Perancangan Pipa Ekonomis Pompa Sentrifugal

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LE-1 1. Pompa Air Sungai (P-201)

Fungsi : Memompa air dari sungai ke bak penampungan air

Jenis : Pompa sentrifugal

Jumlah : 2 unit (1 standby) Bahan Konstruksi : Commercial Steel Data :

Laju alir massa (F) = 742.478,9914 kg/jam = 454,6906 lbm/s Densitas air (ρ) = 995,72 kg/jam = 62,16 lbm/ft³ Viscositas (μ) = 0,8007 cP = 0,0005 lbm/ft³.s Perhitungan :

Laju alir volume (Q) = ^454,6906

lbms 62,16 ^lbm

ft3

= 7,3148 ^ft

3 s

Diameter pipa ekonomis untuk aliran turbulen :

Dopt = 3,9 (Q)^0,45(ρ)^0,13 (Timmerhaus, 2004)

= 3,9 (7,3148)^0,45(62,16)^0,13

= 16,3348 in

Dari Appendix E.19 Mc Graww Hill, dipilih pipa commercial steel:

Ukuran nominal = 6 in Schedule number = 80

Diameter dalam (ID) = 5,826 in = 0,3188 ft Diameter luar (OD) = 6 in

Inside sectional area = 0,0798 ft²

Kecepatan rata-rata fluida dalam pipa :

V = ^Q

A

= ^7,3142

0,0798

= 91,6647 ft/s

(2)

Sehingga :

NRe = ^ρVD

μ

= ^62,16

lbm

ft3 × 91,6647 ^ft

s × 0,3188 ft 0,0005 ^lbm

ft3s

= 3.632.966,4982

Untuk pipa commercial steel diperoleh harga ε = 0,00015 dan ε/D = 0,0004705 Maka harga f = 0,0045 (Gambar 2.10-3, Geankoplis, 1997)

Kehilangan karena gesekan (friction loss):

1. 1 sharp edge enterance (hc) = 0,55 (1-^A²

A₁)( ^V

2 2a.gc)

= 0,55 (1-0)( ^91,6647

2 2(1)(32,714))

= 70,6323 ft.lbf/lbm

2. 2 elbow 90^o (hf) = n.Kf.(^V

2 2.g_c)

= 2(0,75)( ^91,6647

2 (2)(32,714))

= 192,6335 ft.lbf/lbm

3. 1 check valve (hf) = n.Kf.(^V

2 2.g_c)

= 1(0,75)( ^91,6647

2 (2)(32,714))

4. Pipa lurus 50 ft (Ff) = 4f.(^∆L.V

D.2.g_c)

= 4(0,0045)( 50(91,6647) (0,3188)(2)(32,714))

= 3,9551 ft.lbf/lbm

5. 1 sharp edge exit (hex) = (1-^A²

A₁)( ^V

2 2a.gc)

= (1-0)( ^91,6647

2 2(1)(32,714))

= 128,4223 ft.lbf/lbm

Total friction loss ∑F = 491,9600 ft.lbf/lbm

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Dari persamaan Bernoulli

1

2α ((v2)² – (v1)²) + g(z2 – z1) + ^P

2- P¹

P + ∑F + Ws = 0 Dimana : v1 = v2

P1 = 101,325 kPa = 2116,22807 lbf/ft² P2 = 101,325 kPa = 2116,22807 lbf/ft²

∆P = 0 lbf/ft²

∆z = 50 ft 0 + ^32,714

ft s2 32,714 ^ft s2

x (50 ft) + 0 + 491,9600 ft.lbf/lbm + Ws = 0 Ws = -541,9600 ft.lbf/lbm

Efisiensi pompa (𝜂) = 80%

Ws =- 𝜂 x Wp

Wp = -0,8 x -541,9600 ft.lbf/lbm

Wp = 433,5680 ft.lbf/lbm

Daya pompa : P = m x Wp

= 454,6906 lbm/s x 433,5680 ft.lbf/lbm x ^{1 hp}

550 ft.lb_f/s

= 358,4349 hp

Maka dipilih pompa dengan daya motor = 358,4349 hp

2. Bak Penampungan Air (B-201)

Fungsi : Menampung air sungai untuk mengendapkan partikel- partikel yang besar tanpa bantuan bahan kimia

Jenis : Bak dengan permukaan bentuk persegi Junlah : 2 Unit (1 standby)

Bahan konstruksi : Beton Data :

Laju alir massa (F) = 742.478,9914 kg/jam = 454,6909 lbm/s

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Densitas air (ρ) = 995,72 kg/jam = 62,16 lbm/ft³ Laju alir volumetrik (Q) = 7,3148 ft³/s = 745,6798 m³/jam Bak pengendap dirancang untuk penampung air selama 1 hari

Volume air buangan= (745,6798 m³/jam x 1 hari x 24 jam) = 17.896,3144 m³ Bak terisi 90 % maka volume bak = 17.896,3144 m³/0,9 = 19.884,7938 m³ Direncanakan ukuran bak sebagai berikut :

 Panjang bak = 2 x lebar bak (l)

