indrianti susanto_21080114120038_a (responsi 1 dan 2)
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7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
RESPONSI 1 MK SATUAN OPERASI
1. Apabila diketahui: Vs =1
18. g (Ss−1 ) . dp2
υ (huku St!ke"# aka $aba%ka&
ku%'a hubu&$a& a&ta%a keepata& e&$e&dap pa%tikel disk%it te%hadap tepe%atu%
ai% pada ali%a& lai&e% N%e ) 1. *p = +#, # Ss = ,#-.
/atata&: buatlah t%ial de&$a& %a&$e suhu a&ta%a 0 !/ 23 !/.
Pe&4elesaia& :
Untu" a#iran #amin$r Nr$ % 1& ma"a "o$'isi$n dra )* + 24,Nr$- s$.ina:/aunan $r.itunan aa raitasi d$nan aa dra 'ri"si untu" arti"$# s.$ria#&
m$n.asi#"an 'ormu#a s: + ((4,3)*!5((6 7 6!,6!5d!1,2 atau + ((4,3)*!5(s 91!5d!1,2
s +1
18. g (Ss−1 ) . dp2
υ
s +;4.
3
g (Ss−1 )dpCd <
s2 +4
3
g ( Ss−1 ) dp
24 Nre
s2 +1
18
g (Ss−1 )dpv Vs.dp
s +1
18
g (Ss−1 )v
dp2
5555555 (=u"um Sto"$!
>i"a t$m$ratur nai"& ma"a is"ositas "in$matis a"an turun& s$.ina daat diuat "ura
antara t$m$ratur dan "$$atan m$n$nda arti"$# d$nan $rsamaan :
s +1
18
9,81 (2,65−1 )v
(0,0002m)2
s + 2,18 x10
−9
(2,65−1)
1
v
T$m$ratur $
is"ositas*$nsita
ss (m,s!
4 1&?@@ 1 2&2B4?BE90B@ 1&42@ 0&BBBB 2&442@2E90B
10 1&3101 0&BBB3 2&4??BE90B14 1&1?@ 0&BBB2 3&0?B1E90B18 1&0@18 0&BB8@2 3&38@4E90B
22 0&B@2B 0&BB8 3&3??BE90B2@ 0&84 0&BB@81 4&0BB@1E90B30 0&803B 0&BB?@8 4&4444E90B
7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
0 5 10 15 20 25 30 35
0
0
0
0
0
0
Hubungan Temperature dan Vs
Vs
Temperature
Semakin tinggi temperature, maka semakin cepat pengendapan partikel diskritnya5
,. Apabila diketahui: Vs =1
18. g (Ss−1 ) . dp2
υ (huku St!ke"# aka $aba%ka&
ku%'a hubu&$a& a&ta%a keepata& e&$e&dap pa%tikel disk%it te%hadap diaete%
pa%tikel pada ali%a& lai&e% N%e ) 1# pada tepe%atu% ,5 !/# Ss = ,#-.
Pe&4elesaia&:
Untu" a#iran #amin$r Nr$ % 1& ma"a "o$'isi$n dra )* + 24,Nr$- s$.ina:/aunan $r.itunan aa raitasi d$nan aa dra 'ri"si untu" arti"$# s.$ria#&
m$n.asi#"an 'ormu#a s: + ((4,3)*!5((6 7 6!,6!5d!1,2 atau + ((4,3)*!5(s 91!5d!1,2
s +1
18. g (Ss−1 ) . dp2
υ
s +;
4.
