SELY OKTAV OKTAVIOLITA ASRI
KELAS A TEKNIK LINGKUNGAN LINGKUNG AN
21080114120029
Nama : Sely Oktaviolita Asri Problem 8.1 Problem: A Problem: A rapid-mixing basin is to be designed for a water coagulation plant, and
the
design
ow for the basin is 4.0 MGD. he basin is to be s!uare with depth e!ual to ".#$ times
the
width. he %elocit& gradient is to be '00 s -" (at $0) *+, and the detention time is 0
sec.
Determine a+ he basin basin dimensions dimensions if incremen increments ts of " in. are are used. b+ he inpu inputt horse horsepow power er.. c+ he impeller impeller speed speed if a %ane-disc %ane-disc impeller impeller with six at at blades is emplo&e emplo&ed d and the
tan
is
ba/ed. he impeller diameter is to be $0 of the basin width. Approach Approach *irst, we sol%e for the %olume of the tan b& using the ow rate and the
detention
time. 1e 1e then use this %olume %olume to compute basin dimensions. dimensions. hen, we use the %elocit&
gradient
and the inematic %iscosit& to obtain power. 2astl&, we use the nowledge of impeller
relations
to obtain impeller speed. Variables:
3 detention time
7 absolute %iscosit& of water
x length and width of basin
D diameter of impeller
G %elocit& gradient
n impeller %elocit&
d depth of basin
8 9olume of basin
5 ow rate
1 6ower imparted
6 6ower input
: water densit&
Solution: a+ *irst irst,, we need need to calc calcul ulat ate e the the %olum %olume e of the the basi basin. n. 1e acco accomp mplis lish h this this through the following 6 3 4 x 10 gals 1 day 1 hour 1 min 1 ft 3 x x x x x 30 seconds =185.68 ft ∀ =Q x θ = 1 day 24 hours 60 min 60 sec 7.48 gal 1e now tae this %olume and e!uate it to the following ∀ = x
2
x d d =1.25 x ∴ ∀ =1.25 x
3
SELY OKTAVIOLITA ASRI
KELAS A TEKNIK LINGKUNGAN
21080114120029
x =
√ 3
∀
1.25
=
√ 3
3
185.68 ft 1.25
=5.3 feet ≈ 5 ft −4 ∈¿
;ow, sol%ing for depth
d =1.25 x =1.25 x 5.3 feet =6.675 feet ≈ ft −8 ∈¿
ft-?in b+ *irst, we need to calculate the power imparted ft −lb −5 2.73 x 10 = 22.113 3 sec − ft −1 2 2 W =G x μ =( 900 s ) x ¿ ;ow, we can sol%e for the total power
P =W x ∀ =22.113
ft − lb
ft − lb ' 3 ' = ( ) ( ) 5.33 6.67 4193 x x 3 sec sec −ft
2astl&, we con%ert to horsepower
P = 419 as 3
ft −lb sec x x hp =7.62 hp sec 550 ft −lb
c+ *or this, we simpl& need the following e!uation 2.667 ft
¿ ¿
ft −lb sec rev 60 sec = = 84.5 rpm x 1.41 1 /3 sec 1 min 5.75 x (¿ 5 x 1.936 ¿¿ ) 4193
n=
(
¿
P 5
K t D
Problem 8.2 a+ he =asin Dimensions 3
15140 m 1 min ! = x x30s 1440 min 60 s
! =5.26 m
3
he dimensions as gi%en b& " x " x ( 1.25 " )= 5.26 m
3
1/ 3
)
=¿
SELY OKTAVIOLITA ASRI
KELAS A TEKNIK LINGKUNGAN
21080114120029
1.25 "
3
= 5.26 m3
3
" =4.208 m " =1.61 m
;ow, sol%ing for length # =1.25 " =1.25 ( 1.61 ) =2.0125 m
he basin dimensions " =1.61 m # =2.0125 m b+ *irst, we need to calculate the power imparted $ −m 2 −1 2 W = G x μ =( 900 s ) x 0.00131 =1061.1 3 s −m ;ow, we can sol%e for the total power 1061.1
P= W x ∀ =
$ −m 3
s −m 3 m
3
x 5.26 m
$ −m % =5581 =5581 " s s
=5581.38
c+ he impeller speed is a %ane-disc impeller with six at blade is emplo&ed and the tan is ba/ed. he impeller diameter is to be 0 of the bacin width Kr =5.75 d =50 "= 0.5 x 1.61=0.805 m *or this, we simpl& need the following e!uation 1/ 3 1/ 3 5581 " P rev 60 sec = = = 72 rps 1.20 n= x 5 2 1 sec min / 5.75 x 0.806 m x 999.7 &g m K t D
(
) (
)
Problem 8.3 Problem A occulation basin is to be designed for a water coagulation plant,
and
the
design
ow is ".0 MGD. he basin is to be a cross-ow hori@ontal-shaft, paddlewheel
