Appendix.pdf

LAMPIRAN A

PERHITUNGAN NERACA MASSA

Kapasitas Produksi : 40.000 ton / tahun

1 tahun operasi : 330 hari

1 hari produksi : 24 jam

Dasar Perhitungan : 1 jam operasi

Satuan : Kg / jam

Kapasitas produksi dalam 1 jam operasi :

jamHarix

TonKgx

hariTahunx

TahunTon

241

11000

3301

1000.40

=

Heptena = 5.050,5050 Kg/jam

Kemurnian Produk : 97 %

Heptena % = 5.050,5050 x 97 %

= 4.898,9898 Kg / jam

= 4.899 Kg / jam

Perhitungan neraca massa dilakukan dengan alur mundur, dimana

perhitungan dimulai dari alur produk sampai ke alur bahan baku. Adapun kemurnian

heptena yang dihasilkan adalah 97 % .

Universitas Sumatera Utara

LA.1 Mixing Point ( M-101 )

Fungsi : Untuk mencampur gas propena segar dan gas n-butena segar dengan

gas n-butena recycle.

Bahan A B C Propena 13,8782 1875,25 -22,9101 Butena 13,8817 2189,45 -30,5161

Sumber : Reaklaitis, 1983.

Bahan baku Propena pada tekanan 13 atm dan temperatur 30 0C, adalah

berupa fasa cair. Ini dapat didapat dan dibuktikan dengan menggunakan rumus

Antoine.

CT

TKpa

CTBAP

014,32)9101,22(

25,18758782,13225,1317ln)(

ln

=

−+−==

+−=

T hitung > T referensi.

Pada bahan baku n-butena pada tekanan 4 atm dan temperatur 30 0C, adalah

berupa fasa cair. Ini didapat dengan menggunakan rumus Antoine.

CT

TKpa

CTBAP

0471,33)5161,30(

45,21898817,133,405ln)(

ln

=

−+−==

+−=

T hitung < T referensi.

(2)

Berdasarkan perhitungan mundur, didapat hasil bawah dari destilasi 4 (MD -

104) yang direcycle kemixing point, dan propena segar (1) dengan butena segar (2)

C3H6 C3H8

M-101 (1)

(19)

(5)

C3H8 i-C4H8 n-C4H8 n-C4H10

C3H8 C3H6 n-C4H8 i-C4H8 n-C4H10

i-C4H8 n-C4H8 n-C4H10


Alur 19

Propana : 19C3H8F = 47,1348 kg /jam

i-butena : 19C4H8-IF = 660,1796 kg /jam

n-butena : 19C4H8-nF = 7.922,1563 kg /jam

n-butana : 19C4H10-nF = 801,6677 kg /jam

Total : F19 = 9.431,1385 kg/jam

Alur 1

Propena : 1C3H6F = 2.753,5534 kg /jam

Propana : 1C3H8F = 371,1265 kg /jam

Total : F1 = 3.124,6799 kg/jam

Alur 2

i-butena : 2C4H8-IF = 315,7633 kg /jam

n-butena : 2C4H8-nF = 8.668,8747 kg /jam

n-butana : 2C4H10-nF = 2.858,1185 kg /jam

Total : F2 = 11.842,7565 kg/jam

Alur 5

Total : F5 = F1+ F2 + F19 = 24.398,5749 kg/jam

Propena : 5C3H6F = 1

C3H6F = 2.753,5534 kg/jam

Propana : 5C3H8F = 1

C3H8F + 19C3H8F = 418,2613 kg/jam

i-butena : 5C4H8-iF = 2

C4H8-IF + 19C4H8-IF = 975,9429 kg/jam

n-butena : 5C4H8-nF = 2

C4H8-nF + 19C4H8-nF = 16.591,031 kg/jam

n-butana : 5C4H10-nF = 2

C4H10-nF + 19C4H10-nF = 3.659,7862 kg/jam


Tabel LA.1 Neraca Massa Mixing Point (M-101)

Komponen Masuk (kg/jam) Keluar (kg/jam)

Alur 1 Alur 2 Alur 24 Alur 3 C3H6 2.753,5534 - - 2.753,5534 C3H8 371,1265 - 47,1348 418,2613 n-C4H8 - 8.668,8747 7.922,1563 16.591,031 i-C4H8 - 315,7633 660,1796 975,9429 n-C4H10 - 2.858,1185 801,6677 3.659,7862

Total 3.124,6799 11.842,7565 9.431,1385 24.398,5749 kg/jam 24.398,5749 kg/jam

LA.2 Reaktor ( R-101 )

Fungsi : Tempat mereaksikan n-butena dan propena menjadi heptena

5

C3H6 C3H8 n-C4H8 i- C4H8 6 C3H8 n-C4H10 n-C4H8

i-C4H8 n-C4H10

C6H12 C7H14 C8H16 Komposisi pada alur produk:

XC6H12 = 0,14 (Destinigsih, 2003)

XC8H16 = 0,09 (Destinigsih, 2003)

Dalam reaktor ini terjadi reaksi, tetapi tidak ada data / referensi yang rinci

untuk menunjukkan persamaan reaksi yang terjadi pada reaktor. Pada reaktor terjadi

konversi n-butena menjadi heptena sebanyak 50%, sedangkan propena habis

bereaksi. (Destinigsih, 2003)

Total : F6 = F5 = 24.398,5748 kg/jam

Alur 6

n-butena : 5C4H8-nF Χ 50%

16.591,031kg/jam x 50% = 8.295,5155 kg/jam

i-butena : 6C4H8-IF = 5

C4H8-IF = 975,9429 kg /jam


n-butana : 6C4H10-nF = 5

C4H10-nF = 3.659,7862 kg /jam

Propana : 6C3H8F = 6

C3H8F = 418,2613 kg/jam

Heksena : 6TotalF x (XC6H12)

24.398,5748 kg/jam X 0,14 = 3.415,8005 kg/jam

Oktena : 6TotalF x (XC8H16)

24.398,5748 kg/jam X 0,09 = 2.195,8717 kg/jam

Heptena : F6Total – ( 6

C4H8-IF + 6C4H10-nF + 6

C3H8F + 6C6H12F + 5

C4H8-nF ) = 5.437,3967

kg/jam

Tabel LA.2 Neraca Massa Reaktor (R-101)


Alur 5 Alur 6 C3H6 2.753,5534 - C3H8 418,2613 418,2613 n-C4H8 16.591,031 8.295,5155 i-C4H8 975,9429 975,9429 n-C4H10 3.659,7862 3.659,7862 C6H12 - 3.415,8005 C7H14 - 5.437,3967 C8H16 - 2.195,8717

Total 24.398,5748 kg/jam 24.398,5748 kg/jam

LA.3 Kolom Distilasi 1 (MD-101)

Fungsi : Memisahkan campuran C4 dari C6,C7 dan C8

Destilasi 01T= 132,5 oCP= 3,7 atm

F7C3H8n-C4H8i-C4H8n-C4H10C6H12T = 34 0CP = 3,7 atm

F6C3H8n-C4H8i-C4H8n-C4H10C6H12C7H14C8H16T = 40 0CP = 3,75 atm

F8n-C4H10C6H12C7H14C8H16T = 132,5 0CP = 3,7 atm


Komposisi pada alur 7 :

- Xn-C4H8 = 0,595 (Destinigsih, 2003)


- XC6H12 = 0,05 (Destinigsih, 2003)

Alur 7

C3H8 : 7C3H8F = 6

C3H8F = 418,2613 kg/jam

i-C4H8 : 7C4H8-iF = 6

C4H8-iF = 975,9429 kg/jam

n-C4H8 : 7C4H8-nF = 6

C4H8-nF = 8.295,5155 kg/jam

Total : F7 = 7C4H8-nF / Xn-C4H8 = 13.942,0427 kg/jam

n-C4H10 : F7 x( X n-C4H10) = 3.555,2209 kg/jam

C6H12 : F7 x (X C6H12 ) = 697,1021 kg/jam

Alur 8

C7H14 : 8C7H14F = 6

C7H14F = 5.437,3967 kg/jam

C8H16 : 8C8H16F = 6

C8H16F = 2.195,8717 kg/jam

n-C4H10 : 6C4H10-nF - 7

C4H10-nF = 104,5653 kg/jam

C6H12 : 6C6H12F - 7

C6H12F = 2.718,6984 kg/jam

Tabel LA.3 Neraca Massa Kolom Destilasi (MD-101)

Komponen Masuk (kg/jam) Keluar (kg/jam) Alur 6 Alur 7 Alur 8

C3H8 418,2613 418,2613 - i-C4H8 975,9429 975,9429 - n-C4H8 8.295,5155 8.295,5155 - n-C4H10 3.659,7862 3.555,2209 104,5653 C6H12 3.415,8005 697,1021 2.718,6984 C7H14 5.437,3967 - 5.437,3967 C8H16 2.195,8717 - 2.195,8717

Total 13.942,0427 10.456,5321 24.398,5748 kg/jam 24.398,5748 kg/jam


LA.3.1 Kondensor (E-102)

Fungsi : Mengkondensasikan uap dari hasil atas kolom destilasi.

Kondensor I

Vd

Ld D

7

Untuk menghitung harga (L/D)min perlu dihitung harga θ. Harga θ dihitung dengan

rumus:

∑ −=−

θαα

i

ifi xq1

Karena umpan dimasukkan di kondensor adalah zat cair jenuh, maka q = 1

Maka : ∑ −=−

θαα

i

ifi x11 = ∑ −=

θαα

i

ifi x0

Nilai θ dicari dengan cara trial dan erorr.

Tabel LA 4. nilai trial erorr θ

Asumsi (θ) C3H8 i-C4H8 n-C4H8 n-C4H10 C6H12 C7H14 C8H16 Total

0.56112 0.0309 0.147 3.106 -1.496 0.318 -0.902 -0.058 1.1470

0.549 0.0303 0.139 2.641 -1.961 0.310 -1.013 -0.060 0,0873 0.5481 0.030 0.138 2.612 -2.007 0.309 -1.022 -0.060 0.00125

Oleh karena itu ∑ −=

θαα

i

ifi x0 , sehingga trial θ = 0.5481dapat diterima

∑ −=+

θαα

i

idiDm

xR 1

RDm + 1 = 1,465

RDm = 0,465

RD = 1,5 RDm

RD = 0,697

Alur Ld

Total : NLd = 0,697 x N7 = 0,697 x 13.942,0427

= 9.717,6037 kg/Jam

C3H8 : NLdC3H8 = X7

C3H8 x NLd = 0,03 x 9.717,6037

= 291,5281 kg/Jam


i-C4H8 : NLd i-C4H8 = X7 i-C4H8 x NLd = 0,07 x 9.717,6037

= 680,2322 kg/Jam

n-C4H8 : NLd n-C4H8 = X7 n-C4H8 x NLd = 0,59 x 9.717,6037

= 5.733,3861 kg/Jam


= 2.526,5769 kg/Jam

C6H12 : NLdC3H8 = X7

C6H12 x NLd = 0,05 x 9.717,6037

= 485,8801 kg/Jam

Alur Vd

Total : NVd = NLd + N7 = 13.942,0427+ 9.717,6037

= 23.659,6464 kg/Jam

C3H8 : NVdC3H8 = X7

C3H8 x NVd = 0,03 x 23.659,6464

= 709,7893 kg/Jam

i-C4H8 : NVd i-C4H8 = X7 i-C4H8 x NVd = 0,07 x 23.659,6464

= 1.656,1752 kg/Jam

n-C4H8 : NVd n-C4H8 = X7 n-C4H8 x NVd = 0,59 x 23.659,6464

= 13.959,1913 kg/Jam


= 6.151,5080 kg/Jam

C6H12 : NVd C6H12 = X7

C6H12 x NVd = 0,05 x 23.659,6464

= 1.182,9823 kg/Jam

Tabel LA 5. Neraca Massa Kondensor I


Alur Vd Alur Ld Alur 7 C3H8 709,7893 291,5281 418,2613 i-C4H8 1.656,1752 680,2322 975,9429 n-C4H8 13.959,1913 5.733,3861 8.295,5155 n-C4H10 6.151,5080 2.526,5769 3.555,2209 C6H12 1.182,9823 485,8801 697,1021

Total 9.717,6034 13.942,0427

23.659,6461 kg/Jam 23.659,6461 kg/jam


LA.3.2 Reboiler (E-103)

Fungsi : Menguapkan cairan dari hasil bawah kolom destilasi

Vb

Reboiler I

Lb 8

Lb = Ld + q.F

NLb = NLd + N6

Alur Lb

Total : NLb : NLd + N6 =9.717,6037 + 24.398,5748

= 34.116,1785 kg/Jam

n-C4H10 : NLb n-C4H10 : X8 n-C4H10 x NLb = 0,01 x 34.116,1785

= 341,1617 kg/Jam

C6H12 : NLb C6H12 = X8

C6H12 x NLb = 0,26 x 34.116,1785

= 8.870,2064 kg/Jam

C7H14 : : NLbC7H14 = X8

C7H14 x NLb = 0,52 x 34.116,1785

= 17.740,4128 kg/Jam

C8H16 : : NLbC8H16 = X8

C8H16 x NLb = 0,21 x 34.116,1785

= 7164,3974 kg/Jam

Alur Vb

Total : NVb : NLb – N8 = 34.116,1785 - 10.456,5321

= 23.659,6464 kg/Jam

n-C4H10 : NVb n-C4H10 : X8 n-C4H10 x NVb = 0,01 x 23.659,6464

= 236,5964 kg/Jam

C6H12 : NVb C6H12 = X8

C6H12 x NVb = 0,26 x 23.659,6464

= 6.151,5080 kg/Jam

C7H14 : : NVbC7H14 = X8

C7H14 x NVb = 0,52 x 23.659,6464

= 12.303,0161 kg/Jam

C8H16 : : NVbC8H16 = X8

C8H16 x NVb = 0,21 x 23.659,6464

= 4.968,5257 kg/Jam


Tabel LA 6. Neraca Massa Reboiler I


Alur Lb Alur Vb Alur 8 n-C4H10 341,1617 236,5964 104,5653 C6H12 8.870,2064 6.151,5080 2.718,6984 C7H14 17.740,4128 12.303,0161 5.437,3967 C8H16 7164,3974 4.968,5257 2.195,8717

Total 23.659,6462 10.456,5321

34.116,1783 Kg/Jam 34.116,1783 Kg/Jam


Fungsi : Memisahkan campuran hasil utama Heptena dari hasil

samping Heksena dan Oktena

Destilasi 03T= 105,5 oC

P= 1 atm

F15n-C4H10C6H12C7H14T = 63,5 0CP = 1 atmF8

n-C4H10C6H12

C7H14C8H16T = 83,5 0CP = 1,1 atm

F17C6H12C7H14C8H16T = 105,5 0CP = 1 atm



- XC6H12 = 0,84334 (Destinigsih, 2003)

Alur 15

n-C4H10 : 15C4H10-nF = 8

C4H10-nF = 104,5653 kg/jam


C6H12 : F15 x(X C6H12) = 2.645,5235 kg/jam

C7H14 : F15 - 15C4H10-nF - 15

C6H12F = 386,8917 kg/jam


Alur 17

C6H12 : 8C6H12F - 15

C6H12F = 73,1749 kg/jam

C7H14 : 8C7H14F - 15

C7H14F = 5.050,5050 kg/jam

C8H16 : 17C8H16F = 8

C8H16F = 2.195,8717 kg/jam

Tabel LA.7 Massa Kolom Destilasi (MD-103)


Alur 10 Alur 15 Alur 17 n-C4H10 104,5653 104,5653 - C6H12 2.718,6984 2.645,5235 73,1749 C7H14 5.437,3967 386,8917 5.050,5050 C8H16 2.195,8717 - 2.195,8717




Kondensor III

Vd

Ld D

15


rumus:

∑ −=−

θαα

i

ifi xq1


Maka : ∑ −=−

θαα

i

ifi x11 = ∑ −=

θαα

i

ifi x0




Asumsi (θ) n-C4H10 C6H12 C7H14 C8H16 Total

1,68 0.0150 0.9023 -0.7647 -0.0889 0.0637

1,67 0.01501 0.8892 -0.7761 -0.0897 0.0384 1.66 0.0112 0.876 -0.7878 -0.090 0.00939


θαα

i

ifi x0 , sehingga trial θ = 1,66 dapat diterima

∑ −=+

θαα

i

idiDm

xR 1

RDm + 1 = 1,321

RDm = 0,321

RD = 1,5 RDm

RD = 0,4815

Alur Ld

Total : NLd = 0,4815 x N15 = 0,4815 x 3.136,9805

= 1.510,4561 kg/Jam


= 49,8450 kg/Jam

C6H12 : NLd C6H12 = X15

C6H12 x NLd = 0,8433 x 1.510,4561

= 1.273,7676 kg/Jam

C7H14 : NLd C7H14 = X15

C7H14 x NLd = 0,123 x 1.510,4561

= 185,7861 kg/Jam

Alur Vd

Total : NVd = NLd + N15 = 1.510,4561+ 3.136,9805

= 4.647,4366 kg/Jam


= 153,36455 kg/Jam

C6H12 : NVd C6H12 = X15

C6H12 x NVd = 0,8433 x 4.647,4366

= 3.919,1835 kg/Jam

C7H14 : NVd C7H14 = X15

C7H14 x NVd = 0,123 x 4.647,4366

= 571,6346 kg/Jam


Tabel LA 9. Neraca Massa Kondensor III


Alur Vd Alur Ld Alur 15 n-C4H10 153,3655 49,8450 104,5653 C6H12 3.919,1835 1.273, 7676 2.645,5235 C7H14 571,6346 185,7861 386,8917 C8H16 - - -

Total 1.510,4561 3.136,9805

4.646,4366 kg/Jam 4.646,4366 kg/Jam



Vb

Reboiler III

Lb 17

Lb = Ld + q.F

NLb = NLd + N10

Alur Lb

Total : NLb : NLd + N10 =1.510,4561+ 10.456,5321

= 11.966,98`2 kg/Jam

C6H12 : NLb C6H12 = X17

C6H12 x NLb = 0,01 x 11.966,9882

= 119,6698 kg/Jam

C7H14 : NLbC7H14 = X17

C7H14 x NLb = 0,69 x 11.966,9882

= 8.257,2218 kg/Jam

C8H16 : NLbC8H16 = X17

C8H16 x NLb = 0,3 x 11.966,9882

= 3.590,0906 kg/Jam

Alur Vb

Total : NVb : NLb – N17 = 11.966,9812 - 7.319,5516

= 4.647,4296 kg/Jam

C6H12 : NVb C6H12 = X17

C6H12 x NVb = 0,01 x 4.647,4294

= 46,4742 kg/Jam

C7H14 : : NVbC7H14 = X17

C7H14 x NVb = 0,69 x 4.647,4294 4.647,4056

= 3.206,7264 kg/Jam


C8H16 : : NVbC8H16 = X17

C8H16 x NVb = 0,3 x 4.647,4056

= 1.394,2288 kg/Jam

Tabel LA 10. Neraca Massa Reboiler III


Alur Lb Alur Vb Alur 17 n-C4H10 - - - C6H12 119,6698 46,4742 73,1749 C7H14 8.257,2218 3.206,7264 5.050,5050 C8H16 3.590,0906 1.394,2288 2.195,8717

Total 4.647, 4296 7.319,5516

11.966,9812 kg/Jam 11.966,9812 kg/Jam




Destilasi 05T= 74,5 oCP= 10 atm

F20C6H12C7H14C8H16T = 30 0CP = 10 atm

F17C6H12C7H14

C8H16T = 50 0CP = 10 atm

F21C7H14C8H16T = 74,5 0CP = 10 atm


- XC7H14 = 0,97 (Destinigsih, 2003)

Kemurnian produk C7H14 yang diset pada alur 20 adalah 97%, sedangkan C7H14 pada

alur 21 adalah 3%.


Alur 20

C7H14 : 17C7H14F x 97% = 4.899,0001 kg/jam

Total : F20 = 20C7H14F / XC7H14 = 5.050,5050 kg/jam

C6H12 : 20C6H12F = 17

C6H12F = 73,1749 kg/jam

C8H16 : F20 - 20C7H14F - 20

C6H12F = 78,331 kg/jam

Alur 21

C7H14 : 17C7H14F x 3% = 219,5865 kg/jam

Total : F21 = F17 – F20 = 2.269,0466 kg/jam

C8H16 : F21 - 21C7H14F = 2.049,4601 kg/jam



Alur 18 Alur 20 Alur 21 C6H12 73,1749 73,1749 - C7H14 5.050,5050 4.899,0001 219,5865 C8H16 2.195,8717 78,331 2.049,4601




Kondensor V

Vd

Ld D

20


rumus:

∑ −=−

θαα

i

ifi xq1


Maka : ∑ −=−

θαα

i

ifi x11 = ∑ −=

θαα

i

ifi x0




Asumsi (θ) C6H12 C7H14 C8H16 Total

1.244 0.012109 1.22447 -1.22950 0.007072 1.2435 0.012108 1.224089 -1.23203 0.004165 1.243 0.012107 1.22370 -1.23456 0.001248


θαα

i

ifi x0 , sehingga trial θ = 1.243 dapat diterima

∑ −=+

θαα

i

idiDm

xR 1

RDm + 1 = 1,438 RDm = 0,438

RD = 1,5 RDm

RD = 0,657

Alur Ld

Total : NLd = 0,657 x N20 = 0,657 x 5.050,5050

= 3.318,1817 Kg/Jam

C6H12 : NLd C6H12 = X20

C6H12 x NLd = 0,013 x 3.702,9257

= 43,1365 Kg/Jam

C7H14 : NLd C6H12 = X15

C6H12 x NLd = 0,8870 x 3.702,9257

= 2.943,2271 Kg/Jam

C8H16 : NLd C7H14 = X15

C7H14 x NLd = 0, 1 x 3.702,9257

= 331,8181 Kg/Jam

Alur Vd

Total : NVd = NLd + N20 = 3.318,1817 + 5.050,5050

= 8.368,6867 kg/Jam

C6H12 : NVd C6H12 = X20

C6H12 x NVd = 0,013 x 8.368,6867

= 108,7929 kg/Jam

C7H14 : NVd C6H12 = X15

C6H12 x NVd = 0,8870 x 8.368,6867

= 7.423,0252 kg/Jam

C7H14 : NVd C7H14 = X15

C7H14 x NVd = 0,1 x 8.368,6867

= 836,8686 kg/Jam


Tabel LA 13. Neraca Massa Kondensor V


Alur Vd Alur Ld Alur 20 C6H12 108,7929 43,1365 73,1749 C7H14 7.423,0252 2.943,2271 4.899,0001 C8H16 836,8686 331,8181 78,331

Total 3.318,1817 5.5050,5050

8.368,6867 Kg/Jam 8.368,6867 Kg/Jam



Vb

Reboiler V

Lb 21

Lb = Ld + q.F

NLb = NLd + N18

Alur Lb

Total : NLb : NLd + N18 = 3.318,1817+ 7.319,5516

= 10.637,7333 Kg/Jam

C7H14 : NLbC7H14 = X21

C7H14 x NLb = 0,03 x 10.637,7333

= 319,1319 Kg/Jam

C8H16 : NLbC8H16 = X21

C8H16 x NLb = 0,97x 10.637,7333

= 10.318,6014 Kg/Jam

Alur Vb

Total : NVb : NLb – N21 = 10.637,7333 – 2.269,0466

= 8.368,6867 Kg/Jam

C7H14 : : NVbC7H14 = X21

C7H14 x NVb = 0,03 x 8.368,6867

= 251,0607 Kg/Jam

C8H16 : : NVbC8H16 = X21

C8H16 x NVb = 0,97 x 8.368,6867

= 8.117,6260 Kg/Jam


Tabel LA 14. Neraca Massa Reboiler V


Alur Lb Alur Vb Alur 21 C7H14 319,1319 251,0607 219,5865 C8H16 10.318,6014 8.117,6260 2.049,4601

Total 8.368,6867 2.269,0466

10.637,7333 Kg /Jam 10.637,7333 Kg/Jam


Fungsi : Memisahkan campuran hasil propena, butena dan heksena

Destilasi 02T= 50 oCP= 5 atm

F11C3H8i-C4H8n-C4H8n-C4H10T = 37 0CP = 4,5 atm

F7C3H8i-C4H8n-C4H8

n-C4H10C6H12T = 42 0CP = 5 atm

F13N-C4H8N-C4H10C6H12T = 50 0CP = 5 atm


- Xi-C4H8 = 0,093333 (Destinigsih, 2003)


Alur 11

C3H8 : 11C3H8F = 7

C3H8F = 418,2613 kg/jam

i-C4H8 : 11C4H8-iF = 7

C4H8-iF = 975,9429 kg/jam

Total : F11 = 11C4H8-iF / Xi-C4H10 = 10.456,5684 kg/jam

n-C4H8 : F11 x (Xn- C4H8) = 8.260,6890 kg/jam

n-C4H10 : F11 - 11C3H8F - 11

C4H8-iF - 11C4H8-nF = 801,6752 kg/jam


Alur 13

n-C4H8 : 7C4H8-nF - 11

C4H8-nF = 34,8265 kg/jam

n-C4H10 : 7C4H10-nF - 11

C4H10-nF = 2.753,5457 kg/jam

C6H12 : 13C6H12F = 7

C6H12F = 697,1021 kg/jam



Alur 7 Alur 11 Alur 13 C3H8 418,2613 418,2613 - i-C4H8 975,9429 975,9429 - n-C4H8 8.295,5155 8.260,6890 34,8265 n-C4H10 3.555,2209 801,6752 2.753,5457 C6H12 697,1021 - 697,1021



Fungsi : Mengkondensasikan uap dari hasil atas kolom destilasi

Kondensor II

Vd

Ld D

11


rumus:

∑ −=−

θαα

i

ifi xq1


Maka : ∑ −=−

θαα

i

ifi x11 = ∑ −=

θαα

i

ifi x0




Asumsi (θ) C3H8 i-C4H8 n-C4H8 n-C4H10 C6H12

Total

1.03885 0.04248 0.17883 6.3530 -6.563 -0.0097 0.00082

1.0388495 0.0424804 0.17883 6.352970 -6.56379 -0.00978 0.0007141 1.0388488 0.04248 0.17883 6.35292 -6.56390 -0.009783 0.000554