3
g ( Ss−1 )dp
Cd <
s2 +4
3
g ( Ss−1 ) dp
24 Nre
s2 +1
18
g (Ss−1 )dpv
Vs.dp
s +1
18
g (Ss−1 )v
dp2
5555555 =u"um Sto"$
>i"a t$m$ratur nai"& ma"a is"ositas "in$matis a"an turun& s$.ina daat diuat "ura
antara t$m$ratur dan "$$atan m$n$nda arti"$# d$nan $rsamaan :
s +1
18
9,81 (2,65−1 )0,00000082619
(dp)2
7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
*p (" Vs (6s"
050001 05010B050002 05043?050003 050B80050004 0514105000? 0522105000@ 053B1805000 05?333050008 05@B@@
1E-40.0000
0.2000
0.4000
0.6000
0.8000
Grafk Hubungan Vs dan Dp
Vs (m/s)
Diameter Partikel Dp (m)
Kecepatan pengendapan, Vs (m/s)
Semakin besar diameter partikel, maka semakin cepat pengendapan partikel diskrit
2. 7itu&$lah e8isie&si pe&$e&dapa& pa%tikel disk%it pada bak pe&$e&dap 4a&$
epu&4ai dp = +#, # 9s = ,#- pada k!&disi tepe%atu% ,- !/ da& N%e =
1.++. *ebit bak pe&$e&dap = ,++ ;6detik da& luas pe%ukaa& bak As = 2.+++,.
Pe&4elesaia& :
*i"$ta.ui :
Nr$ + 15?00 à )* + 0&4
T + 2@ o) à 6 + 0&BB@81 ram,m3
2 +(4,3)*!5(s 91!5d (&e<t!&"
+ 0&011 m,d$ti"
C + 200 ,d$ti" + 0&2 m3,d$ti"
A + 3000 m2
7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
M$n.itun o
o + C,A + 0&2 m3,d$ti" , 3000 m2 + @&@5109? m,d$ti" + 0&00@ m,d$ti"
Ma"a s D o- ma"a arti"$# m$n$nda 100 untu" u"uran d + 0&2 mm& s$.ina
$'isi$nsi $n$ndaan arti"$# dis"rit ada#a. 100
0. Suatu bak pe&$e&dap h!%i!&tal de&$a& uku%a& pa&>a&$ ; = -+ # leba% ? = 1+ #
kedalaa& 7 = , # e&$!lah ai% pada debit = 1++ ;6detik# T = ,- !/. 7itu&$lah:
a. ?eba& pe%ukaa&6su%8ae l!adi&$ %ate6OVR@
b. ?ila&$a& Re4&!ld da& bila&$a& %!ude@
. ?a$aia&a uku%a& 4a&$ pali&$ baik u&tuk e&$hasilka& N%e da& N8% 4a&$ lebih
baik@
Pe&4elesaia& :
a! r + ?0 & di$ro#$. s + So + 1 mm,d$ti"
So + =,To + =5C, + =5C,=5A + C,ASo + C,A
1 mm,d$ti" + ?0051093 , A
A + ?00 m2
R (! So (mm , d$ti"! A (m2!?0 1 10@0 0&8 80 0&@ @80 0&48 4&8
B0 0&3? 3&?100 0&2 2
7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
1 2 3 4 5 6 7 8 9 10 110
50
100
150
Hubungan Efsiensi Rem!al dg "uas #ak
$ (m%)
r (&)
OR + C , A
OR + 0&1 m3,d$ti" : 10 m2 + 0&01 m,d$ti" + 5-0 6>a
! K$$atan =oriFonta# Parti"$# +Q
L xH =
0,1m3/ s(10 x2 )m2
=5 x10−3m /s
R + L x H
L+2 H =
10 x 2
10+(2 x2)=1,43m
Nr$ +VHx R
υ +0,005 x 1,43
0,893 x 10−6=8006,72 (tida" m$m$nu.i!
NGr +VH
2
g x R=
0,0052
9,81 x1,43=1,782 x10
−6
(tida" m$m$nu.i!
! P$r#u $nama.an #at$ s$tt#$r ada a" s$dim$ntasi :
K$$atan a#iran masu" #at$
o +Q / Asinα =
0,01m /detik
sin 60 =0,012m /detik
*im$nsi P#at$
# +h
sinα +2
sin60=2,31m
. + 2 m
+ 0&0? m
t + 0&00? m
>um#a. P#at$
7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
>ara" =oriFonta# antar#at$& H +w
sin60=0,05m
sin60=0,06m
>um#a. P#at$& n + P
x +60
0,06=1000buah
>ari 7 >ari .idro#is & R +w
2=
0,05m
2=0,025m
)$" i#anan R$no#ds& & NR$ + Vo x R
υ =
0,012m
s x0,025m
0,893 x 10−6
=335(! )
)$" i#anan Groud& NGr + Vo2
g x R=
0,0122
9,81 x 0,025 + 5,872 x 10−4
(OK!
