Bangunan Pengambilan
KP-02 hal. 118
KP-02 hal. 123
Untuk aliran tidak tenggelam (a) Q= μ ∙ a ∙ B √ 2 ∙ g ∙ hi K
(KP-02 hal. 118)
........................ (1)
V = μ ∙ √ 2 ∙g ∙hi
........................(2)
Q= V ∙ a ∙ B ∙ K
........................(3)
Dimana: Q
= debit (m3/det)
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µ
= koefien debit
b
= leba! b"kaan (m)
a
= tin##i b"kaan (m)
#
= $e!%e$atan #!a&itai (= '8 m/det2)
hi
= ked kedal ala aman man ai! ai! di de$an e$an $int $int" " di di at ata amba amban# n#
= ke ke%e$atan al ali!an ma ma"k (m (m/det)
K
= *akto! ali!an ten##elam
KP-02 hal. 118
KP-02 hal. 108
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V = μ ∙ √ 2 ∙g ∙ z
........................(,)
Q= V ∙ a ∙ b
........................()
Dimana: Q
= debit (m3/det)
µ
= koe koef fie ien n deb debit it " "nt nt"k "k b"ka b"kaan an di baa baah h $e! $e!m" m"ka kaan an ai! ai! den den#a #an n kehilan#an tin##i ene!#i µ = 080
b
= leba! b"kaan (m)
a
= tin##i b"kaan (m)
#
= $e!%e$atan #!a&itai (= '8 m/det2)
= kehilan#an tin##i ene! ene!# #i $a $ada b"kaan (m (m)
= ke ke%e$atan al ali!an ma ma"k (m (m/det)
Debit Debit an# an# di#" di#"na naka kan n dala dalam m $e!en $e!en%a %ana naan an $int $int" " $en# $en#am ambi bila lan n (ka$ (ka$a ait ita a $en#ambilan $en#ambilan)) ek"!an# ek"!an#-k"!a -k"!an#n n#na a 120 da!i debit keb"t"ha keb"t"han. n. (KP-02 (KP-02 hal. 122) Q$en#ambilan = 12 Qkeb"t"han
........................()
Ke%e$atan ali!an ma"k di$e!ki!akan den#an
()
h V ≥ 32 d 2
1/ 3
∙d
(KP-02 hal. 122)
= ke%e$atan !ata-!ata (m/det)
h
= kedalaman ai! (m)
d
= diamete! b"ti! (m)
........................ (8)
dalam kondii biaa !"m" ini da$at diede!hanakan men4adi: 5 10 . d 0,
(KP-02 hal. 122)
........................ (')
6nt"k $e!en%anaan no!mal ke%e$atan ma"k be!kia! 10 7 20 m/det den#an ha!a$an b"ti!-b"ti! be!diamete! 001 7 00+ m da$at ma"k. (KP-02 hal.
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di!en%anakan diata daa! "n#ai den#an ketent"an eba#ai be!ik"t : (KP-02 hal. 12+) • • •
min 0,0 m 4ika "n#ai menan#k"t lana". min 100m bila "n#ai 4"#a menan#k"t $ai! dan ke!ikil. min 1,0m kala" "n#ai menan#k"t bat" 7 bat" bon#kah
;ika di!en%anakan $embila baah maka %!ite!ia ini be!#ant"n# $ada "k"!an al"!an $embila baah dalam hal ini "m"mna amban# $en#ambilan di!en%anakan 0 < P < 20 %m diata "4"n# $en"t"$ al"!an $embila baah.