θαα

i


∑ −=+

θαα

i

idiDm

xR 1

RDm + 1 = 1,264

RDm = 0,264

RD = 1,5 RDm

RD = 0,396

Alur Ld

Total : NLd = 0,396 x N11 = 0,396 x 10.456,5684

= 4.140,801086 Kg/Jam


C3H8 x NLd = 0,04 x 4.140,801086

= 165,6320435 Kg/Jam

i-C4H8 : NLd i-C4H8 = X11 i-C4H8 x NLd = 0,093 x 4.140,801086

= 385,094501 Kg/Jam


= 3.271,232858 Kg/Jam


= 318,84167 Kg/Jam

Alur Vd

Total : NVd = NLd + N11 = 4.140,8010 + 10.456,5684

= 14.597,3694 Kg/Jam

C3H8 : NVdC3H8 = X11

C3H8 x NVd = 0,04 x 14.597,3694

= 583,8947 Kg/Jam



= 1.357,5553 Kg/Jam


= 11.531,9218 Kg/Jam


= 1.123,9973 Kg/Jam

Tabel LA 17. Neraca Massa Kondensor II


Alur Vd Alur Ld Alur 11 C3H8 583,8947 165,6320 418,2613 i-C4H8 1.357,5553 385,0944 975,9429 n-C4H8 11.531,9218 3.271,2327 8.260,6890 n-C4H10 1.123,9973 318,8416 801,6752

Total 4.140,8007 10.456,5684

14.597,3691 Kg/Jam 14.597,3691 Kg/Jam



Vb

Reboiler II

Lb 13

Lb = Ld + q.F

NLb = NLd + N9

Alur Lb

Total : NLb : NLd + N9 = 4.140,8010 + 13.942,0427

= 18.082,8437 Kg/Jam

n-C4H8 : NLb n-C4H8 = X13 n-C4H8 x NLb = 0,01 x 18.082,8437

= 180,828437 Kg/Jam


= 14.285,44652 Kg/Jam

C6H12 : NLb C6H12 = X13

C6H12 x NLb = 0,2 x 18.082,8437

= 3.616,56874 Kg/Jam


Alur Vb

Total : NVb : NLb – N13 = 18.082,8437 - 3.485,4743

= 14.597,3694 Kg/Jam

n-C4H8 : NVb n-C4H8 = X13 n-C4H8 x NVb = 0,01 x 14.597,3694

= 145,973694 Kg/Jam

n-C4H10 : NVb n-C4H10 : X13 n-C4H10 x NVb = 0,79 x 14.597,3694

= 11.531,92183 Kg/Jam

C6H12 : NVb C6H12 = X13

C6H12 x NVb = 0,2 x 14.597,3694

= 2.919,47388 Kg/Jam

Tabel LA 18. Neraca Massa Reboiler II


Alur Lb Alur Vb Alur 13 n-C4H8 180,828437 145,9736 34,8265 n-C4H10 14.285,44652 11.531,9218 2.753,5457 C6H12 3.616,56874 2.919,4738 697,1021

Total 14.597,3694 3.485,4743

18.082,8437 Kg/Jam 18.082,8437 Kg/Jam


Fungsi : Memisahkan hasil utama dari campuran propena dan butena

Destilasi 04T= 90 oC

P= 12 atm

F18C3H8i-C4H8n-C4H8T = 30 0CP = 12 atmF12

C3H8i-C4H8n-C4H8

n-C4H10T = 60 0CP = 12 atm

F19C3H8i-C4H8N-C4H8N-C4H10T = 80 0CP = 12 atm



- Xn-C4H10 = 0,085002 (Destinigsih,. 2003)


- Xi-C4H8 = 0,07 (Destinigsih, 2003)

Alur 19

n-C4H10 : 19C4H10-nF = 11

C4H10-nF = 801,6752 kg/jam


n-C4H8 : F19 x(Xn- C4H8) = = 7.922,250865 kg/jam

i- C4H8 : F19 x (Xi- C4H8) = = 660,187572 kg/jam

C3H8 : F19 - 19C4H10-nF - 19

C4H8-nF - 19C4H8-iF = 47,137392 kg/jam

Alur 18

C3H8 : 11C3H8F - 19

C3H8F = 371,123908 kg/jam

i-C4H8 : 11C4H8-iF - 19

C4H8-iF = 315,755328 kg/jam

n-C4H8 : 11C4H8-nF - 19

C4H8-nF = 338,438135 kg/jam



Alur 11 Alur 18 Alur 19 C3H8 418,2613 371,123908 47,137392 i-C4H8 975,9492 315,755328 660,187572 n-C4H8 8.260,6890 338,438135 7.922,250865 n-C4H10 801,6752 - 801,6752




Fungsi : Mengkondensasikan uap dari hasil atas kolom destilasi

Kondensor IV

Vd

Ld D

18


rumus:

∑ −=−

θαα

i

ifi xq1


Maka : ∑ −=−

θαα

i

ifi x11 = ∑ −=

θαα

i

ifi x0



Asumsi (θ) C3H8 i-C4H8 n-C4H8 n-C4H10

Total

1.211379 0.076470265 2.0148069 -3.737362 1.6465089 0.00042342

1.211378 0.0764702 2.01477261 -3.737380 1.6464808 0.0003432 1.211375 0.0764700 2.014669 -3.737433 1.646396 0.0001026


θαα

i


RDm + 1 = 1,326

RDm = 0,326

RD = 1,5 RDm

RD = 0,489

∑ −=+

θαα

i

idiDm

xR 1


Alur Ld

Total : NLd = 0,489 x N18 = 0,489 x 1.025,317371

= 501,380093Kg/Jam


C3H8 x NLd = 0,361 x 501,3801644

= 180,9981679Kg/Jam

i-C4H8 : NLd i-C4H8 = X18 i-C4H8 x NLd = 0,307 x 501,3801644

= 153,9237105 Kg/Jam

n-C4H8 : NLd n-C4H8 = X18 n-C4H8 x NLd = 0,332 x 501,3801644

= 166,4582146 Kg/Jam

Alur Vd

Total : NVd = NLd + N18 = 501,3801644 + 1.025,3173

= 1.526,697464 Kg/Jam

C3H8 : NVdC3H8 = X18

C3H8 x NVd = 0,361 x 1.526,697464

= 551,1377845 Kg/Jam


= 468,6961214 Kg/Jam


= 506,863558 Kg/Jam

Tabel LA 21. Neraca Massa Kondensor IV


Alur Vd Alur Ld Alur 18 C3H8 551,137785 180,9981679 371,123908 i-C4H8 468,696122 153,9237105 315,755328 n-C4H8 506,863558 166,4582146 338,438135

Total 501,380093 1.025,317371

1.526,697464 kg/Jam 1.526,697464 kg/Jam



Fungsi : Menguapkan cairan dari hasil bawah kolom destilasi 4

Vb

Reboiler IV

Lb 19

Lb = Ld + q.F

NLb = NLd + N12

Alur Lb

Total : NLb : NLd + N12 =501,3801644 + 10.456,5684

= 10.957,94856 Kg/Jam

C3H8 : NLbC3H8 = X18

C3H8 x NLb = 0,005 x 10.957,94856

= 54,78974282 Kg/Jam

n-C4H8 : NLb n-C4H8 = X19 n-C4H8 x NLb = 0,07 x 10.957,94856

= 767,0563992 Kg/Jam


= 9.204,67679 Kg/Jam

C6H12 : NLb C6H12 = X19

C6H12 x NLb = 0,085 x 10.957,94856

= 931,4256276 Kg/Jam

Alur Vb

Total : NVb : NLb – N19 = 10.957,94856 - 9.431,251029

= 1.526,697531 Kg/Jam

C3H8 : NVbC3H8 = X18

C3H8 x NVb = 0,005 x 1.526,697531

= 7,633487653 Kg/Jam


= 106,8688272 Kg/Jam


= 1.282,425926 Kg/Jam

C6H12 : NVb C6H12 = X19

C6H12 x NVb = 0,085 x 1.526,697531

= 129,7692901Kg/Jam


Tabel LA 22. Neraca Massa Reboiler IV


Alur Lb Alur Vb Alur 19 C3H8 54,78974282 7,633487653 47,137392 n-C4H8 767,0563992 106,8688272 660,187572 n-C4H10 9.204,67679 1.282,425926 7.922,250865 C6H12 931,4256276 129,7692901 801,6752

Total 1.526,697531 9.431,251029

10.957,94856 kg/Jam 10.957,94856 kg/Jam


LAMPIRAN B

PERHITUNGAN NERACA PANAS

Basis Perhitungan : 1 Jam Operasi

Satuan Operasi : Kj/jam

Temperatur referensi : 25 0C

LB.1 Heater (E-101)

Fungsi : Menaikkan temperatur bahan yang keluar dari pompa

sebelum dimasukkan kedalam reaktor

Saturated steam (1500C)

P = 4,8 bar

4 5

Neraca energi : )P,(THN )P,(THNdTdW

dTdQ

jjout in

jjiiii∑ ∑−=−

Karena sistem tidak melakukan kerja, maka 0dTdW

=

Sehingga, )P,(THN )P,(THN dTdQ

jjout in

jjiiii∑ ∑−=

C3H6 C3H8 n-C4H8 i- C4H8 n-C4H10

C3H6 C3H8 n-C4H8 i- C4H8 n-C4H10

T= 30 0C P = 9,11 bar

T= 40 0C P = 9,11 bar

Air kondensat (1350C) P = 4,8 bar


Panas masuk T = 303 (30 oC)

Panas masuk pada suhu 30 oC (303 K)

Panas keluar pada alur 4: Qout = ∫∑303

298(g)i dT Cp N

Tabel LB.1 Panas masuk Heater pada Alur 4 (E-101)

Komponen F (Kg/jam) dTCp g∫303

298)(

∫

303

298)(

5 dTCpN g

C3H6 2.753,5534 321,2178 884,4903653 C3H8 418,2613 830,7164 347,565214 n-C4H8 16.591,031 430,9289 7.109,554739 i-C4H8 975,9429 447,8453 437,0714408 n-C4H10 3.659,7862 190,5042 697,2046422

Total 9.475,886401 Kj/Jam

Panas keluar pada suhu 40 oC (313 K)

Panas keluar pada alur 5: Qout = ∫∑313

298(g)i dT Cp N

Tabel LB.2 Panas Keluar Heater pada Alur 5 (E-101)


298)(

∫

313

298)(

5 dTCpN g

C3H6 2.753,5534 976.2216 2.688,078306 C3H8 418,2613 2.578,0384 1.078,293693 n-C4H8 16.591,031 1.310,7484 21.746,66734 i-C4H8 975,9429 1.360,4775 1.327,748357 n-C4H10 3.659,7862 455,2146 1.665,988111

Total 28.506,77581 kJ/Jam

Neraca Energi Total Sistem :

dtdQ = QOut– QIn

= [28.506,77581 – 9.475,88640] kJ/jam

= 19.030,88941 kJ/jam

Media pemanas yang dipakai saturated steam pada 150 0C, kemudian keluar sebagai

air pada 1350C (1atm).

dtdQ = m. Cp. ∆T


Steam yang diperlukan adalah :

)135150( 00KondensatSteamSaturated

reboiler

CCHq

−∆

=

− 8,174.2 4119.030,889

)9,5707,745.2( 4119.030,889

= 8,750638 kg/jam

LB.2 Reaktor 1 (R-101)

Fungsi : Tempat berlangsungnya reaksi pembentukan Heptena

5

C3H6 C3H8 n-C4H8 i- C4H8 6 C3H8 n-C4H10 n-C4H8

i-C4H8 n-C4H10

C6H12 C7H14 C8H16 Panas masuk T = 313 (40 oC)

Panas masuk pada alur 5: Qin = )dT Cp Hf (F313

298(g)i ∫∑ +∆

Contoh perhitungan :

Untuk Propena : ∆H Propena = )dT Cp Hf (F313

298(g)Propena ∫∑ +∆

∆H Propena = 2.753,5534Kg/Jam (20,4179 J/Kg + 976,2216 J/Kg)

2.744,300084 Kj/Jam


Tabel LB.3 Panas masuk reaktor pada Alur 5 (R-101)

Komponen F (Kg/jam) ∆Hf dTCp g∫313

298)(

+∆ ∫

313

298)(

5 dTCpHfF g

C3H6 2.753,5534 20,4179 976,2216 2.744,300084 C3H8 418,2613 -130,8468 2578,0384 1.023,56554 n-C4H8 16.591,031 -0,1255 1310,7484 21.744,58516 i-C4H8 975,9429 -16,9033 1360,4775 1.343,5742 n-C4H10 3.659,7862 -126,1476 455,2146 1.204,314865

Total 28.060,33985 Kj/Jam Panas keluar pada alur 6 pada suhu 400C (313 K)

Panas keluar pada alur 6: Qin = ∫∑ +∆313

298(g)i dT Cp Hf (F

Contoh perhitungan :

Untuk Propena : ∆H Propana = ∫∑ +∆313

298(g)Propana dT Cp Hf (F

∆H Propana = 418,2613 Kg/Jam (-103,8468 J/Kg + 2.578,0384 J/Kg)

1.034,858595 Kj/Jam

Tabel LB.4 Panas keluar reaktor pada Alur 6 (R-101)

Komponen F (Kg/jam) ∆Hf dTCp g∫313

298)(

+∆ ∫

313

298)(

5 dTCpHfF g

C3H8 418,2613 -103,8468 2.578,0384 1.034,858595 n-C4H8 8.295,5155 -0,1255 1.310,7484 10.872,29258 i-C4H8 975,9429 -16,9033 1.360,4775 1.311,250701 n-C4H10 3.659,7862 -126,1476 455,2146 1.204,314865 C6H12 3.415,8005 -41,6726 2742,4815 9.225,424391 C7H14 5.437,3967 -62,2997 4522,0756 24.249,57076 C8H16 2.195,8717 -82,9268 3.489 15.686,9934


Total Neraca Energi :

∆H total = ∆H keluar - ∆H masuk

= 63.584,70529 – 28.060,33985

= 35.524,3654 Kj/Jam


Massa steam yang digunakan :

)135150( 00KondensatSteamSaturated

reboiler

CCHq

−∆

8,174.2 435.524,365

)9,5707,745.2( 435.524,365

=−

= 16,3345 kg/jam

LB.3 Kolom Destilasi 01 ( MD-101 )


Destilasi 01T= 132,5 oCP= 3,7 atm

F7C3H8n-C4H8i-C4H8n-C4H10C6H12T = 34 0CP = 3,7 atm

F6C3H8n-C4H8i-C4H8n-C4H10C6H12C7H14C8H16T = 40 0CP = 3,75 atm

F8n-C4H10C6H12C7H14C8H16T = 132,5 0CP = 3,7 atm

Neraca Panas Total

Q6 = Q7 + Q8

Q = ∑m. Cp. Δt

Tabel LB.5 Hasil Perhitungan Neraca Panas umpan pada alur 6

Komponen F (Kg/jam) dTCp l∫313

298)(

∫

313

298)(

5 dTCpF ll

C3H8 418,2613 2.578,0384 1.078,293693 n-C4H8 8.295,5155 1.310,7484 10.873,33367 i-C4H8 975,9429 1.360,4775 1.327,748357 n-C4H10 3.659,7862 455,2146 1.665,988111 C6H12 3.415,8005 2742,4815 9.367,769679 C7H14 5.437,3967 4522,0756 24.588,31894 C8H16 2.195,8717 3.489 766,139361



Tabel LB.6 Hasil Perhitungan Neraca Panas destilat pada alur 7


298)(

∫

307

298)(

5 dTCpF l

C3H8 418,2613 1.100,8116 460,4268909 n-C4H8 975,9429 1.212,2591 1.183,095662 i-C4H8 8.295,5155 1.181,0031 9.797,029522 n-C4H10 3.555,2209 1.290,7085 4.588,753835 C6H12 697,1021 1.635,4376 1.140,066985


Tabel LB.7 Hasil Perhitungan Neraca Panas produk bawah pada alur 8

Komponen F (Kg/jam) dTCp l∫5,405

298)(

∫

5,405

298)(

5 dTCpF l

n-C4H10 104,5653 18.015,6705 1.883,813991 C6H12 2.718,6984 20.491,6601 55.710,64353 C7H14 5.437,3967 36.140,8060 196.511,8993 C8H16 2.195,8717 25.004,5 54.906,67392


Tabel LB.8 Hasil Perhitungan Neraca Panas pada Condensor Komponen F (Kg/jam) ( )KgJ /λ ( )λ5F

C3H8 418,2613 18.773,1 7.852,061211 n-C4H8 975,9429 22.080,6 21.549,4048 i-C4H8 8.295,5155 22.050,6 182.921,0941 n-C4H10 3.555,2209 22.416,0 79.693,83169 C6H12 697,1021 28.207,4 19.663,43778

Total 311.679,8296 Kj/Jam Panas yang dibutuhkan pada Reboiler adalah :

QB = Q7 + Q8 + QC – Q6

QB = 17.169,37289 Kj/Jam + 309.013,0307 Kj/Jam + 311.679,8296 Kj/Jam –

49.667,59181 Kj/Jam

QB = 588.194,6414 KJ/Jam

Massa air pendingin yang digunakan di condensor yaitu :

m =

−∆ )1525( 00 CCH

q

=

− )9,41(

8296,679.311)9,628,104(

8296,679.311

= 7.438,6594 Kg/Jam


Massa steam yang dibutuhkan pada reboiler yaitu :

m = KondensatteamsaturatedS

reboiler

CCHq

00 135150( −∆

= 8,174.2

6414,194.588)9,5707,745.2(

6414,194.588=

−

= 270,4591 Kg/Jam

LB.4 Cooler (E-105)

Fungsi : Menurunkan temperatur bahan yang keluar dari reboiler pada

destilasi I sebelumdimasukkan kedalam destilasi 3

Persamaan Neraca Panas :

Panas Masuk = Panas Keluar + Akumulasi

Asumsi : Keadaan steady state, akumulasi = 0

Maka :

Panas Masuk = Panas Keluar

Panas alur masuk = Panas alur keluar + Panas yang dibawa air pendingin

• Kondisi Masuk:

- Alur masuk = Alur 8

- Temperatur referensi = 25oC = 298,15 oK

- Temperatur alur = 132,5 oC = 405,5oK

Panas masuk : Qin = 309.013,0307 Kj/Jam

83,5 oC P= 3,85 b

132,5 oC P= 3,85 bar

Air pendingin bekas 25 oC P = 1 bar

8 10 n-C4H10 C6H12 C7H14 C8H16

n-C4H10 C6H12 C7H14 C8H16

Air pendingin (15 oC) P = 1 bar


• Kondisi Keluar:

- Alur keluar = Alur 10

- Temperatur referensi = 25oC = 298,15 K

- Temperatur alur 15 = 83,5 oC = 356,5 K

Panas keluar : Qout = dT Cp Hvl dT Cp NBP

298

356,5

BP(g)(l)i

+∆+∫ ∫∑

Tabel LB.9 Panas Keluar Cooler 1 (E-105) Alur 10

Komponen F (Kg/jam) dTCpBP

l∫298

)( ∆Hvl dTCp

BPg∫

5,356

)( dTCp∫

5,356

298

∫

15,363

15,298

10 dTCpF

n-C4H10 104,5653 -931,2384 22.416 5.994,9699 27.479,7315 2.873,426368 C6H12 2.718,6984 7.237,3147 28.207,4 8.343,3806 43.788,0953 119.046,6246 C7H14 5.437,3967 - - - 18.540,9303 100.814,3932 C8H16 2.195,8717 - - - 13.607,1000 29.879,44581



dtdQ = QOut– QIn

= [252.613,89 – 309.013,0307] kJ/jam

= -56.399,14072 kJ/jam

Air pendingin yang diperlukan adalah :

m =

−∆ )1525( 00 CCH

q

= )9,628,104(kJ/jam) 072(56.399,14

−−

= (41,9)

kJ/jam) 072(56.399,14 −

= 1.346,041545 kg/jam


LB.5 Kolom Destilasi 03 ( MD -103 )

Fungsi : Menaikkan temperatur bahan yang keluar dari kondensor

sebelum dimasukkan kedalam destilasi 2

Destilasi 03T= 105,5 oC

P= 1 atm

F15n-C4H10C6H12C7H14T = 63,5 0CP = 1 atmF10

n-C4H10C6H12

C7H14C8H16T = 83,5 0CP = 1,1 atm

F17C6H12C7H14C8H16T = 105,5 0CP = 1 atm

Neraca Panas Total

Q10 = Q15 + Q17

Q = ∑m. Cp. Δt

Panas masuk = panas yang keluar dari cooler 105

Panas masuk = 252.613,89 Kj/Jam


Komponen F (Kg/jam) dTCp∫5,336

298

∫

5,336

298

15 dTCpF

n-C4H10 104,5653 5.747,5060 600,9896891 C6H12 2.645,5235 7.209,1585 19.071,99823 C7H14 386,8917 11.926,9071 4.614,421364



Komponen F (Kg/jam) dTCp g∫5,405

298)(

∫

5,405

298)(

5 dTCpF g

C6H12 73,1749 15.753,7871 1.152,781796 C7H14 5.050,5050 26.167,2236 132.157,6936 C8H16 2.195,8717 18.724,3 41.116,16047




n-C4H10 153,3655 22.416,0 3.437,841048 C6H12 3.919,1835 28.207,4 110.549,9767 C7H14 571,6346 31.7352 181.409,3836


Panas yang dibuthkan pada Reboiler adalah :

QB = Q15 + Q17 + QC – Q10


252.613,89 Kj/Jam

QB = 241.497,3565 KJ/Jam

Massa air pendingin yang digunakan di kondensor yaitu :

m =

−∆ )1525( 00 CCH

q

=

− )9,41(

13295.397,20)9,628,104( 13295.397,20

= 7.050,0525 Kg/Jam

Massa steam yang dibutuhkan pada reboiler yaitu :

m = )135150( 00

KondensatSteamSaturated

reboiler

CCHq

−∆

= 2.174,8

65241.497,35)9,5707,745.2(

65241.497,35=

−

= 111,0434 Kg/Jam

LB.6 Cooler (E-111)


destilasi III sebelum dimasukkan kedalam tangki Produk

Hexene





Maka :



• Kondisi Masuk:


- Temperatur referensi = 25oC = 298 oK

- Temperatur alur 15 = 63,5 oC = 336,5oK


• Kondisi Keluar:


- Temperatur referensi = 25oC = 298 K

- Temperatur alur 15 = 50 oC = 323 K


298,15

323

BP(g)(l)i

+∆+∫ ∫∑

Tabel LB.13 Panas Keluar Cooler (E-111) Alur 16


l∫298

)( ∆Hvl dTCp

BPg∫

323

)( dTCp∫

323

298

∫

323

15,298

16 dTCpF

n-C4H10 104,5653 -931,2384 22.416 775,0459 22.259,8075 2.327,603449 C6H12 2.645,5235 - - - 4.617,3295 12.215,2537 C7H14 386,8917 - - - 7.624,9301 2.950,022169


50 oC P = 1,013 bar

63,5 oC P = 1,013 bar

15 16 n-C4H10 C6H12 C7H14

n-C4H10 C6H12 C7H14





dtdQ = QOut– QIn

= [17.492,8793 - 24.287,16001] kJ/jam

= -6.794,2806 kJ/jam


m =

−∆ )1525( 00 CCH

q

= )9,41(

kJ/jam) 6(6.794,280 )9,628,104(

kJ/jam) 6(6.794,280 =

−−

= 162,1546 kj/jam

LB.7 Cooler (E-116)

Fungsi : Menurunkan temperatur bahan yang keluar dari kondensor

destilasi III sebelum dimasukkan kedalam destilasi V.




Maka :



• Kondisi Masuk:




50 oC P = 1,21 b

105,5 oC P= 1,21 bar

17 19 C6H12 C7H14 C8H16

C6H12 C7H14 C8H16





• Kondisi Keluar:





298,15

323

BP(g)(l)i

+∆+∫ ∫∑



l∫298

)( ∆Hvl dTCp

BPg∫

323

)( dTCp∫

323

298

∫

323

15,298

16 dTCpF

C6H12 73,1749 - - - 4.617,3295 337,872624 C7H14 5.050,5050 - - - 4.617,3295 23.319,84573 C8H16 2.195,8717 - - - 7.624,9301 16.743,36822



dtdQ = QOut– QIn

= [40.401,0865 - 174.427,1828] kJ/jam

= -134.026,0962 kJ/jam


m =

−∆ )1525( 00 CCH

q

= )9,41(

kJ/jam) 962(134.026,0 )9,628,104(

kJ/jam) 962(134.026,0 =

−

= 3.198,7135 kg/jam




samping Oktena

Destilasi 025T= 74,5 oCP= 10 atm

F20C6H12C7H14C8H16T = 30 0CP = 10 atm

F17C6H12C7H14

C8H16T = 50 0CP = 10 atm

F21C7H14C8H16T = 74,5 0CP = 10 atm

Neraca Panas Total

Q19 = Q20 + Q21

Q = ∑m. Cp. Δt

Panas masuk = panas yang keluar dari cooler 105




298)(

∫

303

298)(

20 dTCpF l

C6H12 73,1749 904,9809 66,221886 C7H14 4.899,0001 1.489,8103 7.298,580809 C8H16 78,331 1163 91,098953




298)(

∫

303

298)(

20 dTCpF l

C7H14 219,5865 15.529,6369 3.410,098613 C8H16 2.049,4601 11.513,7000 23.596,86875




C6H12 108,7929 282.074 30.687,64847 C7H14 7.423,0252 31.735 235.569,7047 C8H16 836,8686 32.867 27.505,36028



QB = Q20 + Q21 + QC – Q19


40.401,0865 Kj/Jam

QB = 287.824,496 KJ/Jam


m =

−∆ )1525( 00 CCH

q

=

− )9,41(

35293.762,71)9,628,104( 35293.762,71

= 7.011,0432 Kg/Jam

Massa steam yang dibutuhkan di reboiler yaitu :

m = )135150( 00


reboiler

CCHq

−∆

= 2.174,8

5157,095.183)9,5707,745.2(

57183.095,51=

−

= 132,3452 Kg/Jam


LB.9 Cooler (E-118)

Fungsi : Menurunkan temperatur bahan yang keluar dari reboiler

destilasi V sebelum dimasukkan kedalam tangki produk

oktene.