. Suatu bak p%asedie&tasi be%uku%a& ; ? 7 = 5+ ,+ 0 . Apabila aka&
di$u&aka& u&tuk e&$!lah ai% de&$a& beba& pe%ukaa& +#, ;6detik.,. B =
+#++5330 ,6detik.
a. 7itu&$lah be%apa debit 4a&$ dapat di!lah@
b. 7itu&$lah N%e da& N8%@
. *e&$a& 6As 4a&$ saa# ubahlah !&e i&let da& !utlet bak pe&$e&dap te%sebut
u&tuk e&dapatka& N%e da& N8% 4a&$ lebih baik@
d. ?uatlah !di8ikasi bak u&tuk e&dapatka& N%e ) ,.+++.
e. *ebit
Pe&4elesaia& :
Asumsi Td + 2 Jam + 200 d$ti"
C + o#um$ , Td + ( H # ! , Td + ( 80 m H 20 m H 4 m ! , ( 200 d$ti" ! + 0&88 m 3,s$"on
'5 NRE dan NGR
K$$atan .oriFonta#
. + C : ( H = ! + ( 0&88 m3,d$ti" ! : ( 20 m H 4 m ! + 0&011 m,s
>ari9Jari .idro#is
R + ( H = ! : ( 2 = ! + ( 20 m H 4 m ! : ( 20 m 2 5 4 m ! + 2&8? m
i#anan R$no#d NRE + ( . 5 R ! :
NRE + ( 0&011 m,s H 2&8? m ! : ( 0&8?8@1 H 109@!
7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
NRE + 3@5@02&18 ( 1039104 s$.ina tida" m$m$nu.i !
i#anan Groud
NGR + ( .2! : ( H R ! + (0&011 m,s !2 : ( B&81 m,s2 H 2&8? m !
NGR + 4&32 H 109@ % 109? ( tida" m$m$nu.i !
5 Lona In#$t
C!&a i&let
In#$t diuat dari u"aan dindin aian aa.5
r$nana 3 m,s ( *armas$tiaan& 2001 !
*iami# i + 051 m,s
uan in'#u$n $r$ntu" $rs$i anJan d$nan Jum#a. 4 ua.5PanJan sa#uran + #$ar a" + 20 m
C$r#uan + C , 4 + 0&88 , 4 + 0&22 m3,dt
uas #uan ( Ai ! + 0522 , 051 + 2&2 m2
Tini #uan + a"ar 2&2 + 1&48 m
$ar sa#uran + tini sa#uran + 05? m
>ara" antar #uan& F
20 + (10 H 05? ! (101! FF + 1&4 m
K$$atan .oriFonta#
. + C , ( H = ! + 0&22 , ( 20 H 0&? ! + 0&022 m,s
>ari9Jari .idro#is
R + m
NRE + ( . H R ! , + (0&022 H 0&1! , ( 0&8?81 H 109@! NRE + 43?8&4@ (1039104 s$.ina m$m$nu.i !
.5 Modi'i"asi a" d$nan Lona In#$t
C!&a i&let
In#$t diuat dari u"aan dindin aian aa.5
r$nana 3 m,s ( *armas$tiaan& 2001 !
*iami# i + 051 m,suan in'#u$n $r$ntu" $rs$i anJan d$nan Jum#a. 40 ua.5
PanJan sa#uran + #$ar a" + 20 m
H L
LxH
2+1A&0
?&02?&0
?&0?&0=
+ x
x
7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
C$r#uan + C , 40 + 0&88 , 40 + 0&022 m3,dt
uas #uan ( Ai ! + 05022 , 051 + 0&22 m2
Tini #uan + a"ar 0&22 + 0&4 m
$ar sa#uran + tini sa#uran + 05? m
>ara" antar #uan& F
20 + (10 H 05? ! (101! F
F + 1&4 m
K$$atan .oriFonta#
. + C , ( H = ! + 0&022 , ( 20 H 0&? ! + 0&0022 m,s
>ari9Jari .idro#isR + m
NRE + ( . H R ! , + (0&0022 H 0&1! , ( 0&8?81 H 109@!