 Tinggi bak = lebar bak (l) Maka, volume bak = p x l x t

19.884,7938 m³ = 2l x l x l, maka l = 21,5028 m sehingga,

Panjang bak = 43,0057 m

Lebar bak = 21,5028 m

Tinggi bak = 21,5028 m

3. Pompa Bak Pengendapan (P-202)

Fungsi : Memompa air dari bak penampungan air ke clarifier

lbm s 62,16 ^lbm

ft3

= 7,3148 ^ft

3 s

= 3,9 (7,3148)^0,45(62,16)^0,13

= 16,3348 in

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V = ^Q

A

= ^7,3142

0,0798

= 91,6647 ft/s Sehingga :

NRe =^ρVD

μ

=

62,16 ^lbm

ft3 × 91,6647 ^ft_s × 0,3188 ft 0,0005 ^lbm

ft3s

= 3.632.966,4982

A₁)( ^V

2 2a.gc)

= 0,55 (1-0)( ^91,6647

2 2(1)(32,714))

2. 2 elbow 90^o (hf) = n.Kf.(^V

2 2.g_c)

= 2(0,75)( ^91,6647

2 (2)(32,714))

= 192,6335 ft.lbf/lbm

2 2.g_c)

= 1(0,75)( ^91,6647

2 (2)(32,714))

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D.2.g_c)

= 4(0,0045)( 50(91,6647) (0,3188)(2)(32,714))

= 3,9551 ft.lbf/lbm

A₁)( ^V

2 2a.gc)

= (1-0)( ^91,6647

2 2(1)(32,714))

= 128,4223 ft.lbf/lbm

1

2α ((v2)² – (v1)²) + g(z2 – z1) + ^P

2- P¹

P + ∑F + Ws = 0 Dimana : v1 = v2

∆P = 0 lbf/ft²

∆z = 50 ft 0 + ^32,714

ft s2 32,714 ^ft s2

Ws =- 𝜂 x Wp

Wp = -0,8 x -541,9600 ft.lbf/lbm

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550 ft.lb_f/s

= 358,4349 hp

4. Tangki Pelarutan Alum [Al2(SO4)3] (T-201)

Fungsi : Membuat larutan alum [Al2(SO4)3]

Bentuk : Silinder tegak dengan alas dan tutup datar Bahan Konstruksi : Carbon steel SA-283 Grade C

Jumlah : 1 Unit

Volume tangki :

Al2(SO4)3 yang digunakan = 50 ppm

Al2(SO4)3yang digunakan berupa larutan 30% (% berat) Laju massa Al2(SO4)3 = 37,123 kg/jam

Densitas Al2(SO4)3 30% = 1.363 kg/m³ = 85,093 lbm/ft³ (Perry, 1997) Kebutuhan perancangan = 30 hari

Volume larutan (Vl) = ^37,123

kg jam ×24 jam

hari x 30 hari 0,5 × 1.363 ^kg

m3

= 39,2211 m³ Faktor kelonggaran = 20 %, Maka:

Volume tangki = 1,2 x 39,2211 m³ = 47,0654 m³ Diameter dan tebal tangki :

Vs = ^{π . Di}

2 . Hs

4 (Brownells & Young, 1959)

Dimana:

Vs = Volume silinder (ft³)

Di = Diameter dalam silinder (ft) Hs = Tinggi tangka silinder (ft)

Ditetapkan: Perbandingan tinggi tangka dengan diameter tangka Hs : Di = 2 : 1

Maka:

Vs = ^{π . Di}

2 . (3/2 Di)

4

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47,0654 m³ = ^{π . Di}

2 . (3 Di) 8

Di = 3,1065 m = 10,1918 ft Hs = 6,2129 m = 20,3837 ft Tinggi cairan dalam tangki:

Tinggi cairan dalam tangki = volume cairan x tnggi silinder volume silinder

= 39,2211 x 10,1918 47,0656

= 5,1774 m Phidrostatis = ρ x g x h

= 1.363 kg/m³ x 9,8 m/s² x 5,1774 m

= 69,1573 kPa

Faktor keamanan untuk tekanan = 5%

Pdesain = (100% + 5%) x (69,1573 + 101,325)

Pdesain = 179,0064 kPa

Joint efficiency = 0,8 (Brownells & Young, 1959) Allowable stress = 12.650 psia = 87.218,714 kPa

Tebal shell tangki:

t = ^{P x D}

(2 x S x E) - (1,2 x P)

= (179,0064) x (10,1918)

(2 x 87.218,714 x 0,8) - (1,2 x 179,0064) = 0,0039 in Faktor korosi = 1/8 in/tahun

Maka tebal shell yang dibutuhkan = 0,0039 in + 1/8 in = 0,1289 in

Tebal tangki standart yang digunakan 7/16 in (Brownells & Young, 1959) Daya Pengaduk

Tipe pengaduk : Flat six turbin imp Jumlah baffle : 4 buah

Da/Dt = 1/3 ; Da = (1/3 x 3,1065) m = 1,0354 m E/Da = 1 ; E = (1 x 1,0354) m = 1,0354 m L/Da = 1/4 ; L = (1/4 x 1,0354) m = 0,2588 m