t&pe
with
a
mean %elocit& gradient of #>. s -" (at $0 0*+, a detention time of 4$ min, and
a
G
%alue
from
$0,000 to "00,000. apered occulation is to be pro%ided, and three compartments
of
e!ual
SELY OKTAVIOLITA ASRI
KELAS A TEKNIK LINGKUNGAN
21080114120029
depth in series are to be used, as shown in *igure ?."'(b+. he compartments
are
to
be
separated
b& slotted, redwood ba/e fences, and the basin oor is le%el. he G %alues are to be $0, #0, and "0s -". he occulation basin is to ha%e a width of '0ft to adBoin the settling basin. he paddle wheels are to ha%e blades with a > in width and a length of "0ft. he outside blades should clear the oor b& " ft and be " ft below the water surface. here are to be six blades per paddle wheel, and the blades should ha%e a spacing of " ft. AdBacent paddle wheels should ha%e a clear spacing of 0 to > in. between blades. he wall clearance is "# to "? in. Determine a+ he basin Dimensions b+ he paddle-wheel design c+ he power to be imparted to the water in each compartment and the total power re!uired for the basin. d+ he range in rotational speed for each compartment if "4 %ariable speed
dri%es
are
emplo&ed. Approach: o begin, weCll sol%e for the G %alue. Also, weCll sol%e for the dimensions
of
the
basin using a %olume e!uation compared with the ow. *rom this, we will assume
s!uare
compartments in prole (depth length for each prole+. his will gi%e us our
dimensions.
Esing those dimensions, we can sol%e for the paddle wheel design relati%el&
easil&
with
the
restrictions gi%en. *rom here, we can use the %iscosit& of the water, the separate G %alues and the %olume of each compartment to sol%e for the power (in hp+ re!uired for each compartment. 1e then simpl& add those up to obtain total power. 2astl&, we can use relations of water speed %ersus rotational speed to obtain the rotations in rpms (the range+. Variables: 3 detention time
G %elocit& gradient
x width of basin
d depth of basin
SELY OKTAVIOLITA ASRI
KELAS A TEKNIK LINGKUNGAN
21080114120029
5 ow rate
8 9olume of basin
7 absolute %iscosit& of water
6 6ower imparted
dC diameter of paddle wheel
2 2ength of =asin
% blade %elocit& relati%e to
G G %alue
water
Solution: a+ o begin, we need to chec that the G %alue will fall in the appropriate range
($0,000-
"00,000+
¿=G x θ =
26.7 60 sec = 72.090 x 45 min x 1 min sec
50.000 72.090 100.000 ( o&
;ext, we compute the %olume of the entire basin ∀ =Qθ =
13.0 #GD 1 hour 1 ft =5.43 x 104 ft 3 x x 45 min x 24 60 min 7.48
1e alread& now the width of the basin, so we can di%ide that out to obtain
the
prole
area
of
the basin 4
∀
3
5.43 x 10 ft d x ) = = 90 ft x
=603.46 10 4 ft 3
o obtain that each of proles of the compartments is a s!uare, we assume
that
per
compartment,
length depth. herefore 4
3
d x 3 s =603.46 10 ft ( d =14 ft −2 ∈¿ )tot = 42 ft − 6 ∈¿
inal basin !imensions: }
∀ =54187.5 {ft} ^ {3} '
''
'
'
¿
)= 42 6 x =90 d =14 2
b+ Ff we assume a six-blade paddle wheel design and that both clearances from
the
top
and
bottom of " ft, we will tr& and get our D " to match accordingl&. D 1=d −2 ( 1 ft ) −2
(
)
1 ' ( d ) =14.17 ft −2 ( 1 ft ) −2 1 x 0.5 ft =11.67 ft ≈ 11.6 ft 2 2
1e also now that each of the other diameters is e!ual to twice the spading
between
blades
plus
the width of each blade multiplied b& two. D 2= D1− 2 ( 1 ft ) − 2 ( d ) =11.6 ft −2 ( 1 ft )−2 ( 0.5 ft ) =8.6 ft '
D 3= D2− 2 ( 1 ft ) −2 ( d ) =8.6 ft −2 ( 1 ft )−2 ( 0.5 ft ) =5.6 ft '