Maka :



• Kondisi Masuk:





• Kondisi Keluar:





298,15

323

BP(g)(l)i

+∆+∫ ∫∑

50 oC P = 10,132 bar

74,5 oC P = 10,132 bar

21 22 C7H14 C8H16

C7H14 C8H16





Komponen F (Kg/jam)

dTCpBP

l∫298

)( ∆Hvl dTCp

BPg∫

323

)( dTCp∫

323

298

∫

323

15,298

16 dTCpF

C7H14 219,5865 - - - 4.617,3295 1.013,903224 C8H16 2.049,4601 - - - 7.624,9301 15.626,99001

Total 16.640,89323 Kj/Jam Neraca Energi Total Sistem :

dtdQ = QOut– QIn

= [16.640,89323 - 19.585,67171] kJ/jam

= -2.944,7784 kJ/jam


m =

−∆ )1525( 00 CCH

q

=

− )9,41(

2.944,7784)9,628,104(

2.944,7784

= 70,28110 Kg/Jam

LB.10 Heater 02 (E-104)




P = 4,8 bar

4 5

Air condensat (T=1350C)

P = 4,8 bar


dTdQ

jjout in


C3H8 n-C4H8 i- C4H8 n-C4H10 C6H12

C3H8 n-C4H8 i- C4H8 n-C4H10 C6H12

T= 34 0C P = 3,749 bar

T= 42 0C P = 3,749 bar



Panas masuk pada alur 7: Qin = ∫∑307

298i dT Cp N



Maka :


• Kondisi Masuk:



- Temperatur alur 7 = 34 oC = 307oK

- Panas masuk = 17.169,37289 Kj/Jam

• Kondisi Keluar:





298

315

BP(g)(l)i

+∆+∫ ∫∑

Tabel LB.19 Panas Keluar Heater (E-104) Alur 5

Komponen F (Kg/jam) dTCp∫315

298

∫

315

298

9 dTCpF

C3H8 418,2613 2.941,8341 1.230,455355 n-C4H8 975,9429 1.489,5687 1.453,733997 i-C4H8 8.295,5155 1.545,6990 12.822,37001 n-C4H10 3.555,2209 518,1653 1.842,192104 C6H12 697,1021 3114,4366 2.171,080294


dtdQ = QOut– QIn

= [19.519,83176 - 17.169,37289] kJ/jam

= 2.350,45887 kJ/jam



m = )135150( 00


reboiler

CCHq

−∆

= 2.174,8

72.350,4588)9,5707,745.2(

72.350,4588=

−

= 1,0807 Kg/Jam


Fungsi : Memisahkan campuran C4 dari C6

Destilasi 02T= 50 oCP= 5 atm

F11C3H8i-C4H8n-C4H8n-C4H10T = 37 0CP = 4,5 atm

F7C3H8i-C4H8n-C4H8

n-C4H10C6H12T = 42 0CP = 5 atm

F13N-C4H8N-C4H10C6H12T = 50 0CP = 5 atm

Neraca Panas Total

Q9 = Q11 + Q13

Q = ∑m. Cp. Δt

Panas masuk = panas yang keluar dari heater E-104




298)(

∫

310

298)(

11 dTCpF g

C3H8 418,2613 1.476,8133 617,693850 i-C4H8 975,9429 1.622,71 1.583,672303 n-C4H8 8.260,6890 1.580,9402 13.059,65532 n-C4H10 801,6752 1.727,5748 1.384,953873





298

∫

323

298

13 dTCpF

n-C4H8 34,8265 3.355,4794 116,859603 n-C4H10 2.753,5457 3.661,8641 10.083,11015 C6H12 697,1021 4.617,3295 3.218,750091



C3H8 418,2613 18.773,1 7.852,061211 i-C4H8 975,9429 22.080,6 21.549,4048 n-C4H8 8.260,6890 22.050,6 182.153,1489 n-C4H10 801,6752 22.416,0 17.970,35128



QB = Q11 + Q13 + QC – Q9


19.519,83176 Kj/Jam

QB = 240.069,8296 Kj/Jam


m =

−∆ )1525( 00 CCH

q

=

− )9,41(

62229.524,96)9,628,104( 62229.524,96

= 5.477,922821 Kg/Jam


m = )135150( 00


reboiler

CCHq

−∆

= 2.174,8

96240.069,82)9,5707,745.2(

37479.996,26=

−

= 110,387083 Kg/Jam


LB.12 Cooler (E-110)

Fungsi : Menurunkan temperatur bahan yang keluar dari reboiler

destilasi II sebelum dimasukkan kedalam tangki produk

butana.




Maka :



• Kondisi Masuk:





• Kondisi Keluar:





298,15

303

BP(g)(l)i

+∆+∫ ∫∑

30 oC P= 5,066 bar

50 oC P= 5,066 bar

13 14 n-C4H8 n-C4H10 C6H12

n-C4H8 n-C4H10 C6H12 Air pendingin bekas 25 oC

P = 1 bar





l∫298

)( ∆Hvl dTCp∫

303

298

∫

303

298

14 dTCpF

n-C4H8 34,8265 - - 430,9289 15,007745 n-C4H10 2.753,5457 - - 190,5042 524,562020 C6H12 697,1021 81,1413 56,563770

Total 596,133535 Kj/Jam Neraca Energi Total Sistem :

dtdQ = QOut– QIn

= [596,133535 - 13.418,71984] kJ/jam

= -12.822,58631 kJ/jam


m =

−∆ )1525( 00 CCH

q

=

− )9,41(

3112.822,586)9,628,104(

725.645,384

= 306,028312 Kg/Jam

LB.13 Heater 03 ( E-108 )




P = 4,8 bar

11 12

Air condensat (T=1350C)

P = 4,8 bar

C3H8 n-C4H8 i- C4H8 n-C4H10

C3H8 n-C4H8 i- C4H8 n-C4H10

T= 37 0C P= 4,559 bar

T= 60 0C P= 4,559 bar



dTdQ

jjout in



Panas masuk pada alur 7: Qin = ∫∑310

298i dT Cp N



Maka :


• Kondisi Masuk:




- Panas masuk = 16.645,97535 Kj/Jam

• Kondisi Keluar:





298

333

BP(g)(l)i

+∆+∫ ∫∑

Tabel LB.24 Panas Keluar Heater (E-108) Alur 12


298

∫

333

298

12 dTCpF

C3H8 418,2613 6.440,3380 2.693,744144 i-C4H8 975,9429 3.252,9157 3.174,659982 n-C4H8 8.260,6890 3.207,6491 26.497,39164 n-C4H10 801,6752 1.106,7300 887,237994


dtdQ = QOut– QIn

= [33.253,03376 - 16.645,97535] kJ/jam

= 16.607,05841 kJ/jam



m = )135150( 00


reboiler

CCHq

−∆

= 2.174,8

4116.607,058)9,5707,745.2(

4116.607,058=

−

= 7,636131 Kg/Jam



Destilasi 04T= 90 oC

P= 12 atm

F18C3H8i-C4H8n-C4H8T = 30 0CP = 12 atmF12

C3H8i-C4H8n-C4H8

n-C4H10T = 60 0CP = 12 atm

F19C3H8i-C4H8N-C4H8N-C4H10T = 80 0CP = 12 atm

Neraca Panas Total

Q12 = Q18 + Q19

Q = ∑m. Cp. Δt

Panas masuk = panas yang keluar dari heater E-108




298)(

∫

303

298)(

18 dTCpF g

C3H8 371,123908 830,7164 308,298716 i-C4H8 315,755328 447,8453 141,409539 n-C4H8 338,438135 430,9289 145,842773

Total 595,551028 Kj/Jam




298

∫

353

298

19 dTCpF

C3H8 47,137392 10.808,7938 509,498350 i-C4H8 660,187572 7.052,0581 4.655,681115 n-C4H8 7.922,250865 5.180,2648 41.039,35729 n-C4H10 801,6752 1.800,5356 1.443,444737



C3H8 371,123908 18.773,1 6.967,146237 i-C4H8 315,755328 22.080,6 6.972,067095 n-C4H8 338,438135 22.050,6 7.462,76394


Panas yang dibutuhkan pada Reboiler adalah :

QB = Q18 + Q19 + QC – Q9

QB = 595,551028 Kj/Jam + 47.647,98149 Kj/Jam + 21.401,97727 Kj/Jam –

33.253,03376 Kj/Jam

QB = 36.392,47603 KJ/Jam


m =

−∆ )1525( 00 CCH

q

=

− )9,41(

2721.401,977)9,628,104( 2721.401,977

= 510,787047 Kg/Jam


m = )135150( 00


reboiler

CCHq

−∆

= 2.174,8

0336.392,476)9,5707,745.2(

272.787,455=

−

= 16,7337 Kg/Jam


LAMPIRAN C

PERHITUNGAN SPESIFIKASI PERALATAN

LC.1 Tangki Bahan Baku Propena - 01 (TK-102)

Fungsi : Untuk penyimpanan bahan baku Propena selama 30 hari

Jumlah : 3 unit

Bentuk : Tangki berbentuk silinder vertical dengan alas dan tutup ellipsiodal

Bahan : Carbon Steel, SA-285 (Brownell, 1959)

Kondisi operasi : -Temperatur = 300C

-Tekanan = 13 atm = 191,0473 psi

Perhitungan:

a. Volume Tangki

Kebutuhan propena = 3.124,6799 kg/jam x 24 jam/hari (Lampiran A)

= 74.992,3176 kg/hari

Kebutuhan 30 hari = 74.992,3176 kg/hari x 30 hari

= 2.249.769,5280 Kg

Densitas propena (ρ) = 0,614 kg/dm3 x 1000 dm3/m3

= 614 kg/m3

Volume propena 31197,664.3614

52802.249.769,3

mkgm

mkg

==ρ

Volume larutan untuk 1 tangki = 1197,664.3 / 3 = 1.221,3732 m3

Faktor kelonggaran (fk) = 20 % (Brownell & Young, 1959)

Volume tangki, VT = (1 + 0,2) x 2.442,7465 m3

= 1.465,6479 m3

b. Diameter dan Tinggi Shell

Volume silinder tangki (Vs)

Vs = 4

2 HsxDtxπ (Brownell & Young, 1959)

Dimana : Vs = Volume silinder (m3)

Dt = Diameter tangki (m)

Hs = Tinggi tangki silinder (m)


Direncanakan perbandingan tinggi tangki dengan diameter tangki

Hs : Dt = 5 : 4 , Maka :

Vs = 41 π Dt

2 Hs (Hs : Dt = 5 : 4)

Vs = 165 π Dt3

= 0,9812 Dt3

Vh = 24π Dt3

= 0,13083 Dt3

Volume tangki (Vt)

Vt = Vs + Vh

1.465,6479 = 0,9812 Dt3 + 0,13083 Dt3

1.465,6479 = 1,11203 Dt3

Dt = 10,9640 m = 431,6522 in

r = ½ x Dt = ½ x (10,9640) = 5,482 m = 215,8261 in

Tinggi silinder (Hs) :

Hs = 45 x Dt =

45 x 10,9640 m = 13,705 m = 539,5653 in

Tinggi cairan dalam tangki (Hc)

Volume tangki (Vt) = 1.465,6479 m3

Volume gas (Vc) = 1.221,3732 m3

Tinggi silinder (Hs) = 13,705 m

Tinggi bahan dalam tangki (Hc) = silindervolume

silindertinggixcairanVolume

=6479,465.1

13,705 1.221,3732 x

= 11,4208 m = 449,6364 in

c. Tebal shell tangki

Direncanakan menggunakan bahan konstruksi Carbon Steel SA –285 Grade C

(Timmerhaus,2004)

- Allowable stress (S) = 13.700 psia

- Joint efficiency (E) = 0,8

- Corrosion allowance (C) = 0,125 in/tahun

- Umur tangki (n) = 10 tahun


Tekanan operasi (Po) = 30 atm

= 191,0473 psi

Pdesign = (1,2) × (191,0473 psi) = 229,2567 psi

Tebal shell tangki: (Walas dkk, 2005)

in 5215,3

in)125,0(1067psi)0,6(229,25psi)(0,8) 2(13.700

in) (215,8261 psi) (229,2567

nC0,6P2SE

PDt

=

+

−

=

+−

=

x

Tebal shell standar yang digunakan = 4 in (Brownell, 1959)

d. Tebal tutup tangki

Tutup atas tangki terbuat dari bahan yang sama dengan shell.

Tebal tutup atas yang digunakan = 4 in

LC.2 Tangki Bahan Baku Butena - 01 (TK-101)

Fungsi : Untuk penyimpanan bahan baku Butena selama 30 hari

Jumlah : 3 unit

Bentuk : Tangki berbentuk silinder vertical dengan alas dan tutup elliopsidal


Kondisi operasi : -Temperatur = 300C

-Tekanan = 3 atm

= 44,0878 psi

Perhitungan:

a. Volume Tangki

Kebutuhan butena = 11.842,7565 kg/jam x 24 jam/hari (Lampiran A)

= 284.226,156 kg/hari


= 8.526.784,68 Kg

Densitas butena (ρ) = 0,630 kg/dm3 x 1000 dm3/m3

= 630 kg/m3

Volume butena 35788,534.13630

688.526.784,3

mkgm

mkg

==ρ


Volume larutan untuk 1 tangki = 5788,534.13 / 4 = 3.383,6447 m3


Volume tangki, VT = (1 + 0,2) x 3.383,6447 m3

= 4.060,3736 m3



Vs = 4






Hs : Dt = 5 : 4 , Maka :

Vs = 41 π Dt

2 Hs (Hs : Dt = 5 : 4)

Vs = 165 π Dt3

= 0,9812 Dt3

Vh = 24π Dt3

= 0,13083 Dt3

Volume tangki (Vt)

Vt = Vs + Vh

4.060,3736 = 0,9812Dt3 + 0,13083Dt3

4.060,3736 = 1,11203 Dt3

Dt = 15,3986 m = 606,2423 in

r = ½ x Dt = ½ x (15,3986) = 7,6993 m = 303,1211 in


Hs = 45 x Dt =

45 x 15,3986m = 19,2482 m = 757,8009in


Volume tangki (Vt) = 4.060,3736m3

Volume gas (Vc) = 3.383,6447 m3

LC-1



Tinggi gas dalam tangki (Hc) = silindervolume


=4.060,3736

19,2482 3.383,6447 x

= 16,0402 m = 631.502in



(Timmerhaus,2004)



- Corrosion allowance (C) = 0,125 in/tahun


Tekanan operasi (Po) = 58,78 psi

Pdesign = (1,2 × 58,784 psi) = 70,536 psi

Tebal shell tangki:

in 3721,1

in)125,0(10psi) 0,6(70,536psi)(0,8) 2(13.700

in) (303,1211 psi) (70,536

nC0,6P2SE

Prt

=

+

−

=

+−

=

x

Tebal shell standar yang digunakan = 2 in (Brownell,1959)



Tebal tutup atas yang digunakan = 2 in

LC.3 Pompa Bahan (L-101)

Fungsi : Memompa bahan dari inline mixing ke heater (E-101)

Jenis : Pompa sentrifugal

Jumlah : 1 unit

Kondisi operasi :

Tekanan = 9,8 atm

Temperatur = 30 oC


Laju alir massa (F) = 24.398,5749 kg/jam

Tabel LC.1 Data pada alur 3

Komponen Fraksi Berat Densitas

ρ campuran Viskositas

μ campuran

C3H6 0,113 614 69,382 0,0437 0,00493 C3H8 0,017 501 9,945 0,0373 0,00063 n-C4H8 0,68 630 428,4 0,1096 0,07452 i-C4H8 0,04 635 25,4 0,1126 0,00450 n-C4H10 0,15 573 86,85 0,03216 0,00482 619,977 0,08940

(Sumber: Geankoplis, 2003; Perry, 1999)

Laju alir volumetrik,

mv = 3kg/m 619,977kg/jam 924.398,574

= 0,010931 m3/s = 0,3860246 ft3/s Desain pipa:

Di,opt = 0,363 (mv)0,45(ρ)0,13 (Timmerhaus, 2004)

= 0,363 (0,010931 m3/s)0,45 (619,977 kg/m3)0,13

= 0,1097297 m = 4,32006 in

Dari Tabel A.5-1 Geankoplis (2003), dipilih pipa dengan spesifikasi:

Ukuran nominal : 5 in

Schedule number : 40

Diameter Dalam (ID) : 5,047 in = 0,4205 ft

Diameter Luar (OD) : 5,563 in = 0,4635 ft

Inside sectional area : 0,139 ft2

Kecepatan linier, v = AQ = 2

3

ft 0,139/sft 0,3860246 = 2,7772 ft/s

Bilangan Reynold:

NRe = μ

Dvρ ××

= lbm/ft.s6.00743.10

ft) 05ft/s)(0,42 )(2,7772lbm/ft(619,9775-

3

= 752.558,3548 (Turbulen)


Untuk pipa Commercial Steel diperoleh harga ε = 0,000046 (Geankoplis, 2003)

Pada NRe = 752.558,3548 dan ε/D =m0,15405m0,000046 = 0,000359

Dari Gambar 2.10-3 Geankoplis (2003) diperoleh harga f = 0,0045 Friction loss:

1 Sharp edge entrance: hc = 0,55α2

12

1

2 vAA

− =

)174,32)(1(2 2,7772)01(55,0

2

−

= 0,0659 ft.lbf/lbm

2 elbow 90o: hf = n.Kf.cg

v.2

2

= 2(0,75))174,32(2

2,7772 2

= 0,17979 ft.lbf/lbm

1 check valve: hf = n.Kf.cg

v.2

2

= 1(2))174,32(2

2,7772 2

= 0,2397 ft.lbf/lbm

Pipa lurus 60 ft: Ff = 4fcgD

vL.2.. 2∆ = 4(0,0045) ( )

( ) ( )174,32.2. 0,50541 2,7772.60 2

= 0.307793 ft.lbf/lbm

1 Sharp edge exit: hex = ncg

vAA

..21

22

2

1

α

− = 1 ( ) ( )( )174,3212

2,7772012

2−

= 0,1198 ft.lbf/lbm

Total friction loss: ∑ F = 0.913103 ft.lbf/lbm

Dari persamaan Bernoulli:

( ) ( ) 021 12

122

12

2 =+∑+−

+−+− sWFPPzzgvvρα

(Geankoplis, 2003)

dimana: v1 = v2

P1 = 9,8 atm

P2 = 9 atm

∆P = 0,8 atm = 1.692,973 lbf/ft2

tinggi pemompaan ∆Z = 30 ft

maka : ( ) 0W 0,8789 138.7049145

1.692,973 3032,17432,1740 s =++++

Ws = - 74,65387 ft.lbf/lbm Efisiensi pompa, η= 75 %,

Wp = -η

Ws (Geankoplis, 2003)


= -0,75

74,65387-

= 99,53849 ft.lbf/lbm. Daya pompa: P = m × Wp

= ( )( ) ft.lbf/lbm 99,53849lbm/s/s.hp)(550ft.lbf36000,45359

14.94142×

= 2.704084 hp

Maka dipilih pompa dengan daya motor 3 hp.

LC.4 Heater 01 (E-101)

Fungsi : Menaikkan temperatur bahan yang keluar dari mixer sebelum

dimasukkan kedalam reaktor.

Jenis : Double Pipe Heat Exchanger

Jumlah : 1 unit

Fluida Panas : Saturated Steam

Flowrate, W = 8,7506 Kg/jam (Lampiran B)

= 8,7506 Kg/jam x 2,2046 lb/kg

= 19,2915 lb/jam

T1 = 150 0C

= (150 0C x 1,8) + 32

= 302 0F

T2 = 135 0C

= (135 0C x 1,8) + 32

= 275 0F

Fluida dingin : Senyawa gas

Flowrate, W = 24.398,5749 kg/jam (Lampiran A)

= 24.398,5749 kg/jam x 2,2046 lb/kg

= 53.789,0982 lb/jam

t1 = 30 0C

= (30 0C x 1,8) + 32

= 86 0F


t2 = 40 0C

= (40 0C x 1,8) + 32

= 104 0F

Perhitungan design sesuai dengan literatur Kern

Panas yang diserap (Q) = 19.030,88941 kJ/jam = 18.037,7319 Btu/jam

(1) ∆t = beda suhu sebenarnya

Fluida Panas Fluida Dingin Selisih T1 = 302 0F Temperatur yang lebih tinggi t2 = 104 0F ∆t1 = 198°F T2 = 275 0F Temperatur yang lebih rendah t1 = 86 0F ∆t2 = 189°F

T1 – T2 = 27 °F Selisih t2 – t1 = 18 °F ∆t2 – ∆t1 = -9 °F

193.4651

198189ln

9-

ΔtΔtln

ΔtΔtLMTD

1

2

12 =

=

−

= °F

(2) Tc dan tc

=+

=+

=2

2753022

TTT 21c 288,5 °F

=+

=+

=2

861042

ttt 21c 95 °F

(3) Dari Tabel 8, hal. 840, Kern, 1965, kondensor untuk fluida panas gas dan

fluida dingin air, diperoleh UD = 2-50, dan faktor pengotor (Rd) = 0,003.

Diambil UD = 50 Btu/jam⋅ft2⋅°F

Luas permukaan untuk perpindahan panas,

2

oo2

D

ft8647,1F4651,193

FftjamBtu50

Btu/jam 918.037,731ΔtU

QA =×

⋅⋅

=×

=

Fluida panas : steam, annulus

(1) Flow Area (aa)

Dari tabel 11 hal. 844 (Kern, 1965) diperoleh :

D1 = 1,66 in = 0,138 ft ; pipa bagian luar (OD)

D2 = 2,067 in = 0,172 ft ; pipa bagian dalam (ID)

4

).( 21

22 DD

aa−

=π


00834,04

)138,0172,0.( 22

=−

=π

aa

Diameter Ekivalen (De)

077,0)(

1

21

22 =−

=D

DDDe

(2) Massa Velocity (Ga)

2./1294,313.200834,0

19,2915 ftjamlbmawG

aa ===

(3) Bilangan Reynold:

Pada Tc = 288,5 °F

µ = 0,023 cP = 0,0556 lbm/ft2⋅jam (Gambar 14, Kern, 1965)

µ

aeea

GDR ×=

4346,203.3

0556,0 1294,313.2 077,0

=×

=eaR

(4) Taksir jH dari Gambar 24 Kern (1965), diperoleh jH = 12 pada Rea = 4346,203.3

(5) Pada Tc = 288,5 °F

c = 0,45 Btu/lbm.°F (Gambar 2, Kern, 1965)

k = 0,0156 Btu/jam lbm ft.°F (tabel 5, Kern, 1965)

=

×

=

3

13

1

0156,00556,045,0.

kc µ 1,1709

114,0

=

=

weD

µµ

8466,211709,1

077,00156,012.

14,03

1

=×××=

×

××=

we kc

DkjHho

µµµ

Fluida dingin : pipa dalam

(1) Flow area (aa)

Dari table 11 hal. 844 (kern, 1965) diperoleh :

D1 = 1,38 in = 0,115 ft


0104,04

14,3115,04

22

=×

=×

=πDaa

(2) Kecepatan massa

a

p awG =

673,028.172.50104,0

lb/jam 253.789,098==pG lbm/jam.ft2

(3) Bilangan Reynold

Pada tc = 95 °F

µ = 0,257 cP = 0,62194 lbm/ft2⋅jam

µ

pa

GD ×=Re

896,331.314.2257,0

673,028.172.50,115Re =×

=p

(4) Taksir JH dari Gambar 24, Kern, diperoleh JH = 1000 pada Res = 896,331.314.2

(5) Pada tc = 95 °F

c = 0,35 Btu/lbm⋅°F

k = 0,0143 Btu/jam lbm ft.°F

8458,10143,0

257,035,0. 31

31

=

×=

kc µ

214,0

31

./5988,14511709,1115,00143,01000. ftjamBtu

kc

DkjHhi

we

=×××=

×

××=

µµµ

Koreksi ftjamBtuODIDhihio ./0399,121

66,138,15988,145 =×=

×=

(6) Clean Overall Coefficient, UC

F.ft.Btu/jam 7811,28466,2121,03998466,2121,0399

hhhh

U 2

oio

oioC °=

+×

=+×

=

(7) Design overall coeficient, Ud

RdUc

+=1

Ud1


FftjamBtuUd 2./7662,2

002,07811,21

Ud1

=

+=

(8) Luas yang diperlukan, A :

2t5699,479,1733615,0 918.037,731

tUdQA f=

×=

∆×=

Dari tabel 11 (Kern,1965) diperoleh :

Luas eksternal (a) = 0,435 ft2/ft

Panjang hairpins yang diperlukan = A/a = 5699,4 /0,435 = 10,5055 ft

Digunakan pipa dengan panjang 12 ft dengan 1 hairpins.

Koreksi Ud :

A = a.(48) = 0,435 x24 = 10,44 ft2

Ud baru = FfBtu 02 .tjam./9416,979,17344,10918.037,731

tAQ

=×

=∆×

00193,07662,27811,27662,27811,2

=×−

=×−

=UdUcUdUcRd

Pressure drop

Fluida panas : Anullus

(1) De = D2 – D1 = 0,172 – 0,138 = 0,03455 ft

7226,474.3023,0

1294,313.203455,0GaDe'' =×

=×

=µeaR

s = 1, ρ = 62,5 x 1 = 62,5

0120,00,2640035,0 42,0 =+=eaR

f

(2)

eD

L

'.2810.18,422

. 2aG 4.f.

Faρ××

=∆

ft0000501,003455,025,62810.18,422

42 21294,313.2 0120,04 Fa =×××

×××=∆

(3) sft /0102,05,623600

1294,313.23600

aG V =

×=

×=

ρ


ftcg

00000483,03,322

20102,032

2V3 1F =×

×=

××=

( ) psiaFFa 0000238,0144

5,62)00000483,00000501,0(144

1 Pa =×+

=×+∆

=∆ρ

∆Pa yang diperbolehkan = 2 psia

Fluida dingin : inner pipe

(1) 0035,00,2640035,0 42,0 =+=eaR

f

s = 0,6045, ρ = 62,5 x 0,6045 = 37,7812

(2) D

L

.2810.18,42

. 2pG 4.f.

Fpρ××

=∆

ft0065,0115,027812,37810.18,42

42 2673,028.172.5 0,00354 Fp =×××

×××=∆

psia00170,0144

FppΔP =

×=

ρ

∆Pp yang diperbolehkan = 2 psia

LC.5 Reaktor ( R-101 )

Fungsi : Tempat berlangsungnya reaksi pembentukan Heptena

Jenis : Reaktor fixed bed multitubular

Bentuk : silinder vertikal dengan alas dan tutup ellipsoidal

Bahan konstruksi : carbon steel SA-285 grade A

Jumlah : 1 unit

Temperatur masuk = 40 oC = 313 K

Temperatur keluar = 40 oC = 313 K

Tekanan operasi = 8,5 atm = 861,2625 kPa

Laju alir massa = 24.398,5749 kg/jam

Waktu tinggal (τ) reactor = 0,00138 jam = 5 detik


Perhitungan

Desain Tangki


mv = 3kg/m 619,977

kg/jam 824.398,574

= 0,0109 m3/s = 0,3849 ft3/s


3

ft 0,0232/sft 0,3849 = 16,5905 ft/s

Bilangan Reynold:

NRe = μ

Dvρ ××

= lbm/ft.s6.00743.10

ft) 85ft/s)(1,31 )(16,5905lbm/ft(619,9775-

3

= 225.750.456,4 (Turbulen)

Cao = Vn =

)313)(/ .314,8(2625,861.3 KmolKmPa

kPaRTPi = = 0,3309 kmol/m3

a. Volume reaktor

V = 33

1

m 8907,1kmol/m 0,3309

)kmol/jam3623,534.(jam00138,0 ==

−

AO

AO

CFτ

Setiap 1 m3 katalis dapat menghasilkan sekitar 15.762,2917 kg heptena per

jam.

Heptena yang dihasilkan dalam reaktor adalah 5.437,3967 kg/jam. Maka

jumlah katalis yang dibutuhkan adalah : 715.762,291

5.437,3967 = 0,34 m3

Volume total reaktor = 8907,1 +0,34 = 2,2307 m3

Faktor kelonggaran = 20%

Volume reaktor = 1,2 x 2,2307 = 2,768 m3

b. Jumlah tube

Direncanakan:

Diameter tube = 0,635/0,15 = 4,233 cm= 0,04233 m= 1,666 in

Panjang tube = 7,2 m


Pitch (PT) = 2,975 triangular pitch

Jumlah tube = 2,7.