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W/Da = 1/5 ; W = (1/5 x 1,0354) m = 0,2070 m J/Dt = 1/12 ; J = (1/12 x 3,1065) m = 0,0862 m Dimana:

Dt = diameter tangki Da = diameter impeller

E = tinggi turbin dari dasar tangki L = panjang blade pada turbin W = lebar blade pada turbin J = lebar blade

Kecepatan pengaduk N = 2 rps

Viskositas Al2(SO4)370% = 0,000672 lbm/ft.s (Othmer, 1967) Bilangan Reynold:

NRe = ^{ρ x N x Da}

2 µ

=(85,093) x (2) x (1,0354)²

0,000672 = 271.548,235

NRe> 10.000 maka perhitungan dengan pengadukan menggunakan rumus:

P = ^K^r^{x n}³^{x Da}

5 x ρ

gc (McCabe, 1999)

Kr = 6.3 (McCabe, 1999)

=(6,3) x (2)³ x (1,0354)⁵ x (85,093) x (1)

(32,714) x (550) = 0,2837 hp Efisiensi motor penggerak = 80%

Daya motor penggerak =^0,2837

0,8 = 0,3457 hp

5. Pompa Alum [Al2(SO4)3] (P-203)

Fungsi : Memompa alum dari tangki pelarutan alum ke clarifier

Jenis : Pompa injeksi

Jumlah : 2 unit (1 standby) Bahan Konstruksi : Commercial Steel

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Data :

Laju alir massa (F) = 37,1239 kg/jam = 0,0227 lbm/s Densitas Al2(SO4)3 30% (ρ) = 1.363 kg/m³ = 85,093 lbm/ft³ Viskositas Al2(SO4)3 30% (μ) = 0,000672 lbm/ft³.s

Perhitungan :

lbm s 85,093 ^lbm

ft3

= 0,00026 ^ft

3 s

= 3,9 (0,00026)^0,45(85,093)^0,13

= 0,1713 in

Dari Appendix A.5-1 Geankoplis, dipilih pipa commercial steel:

Ukuran nominal = 1/8 in Schedule number = 40

Diameter dalam (ID) = 0,269 in = 0,0224 ft Diameter luar (OD) = 0,405 in = 0,0337 ft Inside sectional area = 0,0004 ft²

V = ^Q

A

= ^0,00026

0,0004

NRe = ^ρVD

μ

=

85,093 ^lbm

ft3 × 0,6679 ^ft

s × 0,0224 ft 0,000672 ^lbm

ft3s

= 1.895,9530

Untuk pipa commercial steel dengan Nre 1.895,9530 diperoleh harga f= 0,16 (Gambar 2.10-3, Geankoplis, 1997)

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A₁)( ^V

2 2a.gc)

= 0,55 (1-0)( ^0,6679

2 2(1)(32,714))

= 0,0037 ft.lbf/lbm

2. 2 elbow 90^o (hf) = n.Kf.(^V

2 2.g_c)

= 2(0,75)( ^0,6679

2 (2)(32,714))

= 0,0102 ft.lbf/lbm

2 2.g_c)

= 1(0,75)( ^0,6679

2 (2)(32,714))

= 0,0051 ft.lbf/lbm

D.2.g_c)

= 4(0,16)( ^20(0,6679)

(0,0224)(2)(32,714))

= 5,8291 ft.lbf/lbm

A₁)( ^V

2 2a.gc)

= (1-0)( ^0,6679

2 2(1)(32,714))

= 0,0068 ft.lbf/lbm

1

2α ((v2)² – (v1)²) + g(z2 – z1) + ^P

2- P¹

P + ∑F + Ws = 0 Dimana : v1 = v2

∆P = 0 lbf/ft²

∆z = 20 ft

(12)

0 + ^32,714

ft s2 32,714 ^ft.lbm_lb

f.s2

Ws = - 𝜂 x Wp

Wp =-0,8 x -25,8550 ft.lbf/lbm

550 ft.lb_f/s

= 0,00085 hp

6. Tangki Pelarutan Soda Abu [Na2CO3] (T-202)

Fungsi : Membuat larutan soda abu (Na2CO3) Bentuk : Silinder tegak dengan alas dan tutup datar Bahan Konstruksi : Carbon steel SA-283 Grade C

Jumlah : 1 Unit

Volume tangki :

Na2CO3yang digunakan = 27 ppm

Na2CO3yang digunakan berupa larutan 30% (% berat) Laju massa Na2CO3 = 20,0469 kg/jam

Densitas Na2CO330% = 1.327 kg/m³ = 82,8423 lbm/ft³ (Perry, 1997) Kebutuhan perancangan = 30 hari

Volume larutan (Vl) = ^20,0469

kg jam ×24 jam

hari x 30 hari 0,3 × 1.327 ^kg

m3

= 36,2566 m³ Faktor kelonggaran = 20 %, Maka:

Volume tangki = 1,2 x 36,2566 m³ = 43,5080 m³

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Diameter dan tebal tangki : Vs = ^{π . Di}

2 . Hs

Dimana:

Ditetapkan: Perbandingan tinggi tangki dengan diameter tangki Hs : Di = 2 : 1

Maka:

Vs = ^{π . Di}

2 . (2/1 Di)

4

43,5080 m³ = ^{π . Di}

2 . (2 Di) 4

Di = 3,0261 m = 9,9283 ft Hs = 6,0522 m = 19,8566 ft Tinggi cairan dalam tangki:

Tinggi cairan dalam tangki = volume cairan x tnggi silinder volume silinder

= 36,2566 x 6,0522 43,5080

= 5,0435 m Phidrostatis = ρ x g x h

= 1.327 kg/m³ x 9,8 m/s² x 5,0435 m

= 65,5897 kPa

Faktor keamanan untuk tekanan = 5%

Pdesain = (100% + 5%) x (65,5897 kPa + 101,325 kPa)

Pdesain = 175,2604 kPa

Joint efficiency = 0,8 (Brownells & Young, 1959) Allowable stress = 12.650 psia = 87.218,714 kPa

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t = ^{P x D}

(2 x S x E) - (1,2 x P)

= (175,2604) x (3,0261 )

(2 x 87.218,714 x 0,8) - (1,2 x 175,2604 ) = 0,0038 in Faktor korosi = 1/8 in/tahun

Da/Dt = 1/3 ; Da = (1/3 x 3,0261) m = 1,0087 m E/Da = 1 ; E = (1 x 1,0087) m = 1,0087 m L/Da = 1/4 ; L = (1/4 x 1,0087) m = 0,2521 m W/Da = 1/5 ; W = (1/5 x 1,0087) m = 0,2017 m J/Da = 1/12 ; J = (1/12 x 1,0087) m = 0,0840 m Dimana:

Viskositas Al2(SO4)370% = 0,000369 lbm/ft.s (Othmer, 1967) Bilangan Reynold:

2 µ

= (82,8423 ) x (2) x (1,0087 )²

0,000369 = 456.870,5515

P = ^K^r^{x n}³^{x Da}

5 x ρ

gc (McCabe, 1999)

Kr = 6.3 (McCabe, 1999)

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P = (6,3) x (2)³ x (1,0087 )⁵ x (82,8423) x (1)

(32,714) x (550) = 0,2423 hp Efisiensi motor penggerak = 80%

Daya motor penggerak = ^{0,2423 hp}

0,8 = 0,3029 hp 7. Pompa Soda Abu [Na2CO3] (P-204)

Fungsi : Memompa soda abu dari tangki pelarutan soda abu ke clarifier

Jenis : Pompa injeksi

Laju alir massa (F) = 20,0469 kg/jam = 0,0122 lbm/s Densitas Na2CO3 30% (ρ) = 1.327 kg/m³ = 82,8423 lbm/ft³ Viskositas Na2CO3 30% (μ) = 0,000369 lbm/ft³.s

Perhitungan :

lbm s 82,8423 ^lbm

ft3

= 0,00014 ^ft

3 s

= 3,9 (0,00014)^0,45(82,8423)^0,13

= 0,1310 in

Ukuran nominal = 1/8 in Schedule number = 40

V = ^Q

A

(16)

= ^0,00014

0,0004

NRe = ^ρVD

μ

=

82,8423 ^lbm

ft3 × 0,3704 ^ft_s × 0,0224 ft 0,000369 ^lbm

ft3s

= .864,5080

Untuk pipa commercial steel dengan nilai Nre 62,2798 diperoleh harga f = 0,16 (Gambar 2.10-3, Geankoplis, 1997)

A₁)( ^V

2 2a.gc)

= 0,55 (1-0)( ^0,3704

2 2(1)(32,714))

= 0,0011 ft.lbf/lbm

2. 2 elbow 90^o (hf) = n.Kf.(^V

2 2.g_c)

= 2(0,75)( ^0,3704

2 (2)(32,714))

= 0,0031 ft.lbf/lbm

2 2.g_c)

= 1(0,75)( ^0,3704

2 (2)(32,714))

= 0,0015 ft.lbf/lbm

D.2.g_c)

= 4(0,16)( ^20(0,3704)

(0,0224)(2)(32,714))

= 3,2332 ft.lbf/lbm

A₁)( ^V

2 2a.gc)

= (1-0)( ^0,3704

2 2(1)(32,714))

= 0,0020 ft.lbf/lbm

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1

2α ((v2)² – (v1)²) + g(z2 – z1) + ^P

2- P¹

P + ∑F + Ws = 0 Dimana : v1 = v2

∆P = 0 lbf/ft²

∆z = 20 ft 0 + ^32,714

ft s2 32,714 ^ft.lbm_lb

f.s2

Ws = - 𝜂 x Wp

Wp = -0,8 x -23,2412 ft.lbf/lbm

Wp = 18,5929ft.lbf/lbm

550 ft.lb_f/s

= 0,00041 hp

8. Clarifier (CL-201)

Fungsi : Memisahkan endapan (flok-flok) yang terbentuk karena penambahan alum dan soda abu

