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DATOS

Anillos pall (plástico)

∆P/Z=50 (N/m^2)/m

L=1.8 m

∅=15 cm

Balance de materia

Ls/Gs=(〖YA〗_1-〖YA〗_2)/(〖XA〗_1-〖XA〗_2 )

〖yA〗_1=0.032 (kmol A)/(kmol G1)

〖YA〗_1=〖yA〗_1/(1-〖yA〗_1 )=(0.032 (kmol A)/(kmol G1))/(1-0.032 (kmol A)/(kmol G1))=0.033 (kmol A)/(kmol Gs)

〖YA〗_2=1/8 〖YA〗_1

〖YA〗_2=1/8 (0.033 (kmol A)/(kmol Gs))=0.0042 (kmol A)/(kmol Gs)

Datos de equilibrio para el sistema amoniaco-agua

C (g〖NH〗_3)/(100 g H_2 O) P_NH3 (mmhg) Y_NH3 X_NH3

0 0 0 0

0.105 0.791 0.00135 0.00111

0.244 1.83 0.00314 0.00258

0.32 2.41 0.00414 0.00339

0.38 2.89 0.00496 0.00402

0.576 4.41 0.0076 0.0061

0.751 5.8 0.01001 0.00795

1.02 7.96 0.01379 0.0108

1.31 10.31 0.01794 0.01387

1.53 11.91 0.02078 0.0162

1.71 13.46 0.02355 0.01811

1.98 15.75 0.02767 0.02096

2.11 16.94 0.02982 0.02234

2.58 20.86 0.03698 0.02732

2.75 22.38 0.03978 0.02912

3 23.5 0.04185 0.03176

Y_NH3=P_NH3/(585-P_NH3 )

Y_NH3=((C)(18))/((100)(17))

Grafica de los datos de equilibrio

De grafica se obtiene:

〖XA〗_1^*=0.023 (kmol A)/(kmol Ls)

Ls/〖Gs〗_min =(〖YA〗_1-〖YA〗_2)/(〖XA〗_1^*-〖XA〗_2 )

Ls/〖Gs〗_min =(0.033-0.0042)/(0.023-0)

Ls/〖Gs〗_min =1.2522 (kmol Ls)/(kmol Gs)

〖(Ls/Gs)〗_ope=Ls/〖Gs〗_min ×fd

fd=1.6

〖(Ls/Gs)〗_ope=1.2522 (kmol Ls)/(kmol Gs))(1.6)

〖(Ls/Gs)〗_ope=2 (kmol Ls)/(kmol Gs)

Ls/Gs=(〖YA〗_1-〖YA〗_2)/(〖XA〗_1-〖XA〗_2 )

〖XA〗_1-〖XA〗_2=(〖YA〗_1-〖YA〗_2)/(Ls/Gs)

〖XA〗_2=0

〖XA〗_1=(0.033-0.0042)/2=0.0144 (kmol A)/(kmol Ls)

Para la grafica de caída de presión para lechos empacados:

L^'/G^' (ρ_G/(ρ_L-ρ_G ))^(1/2)

ρ_G=P/RT=(0.7697 atm)/((0.08205 (atm m^3)/(kmol K))(298K))=0.0314 kmol/m^3

〖PM〗_G=yA∙〖PM〗_A+(1-yA)∙〖PM〗_Gs

〖PM〗_G=(0.032)(17)+(1-0.032)(29)=28.616 (kg G)/(kmol G)

¯(ρ_G )=0.0314 (kmol G1)/m^3 ((28.616 kg G1)/(kmol G1))((1 Lb)/(0.4535 kg))((0.0283 m^3)/(1〖ft〗^3 ))=0.0560 Lb/〖ft〗^3

¯ρL=998.08 kg/m^3 ((1 Lb)/(0.4535 kg))((0.0283 m^3)/(1〖ft〗^3 ))=62.283 Lb/〖ft〗^3

(L/G)=(〖Ls/Gs)〗_op

(〖Ls/Gs)〗_op=2 (kmol Ls)/(kmol Gs) ((18 kg Ls)/(kmol Ls))((kmol Gs)/(29 kg Gs))=1.24138 (kg Ls)/(kg Gs)

B=1.24138 (kg Ls)/(kg Gs) ((0.0560 Lb/〖ft〗^3 )/(62.283 Lb/〖ft〗^3 -0.0560 Lb/〖ft〗^3 ))^(1/2)=0.03724