ila $en#amb $en#ambilan ilan mem$"n mem$"nai ai b"kaan b"kaan lebih lebih da!i da!i at" at" maka maka $ila! $ila! ebaik ebaikna na dim"nd"!kan dim"nd"!kan "nt"k men%i$taka men%i$takan n kondii kondii ali!an ma"k an# lebih m"l" (#amba! ,.2 KP 7 02)
KP-02 hal. 12+ 9dala 9d alah h $entin $entin# # "nt"k "nt"k me!en% me!en%ana anakan kan dindin dindin# # aa$ aa$ dan dindin dindin# # $en#a! $en#a!ah ah edemik edemikian ian !"$a !"$a ehin# ehin##a #a ali!an ali!an men4ad men4adii m"l". m"l". Pada Pada "m"mn "m"mna a ini be!a!t be!a!tii baha baha len#k len#k"n# "n#-le -len#k n#k"n# "n# da$at da$at dite!a dite!a$ka $kan n den#an den#an 4a!i-4a 4a!i-4a!i !i minim"m > kali kedalaman ai!. ai!. (KP-02 hal. 121) le&ai le&ai ?9 $en#ambilan $en#ambilan be!ada 010 m di baah ele&ai ele&ai me!%" me!%" bend"n# bend"n# "nt"k men%e#ah kehilan#an ai! $ada bend"n# akibat #elomban#. (KP-02 hal. 123)
Contoh perencanaan: Diketahui: Di!en%anakan $en#ambilan den#an $embila te!b"ka le&ai me!%"
= @ 1+00
le&ai daa! "n#ai
= @ 1++0
le&ai dindin# bend"n#= @ 1,02,
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Keb"t"han $en#ambilan= '+, (12,/08) = 1,1++ m3/det
Penyelesaian: Ka$a a$ait ita a $en# $en#am ambi bila lan n
(Qd) (Qd)
=1 12 2 keb keb"t "t"h "han an $en# $en#am ambi bila lan n
= 12 1,1++ le&ai m"ka ai! h"l" intake
= 1813 m3/det
= ele&ai me!%" 7 010 m
= @1+00 - 010 = @1+'0 Diambil kehilan#an tin##i ene!#i $ada b"kaan () = 02, le&ai ?.9. hili! intake
= ele&. ?.9. h"l" 7 = @1+'0 7 02, = @1+,
Den#an mate!ial han"tan $ai! ke!ikil diambil tin##i amban# ($) = 10 m le&ai amban# intake
= ele&ai daa! "n#ai @ $ = @1+ 1++ +0 0 @ 10 10
= @1 @1+, +,0
Mengambil tipe pengambilan aliran tenggelam
Den#an koefien debit µ = 080 di$e!ki!aan ke%e$atan !ata-!ata ali!an: V = μ ∙ √ 2 ∙g ∙ z V = 0,80 ∙ √ 2 ∙ 9,8 ∙ 0,25
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= @1+, 7 1+,0 - 00, = 0'0 m Aeba! b"kaan $int" (b) Cin##i $int" (B) (B)
=
Qd 1,8173 = =1,1408 5 11, m V∙a 1,77 ∙ 0,90
= ele&ai ?.9. h"l" 7 ele&ai amban# intake = @1+'0 7 1+,0
B
= 1 20 m
= ele&ai dindin# bend"n# 7 ele&ai amban# intake = @1,02, 7 1+,0
= + ,, m
Mengambil tipe pengambilan aliran tak tenggelam
Kedalaman ai! h"l" diata amban# (h1) = le&ai ?.9. h"l" 7 ele&ai amban# intake = @1+'0 7 1+,0 = 120 m Kedalaman ai! hili! diata amban# (h2) = le&ai ?.9. hili! 7 ele&ai amban# intake = @1+, 7 1+,0 = 0', m Cin##i b"kaan $int" (a)
= ele&ai ?.9. hili! 7 ele&ai ele&ai amban# intake @ t = @1+, 7 1+,0 @ 010 = 10, m
Pe!bandin#an h1/a
= 120/10, = 11+
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Den#an h1/a = 11+ dan h2/a = 0'0 da!i #!afk di$e!oleh koefien debit (K) 5 0'
V = μ ∙ √ 2 ∙g ∙ hi V = 0,80 ∙ √ 2 ∙ 9,8 ∙ 1,2 V = 3,88
m3/det
Aeba! b"kaan $int" (b)
=
Qd 1,8173 = = 0,5782 5 00 V ∙a ∙ K 3,88 3,88 ∙ 0,90 0,90 ∙ 0,9
m Cin##i $int" (B) (B)
= ele&ai ?.9. h"l" 7 ele&ai amban# intake = @1+'0 7 1+,0
B
= 1 20 m
= ele&ai dindin# bend"n# 7 ele&ai amban# intake