In relation to the bypass example below, rationalize why there are processes where a small fraction of an enormous flow is separated, treated, and introduced back to the large enormous stream it came from.

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
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In relation to the bypass example below, rationalize why there are processes where a small fraction of an enormous flow is separated, treated, and introduced back to the large enormous stream it came from.

In a water treatment plant, water is split into two: a major and minor
stream. The minor stream is treated with chlorine as sodium
hypochlorite and is again mixed with the major stream. The sodium
hypochlorite content of the mixed major and minor stream shall be 5
ppm per PNSDW or Philippine National Safe Drinking Water Standard.
If the minor stream is 5 m3/h and is 2% of the total flow of the
treated water, how much sodium hypochlorite solution is to be used
and how much is the overall treated water. The sodium hypochlorite
solution is a 5w/v% strength, assume all densities to be 1000 kg/m3.
Transcribed Image Text:In a water treatment plant, water is split into two: a major and minor stream. The minor stream is treated with chlorine as sodium hypochlorite and is again mixed with the major stream. The sodium hypochlorite content of the mixed major and minor stream shall be 5 ppm per PNSDW or Philippine National Safe Drinking Water Standard. If the minor stream is 5 m3/h and is 2% of the total flow of the treated water, how much sodium hypochlorite solution is to be used and how much is the overall treated water. The sodium hypochlorite solution is a 5w/v% strength, assume all densities to be 1000 kg/m3.
Еxample I - Вурass
OVERALL CHLORINE BALANCE
mass in = mass out
(5000mg
+ X
(5mg`
= Z
L
250,0
ka.
X m3/h
0.1L
5 w/v% = 5 g/100mL
NaOCI
(kg
50,000X
h
kg
= 5Z
h
2
→ Z = 10,000x
Solving 1 and 2 simultaneously
5000
kg/h
Y + X = Z
1
kg
250,000
+ X = Z
h
L = 245,000 kg/h
Y =
Z m3/h
98% of
kg
= 5 mg/L
+ X = 10,000X
h
250,000
5 ppm
250,000-
flow
kg/h
NaOCI
NODE A
kg
X = 25.0025
h
kg
Z = 250,025
h
Transcribed Image Text:Еxample I - Вурass OVERALL CHLORINE BALANCE mass in = mass out (5000mg + X (5mg` = Z L 250,0 ka. X m3/h 0.1L 5 w/v% = 5 g/100mL NaOCI (kg 50,000X h kg = 5Z h 2 → Z = 10,000x Solving 1 and 2 simultaneously 5000 kg/h Y + X = Z 1 kg 250,000 + X = Z h L = 245,000 kg/h Y = Z m3/h 98% of kg = 5 mg/L + X = 10,000X h 250,000 5 ppm 250,000- flow kg/h NaOCI NODE A kg X = 25.0025 h kg Z = 250,025 h
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