4. Qo S₂ x₂ Consider two- VI X₁, S₁ -stage treatment system as shown: V₂ X₂ S₂ 1 RQ, xr, S₂ Given = k₁= 0.05 day' ; K=5 day' ; Y=0.5 ; Ks= 50 mg |L Q₂ 1 = 28000 m³/day ; S₂ = 560 mglL as BOD V/₁ = 14.000 m³ ; BOD ₁ = 0.68 BOD ; kg 0₂ = 1.42 kg Cells X₁ = 10000 mg|l; S₂ = 7.5 mgll as BOD5 (in reactor 2) = 5 days; R=0.6; X₂ = 20mg|L Assuming Moned kinetics apply and steady-state operation; determine: -Qu a) x, b) S₁ c) Px₁ = excess sludge production in the first reactor d) amount of oxygen required daily in the first reactor.
4. Qo S₂ x₂ Consider two- VI X₁, S₁ -stage treatment system as shown: V₂ X₂ S₂ 1 RQ, xr, S₂ Given = k₁= 0.05 day' ; K=5 day' ; Y=0.5 ; Ks= 50 mg |L Q₂ 1 = 28000 m³/day ; S₂ = 560 mglL as BOD V/₁ = 14.000 m³ ; BOD ₁ = 0.68 BOD ; kg 0₂ = 1.42 kg Cells X₁ = 10000 mg|l; S₂ = 7.5 mgll as BOD5 (in reactor 2) = 5 days; R=0.6; X₂ = 20mg|L Assuming Moned kinetics apply and steady-state operation; determine: -Qu a) x, b) S₁ c) Px₁ = excess sludge production in the first reactor d) amount of oxygen required daily in the first reactor.
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
Related questions
Question
![4.
Qo
S₂₂
Consider a two-stage treatment system as shown:
VI
X₁, S₁
( )
Given :
V₂
X₂ S₂
1
RQ, Xr, S₂
k₁= 0.05 day; K=5 day¹; Y=0.5 ; K₂ = 50 mg/L
Q₁ = 28000 m³/day ; S₂ = 560 mglL as BOD
= 14000 m³ ; BOD ₁ = 0.68 BOD₁;
V₁
kg 0₂ = 1.42
kg Cells
X₁ = 10000 mg/l ; S₂ = 7.5 mgll as BOD5
(in reactor 2) = 5 days; R=0.6
; x₂ = 20 mg/L
Assuming Monod kinetics apply and steady-state operation;
determine:
uxy
h) Px z
= excess sludge production in the (second reactor; and
i) amount of oxygen required daily in the second reactor.
Qw
+く
Qu
a) x₁
b) S₁
c) Px₁ = excess sludge production in the first reactor
d) amount of oxygen required daily in the first reactor.
f) V/₂
g) Qw
e) X₂](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb0cbfe89-fbe8-48b4-af6e-d48c26d5240b%2F77078cae-54a1-46db-b4a6-7da044946f9d%2F08sbu88_processed.png&w=3840&q=75)
Transcribed Image Text:4.
Qo
S₂₂
Consider a two-stage treatment system as shown:
VI
X₁, S₁
( )
Given :
V₂
X₂ S₂
1
RQ, Xr, S₂
k₁= 0.05 day; K=5 day¹; Y=0.5 ; K₂ = 50 mg/L
Q₁ = 28000 m³/day ; S₂ = 560 mglL as BOD
= 14000 m³ ; BOD ₁ = 0.68 BOD₁;
V₁
kg 0₂ = 1.42
kg Cells
X₁ = 10000 mg/l ; S₂ = 7.5 mgll as BOD5
(in reactor 2) = 5 days; R=0.6
; x₂ = 20 mg/L
Assuming Monod kinetics apply and steady-state operation;
determine:
uxy
h) Px z
= excess sludge production in the (second reactor; and
i) amount of oxygen required daily in the second reactor.
Qw
+く
Qu
a) x₁
b) S₁
c) Px₁ = excess sludge production in the first reactor
d) amount of oxygen required daily in the first reactor.
f) V/₂
g) Qw
e) X₂
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