∆P/Z=50 (N/m^2)/m

De grafica de lechos empacados se obtiene:

(G^'2∙Cf∙μ_L^0.1∙J)/(ρ_G (ρ_L-ρ_G )gc)=0.012

G^'=√((0.012(ρ_G (ρ_L-ρ_G )gc)/(Cf∙μ_L^0.1∙J))

¯(ρ_G )=0.0560 Lb/〖ft〗^3

¯(ρ_L )=62.283 Lb/〖ft〗^3

gc=4.18×〖10〗^8 〖ft〗^3/hr

μ_L=0.9648 cp

J=1.502

Cf=52

G^'=√((0.012(0.0560 Lb/〖ft〗^3 (62.283 Lb/〖ft〗^3 -0.0560 Lb/〖ft〗^3 )4.18×〖10〗^8 〖ft〗^3/hr)/(52∙〖0.9648 cp〗^0.1∙1.502))=473.919 Lb/(〖ft〗^2 hr)

G^'=Gs

Ls/Gs=(〖YA〗_1-〖YA〗_2)/(〖XA〗_1 〖-XA〗_1 )

Ls=(〖YA〗_1-〖YA〗_2)/(〖XA〗_1 〖-XA〗_1 )×Gs

G'=473.919 lb/(〖ft〗^2 hr) ((0.4535 kg)/1lb)(〖1ft〗^2/(.0929m^2 ))((1 Kmol)/29Kg)

G'=79.775 Kmol/(m^2 hr)

S=0.785 D^2

S=0.785 〖(0.15m)〗^2

S=0.01766m^2

Gs=79.775 Kmol/(m^2 hr)×〖(0.01766m〗^2)

Gs=1.4088 Kmol/hr

Ls=(0.033-0.0042)/(0.0144-0)×(1.4088)

Ls=2.8176 Kmol/hr

Ls=2.8176 kmol/hr ((18 kg Ls)/(1kmol Ls))(〖1m〗^3/(998.08 kg))((1 L)/(0.001m^3 ))((1 hr)/(60 min))=0.847 L/min

Porcentaje de rotámetro para el agua:

100% rotámetro ----> 2.1 L/min

X ----->0.847 L/min

X=40.56%≈41%

Cantidad de NH3:

G_NH3=〖YA〗_1∙Gs

G_NH3=(0.033 (kmol A)/(kmol Gs))(1.4088 Kmol/hr)=0.04649 Kmol/hr

G_1=〖GA〗_1+Gs

G_1=(0.04649 Kmol/hr)+(1.4088 Kmol/hr)=1.45529 Kmol/hr

G_2=〖GA〗_2+Gs

〖YA〗_2=〖GA〗_2/Gs

〖GA〗_2=〖YA〗_2×Gs

G_2=Gs(〖YA〗_2+1)

G_2=1.4088 Kmol/hr (0.0042 (kmol A)/(kmol Gs)+1)=1.41472 Kmol/hr

G_2=〖GA〗_2+Gs

〖GA〗_2=G_2-Gs

〖GA〗_2=1.41472 Kmol/hr-1.4088 Kmol/hr=0.00592 Kmol/hr

Ecuación para placa de orificio:

Q=104√(∆h/ρ)

Q=G=Gs

Gs=1.4088 Kmol/hr (〖1m〗^3/(0.0314 kmol))((1 L)/(0.001m^3 ))((1 hr)/(60 min))=747.777 L/min

ρ_(G=) 0.0314 kmol/( m^3 ) ((28.616 kg)/kmol)((1000 gr)/(1 kg))((0.001m^3)/1L)=0.8985 ( gr)/L

747.777 L/min=104√(∆h/(0.8985 ( gr)/L))

∆h=46 cm H2O

Q1=41√(Pt/Tt)

Q1=41√(760/298)=65.4761 L/min

Q1∙P1∙T1=Q2∙P2∙T2

Q2=(Q1∙P1∙T1)/(P2∙T2)

Q2=(65.4761 L/min)(760 mm Hg)(298K)/((585 mm Hg)(294 K))=86.2203 L/min

G_NH3=0.04649 Kmol/hr ((17 kg A)/(kmol A))(〖1m〗^3/(0.717 kg))((1 L)/(0.001m^3 ))((1 hr)/(60 min))=18.3712 L/min

100% ----> 86.2203 L/min

X ----->18.3712

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