Ejercicio de Balance (1)
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PROBLEMA 2 – PAG PAG 21 Convertir 0.5 hp a cal/s 1hp = 178.15 cal/s
178. 1 5 0.5ℎ 1 ℎ =89.075 PROBLEMA 6 – PAG PAG 22 Un gas contiene 30% de CH4, 60% de C2h6 y 10% de C3H8, en volumen, a 60°C y 1 atmosfera. Calcule el peso molecular medio y la densidad de la mezcla. Se asume que el comportamiento de la mezcla es el de un gas ideal ya que trabajamos a bajas temperaturas y presión. Entonces fracción molar = fracción volumétrica. BASE DE CÁLCULO 100 GRAMOS:
PMmedio=0.3(16) + 0.6(30) + 0.1(44) PMmedio= 27.2 g/mol P=1atm V=? T=60°C=333 K R=0.082 Lt –atm/mol°K nmedio = 100/27.2 = 3.67 mol
= = = (3.70)(0.0182)(333) = 101.03 .03 100 =0.989 = = 101. 03
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PROBLEMA 8 – PAG PAG 22 Tiene 25 lb de un gas con la siguiente composición: CH=80%, C2H4=10%, C2H6=10%. ¿Cuál es el peso molecular medio de la mezcla? Cuál es la fracción en peso (masa) de cada uno de los componentes de la mezcla? Masa y fracción en peso
:25 (0.8) = 20.0 20.0 . = 9071 9071.8.8 24:25(0.1) = 2.5 5 . =1134.0 26:25(0.1) = 2.5 5 . =1134.0
PM : 13 PM : 28 PM: 30
. =0.8 = . . =0.1 = . . =0.1 = .
Moles y fracción molar
=697.8 = . =0.89 ∶ . . . =0.052 24: . =40.5 = . . =0.049 26: =37.8 = .
PMmedio=f CH CH PMCH + f C2H4 C2H4 PMC2H4 + f C2H6 C2H6 PMC2H6 PMmedio=0.899(13) + 0.052(28) + 0.049(30) PMmedio=14.61 g/mol PROBLEMA 10 – PAG PAG 22
Un medidor de vacío conectado a un tanque marca 315 kPa. ¿Cuál es la presión atmosférica correspondiente si el barómetro indica 98.2 kPa? Pabs = Pman + Pbar Pabs=-315 kPa + 98.2kPa
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PROBLEMA 11 – PAG PAG 22 Preparar un diagrama de flujos de bloques y de procesos pr ocesos para producir etilenglicol. Sabiendo que se mezcla agua y óxido de etileno en un reactor para generar etilenglicol, esta mezcla luego se separa en en primer lugar óxido de etileno etileno no reaccionante y se recircula para reusarse de materia prima, se separa el agua que también se recircula como parte de materia prima. prima. Se destila obteniendo como parte liviana etilenglicol etilenglicol y parte pesada como di y tri glicoles
agua
agua
oxido de etileno
reactor
Oxido de etileno
etilenglicol
destilacion
di y tri glicoles
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PROBLEMA 2: PAG 32 En una operación de secado de pieles, se determino que un lote de piel previamente pasado ´por el secador pesaba 900 lb y que contenia 7% de su peso en humedad. Se sabe que durante el secado la piel lavada perdió 59.1% de su peso inicial cuando se encontraba humeda. Determinar: (a) el peso de la piel totalmente seca o exenta de humedad, en la carga de alimentación inicial; (b) las libras de agua eliminadas durante el proceso de secado por libra de piel totalmente seca; (c) el porcentaje de agua eliminada respecto de la cantidad de agua presente inicialmente en el lote de piel.