Bahan kontruksi : Carbon stell SA-283 grade C

Tipe : Ekternal solid recirculation clarifier

Jumlah : 1 unit

Data:

Laju massa air (F) : 742.478,914 kg/jam

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Laju massa Al2(SO4)3 : 37,1239 kg/jam Laju massa Na2CO3 : 20,0469 kg/jam Laju massa total : 742.536,1623 kg/jam

Densitas Al2(SO4)3 : 2.710 kg/m³ (Perry, 1997) Densitas Na2CO3 : 2.533 kg/m³ (Perry, 1997) Densitas air : 995,72 kg/m³

Reaksi koagulasi:

Al2(SO4)3 + 3Na2CO3 + 3H2O → 2Al(OH)3 + 3Na2SO4 + 3CO2

Perhitungan:

Dari Metcalf & Eddy, 1984 diperoleh Untuk clarifier tipe upflow (radial):

Kedalaman = 5 – 10 m Settling time = 1 – 3 jam

Dipilih kedalaman air (H) = 8 m dan settling time = 2 jam Diameter dan tinggi clarifier:

Densitas larutan (ρ) = 742.536,1623

742.478,9914

995,72 + ^37,1239_2.710 + ^20,046_2.533

= 995,7478 kg/m³ Volume cairan v =

742.536,1623 ^kg jam 995,7478^kg

m3

x 2 jam = 1.491,2992 m³ D = (^{4 x v}

π . H )^1/2 =(4 ×1.491,2992

3,14 × 8 )^1/2 = 15,4099 m

Tinggi clarifier = 1,5 x 15,4099 m = 23,1149 m

Phidrostatis = ρ x g x h

= 995,7478 kg/m³ x 9,8 m/s² x 8 m

= 78,0666 kPa Tekanan operasi = 1 atm = 101,325 kPa

P = 78,0666 kPa + 101,325 kPa = 179,3916 kPa Faktor keamanan untuk tekanan = 5%

Pdesain = 1,05 x 179,3916 kPa = 188,3612 kPa Joint efficiency = 0,8

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Allowable stress = 12.650 psia = 87.218,714 (Brownells & Young, 1959) Tebal shell tangki:

t = ^{P x D}

(2 x S x E) - (1,2 x P)

= (188,3612 ) x (15,4099)

(2 x 87.218,714 x 0,8) - (1,2 x 188,3612 ) = 0,0208 m = 0,8202 in Faktor korosi = 1/8 in/tahun

Tebal shell standart yang digunakan = ¼ in (Brownells & Young, 1959) 9. Pompa Clarifier (P-205)

Fungsi : Memompa air dari clarifier air ke sand filter

lbms 62,16 ^lbm

ft3

= 7,3148 ^ft

3 s

= 3,9 (7,3148)^0,45(62,16)^0,13

= 16,3348 in

(20)

V = ^Q

A

= ^7,3142

0,0798

NRe =^ρVD

μ

=

62,16 ^lbm

ft3 × 91,6647 ^ft_s × 0,3188 ft 0,0005 ^lbm

ft3s

= 3.632.966,4982

A₁)( ^V

2 2a.gc)

= 0,55 (1-0)( ^91,6647

2 2(1)(32,714))

2. 2 elbow 90^o (hf) = n.Kf.(^V

2 2.g_c)

= 2(0,75)( ^91,6647

2 (2)(32,714))

= 192,6335 ft.lbf/lbm

2 2.g_c)

= 1(0,75)( ^91,6647

2 (2)(32,714))

D.2.g_c)

= 4(0,0045)( 50(91,6647) (0,3188)(2)(32,714))

= 3,9551 ft.lbf/lbm

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A₁)( ^V

2 2a.gc)

= (1-0)( ^91,6647

2 2(1)(32,714))

= 128,4223 ft.lbf/lbm

1

2α ((v2)² – (v1)²) + g(z2 – z1) + ^P

2- P¹

P + ∑F + Ws = 0 Dimana : v1 = v2

∆P = 0 lbf/ft²

∆z = 50 ft 0 + ^32,714

ft s2 32,714 ^ft s2

Ws =- 𝜂 x Wp

Wp = -0,8 x -541,9600 ft.lbf/lbm

550 ft.lb_f/s

= 358,4349 hp

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10. Sand Filter (V-201)

Fungsi : Tempat menyaring partikel-partikel yang masih terbawa dalam air yang keluar dari clarifier

Bentuk : Silinder tegak dengan alas dan tutup datar Bahan kontruksi : Carbon steel SA-283 grade C

Jumlah : 1 unit

Volume tangki

Laju alir massa (F) = 742.478,9914 kg/jam = 454,6904 lbm/s Densitas air (ρ) = 995,72 kg/m³ = 62,16 lbm/ft³ Viskositas air (µ) = 0,8007 cP = 0,0005 16 lbm/ft³.s Faktor kelonggaran = 20%

