Technology : Exercises
Part I :
Policy
2 Concepts a
terminology
1
S11
Given : It =
250 000i .
e
.; Whprod = 100L i.e xo ; .
bCOD =
500mg L
; V 10 000m=
A Lond J
·
Wastewater flow :
Q =
WWpeod .
XIE = 100L x 250000 i . e .
i .
2 .
xd
=
25x18"L = 25 000 m3
O d
Load J QXbCOD 25
000m X 05l8 12500
·
: = = =
B Volumetric loading rate Bu
Bu 25
- 12500kg1000 m
·
= = =
1 ,
M3x@
2
S16 GiveniP =
1 MW =
1x10W ; time =
250d ; 1kWh =
3,
61103)
Energy E
1x10"W
E Pxtime 250d
x864002
9 , 16 Mo
3
· = = + =
16x1831X1 HWN 600O
=> E =
2, =
1 2
. P .
1
,Part : Water
1 Natural purification of surface water
&
S20
-
Guien :
It =
100000 i C .
.; Qu =
20 000 M3 d ; Tw 25 %; =
BOD5, W
30MGL ; DOW
Omg L ; QR 300 000 M3 d;
= = =
30 °;
10 MGL
-
Vr =
0 , 25 MS ; Tr = BODs , R =
;
; Up
&°
DOR 4, 5 0 , 23 or ; Kr 345 d
mg L 0
= = =
,
A Location of maximal oxygen duficit X
·
TA =
QRXTr + Qw x Tw =
300000 x 30 + 20000X25 =
29 , 7
QR + Qu 300 000 + 20008
Kp20
7 20
1 , 04725
TA-20
-
Kn 0, 25 d
,
1 , 04 0 . 359
·
= x = x =
* Ta 20 29, 7 -
28
0 , 315 X 1 , 020 d
=
·
Kr =
Kr x 1 , 020 = = 0, 418
DOA QRXDOR QuxDOw 300000 x 4 , 5 + 20000X0 4 , 22
m
·
=
+ = =
QR + Qu 300 000 + 20008
·
DOsat =
14 , 65 -
Q41 TA 99x183xTA 7 781185x +
+7,
*
-
,
3
041x29 7 + 7 , 994183x29,72 7 781829 73
48 mg
=
14 , 65 - -
=
7
, , , ,
·
Da =
Dosa -
Don = 7 , 48 -
4 ,
22 =
3 26 ,
CBODR 14 , 6
BODsir
12 mg MG.
·
= X 1 , 46 =
X 1 , 46 =
CBODW BOD5 , W + 1 , 46 43 , 8
30mg 46
·
my
=
x1
= =
,
L
·
CBODA =
QRX CBODR + Qw XcBODw
QR + Qu
300000 x 14
, 6 + 28000x43 , % 16 43 Mg
= =
,
300 000 + 20 ord L
3.
1 P L
.
, tc Esmax
&n [kr x1 kn kn
xDA]
·
= = 1 x - -
KR-KD KD Ka bCODA
(n[0418 0 , 418 0 ,
359x39] 20
- x x 1 =
-
= -
0, 418 0
, 359 0 , 359
-
, 359
0
Xc (c 2 , 01 d 025M
Vi
864002 43480M
·
= x = x x =
3
=
43 , 4 km
B Minimal dissolved concentration
oxygen Domin
·
Dc =
Dmax = KD x Laxe-Koxto
KR
-
. 359x2 1 01
0
0 , 359 16 , 43x2 6 86
mg
= =
x
,
0 , 418
6 , 86
Domin DOsat-Dmax 7 48 0 . 62
mg/L
·
= = - =
,
C problem ?
=> Yes : Do 1 , 5
mg/L is detrimental for aquatic life
⑰
S30
-
EXAM
-
Given : Qu 50m3/h ; COD
°
60000mg/L ; fb 0 . 87 ; Tw 20 C
= = = =
,
0 , 5 km/h QR 1036800 M3/d
Dow
Omg/L ; Vi ;
=
;
= =
°
Tr =
10 C ; CBODR =
Omg/L ;
25
Dor =
11 ,
27 mg/L ;
K =
0 , 804/d ; KD =
0, 304/d
MaximumOxygenPictAm m
A
h d O
3. 1
p .