40,04233.

2,768tubeV

reaktorV2

π= = 273

c. Tebal tube

Tekanan operasi = 8,5 atm = 124,95 psi

Faktor kelonggaran = 5 %

Maka, Pdesain = (1,05) (124,95) = 131,1975 psi

Joint efficiency = 0,8 (Brownell,1959)

Allowable stress = 11200 psia (Brownell,1959)

psi) 750,6(131,19psi)(0,8) (11200

in 2

1,666si)(131,1975p

P6,0SEPRt

−

=

−=

= 0,01230 in

Faktor korosi = 0,0098 in/tahun

Umur alat = 1 tahun

Maka tebal tube yang dibutuhkan = 0,01230 in + (0,0098)(1) in = 0,0221 in

Tebal tube standar yang digunakan = 1/4 in (Brownell,1959)

d. Diameter dan tinggi shell

q PT = jarak antara 2 pusat pipa PT

= 1,25 OD = 0,053 m

C’ = Clearance = PT-OD

= 0,01 m = 0,39 inchi

CD = PT sin 60O

IDs = 5,02866,04

⋅⋅⋅π

TPNt

= 5,02053,0561866,04

⋅⋅⋅π

= 1,3185 m = 51,9094 in

PT

C'

60o

60o

60o A B

C

D


Tinggi shell (H) = panjang tube = 7,2 m

e. Diameter dan tinggi tutup

Diameter tutup = diameter tangki = 1,3185 m

Rasio axis = 2 : 1 (Brownell,1959)

Tinggi tutup =

2 1,3185

21 = 0,3296 m

f. Tebal shell dan tebal tutup

Tekanan operasi = 8,5 atm = 124,95 psi


Maka, Pdesain = (1,05) (124,95) = 131,1975 psi



psi) 750,6(131,19psi)(0,8) (11200

in 2

51,9094si)(131,1975p

P6,0SEPRt

−

=

−=

= 0,3834 in

Faktor korosi = 0,0098in/tahun

Umur alat = 1tahun

Maka tebal shell yang dibutuhkan = 0,3834 in + (0,0098)(1) in = 0,3932 in

Tebal shell standar yang digunakan = 1/2 in (Brownell,1959)

Tutup atas tangki terbuat dari bahan yang sama dengan shell. Maka tebal

tutup atas yang digunakan = 1/2 in

LC.6 Pompa bahan ( L-102 )

Fungsi : Memompa bahan dari reaktor ke menara destilasi I (MD-101)


Jumlah : 1 unit

Kondisi operasi :

Tekanan = 8,5 atm

Temperatur = 40 oC






μ campuran

C3H8 0,0171 501 8,5671 0,04197 0,00071

n-C4H8 0,34 630 214,3 0,1186 0,04032 i-C4H8 0,04 630 25,2 0,1297 0,00518 n-C4H10 0,15 573 85,95 0,01236 0,00185 C6H12 0,14 675 94,5 0,1765 0,02471 C7H14 0,2228 698 155,5144 0,1268 0,02825 C8H16 0,09 716 64,44 0,1136 0,01022 648,4715 0,11124



mv = 3kg/m648,4715kg/jam 824.398,574



= 0,363 (0,01045 m3/s)0,45 (648,4715 kg/m3)0,13

= 0.108163173 m = 4,2583 in








3

ft 0,139/sft 0,36907 = 2,6551 ft/s

Bilangan Reynold:


NRe = μ

Dvρ ××

= lbm/ft.s7,4750.10

ft) 05ft/s)(0,42 )(2,6551lbm/ft(648,47155-

3

= 604.806,8763 (Turbulen)


Pada NRe = 604.806,8763 dan ε/D =m0,15405m0,000046 = 0,0003588



12

1

2 vAA

− =

)174,32)(1(22,6551)01(55,0

2

−



v.2

2

= 2(0,75))174,32(2

2,65512



v.2

2

= 1(2))174,32(2

2,65512

= 0,2191 ft.lbf/lbm


vL.2.. 2∆ = 4(0,0036) ( )( )

( ) ( )174,32.2. 0,505412,6551.40 2

= 0,179222 ft.lbf/lbm


vAA

..21

22

2

1

α

− = 1 ( ) ( )( )174,3212

2,6551012

2−


Total friction loss: ∑ F = 0,732506 ft.lbf/lbm


( ) ( ) 021 12

122

12

2 =+∑+−


(Geankoplis, 2003)

dimana: v1 = v2

P1 = 8,5 atm

P2 = 3,7 atm

∆P = 4,8 atm = 10.157,9 lbf/ft2



maka : ( ) 0W 0,73250640,483810.157,9 10

32,17432,1740 s =++++

Ws = - 261,645 ft.lbf/lbm Efisiensi pompa, η= 75 %

Wp = -η


= -0,75

261,645-



824.398,574×

= 9,477 hp

Maka dipilih pompa dengan daya motor 10 hp.

LC.7 Kolom Destilasi 1 (MD-101)


Jenis : Sieve – tray


Bahan konstruksi : Carbon steel SA – 285 grade A

Jumlah : 1 unit

Bahan Heptena pada kondisi tekanan 3,7 atm dan temperatur 132,5 0C, adalah

berupa fasa cair dan keluar pada alur bottom. Ini dapat didapat dengan menggunakan

rumus Antoine.

CT

TKpa

CTBAP

0666,143)9388,51(

90,28958748,139025,374ln)(

ln

=

−+−==

+−=


Tabel LC 3. data fraksi berat dan harga K

Komponen Xif Xid Ki (83,250C) αi (83,250C) Xiw

C3H8 0,0171 0,03 6,1 1,24 0

i-C4H8 0,04 0,07 3,8 0,77 0

n-C4H8 0,34 0,59 3,1 0,63 0

n-C4H10 (L) 0,15 0,26 2,5 0,51 0,01

C6H12 (H) 0,14 0,05 4,9 1 0,26

C7H14 0,223 0 2,2 0,45 0,52

C8H16 0,09 0 1,1 0,22 0,21

Total 1,000 1,000 1,000


rumus:

∑ −=−

θαα

i

ifi xq1


Maka : ∑ −=−

θαα

i

ifi x11 = ∑ −=

θαα

i

ifi x0


Tabel LC 4. nilai trial erorr θ

Asumsi (θ) C3H8 i-C4H8 n-C4H8 n-C4H10 C6H12 C7H14 C8H16 Total

0.56112 0.0309 0.147 3.106 -1.496 0.318 -0.902 -0.058 1.1470

0.549 0.0303 0.139 2.641 -1.961 0.310 -1.013 -0.060 0,0873 0.5481 0.030 0.138 2.612 -2.007 0.309 -1.022 -0.060 0.00125


θαα

i

ifi x0 , sehingga trial θ = 0.5481 dapat diterima

∑ −=+

θαα

i

idiDm

xR 1

RDm + 1 = 1,465

RDm = 0,465

RD = 1,5 RDm


RD = 0,697

4107,0)1697,0(

697,0)1(

=+

=+RR

317,0)1465,0(

465,0)1(

=+

=+m

m

RR

Dari Fig 11.7 – 3, Geankoplis, untuk 4107,01=

+RR dan 317,0

1=

+m

m

RR diperoleh

NN m = 0,56; maka:

( ) ( ) 3905,11,5671,234.ααα LWLDavL, =×==

(Geankoplis,2003)

)log()]/)(/log[(

,avL

LWHWHDLDm

XXXXN

α=

(Geankoplis,2003)

)3905,1log(

)]01,0/26,0)(05,0/26,0log[=

= 14,891

N = 56,0891,14

56,0=mN = 26,59

Jumlah piring teoritis = 26,59

Efisiensi piring = 80 % (Geankoplis,2003)

Maka jumlah piring yang sebenarnya =8,0

26,59 = 33,23 piring ≈ 34 piring.

Jumlah piring total = 34 + 1 = 35 piring

Penentuan lokasi umpan masuk

=

2

log206,0logHD

LW

LF

HF

s

e

XX

DW

XX

NN

(Geankoplis,2003)

=2

05,001,0

210,512882,8023

15,014,0log 206,0log

s

e

NN

42,0=s

e

NN

Ne = 0,42 Ns


N = Ne + Ns

35 = 0,42 Ns + Ns

Ns = 24,647 ≈ 25

Ne = 35 – 25 = 10

Jadi, umpan masuk pada piring ke – 10 dari atas.

Design kolom direncanakan : Tray spacing (t) = 0,4 m

Hole diameter (do) = 4,5 mm (Treybal, 1984)

Space between hole center (p’) = 12 mm (Treybal, 1984)

Weir height (hw) = 5 cm

Pitch = triangular ¾ in

Data :

Suhu dan tekanan pada destilasi MD-101 adalah 132,5 0C dan 3,7 atm.

Tabel LC.5 Komposisi Bahan pada Alur Vd Kolom Destilasi MD-101

Komponen Alur Vd (kmol/jam) Fraksi mol BM

(g/mol) Fraksi mol

x BM C3H8 16,13 0.03 44 1,32 i-C4H8 29,57 0.07 56 3,92 n-C4H8 249,27 0,59 56 33,04 n-C4H10 106,06 0,26 58 15,08 C6H12 14,08 0,05 84 4,2 Total 415,11 kmol/Jam 1 57,56

Laju alir massa gas (G’) = 415,11 Kmol/jam

= 415,11 /3600 = 0,11 kmol/detik

ρv = K) K)(405,16atm/kmolm (0,082

kg/kmol) (57,56 atm) (3,7RT

BMP3

av = = 6,41 kg/m3

Laju alir volumetrik gas (Q) = 15,27316,4054,22 0,11 ×× = 3,65 m3/s


Tabel LC.6 Komposisi Bahan pada Alur Lb Kolom Destilasi MD-101

Komponen Alur Lb (kg/jam)

Fraksi massa

Lρ (kg/m3)

Fraksi massa x Lρ

n-C4H10 341,1617 0,01 635 6,35

C6H12 8.870,2064 0,26 675 175,5

C7H14 17.740,4128 0,52 698 362,96

C8H16 7164,3974 0,21 716 150,36

Total 34.116,1783 1 695,17

Laju alir massa cairan (L’) = 9,47 kg/s

Laju alir volumetrik cairan (q) = 17,695 9,47 = 0,0136 m3/s

Surface tension (σ) = 0,04 N/m (Lyman, 1982) 2

o

a

o

p'd

907,0AA

=

2

a

o

0,01200,0045907,0

AA

= = 0,1275

2/12/1

V

L

6,4117,695

3,65 0,0136

ρρ

Q'q

=

= 0,0388

α = 0,0744t + 0,01173 = 0,0744(0,4) + 0,01173 = 0,0415

β = 0,0304t + 0,05 = 0,0304(0,4) + 0,05 = 0,0621

CF = 2,0

2/1VL 0,02

σβ)ρ/(q/Q)(ρ

1αlog

+

= 2,0

0,020,040,0621

0,03881log 0,0415

+

= 0,1385

VF = 5,0

V

VLF ρ

ρρC

−

= 5,0

6,416,4117,6950,1385

−

= 1,4356 m/s


Asumsi 80 % kecepatan flooding (Treybal, 1984)

V = 0,8 x 1,4356 = 1,1484 m/s

An = 1,1484

3,65VQ= = 3,1783 m2

Untuk W = 0,80 T dari Tabel 6.1. Treybal, hal.162, diketahui bahwa luas

downspout sebesar 14,145%.

At = =− 14145,01

3,1783 3,7019 m2

Column Diameter (T) = [4(3,7019)/π]0,5 = 2,1715 m

Weir length (W) = 0,8(2,1715) = 1,7372 m

Downspout area (Ad) = 0,14145(3,1783) = 0,4495 m2

Active area (Aa) = At – 2Ad = 3,7019– 2(0,4495) = 2,8029 m2

Weir crest (h1)

Misalkan h1 = 0,03452 m

h1/T = 0,03452/2,1715 = 0,01589

== 1,7372 2,1715

WT 1,25

2

1

5,0222eff

WT

Th21

WT

WT

WW

+

−

−

=

( ) ( )[ ] ( )( ){ }25,0222

eff 25,10,015892125,125,1W

W+−−=

=

WWeff 0,9389

3/2eff

3/2

1 WW

Wq666,0h

=

( ) 3/23/2

1 0,9389 1,7372 0,0136 666,0h

=

h 1 = 0,02516

Perhitungan diulangi dengan memakai nilai h1 = 0,02516m hingga nilai h1

konstan pada nilai 0,02516m.


Perhitungan Pressure Drop

Dry pressure drop

Ao = 0,1275 x 2,8029 = 0,3573 m2

uo = ==0,3573

3,65AQ

o

10,2155 m/s

Co = 1,346 Dari gambar 18.29, Mc.Cabe

=

695,176,41

346,1 10,21550,51h 2

2

d

hd = 27,0875 mm = 0,0270875 m

Hydraulic head

2,80293,65

AQV

aa == = 1,302 m/s

21,7372 2,1715

2 W Tz +

=+

= = 1,9543

+−+=

zq225,1ρVh 238,0h 725,00061,0h 5,0

vawwL

+−+=

1,95430,0136225,102)(6,41)(0,05)(1,3 238,0(0,05) 725,00061,0h 5,0

L

h L = 0,01167 m

Residual pressure drop

gdρg σ 6

hoL

cR =

,8)(0,0045)(9 695,17(1) (0,04) 6h R = = 0,0078 m

Total gas pressure drop

hG = hd + hL + hR

hG = 0,0270875 +0,01167 + 0,0078

hG = 0,0465 m

Pressure loss at liquid entrance

Ada = 0,025 W = 0,025(1,7372) = 0,04343 m2 2

da2 A

qg23h

=


2

2 0,043430,0136

g23h

= = 0,015 m

Backup in downspout

h3 = hG + h2

h3 = 0,0465 + 0,015

h3 = 0,0615 m

Check on flooding

hw + h1 + h3 = 0,05 +0,02516+ 0,0615

hw + h1 + h3 = 0,1366 m

t/2 = 0,4/2 = 0,2 m

karena nilai hw + h1 + h3 lebih kecil dari t/2, maka spesifikasi ini dapat

diterima, artinya dengan rancangan plate seperti ini diharapkan tidak terjadi

flooding.

Spesifikasi kolom destilasi

Tinggi kolom = 35 x 0,4 m = 14 m

Tinggi tutup = ( ) 2,171541 = 0,5428 m

Tinggi total = 14 + 2(0,5428) = 15,0856 m

Tekanan operasi = 3,7 atm = 374,9025 kPa


P design = (1+0,05) x 374,9025 kPa = 393,6476 kPa = 57,0937 psi

Joint efficiency (E) = 0,8 (Brownell,1959)

Allowable stress (S) = 11200 psi (Brownell,1959)


Umur = 10 tahun

Tebal shell tangki:

0,6P-SEPRt =

7)0,6(57,093-8)(11200)(0,2,1715/2)(57,0937)(t =

= 0,007 in


Maka tebal shell yang dibutuhkan = 0,007 in + (0,0098).(10) in = 0,105 in


Tebal tutup tangki = tebal shell = 1/4 in







Jumlah : 1 unit

Bahan Hexena pada tekanan 1 atm dan temperatur 105,5 0C, adalah berupa

fasa gs dan keluar pada alur top. Ini dapat didapat dengan menggunakan rumus

Antoine.

CT

TKpa

CTBAP

091,62)1749,47(

34,26577987,13325,101ln)(

ln

=

−+−==

+−=



n-C4H10 0,01 0,033 110 15,27 -

C6H12 (L) 0,26 0,8433 17 2,36 0,01

C7H14 (H) 0,52 0,123 7,2 1 0,69

C8H16 0,21 0 3,6 0,5 0,3

Total 1,000 1,000 1,000


rumus:

∑ −=−

θαα

i

ifi xq1



Maka : ∑ −=−

θαα

i

ifi x11 = ∑ −=

θαα

i

ifi x0



Asumsi (θ) n-C4H10 C6H12 C7H14 C8H16 Total

1,68 0.0150 0.9023 -0.7647 -0.0889 0.0637

1,67 0.01501 0.8892 -0.7761 -0.0897 0.0384 1.66 0.0112 0.876 -0.7878 -0.090 0.00939


θαα

i


∑ −=+

θαα

i

idiDm

xR 1

RDm + 1 = 1,321

RDm = 0,321

RD = 1,5 RDm

RD = 0,4815

3250,0)14815,0(

4815,0)1(

=+

=+RR

2429,0)1321,0(

321,0)1(

=+

=+m

m

RR


+RR dan 2429,0

1=

+m

m

RR diperoleh

NN m = 0,44 ; maka:

( ) ( ) 7824,01,570,39.ααα LWLDavL, =×== (Geankoplis,2003)

)log()]/)(/log[(

,avL

LWHWHDLDm

XXXXNα

= (Geankoplis,2003)

)7824,0log(

)]01,0/69,0)(123,0/8433,0log[=

= 9,4150

N = 44,0

415,944,0

=mN = 21,39





21,39 = 26,73 piring ≈ 27 piring.



=

2

log206,0logHD

LW

LF

HF

s

e

XX

DW

XX

NN

=

2

123,001,0

74,4895144,0261

26,052,0log 206,0log

s

e

NN (Geankoplis,2003)

7879,0=s

e

NN

Ne = 0,7879 Ns

N = Ne + Ns

28 = 0,7879 Ns + Ns

Ns = 15,66 ≈ 16

Ne = 28 – 16 = 12







Data :

Suhu dan tekanan pada destilasi MD-103 adalah 378 K dan 1 atm.




(g/mol) Fraksi mol

x BM n-C4H10 2,6442 0,047 58 2,726

C6H12 46,6566 0,8462 84 71,0808

C7H14 5,8329 0,1057 98 10,3586

Total 55,1337 kmol/Jam 1 84,1654 Laju alir massa gas (G’) = 55,1337 Kmol/jam

= 55,1337 /3600 = 0,0153 kmol/detik

ρv = K) K)(336,65atm/kmolm (0,082

kg/kmol) (84,1654 atm) (1RT

BMP3

av = = 3,048 kg/m3




Fraksi massa

Lρ (kg/m3)

Fraksi massa x Lρ

C6H12 119,6697 0,01 675 6,75

C7H14 8.257,2148 0,69 698 481,62

C8H16 3.590,0934 0,3 716 214,3

Total 11.966,9779 Kg/Jam 1 703,17 Laju alir massa cairan (L’) = 3,3241 kg/s

Laju alir volumetrik cairan (q) = 17,703

3,3241 = 0,00472 m3/s


o

a

o

p'd

907,0AA

=

2

a

o

0,01200,0045907,0

AA

= = 0,1275

2/12/1

V

L

3,04817,703

0,4223 0,00472

ρρ

Q'q

=

= 0,1697


α = 0,0744t + 0,01173 = 0,0744(0,4) + 0,01173 = 0,0415

β = 0,0304t + 0,05 = 0,0304(0,4) + 0,05 = 0,0621

CF = 2,0

2/1VL 0,02

σβ)ρ/(q/Q)(ρ

1αlog

+

= 2,0

0,020,040,0621

0,16971log 0,0415

+

= 0,1079

VF = 5,0

V

VLF ρ

ρρC

−

= 5,0

3,0483,04817,7030,1079

−

= 1,6353 m/s


V = 0,8 x 1,6353 = 1,3082 m/s

An = 1,3082

0,4223VQ= = 0,3228 m2

Untuk W = 0,80T dari Tabel 6.1. Treybal, hal.162, diketahui bahwa luas


At = =− 14145,01

0,3228 0,3759 m2


Weir length (W) = 0,8(0,6919) = 0,5535 m


Active area (Aa) = At – 2Ad = 0,3759 – 2(0,0531) = 0,2697 m2

Weir crest (h1)


h1/T = 0,025/0,6919 = 0,03613

==0,55350,6919

WT 1,25


2

1

5,0222eff

WT

Th21

WT

WT

WW

+

−

−

=

( ) ( )[ ] ( )( ){ }25,0222

eff 25,10,036132125,125,1W

W+−−=

=

WWeff 0,8569

3/2eff

3/2

1 WW

Wq666,0h

=

( ) 3/23/2

1 0,8569 0,5535

0,00472666,0h

=

h 1 = 0,0250 Perhitungan diulangi dengan memakai nilai h1 = 0,0250m hingga nilai h1



Dry pressure drop

Ao = 0,1275 x 0,2697 = 0,0343 m2

uo = == 0,0343 0,4223

AQ

o

12,3119 m/s


=

L

v2

o

2o

d ρρ

Cu

0,51h

=

703,173,048

35,13119,120,51h 2

2

d

hd = 18,3868 mm = 0,01838 m

Hydraulic head

0,26970,4223

AQV

aa == = 1,5658 m/s

2 0,5535 0,6919

2 W Tz +

=+

= = 0,6227


+−+=

zq225,1ρVh 238,0h 725,00061,0h 5,0

vawwL

+−+=

0,6227 0,00472225,1)(3,048) 658(0,05)(1,5 238,0(0,05) 725,00061,0h 5,0

L

h L = 0,0191 m


gdρg σ 6

hoL

cR =

,8)(0,0045)(9 703,17(1) (0,04) 6hR = = 0,0077 m


hG = hd + hL + hR

hG = 0,01838 + 0,0191 + 0,0077

hG = 0,0451 m


Ada = 0,025 W = 0,025( 0,5535) = 0,01383 m2 2

da2 A

qg23h

=

2

2 0,013830,00472

g23h

= = 0,0522 m

Backup in downspout

h3 = hG + h2

h3 = 0,0451 + 0,0522

h3 = 0,0973 m

Check on flooding

hw + h1 + h3 = 0,05 +0,0250+ 0,0973

hw + h1 + h3 = 0,1723 m

t/2 = 0,4/2 = 0,2 m




flooding.


Tinggi kolom = 28 x 0,4 m = 11,2 m

Tinggi tutup = ( ) 0,691941 = 0,1729 m

Tinggi total = 11,2 + 2(0,1729) = 11,5458 m

Tekanan operasi = 1 atm = 101,325 kPa


P design = (1+0,05) x101,325 kPa = 106,3912 kPa = 15,4267 psi




Umur = 10 tahun

Tebal shell tangki:

0,6P-SEPRt =

7)0,6(15,426-8)(11200)(0,0,6919/2)(15,4267)(t =

= 0,000596 in






samping Oktena




Jumlah : 1 Unit


Bahan Oktena pada tekanan 10 atm dan temperatur 74,5 0C, adalah berupa

fasa cair dan keluar pada alur bottom. Ini dapat didapat dengan menggunakan rumus

Antoine.

CT

TKpa

CTBAP

0526,131)39,60(

52,31169630,1525,1013ln)(

ln

=

−+−==

+−=



C6H12 0,01 0,013 0,95 7,14 -

C7H14 (L) 0,69 0,887 0,38 2,85 0,03

C8H16 (H) 0,3 0,01 0,197 1 0,97

Total 1,000 1,000 1,000


rumus:

∑ −=−

θαα

i

ifi xq1


Maka : ∑ −=−

θαα

i

ifi x11 = ∑ −=

θαα

i

ifi x0



Asumsi (θ) C6H12 C7H14 C8H16 Total

1.244 0.012109 1.22447 -1.22950 0.007072 1.2435 0.012108 1.224089 -1.23203 0.004165 1.243 0.012107 1.22370 -1.23456 0.001248


θαα

i


∑ −=+

θαα

i

idiDm

xR 1

RDm + 1 = 1,438


RDm = 0,438

RD = 1,5 RDm

RD = 0,657

396,0)1657,0(

657,0)1(

=+

=+RR

3045,0)1438,0(

438,0)1(

=+

=+m

m

RR


+RR dan 3045,0

1=

+m

m

RR diperoleh

NN m = 0,55; maka:


)log()]/)(/log[(

,avL

LWHWHDLDm

XXXXN

α= (Geankoplis,2003)

)228,1log(

)]03,0/97,0)(01,0/887,0log[=

= 4,921

N = 55,0921,4

55,0=mN = 8,947




8,947 = 11,184 piring ≈ 12 piring.



=

2

log206,0logHD

LW

LF

HF

s

e

XX

DW

XX

NN

(Geankoplis,2003)

=

2

97,003,0

113,831430,1913

69,03,0log 206,0log

s

e

NN

15,0=s

e

NN

Ne = 0,15 Ns


N = Ne + Ns

13 = 0,15 Ns + Ns

Ns = 11,3≈12

Ne = 13 – 12 = 1







Data :

Suhu dan tekanan pada destilasi MD-305 adalah 347,65 K dan 10 atm.


Komponen Alur Vd

(kmol/jam) Fraksi mol BM

(g/mol) Fraksi mol

x BM C6H12 1,4453 0,013 84 1,092 C7H14 84,5278 0,97 98 86,926 C8H16 8,3384 0,01 112 1,12 Total 94,3115 kmol/Jam 1 89,138


= 94,3115 /3600 = 0,0261 kmol/detik

ρv = )K)(347,65K atm/kmolm (0,082

kg/kmol) (89,138 atm) (10RTBM P

3av = = 31,268 kg/m3





Fraksi massa

Lρ (kg/m3)

Fraksi massa x Lρ

C7H14 330,6749 0,03

698 20,94 C8H16 10.691,8232

0,97 716 694,52

Total 11.022,4981 Kg/Jam 1,000 715,46

Laju alir massa cairan (L’) = 3,0618kg/s


3,0618kg = 0,0042 m3/s


o

a

o

p'd

907,0AA

=

2

a

o

0,01200,0045907,0

AA

= = 0,1275

2/12/1

V

L

31,26846,715

0,74400,0042

ρρ

Q'q

=

= 0,027

α = 0,0744t + 0,01173 = 0,0744(0,4) + 0,01173 = 0,0415

β = 0,0304t + 0,05 = 0,0304(0,4) + 0,05 = 0,0621

CF = 2,0

2/1VL 0,02

σβ)ρ/(q/Q)(ρ

1αlog

+

= 2,0

0,020,040,0621

0,0271log 0,0415

+

= 0,1459

VF = 5,0

V

VLF ρ

ρρC

−

= 5,0

31,268268,1346,7150,1459

−

= 0,6824 m/s


V = 0,8 x 0,6824 = 0,5459 m/s


An = 0,5459

0,7440VQ= = 1,3628 m2



At = =− 14145,011,3628 1,5873 m2


Weir length (W) = 0,8(1,4219) = 1,1375 m

Downspout area (Ad) = 0,14145(1,5844) = 0,2241m2


Weir crest (h1)


h1/T = 0,034/1,4219 = 0,0239

==1,13751,4219

WT 1,25

2

1

5,0222eff

WT

Th21

WT

WT

WW

+

−

−

=

( ) ( )[ ] ( )( ){ }25,0222

eff 25,10,02392125,125,1W

W+−−=

=

WWeff 0,9069

3/2eff

3/2

1 WW

Wq666,0h

=

( ) 3/23/2

1 0,90691,13640,0042666,0h

=

h 1 = 0,01492 Perhitungan diulangi dengan memakai nilai h1 = 0,01492 m hingga nilai h1

konstan pada nilai 0,01492 m.