A=? piel lavada
C=900 lb piel tratada SECADOR
XA, agua lb agua)lb Xa, piel lb piel/lb B=? agua extraida
Xc, agua =0.07 lbagua/lb Xc, piel= 0.93 lb piel/lb XB, agua = 1,00 lb agua/lb
BASE DE CALCULO = 400 LB PIEL TRATADA E=S A(XA. Piel) = 900 lb(0.93) = 837 Balance de agua: A(XA, agua)=900(0.07agua) + B(1) Balance total A = 900 +B Relación
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Del balance total
= 900 + 0591 = 2200.5 = 1300 2 837 = % 0.38 = 2200. 5 38 = 1 0.38 = 0.62 1900.5 =0.953 %= (0.62)(2200. 5) PROBLEMA 3 – PAG PAG 32 Una corriente A (en kg/min) que contiene 30%m de etanol y 70% m de agua se mezcla con otra corriente B (en kg/min) que contiene 60%m de etanol y el resto de agua. La corriente de mezcla C (a la salida de la unidad mezcladora) contiene 35%m de etanol. Calcular: (A) la proporción entre las corrientes A y B, esto es (A/B); (b) si la corriente de salida C es 4500 kg/h, ¿Cuál es es la relación entre las corrientes corrientes A y B? ¿esta relación ha cambiado o no? (b) la proporción es la misma, cambian los valores de las corrientes pero la relación AIB es igual.
A=? Kg/min XA, etanol=0.3 XA, agua=0.7
MEZCLADOR
C=? Xc,etanol=0.35 Xc, agua=0.65
B=? Kg/min XB, agua = 0.6 XB, agua=0.4 BASE DE CALCULO: C= 100 Kg/min
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Reemplazar (2) en (1) (0.3)(100)=0.3(B) + 0.6(B) =35 B=16.67 Luego: A=83.33 Proporción: A y B (A/B)
= 83.33 =4.999=500 16.67 Corriente se salida
C=4500 Kg/min A=12.5 B=62.5
= 62.5 = 5 12.5 PROBLEMA 1 – PAG 51 Dibuja un diagrama de flujo para el proceso que se describe a continuación, continuación, etiquetando todos los flujos de alimentación, entre las unidades: la deshidrogenacion catalítica del propano se lleva a cabo en un reactor continuo de cama empacada. Un precalentador se alimenta con 100 libras por hora de propano puro, en el que se calienta a una temperatura de 670 grados centígrados antes de pasar al reactor. El gas de salida del reactor Qué contiene el propano propileno, metano, hidrógeno se enfría de este 800 grados centígrados hasta 110 grados centígrados y alimenta una torre de absorción en la que el propano y el propileno se disuelven en aceite.el aceite entra a una torre de desorción en la que se calienta, permitiendo la salida de los gases disueltos; estos gases se comprimen y se transfieren a una columna de destilación de alta presión, en donde el propano y el propileno se separan. El flujo del producto de la columna de destilación contiene 98% de propileno, y el flujo de recirculación coma 97% de propano. Ell aceite desorbido se hace recircular a la torre de absorción.
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PROBLEMA 8 -PAG 52 Una mezcla que contiene 20% molar de butano, 35% molar de pentano y 45% 45 % molar de hexano se separa por destilación fraccionada. El destilado contiene 95% molar de butano, 4% de pentano y 1% de hexano. El destilado debe contener 90% del butano cargado en el alimento. Determínese la composición de los productos de fondo. fondo . DESTILADO
ALIMENTACIÓN
= 0.20 20
= 0.95 95 = 0.04 04 = 0.01 01
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=90%( =90%( ) 0.95∗ =0.9∗ [0.2∗ (100) 100)] = 18.9 18.947 4736 3688 ⁄ℎ 100=18.947368+ = 81.052632 ⁄ℎ
= + Balance del butano:
= + 0.2(100)=0.85(18.947368)+ (81.052632) =0.024675=2.4675 % Balance del pentano:
= + 0.35(100)=0.04(18.947368)+(81.052632) =0.422468 =42.2468 % Balance del hexano:
+ + = 1 0.024575+0.422468+ = 1 =0.552837 = 55.2837% 2837% ℎ ℎ
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(c) En cada etapa, la concentración de sal en su solución de desperdicio es la misma que la concentración de sal acarreada con la lechada de salida de la etapa. etapa.
1000lb/h
20% TiO2 30% Sal
= TiO2 = sal = agua
a)
()= 0.8 ()=0.8(0.3)(1000) ()= 240 lb/h sol ()= 0.2 () =0,2(0.3)(1000) ()=60
()= 0.8 ()=0.8(12) ()= 9.6 lb/h sol ()= 0.2 () =0,2(12) ()=2.4
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PROBLEMA 14 – PAG PAG 53 Un tanque contiene 100 galones de una solución salina, donde se encuentran disueltas 4 lb de sal. El flujo de agua fresca que entra al tanque es de 5 gal/min, saliendo un mismo flujo de solución del estanque. Si el tanque se encuentra perfectamente agitado. agitado. ¿Cuánta sal queda en el tanque a los 50 minutos? Nota: asuma que la densidad de la solución es igual a la densidad del agua. Fe= 5gal/min Ce=0
V=100 gal C0=4lb/100 gal = 0.04 lb/gal Fs=5gal/min Cs=?