NRE + 43?&84@ ( % 2000 s$.ina m$m$nu.i !
-. Re&a&aka& die&si bak p%asedie&tasi de&$a& kapasitas = ;6detik = +#+
26detik. Kete&tua& lai& adalah sbb:
?e&tuk rectangle de&$a& ;:? = 0:1
Kedalaa& bak = 0
Td = 1D2 >a
Vs = +#1 6detik
V!:6A = ,#1 (& = 162 da& 6! = 5+F"
N%e pa%tikel ) 1
N%e ali%a& ) ,.+++
Pe&4elesaia& :
M$n$ntu"an *im$nsi a"
Asumsi& ?0 + 3@ m,.ari + 0&1@ H 1094m,d$ti"
uas P$rmu"aan (As! + C , OR + 0&0?? m3,d$ti" , 0&1@ H 1094m,d$ti" + 32B3&41 m2
P$randinan PanJan dan $ar a" 4 : 1
P H + 32B3&41 m2
4 H + 32B3&41 m2
+ 2B m
H L
LxH
2+1A&0
?&02?&0
?&0?&0=
+ x
x
7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
P + 2B H 4 + 11@ m
)$" OR + C , ( H P! + 0&0?? m3,d$ti" , (2B H 11@! + 1&@3 H 109? m ,d$ti" + 1&41
)$" a"tu d$t$nsi
o#um$ a" + 11@ H 2B H 4 + 1354?@ m 3
Td + , C + 1354?@ m3 , 0&0?? m3,d$ti" + 2445@?4 d$ti" + @8 Jam
RESPONSI SE*IMENTASI
/ase 1
a! *$t$ntion $riod: 'or #ain s$dim$ntation: 3 to 4 .& and 'or oau#at$d s$dim$ntation:2 to 25? .5
! $#oit o' '#o: Not r$at$r t.an 30 m,min (.oriFonta# '#o!5
! Tan" dim$nsions: : + 3 to ?:15 /$n$ra## + 30 m (ommon! maHimum 100 m5r$adt.+ @ m to 10 m5 )iru#ar: *iam$t$r not r$at$r t.an @0 m5 $n$ra## 20 to 40m5
d! *$t. 25? to ?50 m (3 m!5
$! Sur'a$ O$r'#o Rat$: Gor #ain s$dim$ntation 12000 to 18000 ,d,m2 tan" ar$a- 'or t.orou.# '#ou#at$d at$r 24000 to 30000 ,d,m2 tan" ar$a5
'! S#o$s: R$tanu#ar 1 toards in#$t and iru#ar 85
P!pulati!& !%east b4 *i88e%e&t Meth!ds
P%!ble: Pr$dit t.$ ou#ation 'or t.$ $ars 1B81& 1BB1& 1BB4& and 2001 'rom t.$ 'o##oin$nsus 'iur$s o' a ton di''$r$nt m$t.ods5
$ar 1B01 1B11 1B21 1B31 1B41 1B?1 1B@1 1B1
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
Pou#ation:(t.ousands
!
@0 @? @3 2 B 8B B 120
S!luti!&:
$ar Pou#ation:(t.ousands!
Inr$m$nt $r *$ad$
Inr$m$nta#Inr$as$
P$r$nta$ Inr$m$nt $r *$ad$
1B01 @0 9 9 9
1B11 @? ? 9 (?@0! H100+8533
1B21 @3 92 93 (2@?! H100+9350
1B31 2 B (B@3! H100+14528
1B41 B 92 (2! H100+B52
1B?1 8B 10 3 (10B! H100+125@@
1B@1 B 8 92 (88B! H100+85B8
1B1 120 23 1? (23B! H100+2351
N$t a#u$s1
@0 18 45@1
A$ra$s9
85? 350 105@@
+inr$as$- 9 + d$r$as$
Arithmetical Progression Method:
7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
Pn + P ni
A$ra$ inr$as$s $r d$ad$ + i + 85?
Pou#ation 'or t.$ $ars&
1B81+ ou#ation 1B1 ni& .$r$ n+1 d$ad$
+ 120 85? + 1285?