Dry pressure drop

Ao = 0,1275 x 1,136 = 0,14484 m2


uo = ==0,14484

0,7440AQ

o

5,1367 m/s


=

L

v2

o

2o

d ρρ

Cu

0,51h

=

715,46268,13

35,11367,50,51h 2

2

d

hd = 32,2678 mm = 0,03226 m

Hydraulic head

1,1360,7440

AQV

aa == = 0,6549 m/s

2 1,1375 1,4219

2 W Tz +

=+

= = 1,2797

+−+=

zq225,1ρVh 238,0h 725,00061,0h 5,0

vawwL

+−+=

1,27970,0042225,1)268,1549)(3(0,05)(0,6 238,0(0,05) 725,00061,0h 5,0

L

h L = 0,00285 m


gdρg σ 6

hoL

cR =

,8)(0,0045)(9 715,46(1) (0,04) 6h R = = 0,0076 m


hG = hd + hL + hR

hG = 0,03266 - 0,00285 + 0,0076

hG = 0,03741 m


Ada = 0,025 W = 0,025(1,1375) = 0,0284 m2 2

da2 A

qg23h

=


2

2 0,02840,0042

g23h

= = 0,003346 m

Backup in downspout

h3 = hG + h2

h3 = 0,03741 + 0,003346

h3 = 0,040756 m

Check on flooding

hw + h1 + h3 = 0,05 +0,01492+ 0,040756

hw + h1 + h3 = 0,1056 m

t/2 = 0,4/2 = 0,2 m



flooding.



Tinggi tutup = ( ) 1,421941 = 0,3554 m

Tinggi total = 5,2 + 2(0,3551) = 5,9108 m

Tekanan operasi = 10 atm = 1.013,25 kPa


P design = (1+0,05) x 1.013,25 kPa = 1.063,9125 kPa = 154,35 psi




Umur = 10 tahun

Tebal shell tangki:

0,6P-SEPRt =

)0,6(154,35-8)(11200)(0,,4219/2)(154,35)(1t =


= 0,0123 in









Jumlah : 1 unit


Komponen Xif Xid Ki (500C) αi (500C) Xiw

C3H8 0,03 0,04 2,9 3,536 -

i-C4H8 0,07 0,093 1,4 1,707 -

n-C4H8 (L) 0,594 0,79 0,94 1,146 0,01

n-C4H10 (H) 0,255 0,076 0,82 1 0,79

C6H12 0,05 - 0,14 0,170 0,2

Total 1,000 1,000 1,000


rumus:

∑ −=−

θαα

i

ifi xq1


Maka : ∑ −=−

θαα

i

ifi x11 = ∑ −=

θαα

i

ifi x0




Asumsi

(θ) C3H8 i-C4H8 n-C4H8 n-C4H10 C6H12

Total

1.03885 0.04248 0.17883 6.3530 -6.563 -0.0097 0.00082

1.0388495 0.0424804 0.17883 6.352970 -6.56379 -0.00978 0.0007141 1.0388488 0.04248 0.17883 6.35292 -6.56390 -0.009783 0.000554


θαα

i


∑ −=+

θαα

i

idiDm

xR 1

RDm + 1 = 1,264

RDm = 0,264

RD = 1,5 RDm

RD = 0,396

2836,0)1396,0(

396,0)1(

=+

=+RR

2088,0)1264,0(

264,0)1(

=+

=+m

m

RR


+RR dan 2088,0

1=

+m

m

RR diperoleh

NN m = 0,47; maka:


)log()]/)(/log[(

,avL

LWHWHDLDm

XXXXN


)5049,1log(

)]01,0/79,0)(076,0/79,0log[=

= 12,416

N = 56,0416,12

56,0=mN

= 22,1714





22,1714 = 27,714 piring ≈ 28 piring.



=

2

log206,0logHD

LW

LF

HF

s

e

XX

DW

XX

NN

(Geankoplis,2003)

=

2

076,001,0

376,532112,791

594,0255,0log 206,0log

s

e

NN

284,0=s

e

NN

Ne = 0,284 Ns

N = Ne + Ns

29 = 0,284 Ns + Ns

Ns = 22,5856 ≈ 23

Ne = 29 – 23 = 6







Data :




(g/mol) Fraksi mol

x BM C3H8 13,2703 0,05 44 2,2 i-C4H8 24,2420 0,093 56 5,208 n-C4H8 205,9271 0,79 56 44,24 n-C4H10 19,3792 0,076 58 4,408 Total 262,9196 kmol/Jam 1 56,05



= 262,9196/3600 = 0,073 kmol/detik

ρv = K)(323K) atm/kmolm (0,082


3av = = 10,581 kg/m3

Laju alir volumetrik gas (Q) = 15,273

3234,22 0,073 ×× = 1,9336 m3/s

Tabel LC.18. Komposisi Bahan pada Alur Lb Kolom Destilasi MD-102


Fraksi massa

Lρ (kg/m3)

Fraksi massa x Lρ

n-C4H8 180,828437 0,01 630 6,30

n-C4H10 14.285,44652 0,79 635 501,65

C6H12 3.616,56874 0,2 675 135

Total 18.082,8437 Kg/Jam 1 679,2



5,0230 = 0,00739 m3/s


o

a

o

p'd

907,0AA

=

2

a

o

0,01200,0045907,0

AA

= = 0,1275

2/12/1

V

L

10,5812,679

1,9336 0,00739

ρρ

Q'q

=

= 0,0306

α = 0,0744t + 0,01173 = 0,0744(0,4) + 0,01173 = 0,0415

β = 0,0304t + 0,05 = 0,0304(0,4) + 0,05 = 0,0621

CF = 2,0

2/1VL 0,02

σβ)ρ/(q/Q)(ρ

1αlog

+


= 2,0

0,020,040,0621

0,03061log 0,0415

+

= 0,1434

VF = 5,0

V

VLF ρ

ρρC

−

= 5,0

10,58110,5812,6790,1434

−

= 1,1399 m/s


V = 0,8 x 1,1399 = 0,9119 m/s

An = 0,91191,9336

VQ= = 2,12 m2



At = =− 14145,01

2,12 2,4692 m2


Weir length (W) = 0,8(1,7735) = 1,4188 m



Weir crest (h1)


h1/T = 0,025/1,7735 = 0,0443

==1,41881,7735

WT 1,25

2

1

5,0222eff

WT

Th21

WT

WT

WW

+

−

−

=

( ) ( )[ ] ( )( ){ }25,0222

eff 25,10,04432125,125,1W

W+−−=


=

WWeff 0,8217

3/2eff

3/2

1 WW

Wq666,0h

=

( ) 3/23/2

1 0,82171,41880,00739666,0h

=

h 1 = 0,017

Perhitungan diulangi dengan memakai nilai h1 = 0,017m hingga nilai h1



Dry pressure drop

Ao = 0,1275 x 1,7708 = 0,2257 m2

uo = == 0,2257

1,9336AQ

o

8,5671 m/s


=

L

v2

o

2o

d ρρ

Cu

0,51h

=

679,2 10,581

35,1 8,56710,51h 2

2

d

hd = 31,996 mm = 0,03199 m

Hydraulic head

1,77081,9336

AQV

aa == = 1,0919 m/s

21,4188 1,7735

2 W Tz +

=+

= = 1,5961

+−+=

zq225,1ρVh 238,0h 725,00061,0h 5,0

vawwL

+−+=

1,59610,00739225,11)919)(10,58(0,05)(1,0 238,0(0,05) 725,00061,0h 5,0

L

h L = -0,00575 m


gdρg σ 6

hoL

cR =


,8)(0,0045)(9 679,2(1) (0,04) 6h R = = 0,0080 m


hG = hd + hL + hR

hG = 0,03199 – 0,00575 + 0,0080

hG = 0,03424 m


Ada = 0,025 W = 0,025(1,4188) = 0,0354m2 2

da2 A

qg23h

=

2

2 0,03540,00739

g23h

= = 0,00665 m

Backup in downspout

h3 = hG + h2

h3 = 0,03424 + 0,00665

h3 = 0,0408 m

Check on flooding

hw + h1 + h3 = 0,05 +0,017 +0,0408 = 0,1078

hw + h1 + h3 = 0,1078 m

t/2 = 0,4/2 = 0,2 m



flooding.



Tinggi tutup = ( ) 1,773541 = 0,4433 m

Tinggi total = 11,6 + 2(0,4433) = 12,4866 m

Tekanan operasi = 5 atm = 506,625 kPa


P design = (1+0,05) x 506,625 kPa = 531,9562 kPa = 77,175 psi





Umur = 10 tahun

Tebal shell tangki:

0,6P-SEPRt =

)0,6(77,175-8)(11200)(0,,7735/2)(77,175)(1t =

= 0,00764 in









Jumlah : 1 unit


Komponen Xif Xid Ki (900C) αi (900C) Xiw

C3H8 0,04 0,361 2,8 2,54 0,005

i-C4H8 (L) 0,093 0,307 1,4 1,27 0,07

n-C4H8 (H) 0,79 0,330 1,1 1 0,84

n-C4H10 0,076 - 1,4 1,27 0,085

Total 1,000 1,000 1,000


rumus:


∑ −=−

θαα

i

ifi xq1


Maka : ∑ −=−

θαα

i

ifi x11 = ∑ −=

θαα

i

ifi x0

Nilai θ dicari dengan cara trial dan eror


Asumsi

(θ) C3H8 i-C4H8 n-C4H8 n-C4H10

Total

1.211379 0.076470265 2.0148069 -3.737362 1.6465089 0.00042342

1.211378 0.0764702 2.01477261 -3.737380 1.6464808 0.0003432 1.211375 0.0764700 2.014669 -3.737433 1.646396 0.0001026


θαα

i


∑ −=+

θαα

i

idiDm

xR 1

RDm + 1 = 1,326

RDm = 0,326

RD = 1,5 RDm

RD = 0,489

3284,0)1489,0(

489,0)1(

=+

=+RR

2458,0)1326,0(

326,0)1(

=+

=+m

m

RR


+RR dan 2458,0

1=

+m

m

RR

diperoleh

NN m = 0,44; maka:


)log()]/)(/log[(

,avL

LWHWHDLDm

XXXXN



)020,1log(

)]07,0/84,0)(330,0/307,0log[=

= 10,944

N = 56,0944,10

56,0=mN

= 19,5441




19,5441= 24,430 piring ≈ 25 piring.



=

2

log206,0logHD

LW

LF

HF

s

e

XX

DW

XX

NN

(Geankoplis,2003)

=

2

330,0307,0

40,2359336,2916

093,079,0log 206,0log

s

e

NN

3684,0=s

e

NN

Ne = 0,3684 Ns

N = Ne + Ns

26 = 0,3684 Ns + Ns

Ns = 19,0002 ≈ 19

Ne = 26 – 19 = 7







Data :





(g/mol) Fraksi mol

x BM C3H8 12,5258 0,361 44 15,884 i-C4H8 8,3695 0,307 56 17,192 n-C4H8 9,0511 0,332 56 18,592 Total 29,9464 kmol/Jam 1 51,668


= 29,9464 /3600 = 0,0083 kmol/detik

ρv = K)(303K) atm/kmolm (0,082


3av = = 24,9543 kg/m3

Laju alir volumetrik gas (Q) = 15,273

3034,22 0,0083 ×× = 0,2062 m3/s



Fraksi massa

Lρ (kg/m3)

Fraksi massa x Lρ

C3H8 54,78974282 0,005 597 2,985

i-C4H8 767,0563992 0,07 630 44.1

n-C4H8 9.204,67679 0,84 630 529,2

n-C4H10 931,4256276 0,085 573 48,705

Total 10.957,94856 Kg/Jam 1,000 624,99



3,0438 = 0,0048 m3/s


o

a

o

p'd

907,0AA

=

2

a

o

0,01200,0045907,0

AA

= = 0,1275


2/12/1

V

L

24,954399,624

0,20620,0048

ρρ

Q'q

=

= 0,1164

α = 0,0744t + 0,01173 = 0,0744(0,4) + 0,01173 = 0,0415

β = 0,0304t + 0,05 = 0,0304(0,4) + 0,05 = 0,0621

CF = 2,0

2/1VL 0,02

σβ)ρ/(q/Q)(ρ

1αlog

+

= 2,0

0,020,040,0621

0,11641log 0,0415

+

= 0,1157

VF = 5,0

V

VLF ρ

ρρC

−

= 5,0

24,95439543,4299,6240,1157

−

= 0,5673 m/s


V = 0,8 x 0,5673 = 0,4538 m/s

An = 0,45380,2062

VQ= = 0,4543 m2

Untuk W = 0,80T dari Tabel 6.1. Treybal, hal.162, diketahui bahwa luas downspout

sebesar 14,145%.

At = =− 14145,010,4543 0,5291 m2


Weir length (W) = 0,8(0,8209) = 0,6567 m



Weir crest (h1)


h1/T = 0,025/0,8209 = 0,0304

==0,65670,8209

WT 1,25


2

1

5,0222eff

WT

Th21

WT

WT

WW

+

−

−

=

( ) ( )[ ] ( )( ){ }25,0222

eff 25,10,03042125,125,1W

W+−−=

=

WWeff 0,8803

3/2eff

3/2

1 WW

Wq666,0h

=

( ) 3/23/2

1 0,88030,65670,0048666,0h

=

h 1 = 0,0229 Perhitungan diulangi dengan memakai nilai h1 = 0,0229 m hingga nilai h1 konstan

pada nilai 0,0229 m.


Dry pressure drop

Ao = 0,1275 x 0,3795 = 0,0483 m2

uo = ==0,0483

0,2062AQ

o

4,2691 m/s


=

L

v2

o

2o

d ρρ

Cu

0,51h

=

624,99 24,9543

35,1 4,26910,51h 2

2

d

hd = 20,3635 mm = 0,02036 m

Hydraulic head

0,37950,2062

AQV

aa == = 0,5433 m/s

2 0,6567 0,8209

2 W Tz +

=+

= = 0,7388

+−+=

zq225,1ρVh 238,0h 725,00061,0h 5,0

vawwL


+−+=

0,73880,0048225,143)433)(24,95(0,05)(0,5 238,0(0,05) 725,00061,0h 5,0

L

h L = 0,01805 m


gdρg σ 6

hoL

cR =

,8)(0,0045)(9 624,99(1) (0,04) 6h R = = 0,0087 m


hG = hd + hL + hR

hG = 0,02036 + 0,01805 + 0,0087

hG = 0,04711 m


Ada = 0,025 W = 0,025(0,6567) = 0,0164 m2 2

da2 A

qg23h

=

2

2 0,01640,0048

g23h

= = 0,0131 m

Backup in downspout

h3 = hG + h2

h3 = 0,04711 + 0,0131

h3 = 0,06021 m

Check on flooding

hw + h1 + h3 = 0,05 +0,0229 + 0,06021

hw + h1 + h3 = 0,13311 m

t/2 = 0,4/2 = 0,2 m

karena nilai hw + h1 + h3 lebih kecil dari t/2, maka spesifikasi ini dapat diterima,

artinya dengan rancangan plate seperti ini diharapkan tidak terjadi flooding.




Tinggi tutup = ( ) 0,820941 = 0,2052 m

Tinggi total = 10,4 + 2(0,2052) = 10,814 m

Tekanan operasi = 12 atm = 1.215,9 kPa


P design = (1+0,05) x 1.215,9 kPa = 1.276,695 kPa = 185,22 psi




Umur = 10 tahun

Tebal shell tangki:

0,6P-SEPRt =

)0,6(185,22-8)(11200)(0,,8209/2)(185,22)(0t =

= 0,0085 in




LC.12 Heater 02 (E-104)




Jumlah : 1 unit



= 1,0807 Kg/jam x 2,2046 lb/kg

= 2,3825 lb/jam


T1 = 150 0C

= (150 0C x 1,8) + 32

= 302 0F

T2 = 135 0C

= (135 0C x 1,8) + 32

= 275 0F

Fluida dingin : Senyawa campuran


= 13.942,0427 kg/jam x 2,2046 lb/kg

= 30.736,6273 lb/jam

t1 = 34 0C

= (34 0C x 1,8) + 32

= 93,2 0F

t2 = 42 0C

= (42 0C x 1,8) + 32

= 107,6 0F




Fluida Panas Fluida Dingin Selisih T1 = 302 0F Temperatur yang lebih tinggi t2 = 107,6 0F ∆t2 = 194,4°F T2 = 275 0F Temperatur yang lebih rendah t1 = 93,2 0F ∆t1 = 181,8°F

T1 – T2 = 27 °F Selisih t2 – t1 = 14,4 °F ∆t2 – ∆t1 = 12,6°F

188,34

181,8194,42,3log

12,6

ΔtΔt2,3log

ΔtΔtLMTD

1

2

12 =

=

−

= °F

(2) Tc dan tc

=+

=+

=2

2753022

TTT 21c 288,5 °F

=+

=+

=2

2,936,1072

ttt 21

c 100,4 °F


(3) Dari Tabel 8, hal. 840, Kern, 1965, kondensor untuk fluida panas gas dan fluida

dingin air, diperoleh UD = 2-50, dan faktor pengotor (Rd) = 0,003.



2

oo2

D

ft2365,0F34,188

FftjamBtu50


QA =×

⋅⋅

=×

=


(3) Flow Area (aa)




4

).( 21

22 DD

aa−

=π

00834,04

)138,0172,0.( 22

=−

=π

aa


077,0)(

1

21

22 =−

=D

DDDe


2./6714,28500834,0

2,3825 ftjamlbmawG

aa ===


Pada Tc = 288,5 °F


µ

aeea

GDR ×=

624,395

0556,0 285,6714 077,0

=×

=eaR

(4) Taksir jH dari Gambar 24 Kern (1965), diperoleh jH = 4,5 pada Rea = 624,395

(5) Pada Tc = 288,5 °F




=

×

=

3

13

1

0156,00556,045,0.

kc µ 1,1709

114,0

=

=

weD

µµ

0674,111709,1

077,00156,05,4.

14,03

1

=×××=

×

××=

we kc

DkjHho

µµµ


(1) Flow area (aa)


D1 = 1,38 in = 0,115 ft

0104,04

14,3115,04

22

=×

=×

=πDaa

(2) Kecepatan massa

a

p awG =

933,444.955.20104,0

lb/jam 330.736,627==pG

lbm/jam.ft2


Pada tc = 100,4 °F

µ = 0,0129 cP = 0,0313 lbm/ft2⋅jam

µ

pp

GD ×=Re

49,663.858.10313,0

933,444.955.20,115Re =×

=p


(5) Pada tc = 100,4 °F




0134,10089,0

0313,0296,0. 31

31

=

×=

kc µ

214,0

31

./4283,7810134,1115,00089,01000. ftjamBtu

kc

DkjHhi

we

=×××=

×

××=

µµµ


66,138,14283,78 =×=

×=


F.ft.Btu/jam 0502,10674,165,19940674,165,1994

hhhh

U 2

oio

oioC °=

+×

=+×

=


RdUc

+=1

Ud1


002,00502,11

Ud1

=

+=


2t,39762134,1889541,0

2.227,805tUd

QA f=×

=∆×

=



Panjang hairpins yang diperlukan = A/a =12,3976/0,435 = 28,5002 ft


Koreksi Ud :

A = a.(48) = 0,435 x 48 = 20,88 ft2

Ud baru = FfBtu 02 .tjam./9541,034,1883976,12

2.227,805tA

Q=

×=

∆×

0959,09541,00502,19541,00502,1

=×−

=×−

=UdUcUdUcRd


Pressure drop


(1) De = D2 – D1 = 0,172 – 0,138 = 0,03455 ft

1281,294023,0

6714,28503455,0GaDe'' =×

=×

=µeaR

s = 1, ρ = 62,5 x 1 = 62,5

024,00,2640035,0 42,0 =+=eaR

f

(2)

eD

L

'.2810.18,42

. 2aG 4.f.

Faρ××

=∆

ft00000333,003455,025,62810.18,42

84 2285,6714 0,0244 Fa =×××

×××=∆

(3) sft /00126,05,623600

6714,2853600

aG V =

×=

×=

ρ

ftcg

000000073,03,322

200126,032

2V3 1F =×

×=

××=

( ) psiaFFa 00000147,0

1445,62)000000073,000000333,0(

1441 Pa =

×+=

×+∆=∆

ρ



(1) 024,00,2640035,0 42,0 =+=epR

f

s = 0,6, ρ = 62,5 x 0,6 = 37,5

(2) D

L

.2810.18,422

. 2pG 4.f.

Fpρ××

=∆

ft0029,0115,025,37810.18,42

48 2933,444.955.2 0,0244 Fp =×××

×××=∆


psia00125,0144

FppΔP =

×=

ρ


LC.13 Heater 03 (E-108)




Jumlah : 1 unit



= 7,636131 Kg/jam x 2,2046 lb/kg

= 16,8346 lb/jam

T1 = 150 0C

= (150 0C x 1,8) + 32

= 302 0F

T2 = 135 0C

= (135 0C x 1,8) + 32

= 275 0F

Fluida dingin : Senyawa campuran


= 10.456,5684 kg/jam x 2,2046 lb/kg

= 23.052,5506 lb/jam

t1 = 37 0C

= (37 0C x 1,8) + 32

= 98,6 0F

t2 = 60 0C

= (60 0C x 1,8) + 32

= 140 0F


Panas yang diserap (Q) = 16.607,05841 kJ/jam = 15.740,45 Btu/ja



Fluida Panas Fluida Dingin Selisih T1 = 302 0F Temperatur yang lebih tinggi t2 = 140 0F ∆t2 = 162°F T2 = 275 0F Temperatur yang lebih rendah t1 = 98,6 0F ∆t1 = 176,4°F

T1 – T2 = 27 °F Selisih t2 – t1 = 41,4 °F ∆t2 – ∆t1 = -14,4°F

169,4117

176,41622,3log

14,4-

ΔtΔt2,3log

ΔtΔtLMTD

1

2

12 =

=

−

= °F

(2) Tc dan tc

=+

=+

=2

2753022

TTT 21c 288,5 °F

=+

=+

=2

1406,982

ttt 21

c 119,3°F





2

oo2

D

ft8582,1F4117,169

FftjamBtu50


QA =×

⋅⋅

=×

=


(5) Flow Area (aa)




4

).( 21

22 DD

aa−

=π

00834,04

)138,0172,0.( 22

=−

=π

aa


077,0)(

1

21

22 =−

=D

DDDe



2./5371,018.200834,0

16,8346 ftjamlbmawG

aa ===


Pada Tc = 288,5 °F


µ

aeea

GDR ×=

4560,795.2

0556,0 2.018,5371 077,0

=×

=eaR

(4) Taksir jH dari Gambar 24 Kern (1965), diperoleh jH = 11 pada Rea = 8790,590.5

(5) Pada Tc = 288,5 °F



=

×

=

3

13

1

0156,00556,045,0.

kc µ 1,1709

114,0

=

=

weD

µµ

6094,211709,1

077,00156,011.

14,03

1

=×××=

×

××=

we kc

DkjHho

µµµ


(1) Flow area (aa)


D1 = 1,38 in = 0,115 ft

0104,04

14,3115,04

22

=×

=×

=πDaa

(2) Kecepatan massa

a

p awG =

404,591.216.20104,0

lb/jam 623.052,550==pG lbm/jam.ft2



Pada tc = 119,3 °F

µ = 0,0085 cP = 0,02057 lbm/ft2⋅jam

µ

pp

GD ×=Re

24,222.392.1202057,0

404,591.216.20,115Re =×

=p


(5) Pada tc = 119,3 °F



9354,00095,0

02057,0378,0. 31

31

=

×=

kc µ

214,0

31

./2721,7719354,0115,00095,01000. ftjamBtu

kc

DkjHhi

we

=×××=

×

××=

µµµ


66,138,12721,77 =×=

×=


F.ft.Btu/jam 5075,26094,264,23826094,264,2382

hhhh

U 2

oio

oioC °=

+×

=+×

=


RdUc

+=1

Ud1


002,05075,21

Ud1

=

+=


2t2394,734117,1694950,2 15.740,45

tUdQA f=

×=

∆×=




Panjang hairpins yang diperlukan = A/a =37,2394 /0,435 = 85,6078 ft


Koreksi Ud :

A = a.(100) = 0,435 x 100 = 43,5 ft2

Ud baru = FfBtu 02 .tjam./1359,24117,1695,4345,740.15

tAQ

=×

=∆×

00199,04950,25075,24950,25075,2

=×−

=×−

=UdUcUdUcRd

Pressure drop


(1) De = D2 – D1 = 0,172 – 0,138 = 0,03455 ft

1937,032.3023,0

5371,018.203455,0GaDe'' =×

=×

=µeaR

s = 1, ρ = 62,5 x 1 = 62,5

0126,00,2640035,0 42,0 =+=eaR

f

(2)

eD

L

'.2810.18,42

. 2aG 4.f.

Faρ××

=∆

ft000182,003455,025,62810.18,42

100 2.018,53712 0,01264 Fa =×××

×××=∆

(3) sft /0089,05,623600

5371,018.23600

aG V =

×=

×=

ρ

ftcg

00000367,03,322

20089,032

2V3 1F =×

×=

××=

( ) psiaFFa 0000805,0144

5,62)00000367,0000182,0(144

1 Pa =×+

=×+∆

=∆ρ




(1) 003777,00,2640035,0 42,0 =+=epR

f

s = 0,6045, ρ = 62,5 x 0,6045 = 37,7812

(2) D

L

.2810.18,42

. 2pG 4.f.