Balance de sal:
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0.05=0= 0 −. = 0 =0−. =0.04−. Para t=50 min => C= 0.0033lb/gal PROBLEMA 2 PAG 82 Una mezcla de combustible(hidrogeno y metano) se quema completamente en una caldera que usa aire. El analisis de los gases de la chimenea son 83.4 % de N2, 11,3% de O2 y 5,3% de CO2 (en base seca, sin agua). Reaccion: CH4. + O2. --> CO2 CO2 + H2O H2 + O2 -> H2O a) Cuales son los porcentajes de la mezcla de combustible (H2 y CH4) b) Cual es el porcentaje de exeso de aire Aire.
% 79 N2 y 21% O2 Productos ( N2 = 0.834 , O2 = 0.113 y CO2 = 0.053)
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A = 105.57 ---> O2 = 22.17 mol Balance de O2 105.57 x (0.21) + O = 11.3 + 5.3 + O(H2O) O(H2O) = 3.4 H2O = 6.8 Balance de H2 3 x C = 6.8 C = 2.27 Balance de C 2.27 x (CH4) = 6.8 CH4 = 0.428 = 42.8% H2 = 0.572 = 57.2%
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Lechada
Estanque de lavado
solución
decantada
agua de lavado
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230+0.98B=0.95(500+B) B=8166.7 D=8666.7
BALANCE DE NaAlO2 1000(0.16)= NaAlO2(8666.7) NaAlO2=0.0187≈1.85%
PROBLEMA 6 - PÁG. 85 En un reactor químico, donde se logra un 80% de conversión del reactivo limitante, se produce la oxidación del amoniaco de acuerdo a la siguiente reacción: NH3 + O2 → N2 + H2O Si se alimenta 300 kg/h de amoniaco y 2500 kg/h de aire (21% O 2 y 79% de N 2, en moles), determine: I. Escriba la reacción química con los coeficientes estequiometricos que corresponden II. Cuál es el reactivo limitante
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2NH3 + 3/2 O2
→
N2
+ 3H2O
Estequiometria: 17.651 + 13.24
8.83
26.48
X=0.8
7.06
21.1
14.12
10.59
Balance de N 2 = 86.61(0.79)+7.06 N2 SALIDA=75.48
Balance de H 2O 0+21.18= H2O(s)+0 H2O(s)=21.18
Balance de O 2 86.61(0.21)+0=10.54+ O2(s) O2(s)=7.6
Balance de NH 3
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Una muestra de 0.322g de un vaporr organico a 100 ªc y 0.974 atm, ocupa un volumen de 62.7 ml. Un análisis de dicho vapor da una composición elemental de C=65.46%, O=29.16%, e H=5.5%. ¿Cuál es su fórmula f órmula molecular?. Masas atómicas C=12; O=16; H=1
ATOMO
PORCENTAJ E
MASA AT.
PORCENT/ MASA
C
65,43
12
5,4525
O H
29,16 5,5
16 1
1,8225 5,5
RELACION AL MENOR 2,9917695 5 1 3,0178326 5
REDONDE O 3 1 3
FORMULA : C3 O H3 PROBLEMA 9 --- PAG 85 En un motor entra dodecano (C12H26) que se quema con aire para dar unos gses de combustión de análisis molar en base seca del 12.1% de CO2, 3.4% de CO, 0.5% de O2,
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79% N2 21% O2 relación = 3.76 12X= 12,1 + 3,4 X=1,29 2Y=24,2+ 3,4+ 1+ Z 2Y= 28,6+ Z 26X= 3 + 2Z
r% = r% = r% =
n aire / n combustible (y+3,76y)/x (4,76y)/1,29
r% =
80,92 mol aire/ mol combustible
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UNIVERSIDAD NACIONAL DE SAN AGUSTIN FACULTAD DE INGENIERIA DE PROCESOS ESCUELA PROFESIONAL DE INGENIERIA QUÍMICA
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