1BB1+ ou#ation 1B1 ni& .$r$ n+2 d$ad$
+ 120 2 H 85? + 13514
2001+ ou#ation 1B1 ni& .$r$ n+3 d$ad$
+ 120 3 H 85? + 14?51
1BB4+ ou#ation 1BB1 (ou#ation 2001 9 1BB1! H 3,10
+ 13514 (85?! H 3,10 + 13B51
Incremental Increase Method:
Pou#ation 'or t.$ $ars&
1B81+ ou#ation 1B1 a$ra$ inr$as$ $r d$ad$ a$ra$ inr$m$nta# inr$as$
+ 120 85? 350 + 1315?
1BB1+ ou#ation 1B81 115?
+ 1315? 115? + 143514
2001+ ou#ation 1BB1 115?
+ 143514 115? + 1?451
1BB4+ ou#ation 1BB1 115? H 3,10
+ 143514 354 + 14@5@1
Geometric Progression Method:
A$ra$ $r$nta$ inr$as$ $r d$ad$ + 105@@
P n + P (1i,100! n
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
Pou#ation 'or 1B81 + Pou#ation 1B1 H (1i,100! n
+ 120 H (1105@@,100!& i = 10.66, n = 1
+ 120 H 1105@@,100 = 13258
Pou#ation 'or 1BB1 + Pou#ation 1B1 H (1i,100! n
+ 120 H (1105@@,100! 2 & i = 10.66, n = 2
+ 120 H 15224? = 14@5B?
Pou#ation 'or 2001 + Pou#ation 1B1 H (1i,100! n
+ 120 H (1105@@,100! 3 & i = 10.66, n = 3
+ 120 H 153?? = 1@25@0
Pou#ation 'or 1BB4 + 14@5B? (1?584 H 3,10! + 1?150
Sedimentation Tank Design
P%!ble: *$sin a r$tanu#ar s$dim$ntation tan" to tr$at 254 mi##ion #itr$s o' ra at$r $rda5 T.$ d$t$ntion $riod ma $ assum$d to $ 3 .ours5
S!luti!&: Ra at$r '#o $r da is 254 H 10@ #5 *$t$ntion $riod is 3.5
o#um$ o' tan" + G#o H *$t$ntion $riod + 254 H 103 H 3,24 + 300 m3
Assum$ d$t. o' tan" + 350 m5
Sur'a$ ar$a + 300,3 + 100 m2
, + 3 (assum$d!5 + 35
32 + 100 m2 i5$5 + ?58 m
+ 3 + ?58 3 + 154 m
=$n$ sur'a$ #oadin (O$r'#o rat$! + 254 H 10@ + 24&000 #,d,m2 % 40&000 #,d,m2 (OK!100
Rapid Sand Filter Design
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
P%!ble: *$sin a raid sand 'i#t$r to tr$at 10 mi##ion #itr$s o' ra at$r $r da a##oin05? o' 'i#t$r$d at$r 'or a"as.in5 =a#' .our $r da is us$d 'or a"as.in5 Assum$n$$ssar data5
S!luti!&: Tota# 'i#t$r$d at$r + 1050? H 24 H 10@ + 0542@@ M# , .24 H 235?
$t t.$ rat$ o' 'i#tration $ ?000 # , . , m 2 o' $d5
Ar$a o' 'i#t$r + 1050? H 10@ H 1 + 8?5? m2 235? ?000
Proid$ to units5 Ea. $d ar$a 8?5?,2 + 4255 , + 153- 1532 + 425
+ ?5? m - + ?5? H 153 + 5? m
Assum$ d$t. o' sand + ?0 to ? m5
Und$rdraina$ sst$m:
Tota# ar$a o' .o#$s + 052 to 05? o' $d ar$a5
Assum$ 052 o' $d ar$a + 052 H 425 + 0508@ m2
100
Ar$a o' #at$ra# + 2 (Ar$a o' .o#$s o' #at$ra#!
Ar$a o' mani'o#d + 2 (Ar$a o' #at$ra#s!