Fpρ××

=∆

ft00169,0115,027812,37810.18,42

100 2242.392.222,1 0,0037774 Fp =×××

×××=∆

psia000443,0144

FppΔP =

×=

ρ

∆Pp yang diperbolehkan = 2 psi

LC.14 Cooler (E-105)


destilasi I sebelumdimasukkan kedalam destilasi 3


Jumlah : 1 unit

Fluida Panas : senyawa campuran

Flowrate, W = 10.456,5321 Kg/jam (Lampiran B)

= 10.456,5321 Kg/jam x 2,2046 lb/kg

= 23.052,4706 lb/jam T1 = 132,5 0C

= (132,5 0C x 1,8) + 32

= 270,5 0F

T2 = 83,5 0C

= (83,5 0C x 1,8) + 32

= 182,3 0F

Fluida dingin : air pendingin


= 1.346,041545 kg/jam x 2,2046 lb/kg

= 2.967,4831 lb/jam


t1 = 15 0C

= (15 0C x 1,8) + 32

= 59 0F

t2 = 25 0C

= (25 0C x 1,8) + 32

= 77 0F




Fluida Panas Fluida Dingin Selisih T1 = 270,50F Temperatur yang lebih tinggi t2 = 77 0F ∆t2 = 193,5°F T2 =182,30F Temperatur yang lebih rendah t1 = 59 0F ∆t1 = 123,3°F

T1 – T2 = 88,2 °F Selisih t2 – t1 = 18 °F ∆t2 – ∆t1 = 70,2°F

9653,155

123,3193,52,3log

70,2

ΔtΔt2,3log

ΔtΔtLMTD

1

2

12 =

=

−

= °F

(2) Tc dan tc

=+

=+

=2

3,1825,2702

TTT 21

c 226,4 °F

=+

=+

=2

77592

ttt 21

c 68 °F





2

oo2

D

ft8476,6F9633,155

FftjamBtu50


QA =×

⋅⋅

=×

=

Fluida dingin : air pendingin, annulus

(7) Flow Area (aa)





4

).( 21

22 DD

aa−

=π

00834,04

)138,0172,0.( 22

=−

=π

aa


077,0)(

1

21

22 =−

=D

DDDe


2./332,813.35500834,0

2.967,4831 ftjamlbmawG

aa ===


Pada tc = 68 °F


µ

aeea

GDR ×=

8790,018.10

7346,2 2355.813,33 077,0

=×

=eaR

(4) Taksir jH dari Gambar 24 Kern (1965), diperoleh jH = 40 pada Rea 8790,018.10

(5) Pada tc = 68 °F



=

×=

3

13

1

345,07346,298,0.

kc µ 1,9804

114,0

=

=

weD

µµ

9288,35419804,1

077,0345,040.

14,03

1

=×××=

×

××=

we kc

DkjHho

µµµ


Fluida panas : pipa dalam

(1) Flow area (aa)


D1 = 1,38 in = 0,115 ft

0104,04

14,3115,04

22

=×

=×

=πDaa

(2) Kecepatan massa

a

p awG =

712,583.216.20104,0

lb/jam 623.052,470 ==pG

lbm/jam.ft2


Pada Tc = 226,4 °F

µ = 0,171 cP = 0,4138 lbm/ft2⋅jam

µ

pp

GD ×=Re

2897,015.6164138,0

712,583.216.20,115Re =×

=p

(4) Taksir JH dari Gambar 24, Kern, diperoleh JH = 500 pada Res = 1.232.030,582

(5) Pada Tc = 226,4 °F



7123,10633,0

4138,0768,0. 31

31

=

×=

kc µ

214,0

31

./2547,47117123,1115,00633,0500. ftjamBtu

kc

DkjHhi

we

=×××=

×

××=

µµµ


66,138,12547,471 =×=

×=


F.ft.Btu/jam 2193,8619288,3547659,3919288,3547659,391

hhhh

U 2

oio

oioC °=

+×

=+×

=



RdUc

+=1

Ud1


002,02193,1861

Ud1

=

+=


2t7803,19653,1553076,192

53.399,2tUd

QA f=×

=∆×

=





Koreksi Ud :

A = a.(24) = 0,435 x 24 = 10,44 ft2

Ud baru = FfBtu 02 .tjam./794,239653,15544,102,399.53

tAQ

=×

=∆×

001929,09863,1362193,1869863,1362193,186

=×−

=×−

=UdUcUdUcRd

Pressure drop

Fluida dingin : Anullus

(1) De = D2 – D1 = 0,172 – 0,138 = 0,03455 ft

4840,495.47346,2

332,813.35503455,0GaDe'' =×

=×

=µeaR

s = 1, ρ = 62,5 x 1 = 62,5

0112,00,2640035,0 42,0 =+=eaR

f

(2)

eD

L

'.2810.18,42

. 2aG 4.f.

Faρ××

=∆

ft00120,003455,025,62810.18,42

42 255.813,3323 0,01124 Fa =×××

×××=∆


(3) sft /5813,15,623600

332,813.3553600

aG V =

×=

×=

ρ

ftcg

075,03,322

5813,132

2V3 1F =×

×=

××=

( ) psiaFFa 0330,0144

5,62)075,000120,0(144

1 Pa =×+

=×+∆

=∆ρ


Fluida panas : inner pipe

(1) 00447,00,2640035,0 42,0 =+=epR

f

s = 0,675, ρ = 62,5 x 0,675 = 42,1875

(2) D

L

.2810.18,42

. 2pG 4.f.

Fpρ××

=∆

ft00123,0115,021875,42810.18,42

42 212.216.583,72 0,004474 Fp =×××

×××=∆

psia000360,0144

FppΔP =

×=

ρ




destilasi III sebelum dimasukkan kedalam tangki Produk Hexene.


Jumlah : 1 unit



= 3.136,9596 Kg/jam x 2,2046 lb/kg

= 6.915,7411 lb/jam

T1 = 63,5 0C


= (63,5 0C x 1,8) + 32

= 146,3 0F

T2 = 50 0C

= (50 0C x 1,8) + 32

= 122 0F


Flowrate, W = 162,158473 kg/jam (Lampiran A)

= 162,158473 kg/jam x 2,2046 lb/kg

= 357,4945 lb/jam

t1 = 15 0C

= (15 0C x 1,8) + 32

= 59 0F

t2 = 25 0C

= (25 0C x 1,8) + 32

= 77 0F




Fluida Panas Fluida Dingin Selisih T1 = 146,30F Temperatur yang lebih tinggi t2 = 77 0F ∆t2 = 69,3°F T2 =1220F Temperatur yang lebih rendah t1 = 59 0F ∆t1 = 63°F

T1 – T2 = 24,3 °F Selisih t2 – t1 = 18 °F ∆t2 – ∆t1 = 6,3°F

1743,66

6369,32,3log

6,3

ΔtΔt2,3log

ΔtΔtLMTD

1

2

12 =

=

−

= °F

(2) Tc dan tc

=+

=+

=2

1223,1462

TTT 21

c 134,15 °F

=+

=+

=2

77592

ttt 21

c 68 °F






2

oo2

D

ft9463,1F1743,66

FftjamBtu50


QA =×

⋅⋅

=×

=


(9) Flow Area (aa)




4

).( 21

22 DD

aa−

=π

00834,04

)138,0172,0.( 22

=−

=π

aa


077,0)(

1

21

22 =−

=D

DDDe


2./0479,865.4200834,0

357,4945 ftjamlbmawG

aa ===


Pada tc = 68 °F


µ

aeea

GDR ×=

9804,206.1

7346,2 942.865,047 077,0

=×

=eaR

(4) Taksir jH dari Gambar 24 Kern (1965), diperoleh jH = 7 pada Rea= 9804,206.1

(5) Pada tc = 68 °F




=

×=

3

13

1

345,07346,298,0.

kc µ 1,9804

114,0

=

=

weD

µµ

1125,6219804,1

077,0345,07.

14,03

1

=×××=

×

××=

we kc

DkjHho

µµµ


(1) Flow area (aa)


D1 = 1,38 in = 0,115 ft

0104,04

14,3115,04

22

=×

=×

=πDaa

(2) Kecepatan massa

a

p awG =

1058,975.6640104,0

lb/jam 6.915,7411 ==pG lbm/jam.ft2


Pada Tc = 134,15 °F

µ = 0,14 cP = 0,3388 lbm/ft2⋅jam

µ

pp

GD ×=Re

6906,714.2253388,0

1058,975.6640,115Re =×

=p

(4) Taksir JH dari Gambar 24, Kern, diperoleh JH = 310 pada Res = 6906,714.225

(5) Pada Tc = 134,15 °F



8374,200835,0

3388,0563,0. 31

31

=

×=

kc µ


214,0

31

./8661,6318374,2115,0

00835,0310. ftjamBtuk

cDkjHhi

we

=×××=

×

××=

µµµ


66,138,18661,63 =×=

×=


F.ft.Btu/jam 6249,821125,620935,531125,620935,53

hhhh

U 2

oio

oioC °=

+×

=+×

=


RdUc

+=1

Ud1


002,028,6249

1Ud1

=

+=


2t3,59101743,661002,27 6.439,887

tUdQA f=

×=

∆×=





Koreksi Ud :

A = a.(24) = 0,435 x 24 = 10,44 ft2


887,439.6tA

Q=

×=

∆×

072,032154,96249,2832154,96249,28

=×−

=×−

=UdUcUdUcRd

Pressure drop


(1) De = D2 – D1 = 0,172 – 0,138 = 0,03455 ft

5736,5417346,2

0479,865.4203455,0GaDe'' =×

=×

=µeaR

s = 1, ρ = 62,5 x 1 = 62,5


0222,00,2640035,0 42,0 =+=eaR

f

(2)

eD

L

'.2810.18,42

. 2aG 4.f.

Faρ××

=∆

ft0347,003455,025,62810.18,42

42 22.865,04794 0,02224 Fa =×××

×××=∆

(3) sft /1905,05,623600

0479,865.423600

aG V =

×=

×=

ρ

ftcg

00912,03,322

1905,032

2V3 1F =×

×=

××=

( ) psiaFFa 0190,0144

5,62)00912,00347,0(144

1 Pa =×+

=×+∆

=∆ρ



(1) 00498,00,2640035,0 42,0 =+=epR

f

s = 0,66, ρ = 62,5 x 0,66 = 41,25

(2) D

L

.2810.18,42

. 2pG 4.f.

Fpρ××

=∆

ft1488,0115,0225,41810.18,42

42 206225.714,69 0,004984 Fp =×××

×××=∆

psia0426,0144

FppΔP =

×=

ρ




Fungsi : Menurunkan temperatur bahan yang keluar dari kondensor destilasi

III sebelum dimasukkan kedalam destilasi V.


Jumlah : 1 unit



= 7.319,5725 Kg/jam x 2,2046 lb/kg

= 16.136,7295 lb/jam

T1 = 105,5 0C

= (63,5 0C x 1,8) + 32

= 221,9 0F

T2 = 50 0C

= (50 0C x 1,8) + 32

= 122 0F



= 3.136,9596 kg/jam x 2,2046 lb/kg

= 6.915,7411 lb/jam

t1 = 15 0C

= (15 0C x 1,8) + 32

= 59 0F

t2 = 25 0C

= (25 0C x 1,8) + 32

= 77 0F




Fluida Panas Fluida Dingin Selisih T1 = 221,90F Temperatur yang lebih tinggi t2 = 77 0F ∆t2 = 144,9°F T2 =1220F Temperatur yang lebih rendah t1 = 59 0F ∆t1 = 63°F

T1 – T2 = 99,9 °F Selisih t2 – t1 = 18 °F ∆t2 – ∆t1 =81,9°F


4406,98

63144,92,3log

81,9

ΔtΔt2,3log

ΔtΔtLMTD

1

2

12 =

=

−

= °F

(2) Tc dan tc

=+

=+

=2

1229,2212

TTT 21

c 171,95 °F

=+

=+

=2

77592

ttt 21

c 68 °F





2

oo2

D

ft8088,25F4406,98

FftjamBtu50


QA =×

⋅⋅

=×

=


(11) Flow Area (aa)




4

).( 21

22 DD

aa−

=π

00834,04

)138,0172,0.( 22

=−

=π

aa


077,0)(

1

21

22 =−

=D

DDDe


2./8193,221.83300834,0

6.915,7411 ftjamlbmawG

aa ===



Pada tc = 68 °F


µ

aeea

GDR ×=

5958,461.23

7346,2 93833.221,81 077,0

=×

=eaR

(14) Taksir jH dari Gambar 24 Kern (1965), diperoleh jH = 85 pada

Rea= 5958,461.23

(15) Pada tc = 68 °F



=

×=

3

13

1

345,07346,298,0.

kc µ 1,9804

114,0

=

=

weD

µµ

2237,75419804,1

077,0345,085.

14,03

1

=×××=

×

××=

we kc

DkjHho

µµµ


(1) Flow area (aa)


D1 = 1,38 in = 0,115 ft

0104,04

14,3115,04

22

=×

=×

=πDaa

(2) Kecepatan massa

a

p awG =

609,608.551.10104,0



Pada Tc = 171,95 °F


µ = 0,21 cP = 0,5082 lbm/ft2⋅jam

µ

pp

GD ×=Re

7474,000.3515082,0

609,608.551.10,115Re =×

=p


(5) Pada Tc = 171,95 °F



0746,30107,0

5082,0612,0. 31

31

=

×=

kc µ

214,0

31

./289,11710746,3115,00107,0410. ftjamBtu

kc

DkjHhi

we

=×××=

×

××=

µµµ


66,138,1289,117 =×=

×=


F.ft.Btu/jam 3429,682237,7545053,972237,7545053,97

hhhh

U 2

oio

oioC °=

+×

=+×

=


RdUc

+=1

Ud1


002,086,3429

1Ud1

=

+=


2t4210,714406,980740,74

127.032,1tUd

QA f=×

=∆×

=






Koreksi Ud :

A = a.(40) = 0,435 x 40 = 17,4 ft2


1,032.127tA

Q=

×=

∆×

0153,01250,373429,861250,373429,86

=×−

=×−

=UdUcUdUcRd

Pressure drop


(1) De = D2 – D1 = 0,172 – 0,138 = 0,03455 ft

2485,527.107346,2

8193,221.83303455,0GaDe'' =×

=×

=µeaR

s = 1, ρ = 62,5 x 1 = 62,5

00889,00,2640035,0 42,0 =+=eaR

f

(2)

eD

L

'.2810.18,42

. 2aG 4.f.

Faρ××

=∆

ft00175,003455,025,62810.18,42

08 2333.221,8198 0,008894 Fa =×××

×××=∆

(3) sft /7032,35,623600

8193,221.8333600

aG V =

×=

×=

ρ

ftcg

1719,03,322

7032,332

2V3 1F =×

×=

××=

( ) psiaFFa 0753,0144

5,62)1719,000175,0(144

1 Pa =×+

=×+∆

=∆ρ



(1) 00473,00,2640035,0 42,0 =+=epR

f

s = 0,68, ρ = 62,5 x 0,68 = 42,5


(2) D

L

.2810.18,42

. 2pG 4.f.

Fpρ××

=∆

ft00209,0115,025,42810.18,42

80 2609,608.551.1 00473,04 Fp =×××

×××=∆

psia000616,0144

FppΔP =

×=

ρ



Fungsi : Menurunkan temperatur bahan yang keluar dari reboiler destilasi

V sebelum dimasukkan kedalam tangki produk oktene.


Jumlah : 1 unit



= 1.683,4603 Kg/jam x 2,2046 lb/kg

= 3.711,3565 lb/jam

T1 = 74,5 0C

= (74,5 0C x 1,8) + 32

= 166,1 0F

T2 = 50 0C

= (50 0C x 1,8) + 32

= 122 0F



= 164,7000 kg/jam x 2,2046 lb/kg

= 363,0976 lb/jam

t1 = 15 0C

= (15 0C x 1,8) + 32

= 59 0F


t2 = 25 0C

= (25 0C x 1,8) + 32

= 77 0F




Fluida Panas Fluida Dingin Selisih T1 = 166,1,90F Temperatur yang lebih tinggi t2 = 77 0F ∆t2 = 144,9°F T2 =1220F Temperatur yang lebih rendah t1 = 59 0F ∆t1 = 63°F

T1 – T2 = 44,1 °F Selisih t2 – t1 = 18 °F ∆t2 – ∆t1 =26,1°F

3822,75

6389,12,3log

26,1

ΔtΔt2,3log

ΔtΔtLMTD

1

2

12 =

=

−

= °F

(2) Tc dan tc

=+

=+

=2

1221,1662

TTT 21

c 144,05 °F

=+

=+

=2

77592

ttt 21

c 68 °F





2

oo2

D

ft7354,1F3822,75

FftjamBtu50


QA =×

⋅⋅

=×

=


(13) Flow Area (aa)




4

).( 21

22 DD

aa−

=π


00834,04

)138,0172,0.( 22

=−

=π

aa


077,0)(

1

21

22 =−

=D

DDDe


2./8824,536.4300834,0


aa ===


Pada tc = 68 °F


µ

aeea

GDR ×=

8977,225.1

7346,2 3.536,88244 077,0

=×

=eaR

(4) Taksir jH dari Gambar 24 Kern (1965), diperoleh jH = 5,9 pada Rea= 8977,225.1

(5) Pada tc = 68 °F



=

×=

3

13

1

345,07346,298,0.

kc µ 1,9804

114,0

=

=

weD

µµ

3520,5219804,1

077,0345,09,5.

14,03

1

=×××=

×

××=

we kc

DkjHho

µµµ


(1) Flow area (aa)


D1 = 1,38 in = 0,115 ft

0104,04

14,3115,04

22

=×

=×

=πDaa


(2) Kecepatan massa

a

p awG =

2019,861.3560104,0

lb/jam 3.711,3565 ==pG

lbm/jam.ft2


Pada Tc = 144,05 °F

µ = 0,142 cP = 0,3436 lbm/ft2⋅jam

µ

pp

GD ×=Re

4116,438.1193436,0

2019,861.3560,115Re =×

=p


(5) Pada Tc = 144,05 °F



3938,10812,0

3436,064,0. 31

31

=

×=

kc µ

214,0

31

./6083,3713938,1115,00107,0290. ftjamBtu

kc

DkjHhi

we

=×××=

×

××=

µµµ


66,138,16083,37 =×=

×=


F.ft.Btu/jam 5746,193520,522647,313520,522647,31

hhhh

U 2

oio

oioC °=

+×

=+×

=


RdUc

+=1

Ud1


002,05746,191

Ud1

=

+=



2t5989,43822,758679,18 6.541,166

tUdQA f=

×=

∆×=





Koreksi Ud :

A = a.(40) = 0,435 x 40 = 17,4 ft2


6.541,166tA

Q=

×=

∆×

1494,09869,45746,199869,45746,19

=×−

=×−

=UdUcUdUcRd

Pressure drop


(1) De = D2 – D1 = 0,172 – 0,138 = 0,03455 ft

0619,5507346,2

8824,536.4303455,0GaDe'' =×

=×

=µeaR

s = 1, ρ = 62,5 x 1 = 62,5

0221,00,2640035,0 42,0 =+=eaR

f

(2)

eD

L

'.2810.18,42

. 2aG 4.f.

Faρ××

=∆

ft0594,003455,025,62810.18,42

40 23.536,88244 0,02214 Fa =×××

×××=∆

(3) sft /1934,05,623600

8824,536.433600

aG V =

×=

×=

ρ

ftcg

0017,03,322

03740,032

2V3 1F =×

×=

××=


( ) psiaFFa 0265,0144

5,62)0017,00594,0(144

1 Pa =×+

=×+∆

=∆ρ



(1) 00544,00,2640035,0 42,0 =+=epR

f

s = 0,7, ρ = 62,5 x 0,7 = 43,75

(2) D

L

.2810.18,42

. 2pG 4.f.

Fpρ××

=∆

ft6023,0115,0275,43810.18,42

40 2956.861,2013 00544,04 Fp =×××

×××=∆

psia1829,0144

FppΔP =

×=

ρ



Fungsi : Menurunkan temperatur bahan yang keluar dari reboiler destilasi

II sebelum dimasukkan kedalam tangki produk butana.


Jumlah : 1 unit



= 3.485,4743 Kg/jam x 2,2046 lb/kg

= 7.684,0766 lb/jam

T1 = 50 0C

= (74,5 0C x 1,8) + 32

= 122 0F

T2 = 30 0C

= (50 0C x 1,8) + 32


= 86 0F



= 306,028312 kg/jam x 2,2046 lb/kg

= 674,6700 lb/jam

t1 = 15 0C

= (15 0C x 1,8) + 32

= 59 0F

t2 = 25 0C

= (25 0C x 1,8) + 32

= 77 0F




Fluida Panas Fluida Dingin Selisih T1 = 122 0F Temperatur yang lebih tinggi t2 = 77 0F ∆t2 = 45 °F T2 =86 0F Temperatur yang lebih rendah t1 = 59 0F ∆t1 = 63°F

T1 – T2 = 36 °F Selisih t2 – t1 = 18 °F ∆t2 – ∆t1 =18 °F

2767,35

63452,3log

18

ΔtΔt2,3log

ΔtΔtLMTD

1

2

12 =

=

−

= °F

(2) Tc dan tc

=+

=+

=2

861222

TTT 21

c 104 °F

=+

=+

=2

77592

ttt 21

c 68 °F






2

oo2

D

ft8903,6F2767,35

FftjamBtu50


QA =×

⋅⋅

=×

=


(15) Flow Area (aa)




4

).( 21

22 DD

aa−

=π

00834,04

)138,0172,0.( 22

=−

=π

aa


077,0)(

1

21

22 =−

=D

DDDe


2./6834,895.8000834,0


aa ===


Pada tc = 68 °F


µ

aeea

GDR ×=

8350,277.2

7346,2 0.895,68348 077,0

=×

=eaR

(4) Taksir jH dari Gambar 24 Kern (1965), diperoleh jH = 7 pada Rea= 8350,277.2

(5) Pada tc = 68 °F



=

×=

3

13

1

345,07346,298,0.

kc µ 1,9804


114,0

=

=

weD

µµ

1125,6219804,1

077,0345,07.

14,03

1

=×××=

×

××=

we kc

DkjHho

µµµ


(1) Flow area (aa)


D1 = 1,38 in = 0,115 ft

0104,04

14,3115,04

22

=×

=×

=πDaa

(2) Kecepatan massa

a

p awG =

5192,853.7380104,0



Pada Tc = 104 °F

µ = 0,134 cP = 0,3242 lbm/ft2⋅jam

µ

pp

GD ×=Re

6098,085.2623242,0

5192,853.7380,115Re =×

=p


(5) Pada Tc = 104 °F



4419,20118,0

3242,053,0. 31

31

=

×=

kc µ


214,0

31

./7049,7214419,2115,00107,0320. ftjamBtu

kc

DkjHhi

we

=×××=

×

××=

µµµ


66,138,17049,72 =×=

×=


F.ft.Btu/jam 6327,301125,624414,601125,624414,60

hhhh

U 2

oio

oioC °=

+×

=+×

=


RdUc

+=1

Ud1


002,06327,301

Ud1

=

+=


2t9203,112767,359017,28 12.153,47

tUdQA f=

×=

∆×=



Panjang hairpins yang diperlukan = A/a = 11,9203 /0,435 = 27,4029 ft


Koreksi Ud :

A = a.(48) = 0,435 x 48 = 20,88 ft2


12.153,47tA

Q=

×=

∆×

0279,04999,166327,304999,166327,30

=×−

=×−

=UdUcUdUcRd

Pressure drop


(1) De = D2 – D1 = 0,172 – 0,138 = 0,03455 ft

0675,022.17346,2

6834,895.8003455,0GaDe'' =×

=×

=µeaR

s = 1, ρ = 62,5 x 1 = 62,5


0178,00,2640035,0 42,0 =+=eaR

f

(2)

eD

L

'.2810.18,42

. 2aG 4.f.

Faρ××

=∆

ft01982,003455,025,62810.18,42

48 20.895,69348 0,01784 Fa =×××

×××=∆

(3) sft /3595,05,623600

6934,895.803600

aG V =

×=

×=

ρ

ftcg

006,03,322

1292,032

2V3 1F =×

×=

××=

( ) psiaFFa 0112,0144

5,62)006,001982,0(144

1 Pa =×+

=×+∆

=∆ρ



(1) 00489,00,2640035,0 42,0 =+=epR

f

s = 0,6, ρ = 62,5 x 0,6 = 37,5

(2) D

L

.2810.18,42

. 2pG 4.f.

Fpρ××

=∆

ft790,3115,025,37810.18,42

84 2238.853,5197 00489,04 Fp =×××

×××=∆

psia9869,0144

FppΔP =

×=

ρ



LC.19 Akumulator I (V-101)

Fungsi : Mengumpulkan destilat yang keluar dari kondensor MD-101.

Bahan konstruksi : Carbon Steel SA –285 Grade C

Bentuk : Silinder horizontal dengan tutup ellipsoidal

Jenis sambungan : Single welded butt joints

Jumlah : 1 unit

Kondisi operasi :

Tekanan = 374,9025 kPa

Temperatur = 34 oC = 307 K

Kebutuhan perancangan = 1 jam


Tabel LC.23 Data pada akumulator (V-101)

Komponen F (kg/jam)

Fraksi Berat

ρ (kg/m3)

ρcampuran (kg/m3)

C3H8 418,2613 0,03 585 17,55

n-C4H8 975,9429 0,07 630 44,1

i-C4H8 8.295,5155 0,59 635 374,65

n-C4H10 3.555,2209 0,26 579 150,54

C6H12 697,1021 0,05 675 33,75 Total 13.942,0427 1 620,59

Perhitungan:

a. Volume Tangki

Volume larutan, Vl = 3kg/m 620,59 jam1 x kg/jam 713.942,042 = 22,4657 m3

Volume tangki, Vt = (1,2) x 22,4657 m3 = 26,9588 m3

Fraksi volum = t

l

VV =

9588,264657,22 = 0,8333

Dari gambar 18.15 pada buku Walas dkk, Chemical Process Equipment

diperoleh untuk fraksi volum 0,8333 maka H/D = 0,777

Asumsi L/D = 1,777

Digunakan dua buah tutup ellipsoidal maka volume tutup adalah:


Vh = Vo(V/Vo) = 2 [0,1309D3(2)(H/D)2(1,5-H/D)] (Walas dkk, 2005)

= 2 [0,1309D3(2)(0,777)2(1,5-0,777)]

= 0,2285D3

Kapasitas shell:

θ = 2 arccos (1-2H/D) = 2 arccos (1-2(0,777) = 2 arccos (1-1,554)

= 4,3159 rad

Vs = Vo(V/Vo) = ( )θsin -θ 2π1 LD

4π 2

(Walas dkk, 2005)

= ( )4,3159sin -4,3159 2π1 LD

4π 2

= 0,6548 D2L

Volume tangki = Vh + Vs = 0,2285D3 + 0,6548 D2L

Dimana L/D = 1,777, maka volume tangki adalah:

Vt = 0,2285D3 + 1,1636D3 = 1,3921D3

26,9588 m3 = 1,3921D3

D = 33921,1

26.9588 = 2,6854 m = 105,7241 in.

L = ( )( )2

3

6854,26548,06854,22285,0 26,9588 − = 4,7721 m.

Tinggi cairan = H/D = 0,777 (2,6854) = 2,0865 m.