Sand filter dirancang untuk menampung ¼ jam operasi Direncanakan volume bahan penyaring 1/3 volume tangki Volume air (Va) = 7442.478,9914 _jam^kg x 0,25 jam

995,72 ^kg m3

= 186,4176 m³

Volume tangki (Vt) = (1,2 x 186,4176) m³ = 233,7011 m³ Volume total = (1 + 0,3) x 233,7011 m³ = 290,8114 m³ Diamter dan tebal tangki

Vs = ^{π . Di}

2 . Hs

Dimana:

Ditetapkan: Perbandingan tinggi tangki dengan diameter tangki Hs : Di = 4 : 3

Maka:

Vs = ^{π . Di}

2 . (4/3 Di) 4

290,8114 = ^{π . Di}

2 . (4 Di) 12

(23)

Di = 6,5253 m = 21,4084 ft Hs = 8,7004 m = 28,5446 ft Diameter dan tinggi tutup

Diameter tinggi tutup = diameter tangki = 6,5253 m

Direncanakan perbandingan diameter dengan tinggi tutup D : H = 4 : 1 Tinggi tutup = ¼ x (6,5253 m) = 1,6313 m

Tinggi tangki total = 6,5253 m + (2 x 1,6313 m) = 11,9630 m Tebal shell dan dan tutup tangki

Tinggi penyaring = ¼ x 11,9630 m = 2,9907 m Tinggi cairan dalam tangki = (186,4176 m³

223,7011 m³) x 11,9630 m = 9,9692 m

= 995,72 kg/m³ x 9,8 m/s² x 11,9630 m

Pdesain = (1,05 x 198,6052 kPa) = 208,5354 kPa

Joint efficiency = 0,8 (Brownells & Young, 1959) Allowable stress = 12.650 psia = 87.218,714 kPa (Brownells & Young, 1959) Tebal shell tangki:

t = ^{P x D}

(2 x S x E) - (1,2 x P)

= (208,5354 ) x (6,5253 )

Tebal shell standart yang digunakan = 5/8 in (Brownells & Young, 1959)

11. Pompa Sand Filter (P-206)

Fungsi : Memompa air dari sand filter ke tangki air

(24)

lbm s 62,16 ^lbm

ft3

= 7,3148 ^ft

3 s

= 3,9 (7,3148)^0,45(62,16)^0,13

= 16,3348 in

V = ^Q

A

= ^7,3142

0,0798

NRe =^ρVD

μ

=

62,16 ^lbm

ft3 × 91,6647 ^ft

s × 0,3188 ft 0,0005 ^lbm

ft3s

= 3.632.966,4982

(25)

A₁)( ^V

2 2a.gc)

= 0,55 (1-0)( ^91,6647

2 2(1)(32,714))

2. 2 elbow 90^o (hf) = n.Kf.(^V

2 2.g_c)

= 2(0,75)( ^91,6647

2 (2)(32,714))

= 192,6335 ft.lbf/lbm

2 2.g_c)

= 1(0,75)( ^91,6647

2 (2)(32,714))

D.2.g_c)

= 4(0,0045)( 50(91,6647) (0,3188)(2)(32,714))

= 3,9551 ft.lbf/lbm

A₁)( ^V

2 2a.gc)

= (1-0)( ^91,6647

2 2(1)(32,714))

= 128,4223 ft.lbf/lbm

1

2α ((v2)² – (v1)²) + g(z2 – z1) + ^P

2- P¹

P + ∑F + Ws = 0 Dimana : v1 = v2

∆P = 0 lbf/ft²

(26)

∆z = 50 ft 0 + ^32,714

ft s2 32,714 ^ft s2

Ws =- 𝜂 x Wp

Wp = -0,8 x -541,9600 ft.lbf/lbm

550 ft.lb_f/s

= 358,4349 hp

12. Tangki Air (T-203)

Fungsi : Tempat menampung air sementara untuk didistribusikan ke unit lain

Jenis : Silinder tegak dengan alas dan tutup datar Bahan kontruksi : Carbon steel SA-283 Grade C

Jumlah : 1 unit Data:

Laju alir massa (F) = 742.478,9914 kg/jam = 454,6904 lbm/s Densitas air (ρ) = 995,72 kg/m³ = 62,16 lbm/ft³ Viskositas (µ) = 0,8007 cP = 0,0005 lbm/ft³.s Tangki dirancang untuk kebutuhan selama 1 jam operasi

Volume air (Va) = 742.478,9914 kg/jam x 1 jam

995,72 kg/m³ = 745,6704 m³

Faktor kelonggaran = 20% maka:

Volume tangki = (1,2 x 745,6704 m³) = 849,8045 m³

(27)

Diamter dan tebal tangki Vs = ^{π . Di}

2 . Hs

Dimana:

Ditetapkan: Perbandingan tinggi tangka dengan diameter tangki Hs : Di = 3 : 4

Maka:

Vs = ^{π . Di}

2 . (3/4 Di) 4

894,8045 = ^{π . Di}

2 . (3 Di) 16

Di = 11,4973 m = 37,7210 ft Hs = 8,6230 m = 28,2908 ft

Tinggi cairan dalam tangki = volume cairan x tinggi silinder volume silender

= (745,6704 m³× 8,6230 m

894,8045 m³ ) = 7,1858 m

= 995,72 kg/m³ x 9,8 m/s² x 7,1858 m

Pdesain = (1,05 x 171,4450 kPa) = 180,0173 kPa

t = ^{P x D}

(2 x S x E) - (1,2 x P)

(28)

= (180,0173 ) x (11,4973)

Maka tebal shell yang dibutuhkan = 0, 5848 in + 1/8 in = 0,7098 in

Tebal shell standart yang digunakan = 1 1/8 in (Brownells & Young, 1959)

13. Pompa Tangki Air (P-207)

Fungsi : Memompa air dari tangki air ke kation exchanger

lbm s 62,16 ^lbm

ft3

= 7,3148 ^ft

3 s

= 3,9 (7,3148)^0,45(62,16)^0,13

= 16,3348 in

V = ^Q

A

(29)

= ^7,3142

0,0798

NRe =^ρVD

μ

=

62,16 ^lbm

ft3 × 91,6647 ^ft

s × 0,3188 ft 0,0005 ^lbm

ft3s

= 3.632.966,4982

A₁)( ^V

2 2a.gc)

= 0,55 (1-0)( ^91,6647

2 2(1)(32,714))

2. 2 elbow 90^o (hf) = n.Kf.(^V

2 2.g_c)

= 2(0,75)( ^91,6647

2 (2)(32,714))

= 192,6335 ft.lbf/lbm

2 2.g_c)

= 1(0,75)( ^91,6647

2 (2)(32,714))

D.2.g_c)

= 4(0,0045)( 50(91,6647) (0,3188)(2)(32,714))

= 3,9551 ft.lbf/lbm

A₁)( ^V

2 2a.gc)

= (1-0)( ^91,6647

2 2(1)(32,714))

= 128,4223 ft.lbf/lbm

(30)

1

2α ((v2)² – (v1)²) + g(z2 – z1) + ^P

2- P¹

P + ∑F + Ws = 0 Dimana : v1 = v2

∆P = 0 lbf/ft²

∆z = 50 ft 0 + ^32,714

ft s2 32,714 ^ft s2

Ws =- 𝜂 x Wp

Wp = -0,8 x -541,9600 ft.lbf/lbm

550 ft.lb_f/s

= 358,4349 hp

14. Penukar Kation (Cation Exchanger) (T-204)

Fungsi : Mengikat senyawa logam yang terdapat dalam air umpan ketel (umpan boiler)

Bentuk : Silinder tegak dengan atas dan bawah elipsoidal Bahan kontruksi : Carbon steel SA-283 grade C

Jumlah : 1 unit

(31)

Data:

Laju alir massa air = 19.097,155 kg/jam = 11,6950 lbm/s

Densitas air = 995,72 kg/m³ = 62,16 lbm/ft³ (Perry, 1997) Direncanakan ¼ volume tangki berisi resin

Volume larutan (V1) = 19.097,155 kg/jam x 5/4

995,72 kg/m³ = 23,9740 m³ Faktor kelonggaran = 20%, maka:

Volume tangki = 1,2 x 23,9740 m³ = 28,7688 m³ Diameter dan tebal tangki

Volume silinder tangki

V = ¼ x π x Dt2 x Hs (Hs : Dt = 3 : 2) Vs = ¼ x π x Dt3 = 1,1775 Dt3

Volume tutup tangka ellipsoidal (Vh)

Vh = 1/24 x π x Dt3 = 0,1308 Dt3 (Brownells & Young, 1959) Volume tangki (Vt)

Vt = Vs + Vh

28,7688 = 1,1775 Dt3 + 0,1308 Dt3 21,9889 = 1,30833 Dt3

Dt = 2,8015 m Tinggi silinder (Hs) Hs = 3/2 x Dt

Hs = 3/2 x 2,8015 m = 4,2023 m Tinggi head (Hh) : (Hh : Dt = 1 : 4) Hh = ¼ x Dt

Hh = ¼ x 2,8015 m = 0,7003 m

(32)

Tinggi total tangki (Ht):

Ht = Hs + (2 x Hh)

= 4,2023 m + (2 x 0,7003 m) = 5,6031m Tinggi cairan dalam tangki (Hc):

Tinggi cairan dalam tangki = ^{Vc x Hs}

Vs

= (23,9740 × 14,2023 28,7688 )

= 3,5019 m

= 995,72 kg/m³ x 9,8 m/s² x 3,5019 m

= 41,0068 kPa Tekanan operasi = 1 atm = 101,325 kPa P = (41,0068 + 101,325) kPa = 142,3318 kPa Faktor keamanan untuk tekanan = 5%

Pdesain = (1,05 x 142,3318 kPa) = 149,4484 kPa

t = ^{P x D}

(2 x S x E) - (1,2 x P)

= (149,4484 ) x (2,8015 )

Tebal shell standart yang digunakan = 3/16 in (Brownells & Young, 1959)