So& ar$a o' mani'o#d + 4 H ar$a o' .o#$s + 4 H 0508@ + 05344 + 053? m2 5
*iam$t$r o' mani'o#d + H 053? Q 1,2 + @@ m
Assum$ , o' #at$ra# + 30 m5 Tota# num$rs + 5?, 053 + 2? on $it.$r sid$5
$nt. o' #at$ra# + ?5?,2 9 05@@,2 + 25?4? m5
)5S5 ar$a o' #at$ra# + 2 H ar$a o' $r'orations $r #at$ra#5 Ta"$ dia o' .o#$s + 13 mm
Num$r o' .o#$s: n (153!2 + 0508@ H 104 + 8@0 m2
4
n + 4 H 8@0 + @48& sa @?0 (153!2
Num$r o' .o#$s $r #at$ra# + @?0,?0 + 13
Ar$a o' $r'orations $r #at$ra# + 13 H (153!2 ,4 + 1524 m2
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
Sain o' .o#$s + 25?4?,13 + 1B5? m5
)5S5 ar$a o' #at$ra# + 2 H ar$a o' $r'orations $r #at$ra# + 2 H 1524 + 345? m 25
*iam$t$r o' #at$ra# + (4 H 345?Q 1,2 + @5@3 m
).$": $nt. o' #at$ra# % @0 d + @0 H @5@3 + 35B8 m5 # + 25?4? m (=$n$ a$ta#$!5
Risin as.at$r $#oit in $d + ?0 m,min5
as.at$r dis.ar$ $r $d + (05?,@0! H ?5? H 5? + 053@ m3,s5
$#oit o' '#o t.rou. #at$ra# + 053@ + 053@ H 10 4 + 2508 m,s (o"! Tota# #at$ra# ar$a ?0 H 345?
Mani'o#d $#oit + 053@ + 1504 m,s % 252? m,s (o"!0534?
as.at$r utt$r
*is.ar$ o' as.at$r $r $d + 053@ m3,s5 SiF$ o' $d + 5? H ?5? m5
Assum$ 3 trou.s runnin #$nt.is$ at ?5?,3 + 15B m ,5
*is.ar$ o' $a. trou. + C,3 + 053@,3 + 0512 m3,s5
C +151 H H .3,2
Assum$ +053 m
.3,2 + 0512 + 05234151 H 053
. + 0538 m + 358 m + 40 m
+ 40 ('r$$ oard! ? m + 4? m- s#o$ 1 in 40
)#$ar at$r r$s$roir 'or a"as.in
Gor 4 . 'i#t$r aait& )aait o' tan" + 4 H ?000 H 5? H ?5? H 2 + 12? m3
1000
Assum$ d$t. d + ? m5 Sur'a$ ar$a + 12?,? + 34? m2
, + 2- 22 + 34?- + 13 m + 2@ m5
*ia o' in#$t i$ omin 'rom to 'i#t$r + ?0 m5
7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
$#oit %05@ m,s5 *iam$t$r o' as.at$r i$ to o$r.$ad tan" + @5? m5
Air omr$ssor unit + 1000 # o' air, min, m2 $d ar$a5
Gor ? min& air r$uir$d + 1000 H ? H 5? H ?5 H 2 + 4532 m3 o' air5
Flow in Pipes of a Distribtion !etwork b" #ard" $ross Method
P%!ble: )a#u#at$ t.$ .$ad #oss$s and t.$ orr$t$d '#os in t.$ arious i$s o' a
distriution n$tor" as s.on in 'iur$5 T.$ diam$t$rs and t.$ #$nt.s o' t.$ i$s us$d ar$i$n aainst $a. i$5 )omut$ orr$t$d '#os a't$r on$ orr$tions5
S!luti!&: Girst o' a##& t.$ manitud$s as $## as t.$ dir$tions o' t.$ ossi#$ '#os in $a. i$ ar$ assum$d "$$in in onsid$ration t.$ #a o' ontinuit at $a. Juntion5 T.$ to#os$d #oos& A)* and )*EG ar$ t.$n ana#F$d =ard )ross m$t.od as $r ta#$s 1 2r$s$ti$#& and t.$ orr$t$d '#os ar$ omut$d5
7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
Table 1
/!&side% l!!p A?/*
Pi$ Assum$d '#o *ia o' i$ $nt.o' i$
(m!
