CASCADE AERATION

Dr. A. Saatci

KASKAT HAVALANDIRMA

Kağıthane Kaskat Havalandırma

K. Hane SAT Kaskat Havalandırma

Ömerli (Emirli) SAT

Ömerli (Emirli)

Kırıkkale SAT – Kapalı (!) Kaskat Havalandırma

Kaskat Havalandırma

Cumhuriyet SAT-Kaskat Havalandırma

10/8/2012 Prof. Dr. A. Saatci 9

AERATION Weir Aeration and Cascades

Referans : Pöpel, H.J., “Aeration and Gas Transfer”, Delft University of Technology

Dept. of Civil Engrg., Division of Sanitary Engineering. 1974.

Mechanism of Gas Transfer

During free fall of water certain surface area A is created. (CO2 VOCs and taste-odor producing substances are removed trough A) From the weir height “h” the average time of exposure of the surface area A: Size of A depends on: configuration of the weir portion of the nagge into several jets (will increase A/).

h = 2

2

1Cgt (free fall)

tc = gh /2

Mechanism #2 of Gas Transfer:

When nappe or its jets submerge into the receiving body significant amount of air entrained. The amount of air entrapped depends on the velocity of the nappe

when passing the surface of the water.

h =

g

V

2

2

V = gh2

Depth of receiving water influences the amount of gas transferred Vel of nappe when passing at this point depends on its energy at the ptA which is h = v2/(2g) (potential energy h converted to kinetic energy) . Final velocity of jets within the tail water before reaching its bottom = Rising velocity of the bubbles produced. Empirical estimates recommend a min depth of 2h/3.

Efficiency Coefficient K;

Unpolluted water K = 0.45 (1+0.046 tC).h

Polluted water K = 0.36 (1+0.046 tC).h

Sewage K = 0.29 (1+0.046 tC).h

h = weir height for straight weir.

lm

for rectangular notcher at least 4 jets/m. weir length the proportionality constant

increases up to 0.64 for heights < 0.70 m.

Modest increase of K value above 0.7 m ( Do not have step heights < 0.6-

0.7m)

8

6

K CS,

gO2/m3

4

2

0

0.5 1.0 1.5

length of weir, m

Fig 1: Efficiency coefficient in dependence of

The height of fall over weirs.

s

mgCCK

dt

dCSLa

3/)( ……………………………………………….(1)

aKKA

t

DAaK LL

c

L

2 ……………………………………….(2)

taK

S

S LeCC

CC ..

0

C = C0 @ t = 0 ……………………………….(3)

ln [(CS – C) / (CS – C0)] = (KLa) t St line ……………………………….(4)

taK

S

S LeCC

CC ..

0

= 1 – K ……………………………………………….(5)

K = taK

S

S LeCC

CC ..

0

1

……………………………………………….(6)

If height of weir being divided into (n) equal steps each having Kn = n

K from Eqn (5)

C1 = C0 (1 – Kn) + KnCS ……………………………………………….(7)

Cn = C0 (1 – Kn) + KnCS

Or

Cn = CS – (CS – C0)(1 - n

n

K)

as n (1 - n

n

K) e

-K ……………………………………….(8)

Subdivision into steps of a height of less than 0.6 m will decrease the

oxygenation effic but will promote CO2 desorption … odor … taste prod VOC step

heights of 0.2-0.40 m are quite common. Cascades require little space (~ 50-200

m2/m

3/s).

Example (Application of Eqn 8)

Calculate no of steps for max oxygenation

h = 1.5m, CS = 10 g/m3, C0 = 2 g/m

3

Form Fig 1. h = 1.5 m KCS = 7.0 gO2/m3

K = 7.0/10 = 0.7, K/n = 0.7 (for n = 1)

Form Eqn 8, C1 = CS – (CS – C0) (1 - n

K) = 0.7 CS + 0.3C0 = 7.6 g/m

3

C2 = CS – (CS – C0) (1 - n

K)

2 =0.75 CS + 0.25C0 = 8.0 g/m

3 max value

C3 = CS – (CS – C0) (1 – 0.35)3 = 7.8 g/m

3

Max oxygen concentration is reached at two steps

142.74

142.55

139.92 140.14 140.36 140.07

0.15 20.20

142.40

139.77 140.77

0.60

141.16 141.80

0.60

141.20

140.33 140.96 141.16 Kgiris= 0.50

0.60 TaşmaSavağı Kot= Kçıkış= 1.00

140.35 140.60 Qpipe= 5.79

22.00 m Dpipe= 2.00

Vpipe= 1.84

Qgaleri= 5.79 m3/s Kvalve= 1.50

137.00 Kgiris= 0.50 HL= 0.52

139.00 Kçıkış= 1.00

Agaleri= 6.25 m2 Apenstok=6.25

Vgaler= 0.93 m/s No Penst= 2.00

137.00 137.00 Hlgaler= 0.066 Qpens= 5.79

Vpenstk= 0.93

Kpensk= 2.70

HLpenst 0.118 m

1.20 m

133.92

143.26

142.74

142.55

0.15 20.20

142.40

0.60

141.16 141.80

0.60

141.20

140.96 141.16 Kgiris= 0.50

0.60 TaşmaSavağı Kot= Kçıkış= 1.00

140.60 Qpipe= 5.79 m3/s

22.00 m Dpipe= 2.00 m

Vpipe= 1.84 m/s

Qgaleri= 5.79 m3/s Kvalve= 1.50

137.00 Kgiris= 0.50 HL= 0.52 m

Kçıkış= 1.00

Agaleri= 6.25 m2 Apenstok=6.25

Vgaler= 0.93 m/s No Penst= 2.00

137.00 Hlgaler= 0.066 Qpens= 5.79

Vpenstk= 0.93

Kpensk= 2.70

OZON TEMAS TANKI HİDROLİĞİ

Batık Savak

Batık Savak

H

h

B

a

Batık Savak Formülü:

2/361.0'' )(23

2)(2 hHgBChHghBCQ DD

Perde Hidroliği

%52

A

Ab (% 10’dan fazla olursa kısa devreyi engelleyemez)

Batık orifis denkleminden

hgACQ D 2

gBaC

Qh

D 2

veya

2.3180

dönüşCoK

g

Vh

22.3

2

Perdelerdeki yük kaybı 0.8 cm geçerse yumaklar kırılır. Perde aralığındaki hızlarından aynı

sebepten 0.3 m/s geçmemeleri gerekir.

3/1

2

2

gB

Qdc