3
Part I :
Policy
2 Concepts a
terminology
1
S11
Given : It =
250 000i .
e
.; Whprod = 100L i.e xo ; .
bCOD =
500mg L
; V 10 000m=
A Lond J
·
Wastewater flow :
Q =
WWpeod .
XIE = 100L x 250000 i . e .
i .
2 .
xd
=
25x18"L = 25 000 m3
O d
Load J QXbCOD 25
000m X 05l8 12500
·
: = = =
B Volumetric loading rate Bu
Bu 25
- 12500kg1000 m
·
= = =
1 ,
M3x@
2
S16 GiveniP =
1 MW =
1x10W ; time =
250d ; 1kWh =
3,
61103)
Energy E
1x10"W
E Pxtime 250d
x864002
9 , 16 Mo
3
· = = + =
16x1831X1 HWN 600O
=> E =
2, =
1 2
. P .
1
,Part : Water
1 Natural purification of surface water
&
S20
-
Guien :
It =
100000 i C .
.; Qu =
20 000 M3 d ; Tw 25 %; =
BOD5, W
30MGL ; DOW
Omg L ; QR 300 000 M3 d;
= = =
30 °;
10 MGL
-
Vr =
0 , 25 MS ; Tr = BODs , R =
;
; Up
&°
DOR 4, 5 0 , 23 or ; Kr 345 d
mg L 0
= = =
,
A Location of maximal oxygen duficit X
·
TA =
QRXTr + Qw x Tw =
300000 x 30 + 20000X25 =
29 , 7
QR + Qu 300 000 + 20008
Kp20
7 20
1 , 04725
TA-20
-
Kn 0, 25 d
,
1 , 04 0 . 359
·
= x = x =
* Ta 20 29, 7 -
28
0 , 315 X 1 , 020 d
=
·
Kr =
Kr x 1 , 020 = = 0, 418
DOA QRXDOR QuxDOw 300000 x 4 , 5 + 20000X0 4 , 22
m
·
=
+ = =
QR + Qu 300 000 + 20008
·
DOsat =
14 , 65 -
Q41 TA 99x183xTA 7 781185x +
+7,
*
-
,
3
041x29 7 + 7 , 994183x29,72 7 781829 73
48 mg
=
14 , 65 - -
=
7
, , , ,
·
Da =
Dosa -
Don = 7 , 48 -
4 ,
22 =
3 26 ,
CBODR 14 , 6
BODsir
12 mg MG.
·
= X 1 , 46 =
X 1 , 46 =
CBODW BOD5 , W + 1 , 46 43 , 8
30mg 46
·
my
=
x1
= =
,
L
·
CBODA =
QRX CBODR + Qw XcBODw
QR + Qu
300000 x 14
, 6 + 28000x43 , % 16 43 Mg
= =
,
300 000 + 20 ord L
3.
1 P L
.
, tc Esmax
&n [kr x1 kn kn
xDA]
·
= = 1 x - -
KR-KD KD Ka bCODA
(n[0418 0 , 418 0 ,
359x39] 20
- x x 1 =
-
= -
0, 418 0
, 359 0 , 359
-
, 359
0
Xc (c 2 , 01 d 025M
Vi
864002 43480M
·
= x = x x =
3
=
43 , 4 km
B Minimal dissolved concentration
oxygen Domin
·
Dc =
Dmax = KD x Laxe-Koxto
KR
-
. 359x2 1 01
0
0 , 359 16 , 43x2 6 86
mg
= =
x
,
0 , 418
6 , 86
Domin DOsat-Dmax 7 48 0 . 62
mg/L
·
= = - =
,
C problem ?
=> Yes : Do 1 , 5
mg/L is detrimental for aquatic life
⑰
S30
-
EXAM
-
Given : Qu 50m3/h ; COD
°
60000mg/L ; fb 0 . 87 ; Tw 20 C
= = = =
,
0 , 5 km/h QR 1036800 M3/d
Dow
Omg/L ; Vi ;
=
;
= =
°
Tr =
10 C ; CBODR =
Omg/L ;
25
Dor =
11 ,
27 mg/L ;
K =
0 , 804/d ; KD =
0, 304/d
MaximumOxygenPictAm m
A
h d O
3. 1
p .
3