K + 40 d458
Ca158? =+
K5Ca158?
#=,Ca#
in
#,s$in
um$sd inm
d458
(1! (2! (3! (4! (?! (@! (! (8! (B! (10!
A
)
)*
*A
(!43
(!23
(9! 20
(9! 3?
05043
05023
905020
90503?
0530
0520
0520
0520
258?1093
35B?1094
35B?1094
35B?
1094
?00
300
?00
300
33
1@1?
2@B0
1@1?
3 1093
B541094
521094
2 1093
1512
15?2
915B4
93523
2@
@@
B
B2
925?3 281
V =+ (Ca158?!,(050B4 H 100 158? d458!
or K5Ca158?+ (Ca
158?!,(40 d458!or K +(!,(40 d458!
Gor #oo A)*& $ .a$ +9 = , H5 #= ,Ca#
+(9! 925?3,(158? 281! um$s
+(9! (925?3 1000!,(158? 281! #,s
+458@ #,s +? #,s (sa!
=$n$& orr$t$d '#os a't$r 'irst orr$tion ar$:
Pi$ A ) )* *A
)orr$t$d '#os a't$r 'irst
orr$tion in #,s 48 28 9 1? 9 30
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
Table ,
/!&side% l!!p */E
Pi$ Assum$d '#o *ia o' i$ $nt.o' i$
(m!
K + 40 d458
Ca158? =+
K5Ca158?
#=,Ca#
in
#,s$in
um$sd inm
d458
(1! (2! (3! (4! (?! (@! (! (8! (B! (10!
*)
)G
GE
E*
(! 20
(! 28
(9! 8
(9! ?
05020
05028
905008
90500?
0520
051?
051?
051?
35B? 109
4
B5 109?
B5 109?
B5 109?
?00
300
?00
300
2@B0
@?80
10B40
@?80
52 1094
1534 109
3
1534 109
4
?5@ 109?
15B4
8580
9154
9053
B
314
184
4
85B @@B
Gor #oo A)*& $ .a$ +9 = , H5 #= ,Ca#
+(9! 85B,(158? @@B! um$s
+(9! (85B 1000!,(158? @@B!! #,s
+ 952 #,s
=$n$& orr$t$d '#os a't$r 'irst orr$tion ar$:
Pi$ *) )G GE E*
)orr$t$d '#os a't$r 'irstorr$tion in #,s
1258
2058
91?52
91252
T%ikli&$ ilte% *esi$&
7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
Pro#$m: *$sin a #o rat$ 'i#t$r to tr$at @50 M#d o' s$a$ o' O* o' 210 m,#5 T.$ 'ina#$''#u$nt s.ou#d $ 30 m,# and orani #oadin rat$ is 320 ,m3,d5
So#ution: Assum$ 30 o' O* #oad r$mo$d in rimar s$dim$ntation i5$5& + 210 H 0530 +@3 m,#5 R$mainin O* + 210 9 @3 + 14 m,#5P$r$nt o' O* r$moa# r$uir$d + (14930! H 100,14 + 80
O* #oad a#i$d to t.$ 'i#t$r + '#o H on5 o' s$a$ (",d! + @ H 10 @ H 14,10@ + 882 ",d
To 'ind out 'i#t$r o#um$& usin NR) $uationE2+ 100
10544( F 1.!",15R' 1!1,2
80 + 100 R' 1+ 1& $aus$ no iru#ation510544(##2,1!1,2
1+ 204 m3
*$t. o' 'i#t$r + 15? m& Git$r ar$a + 204,15? + 18025@@ m2& and *iam$t$r + 48 m % @0 m
=drau#i #oadin rat$ + @ H 10@,103 H 1,18025@@ + 3533m3,d,m2 % 4 .$n$ o5"5
Orani #oadin rat$ + 882 H 1000 , 204 + 32@518 ,d,m3 .i. is aroH5 $ua# to 3205
/ase ,
Tim$& min 0 @0 80 100 130 200 240 420
)on5& m, 300 18B 180 1@8 1?@ 111 8 2
.at is t.$ r$moa# $''ii$n in a s$tt#in asin it. a #oadin rat$ o' 2? m3 ,m2 Vd (m,d!W15 )a#u#at$ mass 'ration r$mainin and orr$sondin s$tt#in rat$s5
Tim$(min!