;ow that the wheel dimensions are nown, we need to calculate how man& paddle wheels per arm. o accomplish this, we tae the total
width of the basin and subtract out the minimum wheel spacing between wheels and the minimum spacing between the walls. ;ote the spacing between wheels will be a function of the number of wheels themsel%es, but the wall spacing will remain constant. )nce we ha%e this number, we will round down to the nearest whole integer. 1e cannot round up because this would mean we ha%e exceeded our basin width. x =90 ft =n ( 10 ft ) + 2 ( 1 ft ) +( n−1 )( 2.5 ft ) ∴n
=7 "heels
1e must now chec our answer to mae sure this number of wheels will wor. *irst, attempt to hold the wall clearances constant at "ft and increase wheel spacing, being sure not to cross the > in. clearance limit. x =90 ft =7 ( 10 ft ) + 2 ( 1 ft ) +(7 −1 )( y ) ∴ y
=3 ft
c+ *irst, we must chec that the total paddle blade areas are between "$ and
#0
of
the
total
cross-sectional area of each compartment. 1e start b& calculating out the total cross-sectional area of the blades in each compartment * = ) x d x 6 x 7 = ( 10 ft ) ( 0.5 ft ) ( 6 ) ( 7 )= 210 ft '
2
'
1e now compare this to the area of each compartment that the blades co%er (width times depth+ 2
* blades 210 ft occupied = x 100 = x 100 = 16.5 x x d 90 ft x 14.17 ft
his falls rml& between the bounds of "$-#0. herefore, the following e!uation can be used in sol%ing for the power imparted on each shaft P 1= μ x G
2 ∀
3
=
(
−5 lb
2.73 x 10
−s 2
ft
)( ) ( 50
sec
2
3
54187.5 ft 3
)
=1233
ft −lb s
1e simpl& repeat this step for compartments # and , changing the G %alues to match
P 2= μ x G
P 3= μ x G
2 ∀
3 2 ∀
3
(
−5 lb
= 2.73 x 10
(
)( ) ( − )( ) ( − s 20 2
sec
ft
−5 lb
= 2.73 x 10
2
s
2
ft
10 sec
2
3
54187.5 ft 3
3
54187.5 ft 3
)
− =197 ft lb
)=
s
49
ft −lb s
o obtain the total energ& re!uired, we simpl& sum these "otal po#er $ 1%&'
ft −lb s
Problem 8.% Solution: a+ he =asin Dimensions 3
m 19,200 d 1 min ! = x x 2700 s 1440 min 60 s ! =0.0133 x 0.0166 x 2.700= 5.985
he dimensions as gi%en b& " x " x ( 26.7 " )= 5.985 m 26.7 "
3
3
=5.985 m3
3
" =0,224 m " =0.473 m
;ow, sol%ing for length # =26.7 "=26.7 ( 0.473 )=12.6291 m
he basin dimensions " = 0.473 # =12.6291 m
b+ he paddle wheel design D 1=d −2 ( 1 ft ) −2
1 ' ( d ) =27.43 ft − 2 ( 1 ft ) −2 1 x 3.054 =32.38 ft 2 2
D 2= D1− 2 ( 1 ft ) − 2 ( d ) =32.38 ft − 2 ( 1 ft ) −2 ( 6 ft )=18.38 ft '
D 3= D2− 2 ( 1 ft ) −2 ( d ) =18.38 ft − 2 ( 1 ft ) −2 ( 3.05 ft )=10.28 ft '
x =27.43 ft =n ( 10 ft ) + 2 ( 1 ft )+( n −1 )( 3.05 ft ) n =6 "heels
c+ he power to be imparted to the water in each compartment * = ) x d x 6 x 7 = ( 10 ft ) ( 0.5 ft ) ( 6 ) ( 7 )= 210 ft '
'
2
2
210 ft * blades occupied = x 100 = x 100 =12.38 27.43 x 18.38 +xd
P 1= μ x G
P 2= μ x G
P 3= μ x G
2 ∀
3 2 ∀
3 2 ∀
3
( =( =(
−5 lb
= 3.05 x 10
2
3.05 x 10
s
20 sec
2
s
10 sec
2
2
ft
−5 lb
2
sec
ft
−5 lb
3.05 x 10
)( ) ( − )( ) ( − )( ) ( − s 50
2
ft
otal power "$4.''$
50000 3
)
50000 3
)
50000 3
)
=1270.832
= 203.33
=50.833
ft − lb s
ft −lb s ft −lb s
ft −lb s
Problem 8.( Problem A pneumatic occulation basin is to be designed for a tertiar& treatment plant ha%ing a ow of $.0 MGD. he plant is to emplo& high-p lime coagulation, and the pertinent data for the occulation basin are as follows detention time $min, G "$0s-" (at $0) *+, length # times width, depth 'ft-"0in, diHuser depth 'ft-0in, and air ow 4 cfm per diHuser. Determine a+ he basin dimensions if "in increments are used. b+ he total air ow in ftImin. c+ he number of diHusers.
