21080114120029_sely Oktaviolita Asri_kelas A

7/23/2019 21080114120029_sely Oktaviolita Asri_kelas A

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SELY OKTAVIOLITA ASRI KELAS A TEKNIK LINGKUNGAN

21080114120029

Nama : Sely Oktaviolita Asri

Problem 8.1

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 dimensions if increments of " in. are used.b+ he input horsepower.c+ he impeller speed if a %ane-disc impeller with six at blades is emplo&ed and

the tan is

ba/ed. he impeller diameter is to be $0 of the basin width.

Approach *irst, we sol%e for the %olume of the tan b& using the ow rate and

the detention

time. 1e then use this %olume to compute basin 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

x length and width of basin

G %elocit& gradient

d depth of basin

5 ow rate6 6ower input

7 absolute %iscosit& of water

D diameter of impeller

n impeller %elocit&

8 9olume of basin

1 6ower imparted: water densit&

Solution:

a+ *irst, we need to calculate the %olume of the basin. 1e accomplish this

through the following

∀=Q x θ=4 x 10

6gals

1day x

1day

24 hours x 1 hour

60min x 1min

60sec x

1 ft 3

7.48 gal x 30 seconds=185.68 ft

3

1e now tae this %olume and e!uate it to the following

∀= x2 x d d=1.25 x∴∀=1.25 x3




21080114120029

x=3

√ ∀

1.25=

3

√185.68 ft

3

1.25 =5.3 feet ≈ 5 ft −4∈¿

;ow, sol%ing for depth

d=1.25 x=1.25 x5.3 feet =6.675 feet ≈ ft −8∈¿

<ummar&

=asin Dimensions $ft-4in x $ft-4in x >ft-?in

b+ *irst, we need to calculate the power imparted

2.73 x10−5=22.113

ft −lb

sec−ft 3

W =G2 x μ=(900s

−1)2 x ¿

;ow, we can sol%e for the total power

P=W x ∀=22.113 ft −lb

sec−ft 3 x (5.33' )3 x (6.67' )=4193

ft −lb

sec

2astl&, we con%ert to horsepower

P=419 as3 ft −lb

sec x

sec

550 ft −lb x hp=7.62 hp

c+ *or this, we simpl& need the following e!uation2.667 ft

¿

¿4193

ft −lb

sec

5.75 x (¿5 x 1.936¿¿)1 /3=1.41

rev

sec x

60sec

1min =84.5 rpm

¿

n=( P

K t D5 )

1/3

=¿

Problem 8.2

a+ he =asin Dimensions

! =15140m

3

1440min x

1min

60s x30 s

! =5.26m3

he dimensions as gi%en b&

" x " x (1.25" )=5.26m3




21080114120029

1.25"3=5.26m

3

"3=4.208m

"=1.61m

;ow, sol%ing for length

# =1.25"=1.25 (1.61 )=2.0125m

he basin dimensions

"=1.61m

# =2.0125 m

b+ *irst, we need to calculate the power imparted

W =G2 x μ=(900s

−1)2 x 0.00131=1061.1 $ −m

s−m3


P=W x ∀=

1061.1 $ −m

s−m3 x 5.26m

3

m3

=5581.38 $ −m

s =5581

%

s =5581 "

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

n=( P

K t D5 )

1/3

=( 5581"

5.75 x 0.806m x999.7 &g/m2 )

1/3

=1.20 rev

sec x 60 sec

1min =72rps

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, paddle-

wheel t&pe with a

mean %elocit& gradient of #>. s-" (at $00*+, a detention time of 4$ min,

and a G %alue from

$0,000 to "00,000. apered occulation is to be pro%ided, and three

compartments of e!ual




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 widthof '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 Dimensionsb+ he paddle-wheel designc+ 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 thebasin 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

x width of basin

G %elocit& gradient

d depth of basin




21080114120029

5 ow rate

7 absolute %iscosit& of water

dC diameter of paddle wheel

% blade %elocit& relati%e towater

8 9olume of basin

6 6ower imparted

2 2ength of =asin

G G %alue



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

sec x 45min x

60 sec

1min =72.090

50.000 72.090 100.000 ( o&

;ext, we compute the %olume of the entire basin

∀=Qθ=13.0 #GD

24 x

1hour

60 min x 45 min x

1 ft

7.48=5.43 x 10

4ft

3

1e alread& now the width of the basin, so we can di%ide that out to

obtain the prole area of

the basin

d x )=∀

x =

5.43 x 104

ft 3

90 ft =603.4610

4ft

3

o obtain that each of proles of the compartments is a s!uare, we

assume that per compartment,

length depth. herefore

d x 3 s=603.46104

ft 3

( 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&.

D1=d−2 (1 ft )−21

2( d

' )=14.17 ft −2 (1 ft )−2( 12 x0.5 ft )=11.67 ft ≈11.6 ft

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.