@0 80 100 130 200 240 420
MGr$mainin
05@3 05@0 05?@ 05?2 053 052@ 050B
t H 1092(m,min!
353 25? 250 15?? 150 0583 054
.$r$& mass 'ration (MG! r$mainin + )i ,)o and t + Lo ,t
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
25 P#ot mass 'ration r$mainin s s$tt#in $#oit as s.on in Giur$ 4
35 *$t$rmin$ $#oit (o !& .i. $ua#s sur'a$ #oadin rat$ + 2? m3 ,m2 9d (15 H 1092
m,min!
45 *$t$rmin$ 'rom ra. ro + ?4 5
?5 Int$rat$ ur$
@5 R$moa# $''ii$n (R! + 1 9 ro XInt$rat$d Ar$aY
E#$m$nt Zr t H 1092 Zr[t H 10921 0502 15@ 05032 051? 152? 051B3 0511 05B1 05104 051 05@@ 0511? 050B 0524 0502
Tota# + 054?
R$moa# + (1 9 05?4! 054?,15
+ 054@ 052@
+ 052 or 2
7/23/2019 Indrianti Susanto_21080114120038_a (Responsi 1 Dan 2)
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/ase 2
A sin#$ r$tanu#ar s$dim$ntation asin is to $ d$sin$d to tr$at a at$r '#o o' 15? M/*at t.$ d$sin o$r'#o rat$ o' B00 d,'t2 5
15 *$t$rmin$ t.$ asin dim$nsions (idt.& #$nt.& d$t.! 'or a d$t$ntion tim$ o' 3 .ours anda #$nt. to idt. ratio o' 4 to 15
T.$ d$t. is d$t$rmin$d t.$ ar$a Just a#u#at$d and t.$ o#um$& .i. $ an $t 'romt.$ '#o and r$t$ntion tim$5
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Nama : Indrianti Susanto NIM : 21080114120038SATUAN OPERASI (KEAS A!
25 I' t.$ s$dim$ntation tan" $.a$s #i"$ an id$a# #u '#o r$ator& i## a## arti#$s it. a@? miron (109@ m! diam$t$r (or r$at$r! and a d$nsit o' 1200 ",m3 $ om#$t$#r$mo$d t.$ id$a# s$dim$ntation tan"W (at$r t$m$ratur$ is 10 \)! Pro$ our ans$ruantitati$#5
35 I' t.$ '#o rat$ to t.$ s$dim$ntation tan" inr$as$s& r$su#tin in a .i.$r o$r'#o rat$&ou#d t.$ arti#$ r$moa# $''ii$n d$r$as$& sta t.$ sam$& or inr$as$W ri$'# $H#ain5
Parti#$ r$moa# $''ii$n ou#d d$r$as$& $aus$ t.$ o$r'#o rat$ ou#d inr$as$&narroin t.$ ran$ o' arti#$ s$tt#in $#oiti$s t.at ou#d $ 100 r$tain$d& andd$r$asin t.$ r$t$ntion on t.os$ arti#$s it. s$tt#in $#oiti$s ao$ t.$ o$r'#o$#oit5
45 Suos$ t.$ d$t. o' t.$ s$dim$ntation tan" as r$du$d ?0& and ou ou#d assum$disr$t$ arti#$ s$tt#in5 .at is t.$ $''$t on arti#$ r$moa# $''ii$n i' t.$ '#o rat$is un.an$d ('rom t.$ oriina# as$!W EH#ain our ans$r5
No di''$r$n$& as #on as t.$ tan" as $.ain in an id$a# 'as.ion5 T.is is $aus$o$r'#o rat$ ditat$s s$tt#in $r'orman$& and it is on# a 'untion o' ar$a and '#o&
ut not d$t.5 So ot. o$r'#o rat$ and t.$ sto"$s s$tt#in $#oiti$s i## r$mainun.an$d5
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