Approach: *irst, weCll sol%e for the basin dimensions the same wa& as was sol%ed for in the pre%ious problem. hen, we can sol%e for the total air ow b& using a relation between power Solution: a+ *irst, we need to calculate the %olume of the basin. 1e accomplish this through the following 6
6
5 x 10 gals 1 day 1 hour 5 x 10 gals 3 x x x x 5 minutes= 2321 ft ∀ =Q x θ = 1 day 24 hours 60 min 7.48 gal
1e now tae this %olume and e!uate it to the following ∀=" x ) x d
"=
√
) =2 " d = 9 ft −10 ∈ ( 9.8333 ft ) ∴
∀
2 x 9.8333
=
√
∀
9.833 ft
=2 " 2
3
2321 ft =10.86 feet ≈ 10 ft −10 ∈¿ 2 x 9.8333
;ow, sol%ing for length )=2 " =2 x 10.86 feet = 21.72 feet ≈ 21 ft − 9 ∈¿
b+ o begin, weCll sol%e for the power −1
P= G + ∀= ( 150 s 2
¿ 1423,2
2
) x ( 2.73 x 10−5 ) x ( 10.8333 ft x 21.75 ft x 9.8333 ft )
ft −lb sec
;ow, we use this power in the following e!uation to sol%e for air ow P =, 1 Ga log
( ) h + c2 c2
1here h height of diHuser J" ?".$ (=ritish Enits+
c# 4 (=ritish Enits+ Kearranging and sol%ing 3
P = 1423.2 =171.2 ft Ga = min 9 ft + 34 h +c 2 81.5 x log , 1 log 34 c2
(
( )
)
c+ *or this, we simpl& di%ide the total air ow pro%ided b& the amount re!uired
per
diHuser
cfm+ 3
ft 171.2 min =42.8 diffusers ⇾ 43 diffusers Diffusers= 3 ft 4 per diffuser min
(4
Problem 8.& a+ =asin Dimension 3
875 m 1 min ! = x x 35 s 20 min 60 s 3
! = 43.75 x 0.0166 x 35 = 25.520 m
he dimensions as gi%en b& " x " x ( 1.25 " )=25.520 m 1.25 "
3
3
= 25.520 m3
" =2.247 m
;ow, sol%ing for length # =1.25 " =1.25 ( 2.247 )=2.809 m
he basin dimensions " =2.247 # =2.809 m
b+ Fmpeller Diameter c+ Fmpeller
P=W x ∀ =1.60475 x 25,520 = 40.95322 *or this, we simpl& need the following e!uation
(
n=
40.95322 1.65 x 0.3 x 94600
) ( 1/ 3
=
40.95322 46827
)
1/ 3
3
=√ 8.7456= 0,00295=1.74 rpm
Problem 8.8 a. =asin Dimension
! =
94600 1 min x x 35 s 20 min 60 s
! = 4730 x 0.0166 x 35 =2759.167 m
3
he dimensions as gi%en b& " x " x ( 1.25 " )= 2759.167 m 1.25 "
3
3
= 2759.167 m3
" =6.854 m
;ow, sol%ing for length # =1.25 " =1.25 ( 6.854 ) =8.567 m
he basin dimensions " =6.854 # =8.567 m
b. Fmpeller Diameter c. Fmpeller
P =W x ∀ =8.5675 x 2759.167 =19355.413 *or this, we simpl& need the following e!uation
(
n=
19355.413 1.65 x 0.3 x 94600
) =( 1/ 3
19355.413 46827
)
1 /3
= 0,64 rps=38.57 rpm