Perhitungan tinggi tutup:

Hd = 4D =

46854,2 = 0,6713 m (Walas dkk, 2005)

Perhitungan tinggi shell:

Hs = L – 2Hd = 4,7721 – 2(0,6713) = 3,4295 m.

d. Tebal shell tangki


diperoleh data :

- Allowable stress (S) = 13750 psia



- Corrosion allowance (C) = 0.25 mm/tahun (Timmerhaus dkk,2004)

= 0,0098/tahun


Tekanan Hidrostatik:

PHidrostatik = ρ × g × l

= 620,59 kg/m3 × 9,8 m/det2 × 2,0865 m = 0,1228 atm

Po = 3,7 atm

P = 3,7 atm + 0,1228 atm = 3,8228 atm

Pdesign = 1,2 × 3,8228 = 4,5873atm = 67,4147 psi

Tebal shell tangki:

0,6P-SEPRt = (Walas dkk, 2005)

Dimana :

P = tekanan desain (psig)

R = jari-jari dalam tangki (in)

S = allowable stress (psia)

E = joint efficiency

in 3251,0psi) 70,6(67,414)(0,8)(13750psia

in) 2(105,7241/ psi) (67,41470,6PSE

PRt

=

−

=

−=


Maka tebal shell yang dibutuhkan dengan perkiraan umur alat adalah 10 tahun

= 0,3251 + (10 x 0,0098)

= 0,4231 in


e. Tebal tutup tangki


Tebal tutup atas yang digunakan = 3/4 in


LC.20 Akumulator II (V-102)

Fungsi : Mengumpulkan destilat yang keluar dari kondensor

MD-102.




Jumlah : 1 unit

Kondisi operasi :






Komponen F (kg/jam)

Fraksi Berat

ρ (kg/m3)

ρcampuran (kg/m3)

C3H8 418,2613 0,04 585 23,4

n-C4H8 975,9429 0,093 630 58,59

i-C4H8 8.260,6890 0,79 635 501,65

n-C4H10 801,6752 0,077 579 44,583 Total 10.456,5684 1 628,223

Perhitungan:

a. Volume Tangki



Fraksi volum = t

l

VV =

9735,196446,16 = 0,8333



Asumsi L/D = 1,777




= 2 [0,1309D3(2)(0,777)2(1,5-0,777)]

= 0,2285D3

Kapasitas shell:


= 4,3159 rad


4π 2

(Walas dkk, 2005)

= ( )4,3159sin -4,3159 2π1 LD

4π 2

= 0,6548 D2L



Vt = 0,2285D3 + 1,1636D3 = 1,3921D3

19,9735 m3 = 1,3921D3

D = 33921,1

19,9735 = 2,4299 m = 95,6650 in.

L = ( )( )2

3

4299,26548,04299,22285,0 19,9735 − = 4,3182 m.



Hd = 4D =

44299,2 = 0,6074 m (Walas dkk, 2005)


Hs = L – 2Hd = 4,3128 – 2(0,6074) = 3,098 m.



diperoleh data :





= 0,0098/tahun




= 627,644 kg/m3 × 9,8 m/det2 × 1,8880 m = 0,1123 atm

Po = 4,5 atm

P = 4,5 atm + 0,1123 atm = 4,6123 atm

Pdesign = 1,2 × 4,6123 = 5,5347 atm = 81,3376 psi

Tebal shell tangki:


Dimana :





in 3552,0psi) 60,6(81,337)(0,8)(13750psia

in) (95,6650/2 psi) (81,33760,6PSE

PRt

=

−

=

−=



= 0,3552 + (10 x 0,0098)

= 0,4532 in




Tebal tutup atas yang digunakan = 3/4 in LC.21 Akumulator IV (V-104)



MD-104.




Jumlah : 1 unit

Kondisi operasi :

Tekanan = 1.215,9 kPa





Komponen F (kg/jam)

Fraksi Berat

ρ (kg/m3)

ρcampuran (kg/m3)

C3H8 371,123908 0,361 585 211,185

n-C4H8 315,755328 0,307 630 193,41

i-C4H8 338,438135 0,332 635 210,82 Total 1.025,3173 1 615,415

Perhitungan:

a. Volume Tangki



Fraksi volum = t

l

VV =

9992,16660,1 = 0,8333



Asumsi L/D = 1,777




= 2 [0,1309D3(2)(0,777)2(1,5-0,777)]

= 0,2285D3

Kapasitas shell:


= 4,3159 rad


4π 2

(Walas dkk, 2005)

= ( )4,3159sin -4,3159 2π1 LD

4π 2

= 0,6548 D2L



Vt = 0,2285D3 + 1,1636D3 = 1,3921D3

1,9992 m3 = 1,3921D3

D = 33921,1

1,9992 = 1,1282 m = 44,4171 in.

L = ( )( )2

3

1282,16548,01282,12285,0 1,9992 − = 2,0051 m.



Hd = 4D =

41282,1 = 0,2820 m (Walas dkk, 2005)


Hs = L – 2Hd = 2,0051– 2(0,2820) = 1,4411 m.



diperoleh data :





= 0,0098/tahun




= 614,145 kg/m3 × 9,8 m/det2 × 0,8766 m = 0,0510 atm

Po = 12 atm

P = 12 atm + 0,0510 atm = 12,051 atm

Pdesign = 1,2 × 12,051 = 14,4612 atm = 212,5211 psi

Tebal shell tangki:


Dimana :





in 4341,0psi) 110,6(212,52)(0,8)(13750psia

in) /2(44,4171 psi) (212,52110,6PSE

PRt

=

−

=

−=



= 0,4341 + (10 x 0,0098)

= 0,5321 in




Tebal tutup atas yang digunakan = 3/4 in LC.22 Akumulator III (V-103)



MD-103.




Jumlah : 1 unit

Kondisi operasi :


Temperatur = 105,5 oC = 378,5 K




Komponen F (kg/jam)

Fraksi Berat

ρ (kg/m3)

ρcampuran (kg/m3)

n-C4H10 104,5653 0,033 579 19,107

C6H12 2.645,5235 0,8433 675 569,2275

C7H14 386,8708 0,123 698 85,854 Total 3.136,9596 1 674,1885

Perhitungan:

a. Volume Tangki



Fraksi volum = t

l

VV =

5834,56529,4 = 0,8333



Asumsi L/D = 1,777




= 2 [0,1309D3(2)(0,777)2(1,5-0,777)]

= 0,2285D3

Kapasitas shell:


= 4,3159 rad


4π 2

(Walas dkk, 2005)

= ( )4,3159sin -4,3159 2π1 LD

4π 2

= 0,6548 D2L



Vt = 0,2285D3 + 1,1636D3 = 1,3921D3

5,5834 m3 = 1,3921D3

D = 33921,1

5,5834 = 1,5888 m = 62,5509 in.

L = ( )( )2

3

5888,16548,05888,12285,0 5,5834 − = 2,8235 m.



Hd = 4D =

45888,1 = 0,3972 m (Walas dkk, 2005)


Hs = L – 2Hd = 2,8235 – 2(0,3972) = 2,0291 m.



diperoleh data :





= 0,0098/tahun




= 674,1885 kg/m3 × 9,8 m/det2 × 1,2344 m = 0,0789 atm

Po = 1 atm

P = 1 atm + 0,0789 atm = 1,0789 atm

Pdesign = 1,2 × 1,0789 = 1,2946 atm = 19,0253 psi

Tebal shell tangki:


Dimana :





in 0541,0psi) 30,6(19,025)(0,8)(13750psia

in) (62,5509/2 psi) (19,02530,6PSE

PRt

=

−

=

−=



= 0,0541 + (10 x 0,0098)

= 0,1521 in






LC.23 Akumulator V (V-105)

Fungsi : Mengumpulkan destilat yang keluar dari kondensor MD-105.




Jumlah : 1 unit

Kondisi operasi :

Tekanan = 1.013,25 kPa

Temperatur = 74,5 oC = 346,5 K




Komponen F (kg/jam)

Fraksi Berat

ρ (kg/m3)

ρcampuran (kg/m3)

C6H12 73,1749 0,013 675 8,775

C7H14 5.000,0206 0,887 698 619,126

C8H16 562,9167 0,01 716 7,16 Total 5.636,1122 1 635,061

Perhitungan:

a. Volume Tangki



Fraksi volum = t

l

VV =

6498,108749,8 = 0,8333



Asumsi L/D = 1,777



= 2 [0,1309D3(2)(0,777)2(1,5-0,777)]

= 0,2285D3


Kapasitas shell:


= 4,3159 rad


4π 2

(Walas dkk, 2005)

= ( )4,3159sin -4,3159 2π1 LD

4π 2

= 0,6548 D2L



Vt = 0,2285D3 + 1,1636D3 = 1,3921D3

10,6498 m3 = 1,3921D3

D = 33921,1

10,6498 = 1,9704 m = 77,5745 in.

L = ( )( )2

3

9704,16548,09704,12285,0 10,6498 − = 3,5016 m.



Hd = 4D =

49704,1 = 0,4926 m (Walas dkk, 2005)


Hs = L – 2Hd = 3,5016 – 2(0,4926) = 2,5164 m.



diperoleh data :




= 0,0098/tahun





= 635,061 kg/m3 × 9,8 m/det2 × 1,5310 m = 0,0922 atm

Po = 10 atm

P = 10 atm + 0,0922 atm = 10,0922 atm

Pdesign = 1,2 × 10,0922 = 12,1106 atm = 177,9768 psi

Tebal shell tangki:


Dimana :





in 6337,0psi) 680,6(177,97)(0,8)(13750psia

in) (77,5745/2 psi) (177,97680,6PSE

PRt

=

−

=

−=



= 0,6337 + (10 x 0,0098)

= 0,7317 in

Tebal shell standar yang digunakan = ¾ in (Brownell,1959)



Tebal tutup atas yang digunakan = ¾ in C.24 Pompa refluks Destilasi ( L-103 )

Fungsi : Memompa campuran dari Akumulator (V-101) ke refluks Destilasi


Kondisi operasi :


Tekanan = 3,7 atm

Temperatur = 34 oC


Tabel LC.28 Data pada alur destilat



μ campuran

C3H8 0,03 585 17,55 0,0425 0,00127 i-C4H8 0,07 630 44,1 0,1215 0,00850 n-C4H8 0,59 635 374,65 0,1217 0,07180 n-C4H10 0,26 579 150,54 0,1290 0,03354 C6H12 0,05 675 33,75 0,1956 0,00978 620,59 0,12489



mv = 3kg/m59,206kg/jam 713.942,042



= 0,363 (0,0062 m3/s)0,45 (620,59 kg/m3)0,13

= 0,0852 m = 3,3572 in




Diameter Dalam (ID) : 5,047 in = 0.4205 ft




3

0,1390ft/sft 0,2189 = 1,5854 ft/s


Bilangan Reynold:

NRe = μ

Dvρ ××

=lbm/ft.s 0,00067) x (0,12489

05ft)ft/s)(0,42 )(1,5854lbm/ft0,062) x ((620,59 3

= 615.809,4 (Turbulen)


Pada NRe = 615.661,484 dan ε/D =m0.128195m0,000046 = 0.000359

Dari Gambar 2.10-3 Geankoplis (2003) diperoleh harga f = 0,00375

Friction loss: 13.942,0427


12

1

2 vAA

− =

)174,32)(1(23,1708)01(55,0

2

−

= 0.085935943 ft.lbf/lbm


v.2

2

= 2(2))174,32(2

1708,3 2

= 0.624988 ft.lbf/lbm

1 tee : hf = n.Kf.cg

v.2

2

= 1(1) )174,32(2

1708,3 2

= 0.15624 ft.lbf/lbm


vL.2.. 2∆ = 4(0,00375) ( )( )

( ) ( )174,32.2.4205,03,1708.50 2



vAA

..21

22

2

1

α

− = 1 ( ) ( )( )174,3212

1708,3012

2−

= 0.15624 ft.lbf/lbm



( ) ( ) 021 12

122

12

2 =+∑+−

+−+− sWFPP

zzgvvρα

(Geankoplis,2003)

dimana: v1 = v2

P1 = P2

∆P = 0



maka : ( ) 0W52123,10 1232,17432,1740 s =++++

Ws = - 13,5212304ft.lbf/lbm Efisiensi pompa, η= 75 %

Wp = -η


= -0,75

213,5212304-



827.884,085×

= 0,5597 hp Maka dipilih pompa dengan daya motor 1 hp.

C.25 Pompa Refluks Destilasi (L-104)



Kondisi operasi :

Tekanan = 5 atm

Temperatur = 50 oC


Tabel LC.29 Data pada alurdestilat



μ campuran

C3H8 0,04 585 23,4 0,0429 0,00171 i-C4H8 0,093 630 58,59 0,1213 0,01128 n-C4H8 0,79 635 501,65 0,1215 0,09598 n-C4H10 0,076 579 44,004 0,03559 0,00270 627,644 0,11167




mv = 3kg/m644,276kg/jam 410.456,568



= 0,363 (0,004628 m3/s)0,45 (627,644 kg/m3)0,13

= 0,07465 m = 2,938954 in Dari Tabel A.5-1 Geankoplis (2003), dipilih pipa dengan spesifikasi:







3

0,0513ft/sft 0.163423 = 3,185629 ft/s

Bilangan Reynold:

NRe = μ

Dvρ ××

=lbm/ft.s 0,00067) x (0,11167

ft) 5667ft/s)(0,25 )(3,185629lbm/ft0,06243) x ((627,644 3

= 425.290,2 (Turbulen)


Pada NRe = 425.290,2 dan ε/D =m0,42058m0,000046 = 0.00059


Friction loss:


12

1

2 vAA

− =

)174,32)(1(2 3,185629)01(55,0

2

−




v.2

2

= 2(2))174,32(2

3,1856292

= 0,630834 ft.lbf/lbm

1 tee hf = n.Kf.cg

v.2

2

= 1(1) )174,32(2

3,1856292

= 0,157708 ft.lbf/lbm


vL.2.. 2∆ = 4(0,0033) ( )( )

( ) ( )174,32.2.0,2556673,185629.50 2

= 0,407123 ft.lbf/lbm


vAA

..21

22

2

1

α

− = 1 ( ) ( )( )174,3212

3,185629012

2−

= 0,157709 ft.lbf/lbm

Total friction loss: ∑ F 1,440113 ft.lbf/lbm


( ) ( ) 021 12

122

12

2 =+∑+−

+−+− sWFPP

zzgvvρα

(Geankoplis,2003)

dimana: v1 = v2

P1 = P2

∆P = 0


maka : ( ) 0W440113,10 1232,17432,1740 s =++++


Wp = -η


= -0,75

13,440113-

= 17,920150 ft.lbf/lbm.

Daya pompa: P = m × Wp


410.456,568×

= 0,208638 hp

Maka dipilih pompa dengan daya motor 1/2 hp.





Kondisi operasi :

Tekanan = 12 atm

Temperatur = 90 oC





μ campuran

C3H8 0,361 585 211,185 0,4273 0,1542 i-C4H8 0,307 630 193,41 0,1228 0,0376 n-C4H8 0,330 635 209,55 0,1230 0,04059 614,145 0,23239



mv = 3kg/m614,145kg/jam 711.025,3173



= 0,363 (0,000464 m3/s)0,45 (614,145 kg/m3)0,13

= 0,026437 m = 1,04082 in

Dari Tabel A.5-1 Geankoplis (2003), dipilih pipa dengan spesifikasi: Ukuran nominal : 1,5 in


Diameter Dalam (ID) : 1,610 in = 0.134167 ft

Diameter Luar (OD) : 1,900 in = 0.15833 ft



3

ft 0,01414/sft 0,016377 = 1,158177 ft/s

Bilangan Reynold:


NRe = μ

Dvρ ××

=lbm/ft.s 0,000156) x (0,23239

ft) 4167ft/s)(0,13 )(1.158177lbm/ft0,06243) x ((614,145 3

= 38.151,54 (Turbulen)


Pada NRe = 38.151,54 dan ε/D =m 0,134167m0,000046 = 0,001125


Friction loss:


12

1

2 vAA

− =

)174,32)(1(2 1,158177)01(55,0

2

−

= 0,011465 ft.lbf/lbm


v.2

2

= 2(2))174,32(2

1,158177 2

= 0,083382 ft.lbf/lbm


v.2

2

= 1(1) )174,32(2

1,158177 2

= 0,020845 ft.lbf/lbm


vL.2.. 2∆ = 4(0,001) ( )( )

( ) ( )174,32.2.0.13417 1,158177.50 2

= 0,031074 ft.lbf/lbm


vAA

..21

22

2

1

α

− = 1 ( ) ( )( )174,3212

1,158177012

2−

= 0,020846 ft.lbf/lbm Total friction loss: ∑ F = 0,167612 ft.lbf/lbm Dari persamaan Bernoulli:

( ) ( ) 021 12

122

12

2 =+∑+−


(Geankoplis,2003)

dimana: v1 = v2

P1 = P2

∆P = 0



maka : ( ) 0W 0,1676120 1232,17432,1740 s =++++

Ws = - 12.16761 ft.lbf/lbm Efisiensi pompa, η= 75 %

Wp = -η


= -0,75

12,16761-



711.025,3173×

= 0,018521 hp





Kondisi operasi :

Tekanan = 1 atm

Temperatur = 63,5 oC


Tabel LC.31 data pada alur destilat

Komponen Fraksi Berat Densitas ρ

campuran Viskositas μ

campuran n-C4H10 0,033 579 19,107 0,03489 0,00115 C6H12 0,8433 675 569,2275 0,1983 0,1672 C7H14 0,123 698 85,854 0,1396 0,01717 675,1885 0,18552




mv = 3kg/m675,1885kg/jam 3.136,9596



= 0,363 (0,001291 m3/s)0,45 (675,1885 kg/m3)0,13

= 0,042421 m = 1,670125 in





Diameter Luar (OD) : 2,375 in = 0.197917 ft



3

ft 0,02330/sft 0,045574 = 1,955981 ft/s

Bilangan Reynold:

NRe = μ

Dvρ ××

=lbm/ft.s 0,00067) x (0,18552

ft) 225ft/s)(0,17 )(1,955981lbm/ft0,06243) x ((675,1885 3

= 113.919,3 (Turbulen)


Pada NRe = 113.919,3 dan ε/D =m0,17225m0,000046 = 0,000876


Friction loss:


12

1

2 vAA

− =

)174,32)(1(2 1,955981)01(55,0

2

−

= 0,032701 ft.lbf/lbm


v.2

2

= 2(2))174,32(2

1,955981 2

= 0,237823 ft.lbf/lbm



v.2

2

= 1(1) )174,32(2

1,955981 2

= 0,0594557

ft.lbf/lbm


vL.2.. 2∆ = 4(0,00497) ( )( )

( ) ( )174,32.2.0,20575 1,955981.50 2



vAA

..21

22

2

1

α

− = 1 ( ) ( )( )174,3212

1,955981012

2−

= 0,059456 ft.lbf/lbm Total friction loss: ∑ F = 0,651766 ft.lbf/lbm


( ) ( ) 021 12

122

12

2 =+∑+−


(Geankoplis,2003)

dimana: v1 = v2

P1 = P2

∆P = 0


maka : ( ) 0W 0,6517660 1232,17432,1740 s =++++


Wp = -η


= -0,75

12,65177-



3.136,9596×

= 0,05892 hp






Kondisi operasi :

Tekanan = 10 atm

Temperatur = 30 oC


Tabel LC.32 data pada alur destilat

Komponen Fraksi Berat Densitas ρ

campuran Viskositas μ

campuran C6H12 0,013 675 8,775 0,1876 0,00243 C7H14 0,887 698 619,126 0,4067 0,36074 C8H16 0, 1 716 71,6 0,1327 0,01327 699,501 0,37644



mv = 3kg/m061,356kg/jam 5.636,1122



= 0,363 (0,002238 m3/s)0,45 (699,501 kg/m3)0,13








3

ft 0,05130/sft 0,07903657 = 1,54067379 ft/s


Bilangan Reynold:

NRe = μ

Dvρ ××

=lbm/ft.s 0,00067) x (0,37644

ft) 5667ft/s)(0,25 79)(1,540673lbm/ft0,06243) x ((699,501 3

= 68.001,2023 (Turbulen)


Pada NRe = 68.001,2023 dan ε/D =m255667,0m0,000046 = 0,00059029


Friction loss:


12

1

2 vAA

− =

)174,32)(1(2 1,54067379)01(55,0

2

−

= 0,02028846 ft.lbf/lbm


v.2

2

= 2(2))174,32(2 1,54067379 2



v.2

2

= 1(1) )174,32(2 1,54067379 2

= 0,0368 ft.lbf/lbm


vL.2.. 2∆ = 4(0,005) ( )( )

( ) ( )174,32.2.255667,0 1,54067379.50 2

= 0,23085124 ft.lbf/lbm


vAA

..21

22

2

1

α

− = 1 ( ) ( )( )174,3212

1,54067379012

2−

= 0,0368881 ft.lbf/lbm



( ) ( ) 021 12

122

12

2 =+∑+−

+−+− sWFPP

zzgvvρα

(Geankoplis,2003)

dimana: v1 = v2

P1 = P2

∆P = 0



maka : ( ) 0W 0,472380210 1232,17432,1740 s =++++

Ws = -12,4723802 ft.lbf/lbm Efisiensi pompa, η= 75 %

Wp = -η


= -0,75

12,4723802-



5.636,1122×

= 0,10436 hp


C.29 Pompa tangki penyimpanan (L-113)

Fungsi : Memompa oktena ke tangki produk oktena (TK-107)


Jumlah : 1 unit

Kondisi operasi :

Tekanan = 10 atm






μ campuran

C7H14 0,03 698 20,94 0,1386 0,004158 C8H16 0,97 716 694,52 0,1392 0,135024 715,46 0,139182




mv = 3kg/m 715,46kg/jam 1.683,4603



= 0,363 (0,000654 m3/s)0,45 (715,46 kg/m3)0,13

= 0,03146984 m = 1,238967 in







3

ft 0,01414/sft 0,0230809 = 1,632318 ft/s

Bilangan Reynold:

NRe = μ

Dvρ ××

= lbm/ft.s 9,3526.10

4167ft)ft/s)(0,13 )(1,632318lbm/ft(715,465-

3

= 104.590,204 (Turbulen)


Pada NRe = 104.590,204 dan ε/D =m0,040894m0,000046 = 0,00112485



12

1

2 vAA

− =

)174,32)(1(21,632318)01(55,0

2

−

= 0,0227739 ft.lbf/lbm


v.2

2

= 2(0,75))174,32(2

1,6323182

= 0,0621 ft.lbf/lbm


v.2

2

= 1(2))174,32(2

1,6323182




vL.2.. 2∆ = 4(0,0048) ( )( )

( ) ( )174,32.2.134167,01,632318.100 2

= 0,592558 ft.lbf/lbm


vAA

..21

22

2

1

α

− = 1 ( ) ( )( )174,3212

1,632318012

2−

= 0,0414107 ft.lbf/lbm



( ) ( ) 021 12

122

12

2 =+∑+−


(Geankoplis, 2003)

dimana: v1 = v2

P2 = 10 atm

∆P = 0 atm =


maka : ( ) 0W 0,88447858715,46

0 3032,17432,1740 s =++++

Ws = -30,884478 ft.lbf/lbm Efisiensi pompa, η= 75 %

Wp = -η


= -0,75

30,884478-

= 41,1793048 ft.lbf/lbm.

Daya pompa: P = m × Wp


1.683,4603×

= 0,077187 hp



C.30 Pompa Reboiler Destilasi I (L-108)

Fungsi : Memompa campuran dari alur bawah Destilasi I ke reboiler.


Kondisi operasi :

Tekanan = 3,7 atm






μ campuran

n-C4H10 0,01 635 6,35 0,03537 0,000353 C6H12 0,26 675 175,5 0,2136 0,05553 C7H14 0,52 698 362,96 0,1413 0,07347 C8H16 0,21 716 150,36 0,1406 0,02952 695,17 0,158879



mv = 3kg/m695,17kg/jam 110.456,532



= 0,363 (0,004178 m3/s)0,45 (695,17 kg/m3)0,13








3

ft 0,05130/sft 0,14754803 = 2,87617996 ft/s


Bilangan Reynold:

NRe = μ

Dvρ ××

=lbm/ft.s 0,00067) x (0,158879

ft) 5667ft/s)(0,25 96)(2,876179lbm/ft0,06243) x ((695,17 3

= 298.919,263 (Turbulen)


Pada NRe = 298.919,263 dan ε/D =m0,077928m0,000046 = 0,00059029


Friction loss:


12

1

2 vAA

− =

)174,32)(1(287617996,2)01(55,0

2

−

= 0,07070657 ft.lbf/lbm


v.2

2

= 2(2))174,32(2

87617996,2 2



v.2

2

= 1(1) )174,32(2

87617996,2 2



vL.2.. 2∆ = 4(0,0042) ( )( )

( ) ( )174,32.2.255667,087617996,2.15 2

= 0,14481563 ft.lbf/lbm


vAA

..21

22

2

1

α

− = 1 ( ) ( )( )174,3212

87617996,2012

2−

= 0,12855739 ft.lbf/lbm



( ) ( ) 021 12

122

12

2 =+∑+−

+−+− sWFPP

zzgvvρα

(Geankoplis,2003)

dimana: v1 = v2

P1 = P2

∆P = 0



maka : ( ) 0W 0,9868590532,17432,1740 s =++++


Wp = -η


= -0,75

5,986859-



110.456,532×

= 0,092937 hp


C.31 Pompa Reboiler Destilasi II (L-109)

Fungsi : Memompa campuran dari alur bawah Destilasi II ke reboiler.


Kondisi operasi :

Tekanan = 5 atm

Temperatur = 50 oC





μ campuran

n-C4H8 0,01 630 6,30 0,02356 0,000235 n-C4H10 0,79 635 501,65 0,1218 0,096222 C6H12 0,2 675 135 0,1958 0,03916 642,95 0,135617




mv = 3kg/m642,95kg/jam4743,485.3

= 0,001506 m3/s = 0.053177 ft3/s Desain pipa:


= 0,363 (0,001506m3/s)0,45 (642,95 kg/m3)0,13

= 0,045183 m = 1,778841 in

Dari Tabel A.5-1 Geankoplis (2003), dipilih pipa dengan spesifikasi: Ukuran nominal : 2 in






3

ft 0,02330/sft 0,053177 = 2,282262 ft/s

Bilangan Reynold:

NRe = μ

Dvρ ××

=lbm/ft.s 0,00067) x (0,135617

ft) 225ft/s)(0,17 )(2,282262lbm/ft0,06243) x ((642,95 3

= 173.151,8 (Turbulen)


Pada NRe = 173.151,8 dan ε/D =m052502,0m0,000046 = 0,000876


Friction loss:


12

1

2 vAA

− =

)174,32)(1(2 2,282262)01(55,0

2

−



v.2

2

= 1(2))174,32(2 2,282262 2

= 0,323784 ft.lbf/lbm



v.2

2

= 1(1) )174,32(2 2,282262 2

= 0,0809 ft.lbf/lbm


vL.2.. 2∆ = 4(0,00487) ( )( )

( ) ( )174,32.2.355,0 2,282262.15 2

= 0,107145 ft.lbf/lbm


vAA

..21

22

2

1

α

− = 1 ( ) ( )( )174,3212

2,282262012

2−

= 0,080946 ft.lbf/lbm



( ) ( ) 021 12

122

12

2 =+∑+−

+−+− sWFPP

zzgvvρα

(Geankoplis,2003)

dimana: v1 = v2

P1 = P2

∆P = 0


maka : ( ) 0W 0,6372960532,17432,1740 s =++++


Wp = -η


= -0,75

5,637296-



3.485,4743×

= 0,0291 hp



C.32 Pompa Reboiler Destilasi IV (L-110)

Fungsi : Memompa campuran dari alur bawah Destilasi IV ke reboiler.