Volume resin =

1

4 x 19.097,155 ^kg jam

995,72 kg/m³ = 4,7948 m³

15. Tangki Pelarutan H2SO4 (T-205)

Fungsi : Tempat membuat larutan asam sulfat (H2SO4) Bentuk : Silinder tegak dengan alas dan tutup datar

(33)

Bahan kontruksi : Carbon steel SA-283 grade C

Jumlah : 1 unit

Volume tangki

Laju massa H2SO4 = 2,0554 kg/jam

Densitas H2SO4 50% = 1.387 kg/m³ = 86,5879 lbm/ft³ (Perry, 1997) Kebutuhan perancangan = 30 hari

Volume larutan (V1) = 2,0554 kg/jam x 30 hari x 24 jam/hari

0,5 x 1.387 kg/m³ = 2,1339 m³ Faktor kelonggaran = 20%, maka:

Volume tangki (Vt) = (1,2 x 2,1339 m³) = 2,5607 m³ Diamter dan tebal tangki

Vs = ^{π . Di}

2 . Hs

Dimana:

Ditetapkan: Perbandingan tinggi tangka dengan diameter tangka Hs : Di = 3 : 2

Maka:

Vs = ^{π . Di}

2 . (3/2 Di) 4

2,5607 = ^{π . Di}

2 . (3 Di) 8

Di = 1,2955 m Hs = 1,9433 m

Tinggi cairan dalam tangki:

Tinggi cairan dalam tangki = volume cairan x tinggi silinder volume silinder

= ^{2,1339 m}³ × 1,9433 m

2,5607 m³ = 1,6194 m

= 1.387kg/m³ x 9,8 m/s² x 1,6194 m

(34)

= 26,4157 kPa

Tekanan operasi = 1 atm = 101,325 kPa P = (26,4157 + 101,325) kPa = 127,7407 kPa Faktor keamanan untuk tekanan = 5%

Pdesain = (1,05 x 127,7407 kPa) = 134,1278 kPa

Joint efficiency = 0,8 (Brownells & Young, 1959) Allowable stress =12.650 psia = 87.218.714 kPa

t = ^{P x D}

(2 x S x E) - (1,2 x P)

= (134,1278 ) x (1,2955)

Maka tebal shell yang dibutuhkan = 0, 0490 in + 1/8 in = 0,1740 in

Da/Dt = 1/3 ; Da = (1/3 x 1,2955 m) = 0,4318 m E/Da = 1 ; E = (1 x 0,4318 m) = 0,4318 m L/Da = 1/4 ; L = (1/4 x 0,4318 m) = 0,1079 m W/Da = 1/5 ; W = (1/5 x 0,4318 m) = 0,0863 m J/Dt = 1/12 ; J = (1/12 x 1,2955 m) = 0,1079 m Dimana:

Viskositas H2SO4 50% = 0.00034942 lbm/ft.s (Othmer, 1967)

(35)

Bilangan Reynold:

2 µ

=(86,5879) x (1) x (0,4318 )²

0,00034942 = 415.957,0511

P = ^K^r^{x n}³^{x Da}

5 x ρ

gc (McCabe, 1999)

Kr = 6.3 (McCabe, 1999)

=(6,3) x (1)³ x (0,4318 )⁵ x (86,4879)

(32,174) x (550) = 0,00046 hp

Efisiensi motor penggerak = 80%

Daya motor penggerak =^{0,00046 hp}

0,8 = 0,00057 hp

16. Pompa Penukar Kation (Cation Exchanger) (P-208)

Fungsi : Memompa air dari kation exchanger ke anion exchanger

lbm s 62,16 ^lbm ft3

= 0,1881 ^ft

3 s

= 3,9 (0,1881)^0,45(62,16)^0,13

= 3,1458 in

Ukuran nominal = 1 in

(36)

Schedule number = 40

V = ^Q

A

= ^0,1881

0,006

NRe = ^ρVD

μ

=

62,16 ^lbm

ft3 × 31,3568 ^ft_s × 0,0874 ft 0,0005 ^lbm

ft3s

= 340.779,372

A₁)( ^V

2 2a.gc)

= 0,55 (1-0)( ^31,3568

2 2(1)(32,714))

= 8,2654 ft.lbf/lbm

2. 2 elbow 90^o (hf) = n.Kf.(^V

2 2.g_c)

= 2(0,75)( ^31,3568

2 (2)(32,714))

2 2.g_c)

= 1(0,75)( ^31,3568

2 (2)(32,714))

4. Pipa lurus 20 ft (Ff) = 4f.(^∆L.𝑉

2 D.2.g_c)

(37)

= 4(0,008)( 20(31,3568)² (0,0874)(2)(32,714))

= 3,5087 ft.lbf/lbm

A₁)( ^V

2 2a.gc)

= (1-0)( ^31,3568

2 2(1)(32,714))

1

2α ((v2)² – (v1)²) + g(z2 – z1) + ^P

2- P¹

P + ∑F + Ws = 0 Dimana : v1 = v2

P1 = 101,325 kPa = 2.116,22807 lbf/ft² P2