Environmental Architectural Assignment Article

Civil Engin

neering Depa

artment

CVG2132

2 – FUNDAMENTALS OF ENVIRO

ONMENTA

ALВ ENGINEEERINGВ

HomeworrkВ 5: В В

14 (noon) – C

Cubby « CVG 2

2132В В», В Mezz anineВ AВ (0. 5)В В CBYВ В

DueВ Date: В Nov. В 28, В 201

Problem

nВ 1. В В DetermineВ theВ BOD

D5В ofВ aВ wastewaterВ baseddВ onВ theВ follo

owingВ experrimentalВ resultsВ В

DISSOLV

VEDВ OXYGE NВ (gВ O2/m3))В

Dillution, В mLВ /В 300mLВ

TIMEВ ((d)В

10В

20В

almost eight. 95В

eight. 9В

8.. 8В

eight. 6В

8. 5В

44В

0. 5В

SOLUTIO

ONВ

п‚·

п‚·

The

eВ 1mLВ /300В mLВ dilutionВ can

nnotВ beВ usedВ becauseВ theВ changeВ inВ DO

OВ isВ lessВ thanВ 2В mg/LВ orВ g/m

m3В

The

eВ 20mLВ /В 300В mLВ dilutionВ cannotВ beВ use

edВ becauseВ thheВ finalВ DOВ isВ lessВ thanВ 2В m

mg/LВ

( DOs, we пЂ­ DOs, t ) пЂ­ пЃ›( DO

M b, i actually пЂ­ DOb, t ) * (1 пЂ­ L )пЃќ

BOD

Dt пЂЅ

P

eВ 5В mLВ /В 300

0LВ solution, В P

P=5/300В and

SoВ forВ the

(8. 9mg UNITED KINGDOM / D пЂ­ 6mg O2 as well as L) пЂ­ (9mg O2 / L пЂ­ eight. 5mg

meters O2 as well as L)(1 пЂ­ (5mL / 300mL

m )

пЂЅ 144. five mgO2 as well as L

BOD5 пЂЅ

(5mL / 300mL)

AndВ forВ theВ 10В mlВ /В 3

300В mLВ dilutionВ P=10/300В andВ В

(8. 8mg O2 / L пЂ­ 4mg UNITED KINGDOM / L) пЂ­ (8. 9mg

g O2 / L пЂ­ 8. 5mg O2 / L) 5. (1 пЂ­ (10 cubic centimeters / 300mL))

(10 mL / 300 mL

L)

пЂЅ 129

1 . your five mgO2 / L

BOD5 пЂЅ

NowВ calcculateВ theВ avverageВ ofВ the

eseВ twoВ valu

uesВ

.

ANSWE

ER:

.

В 

QuestionВ #В 2. В TheВ town

nВ ofВ Avepitae

eonmiВ hasВ file

edВ aВ complainntВ withВ theВ staateВ DepartmeentВ ofВ NaturalВ ResourcessВ (DNR)В thatВ ttheВ CityВ ofВ WaatapitaeВ isВ haarmingВ theВ W ashВ RiverВ byВ d dischargingВ raawВ sewage. В

Avepitaeo

onmiВ isВ locate

edВ 15. 55В kmВ d

downstreamВ TTheВ DNRВ wateerВ qualityВ critteriumВ forВ theeВ WashВ RiverВ isВ 5В mg/LВ ofВ DO. В В TheВ follow

wingВ isВ dataВ p

pertaining to tthe 7‐year,  100‐day low flow

wВ conditionsВ atВ Watapitaee. В В

Determine: В

a)

b)

c)

d)

Th

heВ locationВ ofВ theВ criticalВ DOВ deficit. В

Th

heВ DOВ concen

ntrationВ atВ th

heВ criticalВ pointВ

IsВ thisВ satisfacttory? В

IfВ theВ lowestВ D

DOВ levelВ isВ toВ beВ 5В mg/L, В aВ treatmentВ p lantВ willВ haveeВ toВ beВ built. В В WhatВ willВ th heВ

trreatmentВ plan

nt'sВ effluentВ BOD5В concen

ntrationВ haveeВ toВ beВ toВ havveВ suchВ aВ minimumВ DO? В

CVG2132

ASSIGNMENT #5

Fall 2013

Watapitae

Suppose steady-state conditions so Deposition =0

Assume the three fields are thin down and have the same density Presume the river flow speed downstream in the discharge equates to that upstream of the launch Assume the blending is instant and complete. This implies the mixing volume is minimal So there is not any DO generation or consumption in the combining point.

Qr

Qw

Ts

Tww

DOr =

Dow

Dosat

BOD5_ww =

1 ) 08

0. 1507

sixteen

16

six. 95

you

9. 95

128

e (20EC)

zero. 4375

Ls = Upstream BOD

10. 4

ult =

v (m/s) sama dengan

0. 39

H = depth (m) = installment payments on your 8

Е‹ =bed activity coefficient

zero. 2

=

Total mass balance around the mixing point

‫݉ݑܿܿܣ‬. ൌ ܳ௥ ∙ ߩ௥ ൅ ܳ௪ ∙ ߩ௪ െ ܳ௢ ∙ ߩ௢ Steady-state so Build up =0

All the densities are equal because these are thin down solutions

so

3

1 ) 08 m /s

Qo =

&

3

0. 1507 m /s =

1 . 2307

m3/s

Mixed Oxygen (DO) mass equilibrium around the blending point

‫݉ݑܿܿܣ‬. ൌ ܳ௥ ∙ ‫ܱܦ‬௥ ൅ ܳ௪ ∙ ‫ܱܦ‬௪ െ ܳ௢ ∙ ‫ܱܦ‬௢ + Generation - Intake Accumulation =0 because of the steady-state conditions

Due to negligible amount the generation and usage are minimal so

‫ܱܦ‬௢ ൌ

DOo =

DOo =

ܳ௥ ∙ ‫ܱܦ‬௥ ൅ ܳ௪ ∙ ‫ܱܦ‬௪

ЬіаЇў

(

several

1 . '08 m /s x

six. 098968 mg/L

7. 95

mg/L )+(

1 . 2307

3

0. 1507 meters /s x

1

mg/L )

Ultimate BOD from the effluent

‫ ݐ ܦܱܤ‬ൌ ‫ܮ‬௪ ൈ 1 െ ݁ ି௞∙௧

So

‫ܮ‬௪ ൌ ‫ܦܱܤ‬ሺ‫ݐ‬ሻ/ 1 െ ݁ ି௞∙௧ Lw =

-1

0. 4375 g x

128 / [1-exp(-

Langwelle =

a few

d )]

a hundred and forty four. 1761 mg/L

Ultimate BOD remaining (L) Mass stability around the mixing up point ‫݉ݑܿܿܣ‬. ൌ ܳ௥ ∙ ‫ܮ‬௥ ൅ ܳ௪ ∙ ‫ܮ‬௪ െ ܳ௢ ∙ ‫ܮ‬௢ + Generation - Usage Accumulation =0 because of the steady-state conditions

As a result of negligible quantity the technology and intake are minimal so

‫ܮ‬௢...



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