO

QT = EHE + MPHP + MAHA
HP =CpT
HP = (1 Kcal/(Kg°C))(74°C)=74 Kcal/Kg
HA= (1 Kcal/(Kg°C))(64°C)=64 Kcal/Kg

QT = (41.91 kg/h) (628.9 Kcal/Kg) +(120.08 kg/h)(74 Kcal/Kg)+ (164.7 kg/h )(64 Kcal/Kg)
QT=24702.3Kcal/h

QS = MVλ_V

Qs=(48.93 kg/h)(531.3 Kcal/h)=25996.5 Kcal/h

AREA DE TRANSFERENCIA DE CALOR
A= π NT D h
A= π*4(3X〖10〗^(-3) m)(2.54m) = 0.095m

EFICIENCIA TERMICA

ŋ=Q/Qs=(24702.3 Kcal/h)/(25996.5 Kcal/h) =0.95(100) = 95%

CQ=QT/A=(24702.3 Kcal/hr)/(0.095 m2)=260024.21 kcal/hrm2

Coeficiente Global de Transferencia de Calor...

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