D2= D1−2 (1 ft )−2 ( d' )=11.6 ft −2 (1 ft )−2 (0.5 ft )=8.6 ft

D3= 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 howman& 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%ethis 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 x7=(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+

occupied=

* blades

x x d x 100=

210 ft 2

90 ft x 14.17 ft x 100=16.5

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

P1= μ x G2 ∀

3=(2.73 x10

−5 lb−s

ft 2 )( 50

sec )2

( 54187.5 ft 3

3 )=1233 ft −lb

s

1e simpl& repeat this step for compartments # and , changing the G

%alues to match



P2= μ x G2 ∀

3=(2.73 x10

−5 lb−s

ft 2 )( 20sec )

2

( 54187.5 ft 3

3 )=197 ft −lb

s

P3= μ x G2 ∀

3

=

(2.73 x10

−5 lb−s

ft 2

)( 10

sec

)

2

(54187.5 ft

3

3

)=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

! =

19,200m

3

d

1440min x 1min

60 s x2700 s

! =0.0133 x 0.0166 x 2.700=5.985


" x " x (26.7 " )=5.985m3

26.7"3=5.985m

3

"3=0,224m

"=0.473 m


# =26.7 "=26.7 (0.473 )=12.6291m

he basin dimensions

"=0.473

# =12.6291m

b+ he paddle wheel design

D1=d−2 (1 ft )−21

2

( d' )=27.43 ft −2 (1 ft )−2

1

2

x 3.054=32.38 ft



D2= D1−2 (1 ft )−2 ( d' )=32.38 ft −2 (1 ft )−2 (6 ft )=18.38 ft

D3= 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 x7=(10 ft ) (0.5 ft ) (6 ) (7 )=210 ft

2

occupied= * blades

+ x d x100=

210 ft 2

27.43 x18.38 x 100=12.38

P1= μ x G2 ∀3=(3.05 x10−5 lb−s

ft 2 )(

50sec )

2

(50000

3 )=1270.832 ft −lbs

P2= μ x G2 ∀

3=(3.05 x10

−5 lb−s

ft 2 )( 20sec )

2

( 500003 )=203.33ft −lb

s

P3= μ x G2 ∀

3=(3.05 x10

−5 lb−s

ft 2 )( 10sec )

2

( 500003 )=50.833ft −lb

s

otal power "$4.''$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 # timeswidth, 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

∀=Q x θ=5 x 10

6gals

1day x

1day

24 hours x 1hour

60min x 5 x10

6gals

7.48gal x5minutes=2321 ft

3

1e now tae this %olume and e!uate it to the following

∀=" x ) x d )=2" d=9 ft −10∈(9.8333 ft )∴ ∀

9.833 ft =2"

2

"=√ ∀

2 x 9.8333=√

2321 ft 3

2 x 9.8333=10.86 feet ≈10 ft −10∈¿


)=2"=2 x 10.86 feet =21.72 feet ≈21 ft −9∈¿

<ummar&

=asin Dimensions "0ft-"0in x #"ft-'in x 'ft-"0in

b+ o begin, weCll sol%e for the power

P=G2

+ ∀= (150 s−1 )2 x (2.73 x 10

−5 ) x (10.8333 ft x 21.75 ft x 9.8333 ft )

¿1423,2ft −lb

sec

;ow, we use this power in the following e!uation to sol%e for air ow

P=, 1Ga log ( h+c2

c2)

1here

h height of diHuser

J" ?".$ (=ritish Enits+



c# 4 (=ritish Enits+

Kearranging and sol%ing

Ga= P

, 1 log

(h+c2

c2 )=

1423.2

81.5 x log

(9 ft +34

34 )

=171.2 ft

3

min

c+ *or this, we simpl& di%ide the total air ow pro%ided b& the amount

re!uired per diHuser (4

cfm+

Diffusers=171.2

ft 3

min

4 ft

3

min per diffuser

=42.8diffusers⇾ 43diffusers



Problem 8.&

a+ =asin Dimension

! =875 m

3

20 min x

1min

60 s x 35 s

! =43.75 x 0.0166 x35=25.520m3


" x " x (1.25" )=25.520m3

1.25 "3=25.520 m

3

"=2.247 m


# =1.25"=1.25 (2.247 )=2.809m

he basin dimensions

"=2.247

# =2.809m

b+ Fmpeller Diameter

c+ Fmpeller <peed

Kr=1.54 K t =7.0 d=30

*irst, we need to calculate the power imparted

W =G2 x μ=(35)2 x(0.00131)=1.604

$ −m

s−m


P=W x ∀=1.60475 x 25,520=40.95322


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

20min x 1 min

60s x 35s

! =4730 x0.0166 x35=2759.167 m3


" x " x (1.25" )=2759.167m3

1.25"3=2759.167 m

3

"=6.854 m


# =1.25"=1.25 (6.854 )=8.567m

he basin dimensions

"=6.854

# =8.567m

b. Fmpeller Diameter

c. Fmpeller <peed

Kr=1.65 Kt =7.0 d=30

*irst, we need to calculate the power imparted

W =G2 x μ=6.854 x 1.25=8.5675 m


P=W x ∀=8.5675 x 2759.167=19355.413


n=( 19355.413

1.65 x 0.3 x94600 )1/3

=( 19355.41346827 )1 /3

=0,64 rps=38.57rpm

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21080114120029_sely Oktaviolita Asri_kelas A