Kondisi operasi :

Tekanan = 12 atm

Temperatur = 80 oC





μ campuran

C3H8 0,005 505 2,525 0,04275 0,000213 i-C4H8 0,07 630 44,1 0,1236 0,00865 n-C4H8 0,84 630 529,2 0,1236 0,10382 n-C4H10 0,085 573 48,705 0,03483 0,00296 624,53 0,11564



mv = 3kg/m 624,53kg/jam 299.431,2510



= 0,363 (0,004195 m3/s)0,45 ( 624,53kg/m3)0,13








3

ft 0,05130/sft 0,148133 = 2,887589 ft/s


Bilangan Reynold:

NRe = μ

Dvρ ××

=lbm/ft.s 0,00067) x (0,11564

ft) 5667ft/s)(0,25 )(2,887589lbm/ft0,06243) x ((624,53 3

= 370.419,6 (Turbulen)


Pada NRe = 370.419,6 dan ε/D =m0,077928m0,000046 = 0,00059


Friction loss:


12

1

2 vAA

− =

)174,32)(1(2887589,2)01(55,0

2

−

= 0,071269 ft.lbf/lbm


v.2

2

= 1(2))174,32(2

887589,2 2

= 0,259459 ft.lbf/lbm


v.2

2

= 1(1) )174,32(2

887589,2 2

= 0,129579 ft.lbf/lbm


vL.2.. 2∆ = 4(0,0034) ( )( )

( ) ( )174,32.2.255667,0 887589,2.15 2

= 0,103393 ft.lbf/lbm


vAA

..21

22

2

1

α

− = 1 ( ) ( )( )174,3212

887589,2012

2−

= 0,129579 ft.lbf/lbm



( ) ( ) 021 12

122

12

2 =+∑+−

+−+− sWFPP

zzgvvρα

(Geankoplis,2003)

dimana: v1 = v2

P1 = P2

∆P = 0



maka : ( ) 0W 0,6929790532,17432,1740 s =++++


Wp = -η


= -0,75

5,692979-



299.431,2510×

= 0,07971 hp


C.33 Pompa Reboiler Destilasi 3 (L-111)

Fungsi : Memompa campuran dari alur bawah Destilasi III ke reboiler.


Kondisi operasi :

Tekanan = 1 atm






μ campuran

C6H12 0,01 675 6,75 0,2183 0,002183 C7H14 0,69 698 481,62 0,1396 0,09632 C8H16 0,3 716 214,3 0,1872 0,05616 703,17 0,15466




mv = 3kg/m 703,17kg/jam 7.319,5725



= 0,363 (0,002891 m3/s)0,45 (703,17 kg/m3)0,13

= 0,06131 m = 2,413766 in




Diameter Luar (OD) : 2,875in = 0,239583 ft



3

ft 0,03322/sft 0,102109 = 3,073707 ft/s

Bilangan Reynold:

NRe = μ

Dvρ ××

=lbm/ft.s 0,00067) x (0,15466

ft) 575ft/s)(0,20 )(3,073707lbm/ft0,06243) x ((703,17 3

= 267.130,8 (Turbulen)


Pada NRe = 267.130,8 dan ε/D =m0,062713m0,000046 = 0,000733


Friction loss:


12

1

2 vAA

− =

)174,32)(1(23,073707)01(55,0

2

−

= 0,080752 ft.lbf/lbm


v.2

2

= 1(2))174,32(2

3,0737072

= 0,293643 ft.lbf/lbm



v.2

2

= 1(1) )174,32(2

3,0737072

= 0,146821 ft.lbf/lbm


vL.2.. 2∆ = 4(0,00475) ( )( )

( ) ( )174,32.2.0,205753,073707.15 2

= 0,160558 ft.lbf/lbm


vAA

..21

22

2

1

α

− = 1 ( ) ( )( )174,3212

3,073707012

2−

= 0,146822 ft.lbf/lbm



( ) ( ) 021 12

122

12

2 =+∑+−

+−+− sWFPP

zzgvvρα

(Geankoplis,2003)

dimana: v1 = v2

P1 = P2

∆P = 0


maka : ( ) 0W 0,8285960532,17432,1740 s =++++


Wp = -η


= -0,75

5,828596-



7.319,5725×

= 0,06333 hp



C.34 Pompa Reboiler Destilasi V (L-112)

Fungsi : Memompa campuran dari alur bawah Destilasi V ke reboiler.


Kondisi operasi :

Tekanan = 10 atm






μ campuran

C7H14 0,03 698 20,94 0,1286 0,00385 C8H16 0,97 716 694,52 0,1293 0,12542 715,46 0,12927



mv = 3kg/m 715,46kg/jam 1.683,4603

= 1.683,4603 m3/s = 0,023081 ft3/s Desain pipa:


= 0,363 (1.683,4603 m3/s)0,45 ( 715,46 kg/m3)0,13

= 0,03147 m = 1,238967 in







3

ft 0,01414/sft 0,023081 = 1,632318 ft/s

Bilangan Reynold:


NRe = μ

Dvρ ××

=lbm/ft.s 0,00067) x (0,12927

ft) 4167ft/s)(0,13 )(1,632318lbm/ft0,06243) x ((715,46 3

= 112.609,8 (Turbulen)


Pada NRe = 112.609,8 dan ε/D =m0,040894m0,000046 = 0,001125


Friction loss:


12

1

2 vAA

− =

)174,32)(1(21,632318)01(55,0

2

−

= 0,022774 ft.lbf/lbm


v.2

2

= 1(2))174,32(2 1,632318 2

= 0,082814 ft.lbf/lbm


v.2

2

= 1(1) )174,32(2

1,6323182



vL.2.. 2∆ = 4(0,00516) ( )( )

( ) ( )174,32.2.0.17221,632318.15 2



vAA

..21

22

2

1

α

− = 1 ( ) ( )( )174,3212

1,632318012

2−

= 0,041407 ft.lbf/lbm



( ) ( ) 021 12

122

12

2 =+∑+−

+−+− sWFPP

zzgvvρα

(Geankoplis,2003)

dimana: v1 = v2

P1 = P2

∆P = 0



maka : ( ) 0W 0,2578360532,17432,1740 s =++++


Wp = -η


= -0,75

5,257836-



1.683,4603×

= 0,0131 hp


LC.35 Tangki Penyimpanan - 04 (TK-104)

Fungsi : Untuk penyimpanan produk butana dari hasil bawah MD-02

selama 20 hari

Jumlah : 4 unit



Kondisi operasi : -Temperatur = 30 0C

-Tekanan = 2,5 atm = 36,73987 psi

Perhitungan:

a. Volume Tangki

Laju alir = 3.485,4743 kg/jam x 24 jam/hari (Lampiran A)

= 83.651,3832 kg/hari


= 1.673.027,664 Kg

Densitas campuran (ρ) = 642 kg/m3

Volume campuran 39620,605.2642

6641.673.027,3

mkgm

mkg

==ρ


Volume larutan untuk 1 tangki = 9620,605.2 / 4 = 651,4905 m3


Volume tangki, VT = (1 + 0,2) x 651,4905 m3

= 781,7886 m3



Vs = 4






Hs : Dt = 5 : 4 , Maka :

Vs = 41 π Dt

2 Hs (Hs : Dt = 5 : 4)

Vs = 165 π Dt3

= 0,9812 Dt3

Vh = 24π Dt3

= 0,13083 Dt3

Volume tangki (Vt)

Vt = Vs + Vh

781,7886 = 0,9812 Dt3 + 0,13083 Dt3

781,7886 = 1,11203 Dt3

Dt = 8,8918 m = 350,0698 in

r = ½ x Dt = ½ x (8,8918) = 4,4459 m = 175,0349 in


Hs = 45 x Dt =

45 x 8,8918m = 11,1147 m = 437,5853 in


Volume tangki (Vt) = 781,7886 m3

Volume cairan (Vc) = 651,4905m3



Tinggi cairan dalam tangki (Hc) = silindervolume


= 781,7886

1147,11 651,4905 x

= 9,2622 m = 364,6524 in



(Timmerhaus,2004)




= 0,0098 in/tahun




= 642 kg/m3 × 9,8 m/det2 × 9,2622 m = 0,5640 atm

Po = 2,5 atm

P = 2,5 atm + 0,5640 atm = 3,064 atm

Pdesign = 1,2 × 3,064 = 3,6768 atm = 54,0340 psi


in 1022,0

in)0098,0(10psi) 00,6(54,034psi)(0,8) 2(13.700

in) (175,0349 psi) (54,0340

nC0,6P2SE

Prt

=

+

−

=

+−

=

x






LC.36 Tangki Penyimpanan- 03 (TK-103)

Fungsi : Untuk penyimpanan produk propena dari hasil atas MD-104 selama

20 hari

Jumlah : 2 unit




-Tekanan = 11,5 atm = 169,0034 psi

Perhitungan:

a. Volume Tangki


= 24.607,6169 kg/hari


= 492.152,3381 Kg

Densitas campuran (ρ) = 614,145 kg/m3

Volume campuran 33617,801 614,145

3381,152.9243

mkgm

mkg

==ρ

Volume larutan untuk 1 tangki = 3617,801 / 2 = 400,6808 m3



= 480,8170 m3



Vs = 4






Hs : Dt = 5 : 4 , Maka :

Vs = 41 π Dt

2 Hs (Hs : Dt = 5 : 4)


Vs = 165 π Dt3

= 0,9812 Dt3

Vh = 24π Dt3

= 0,13083 Dt3

Volume tangki (Vt)

Vt = Vs + Vh

480,8170 = 0,9812 Dt3 + 0,13083 Dt3

480,8170 = 1,11203 Dt3

Dt = 7,5617 m = 297,7038 in

r = ½ x Dt = ½ x (7,5617) = 3,7808 m = 148,8499 in


Hs = 45 x Dt =

45 x 7,5617m = 9,4521 m = 372,1288 in


Volume tangki (Vt) = 480,8170m3

Volume cairan (Vc) = 400,6808 m3




= 480,8170

4521,9 400,6808 x

= 7,8767 m = 310,1054 in



(Timmerhaus,2004)




= 0,0098 in/tahun





= 614,145 kg/m3 × 9,8 m/det2 × 7,8767 m = 0,4588 atm

Po = 11,5 atm

P = 11,5 atm + 0,4588 atm = 11,9588 atm

Pdesign = 1,2 × 11,9588 = 14,3505 atm = 210,8942 psi


in 1121,0

in)0098,0(10psi) 420,6(210,89psi)(0,8) 2(13.700

in) (148,8499 psi) (210,8942

nC0,6P2SE

Prt

=

+

−

=

+−

=

x






Fungsi : Untuk penyimpanan produk heksena dari hasil atas MD-103

selama 20 hari

Jumlah : 4 unit

Bentuk : Tangki berbentuk silinder vertical dengan alas dan tutup

ellipsiodal



-Tekanan = 1atm = 14,69 psi

Perhitungan:

a. Volume Tangki


= 75.287,0304 kg/hari


= 1.505.740,608 Kg



Volume campuran 31040,230.21885,675

6081.505.740,3

mkgm

mkg

==ρ

Volume larutan untuk 1 tangki = 1040,230.2 / 4 = 557,5260 m3



= 669,0312 m3



Vs = 4






Hs : Dt = 5 : 4 , Maka :

Vs = 41 π Dt

2 Hs (Hs : Dt = 5 : 4)

Vs = 165 π Dt3

= 0,9812 Dt3

Vh = 24π Dt3

= 0,13083 Dt3

Volume tangki (Vt)

Vt = Vs + Vh

669,0312 = 0,9812 Dt3 + 0,13083 Dt3

669,0312 = 1,11203 Dt3

Dt = 8,4419 m = 332,3573 in

r = ½ x Dt = ½ x (8,4419) = 4,2209 m = 166.1767 in


Hs = 45 x Dt =

45 x 8,4419 m = 10,5523 m = 415,4436 in








= 669,0312

10,5523 557,5260 x

= 8,7935 m = 346,1997 in



(Timmerhaus,2004)




= 0,0098 in/tahun




= 675,1885 kg/m3 × 9,8 m/det2 × 8,7935 m = 0,5631 atm

Po = 1 atm

P = 1 atm + 0,5631 atm = 1, 5631 atm

Pdesign = 1,2 × 1, 5631 = 1,8757 atm = 27,5651 psi


in 10,0

in)0098,0(10psi) 10,6(27,565psi)(0,8) 2(13.700

in) (166,1767 psi)(27,5651

nC0,6P2SE

Prt

=

+

−

=

+−

=

x







Fungsi : Untuk penyimpanan produk heptena dari hasil atas MD-105

selama 20 hari

Jumlah : 6 unit


ellipsiodal



-Tekanan = 1,2 atm = 17,635 psi

Perhitungan:

a. Volume Tangki

Laju alir = 5.636,1122 kg/jam x 24 jam/hari (Lampiran A) = 135.266,6928 kg/hari


= 2.705.333,856 Kg



856,333.705.23

mkgm

mkg

==ρ

Volume larutan untuk 1 tangki = 9590,259.4 /6 = 709,993 m3



= 851,9918 m3



Vs = 4






Hs : Dt = 5 : 4 , Maka :


Vs = 41 π Dt

2 Hs (Hs : Dt = 5 : 4)

Vs = 165 π Dt3

= 0,9812 Dt3

Vh = 24π Dt3

= 0,13083 Dt3

Volume tangki (Vt)

Vt = Vs + Vh

851,9918 = 0,9812 Dt3 + 0,13083 Dt3

851,9918 = 1,11203 Dt3

Dt = 9,1503 m = 360.2469in

r = ½ x Dt = ½ x (9,1503) = 4,5751 m = 180,1215in


Hs = 45 x Dt =

45 x 9,1503 m = 11,4378 m = 450,3057 in







= 851,9918

11,4378 709,993 x

= 9,5314 m = 375,2508 in



(Timmerhaus,2004)




= 0,0098 in/tahun





= 635,061 kg/m3 × 9,8 m/det2 × 9,5314 m = 0,5741 atm

Po = 1,2 atm

P = 1,2 atm + 0,5741 atm = 1,7741 atm

Pdesign = 1,2 × 1,7741 = 2,1289 atm = 31,2862 psi


in 1005,0

in)0098,0(10psi) 20,6(31,286psi)(0,8) 2(13.700

in) (180,1215 psi) (31,2862

nC0,6P2SE

Prt

=

+

−

=

+−

=

x






Fungsi : Untuk penyimpanan produk oktena dari hasil bawah MD-105

selama 20 hari

Jumlah : 2 unit


ellipsiodal



-Tekanan = 1,2 atm = 17,635 psi

Perhitungan:

a. Volume Tangki


= 40.403,0472 kg/hari


= 808.060,944 Kg




4808.060,943

mkgm

mkg

==ρ

Volume larutan untuk 1 tangki = 1.129,4285/2 = 564,7142 m3



= 677,6571 m3



Vs = 4






Hs : Dt = 5 : 4 , Maka :

Vs = 41 π Dt

2 Hs (Hs : Dt = 5 : 4)

Vs = 165 π Dt3

= 0,9812 Dt3

Vh = 24π Dt3

= 0,13083 Dt3

Volume tangki (Vt)

Vt = Vs + Vh

677,6571 = 0,9812 Dt3 + 0,13083 Dt3

677,6571 = 1,11203 Dt3

Dt = 8,4780 m = 333,7785 in

r = ½ x Dt = ½ x (8,4780) = 4,239 m = 166,8893 in


Hs = 45 x Dt =

45 x 8,4780 m = 10,5975 m = 417,2232 in


Volume tangki (Vt) = 677,6571m3






= 677,6571

5975,01 564,7142 x

= 8,8312 m = 347,684 in



(Timmerhaus,2004)




= 0,0098 in/tahun




= 715,46 kg/m3 × 9,8 m/det2 × 8,8312 m = 0,5992 atm

Po = 1,2 atm

P = 1,2 atm + 0,5992atm = 1,7992 atm

Pdesign = 1,2 × 1,7992 = 2,1590 atm = 31, 7285 psi


in 1003,0

in)0098,0(10psi) 50,6(31,728psi)(0,8) 2(13.700

in) (166,8893 psi)(31,7285

nC0,6P2SE

Prt

=

+

−

=

+−

=

x




Tebal tutup atas yang digunakan = 1/4 in LC.40 Condensor I (E-102)


Fungsi : Mengkondensasikan bahan yang keluar dari hasil atas destilasi I.

Jenis : Sheal - Tube Heat Exchanger

Jumlah : 1 unit


Flowrate, W = 13.942,0427 Kg/jam (Lampiran A)

= 13.942,0427 Kg/jam x 2,2046 lb/kg

= 30.737,1033 lb/jam T1 = 34 0C

= (340C x 1,8) + 32

= 93,20 0F

T2 = 340C

= (34 0C x 1,8) + 32

= 93,200 0F


Flowrate, W = 7.438,6594 kg/jam (Lampiran B)

= 7.438,6594 kg/jam x 2,2046 lb/kg

= 16.399,2685 lb/jam t1 = 15 0C

= (15 0C x 1,8) + 32

= 59 0F

t2 = 25 0C

= (25 0C x 1,8) + 32

= 77 0F




Fluida Panas Fluida Dingin Selisih T1 = 93,20 0F Temperatur yang lebih tinggi t2 = 77 0F ∆t2 = 16.20 °F T2 =93,20 0F Temperatur yang lebih rendah t1 = 59 0F ∆t1 = 34.20 °F

T1 – T2 = 0 °F Selisih t2 – t1 = 18 °F ∆t2 – ∆t1 = -18,00°F


24,1165

34,2016,202,3log

18,00-

ΔtΔt2,3log

ΔtΔtLMTD

1

2

12 =

=

−

= °F

(2) Tc dan tc

=+

=+

=2

20,9320,932

TTT 21

c 93,20 °F

=+

=+

=2

77592

ttt 21

c 68 °F

Dalam perancangan ini digunakan kondensor dengan spesifikasi:

- Diameter luar tube (OD) = ¾ in

- Jenis tube = 18 BWG

- Pitch (PT) = 1 in triangular pitch

- Panjang tube (L) = 20 ft - Pass = 4 P

a. Dari Tabel 8, hal. 840, Kern, 1965, kondensor untuk fluida panas gas dan




2

oo2

D

ft 9879,442F1165,24

FftjamBtu50


QA =×

⋅⋅

=×

=

Luas permukaan luar (a″) = 0,1963 ft2/ft (Tabel 10, Kern)

Jumlah tube, 4014,26/ftft 1963,0ft20

ft9879,442aL

AN 2

2

"t =×

=×

= buah

b. Dari Tabel 9, hal 842, Kern, 1965, nilai yang terdekat adalah 76 tube dengan

ID shell 12 in.

c. Koreksi UD

2

"t

ft 376,982ft2/ft 0,196367ft 20

aNLA

=

××=××=

Fftjam

Btu41,0535F24,1165x ft376,298

Btu/jam 295.412,6ΔtA

QU 22D °⋅⋅=

°=

⋅=


Fluida dingin : Air, tube

(3) Flow area tube,at’ = 0,334 in2 (Tabel 10, Kern, 1965)

n144aN

a'tt

t ××

= (Pers. (7.48), Kern, 1965)

=×

×=

4144 0,33476

ta 0,0441 ft 2

(4) Kecepatan massa:

t

t awG = (Pers. (7.2), Kern, 1965)

== 0,0441

516.399,268tG 372.129,0950 lbm/jam.ft 2


Pada tc = 68 °F

µ = 1.09 cP = 2,63682 lbm/ft2⋅jam (Gambar 14, Kern, 1965)

Dari tabel 10, Kern, untuk ¾ in OD, 18 BWG, diperoleh :

ID = 0,652 in = 0,0543 ft

µ×

= tt

GIDRe (Pers.(7.3), Kern, 1965)

=×

=63682,2

50372.129,090543,0Re t 7667,9568

Taksir jH dari Gambar 24 Kern (1965), diperoleh jH = 30 pada Ret = 7667,9568

(6) Pada tc = 68 °F


k = 0,347 Btu/jam lbm ft.°F (Tabel 5, Kern, 1965)

=

×=

3

13

1

347,063682,299,0.

kc µ 1,9594


(7) 3

1

t

i

k.c

IDkjHh

µ

××=ϕ

=××= 1,95940,0543

347,030t

ihϕ

375,6418

5579,26375,0

652,06418,753 =×=

×=

t

io

t

i

t

io

hODIDhh

ϕ

ϕϕ

(9) Karena viskositas rendah, maka diambil tϕ = 1 (Kern, 1965)

F ft Btu/jam 5579,26315579,263 o2=×=

×=

io

tt

ioio

h

hh ϕϕ

Fluida panas : shell

(3’) Flow area shell

2

T

's

s ftP144

BCDa×××

= (Pers. (7.1), Kern, 1965)

Ds = Diameter dalam shell = 15,25 in

B = Baffle spacing = 5 in

PT = Tube pitch = 1 in

C′ = Clearance = PT – OD

= 1 – 0,75 = 0,25 in

0,13241144

525,025,15=

×××

=sa ft 2

(4’) Kecepatan massa

s

s awG = (Pers. (7.2), Kern, 1965)

3484,153.2321324,0

30737,1033==sG lbm/jam.ft2

(5’) jamftlbLN

wGt

/6806,857620

30737,10333/23/2

" =×

==

(6’) Asumsi h-=ho = 100


ioh = F ft Btu/jam 5579,263 o2

FtThohio

hott cvcw09068,73)682,93(

1005579,32610068)( =−

++=−

++=

FtT

t wvf

05534,832

9068,732,932

=+

=+

=

Pada tf = 83,5534 °F

kf = 0,05587 Btu/jam lbm ft.°F

sf = 0,60

µ = 0,07784 cP

Dari gambar 12.9 (Kern, 1965), h-=ho = 110 Btu/jam ft2 0F

Pressure drop

Fluida dingin : sisi tube

(1) Untuk Ret = 7667,9568 f = 0,00027 ft2/in2 (Gambar 26, Kern, 1965)

s = 1 (Tabel 6, Kern, 1965)

φt = 1

(2) tφsID10105,22

nL2tGf

tΔP⋅⋅⋅⋅

⋅⋅⋅= (Pers. (7.53), Kern, 1965)

(1)(1)(0,0543))1010(5,22

4)()20(2)50372.129,09((0,00027)tΔP

×××⋅

×××= = 1,1718 psi

(3) Dari Gambar 27, Kern, 1965 diperoleh 2g'

2V = 0,019

psi 3040,0

.0,0171

(4).(4)2g'

2V.s

4nrΔP

=

=

=

∆PT = ∆Pt + ∆Pr

= 1,1718 psi + 3040,0 psi

= 1,4758 psi ∆Pt yang diperbolehkan = 10 psi


Fluida panas : sisi shell

(1′) Untuk Res = 74999.7676

f = 0,0009 ft2/in2 (Gambar 29, Kern, 1965)

φs =1

s = 0,9

(2′) BL x 12 1N =+

520 x 12 1N =+ = 48 (Pers. (7.43), Kern, 1965)

Ds = 15,25/12 = 1,2708 ft

(3′)

sϕ .s. eD . 1010.22,5

1)(N .sD . 2sG f.

sP+

=∆ (Pers. (7.44), Kern, 1965)

(1)(0,9)(0,0608) 1010.22,5

(48) (1,2708) 2484)(232.153,3 0,0009 sP×××

×××=∆ = 1,5686 psi

∆Ps yang diperbolehkan = 10 psi

LC.41 Condensor II (E-106)

Fungsi : Mengkondensasikan bahan yang keluar dari hasil atas

destilasi II.


Jumlah : 1 unit


Flowrate, W = 10.456,5684 Kg/jam (Lampiran A)

= 10.456,5684 Kg/jam x 2,2046 lb/kg

= 7.438,6594 lb/jam

T1 = 37 0C

= (370C x 1,8) + 32

= 98,6 0F

T2 = 370C

= (370C x 1,8) + 32

= 98,6 0F



Flowrate, W = 4.721,188002 kg/jam (Lampiran B)

= 4.721,188002 kg/jam x 2,2046 lb/kg

= 10.408.4923 lb/jam

t1 = 15 0C

= (15 0C x 1,8) + 32

= 59 0F

t2 = 25 0C

= (25 0C x 1,8) + 32

= 77 0F


Panas yang diserap (Q) = 197.817,7773 kJ/jam = 187.494,3390 Btu/jam (1) ∆t = beda suhu sebenarnya

Fluida Panas Fluida Dingin Selisih T1 = 98,6 0F Temperatur yang lebih tinggi t2 = 77 0F ∆t2 = 39,6 °F T2 =98,6 0F Temperatur yang lebih rendah t1 = 59 0F ∆t1 = 21,6 °F

T1 – T2 = 0 °F Selisih t2 – t1 = 18 °F ∆t2 – ∆t1 = 18 °F

29.730

21,639,62,3log

18,00

ΔtΔt2,3log

ΔtΔtLMTD

1

2

12 =

=

−

= °F

(2) Tc dan tc

=+

=+

=2

98,6 98,62

TTT 21c 98,6 °F

=+

=+

=2

77592

ttt 21

c 68 °F







d. Dari Tabel 8, hal. 840, Kern, 1965, kondensor untuk fluida panas gas dan




2

oo2

D

ft 126,1327F29.730

FftjamBtu50

Btu/jam 90187.494,33ΔtU

QA =×

⋅⋅

=×

=



ft 126,1327aL

AN 2

2

"t =×

=×

= buah

e. Dari Tabel 9, hal 842, Kern, 1965, nilai yang terdekat adalah 40 tube dengan

ID shell 10 in.

f. Koreksi UD

2

"t

ft 04,571ft2/ft 0,196340ft 20

aNLA

=

××=××=

Fftjam

Btu40.159F29,696x ft04,157

Btu/jam 90187.494,33ΔtA

QU 22D °⋅⋅=

°=

⋅=



naN

a ttt ×

×=

144

'

(Pers. (7.48), Kern, 1965)

=×

×=

4144 0,33440

ta 0,0232 ft 2


t

t awG = (Pers. (7.2), Kern, 1965)

==0232,0

310.408.492tG 448.749,3667 lbm/jam.ft 2


Pada tc = 68 °F




ID = 0,652 in = 0,0543 ft

µ×

= tt


=×

=63682,2

67448.749,360543,0Re t 9.246,7662

(8) Taksir jH dari Gambar 24 Kern (1965), diperoleh jH = 38pada Ret = 9.246,7662

(9) Pada tc = 68 °F



=

×=

3

13

1

347,063682,299,0.

kc µ 1,9594

(10) 3

1

t

i

k.c

IDkjHh

µ

××=ϕ

=××= 1,95940,0543

347,038t

ihϕ

475,5272

413,391775,0

652,05272,475 =×=

×=

t

io

t

i

t

io

hODIDhh

ϕ

ϕϕ


F ft Btu/jam 413,39171413.3917 o2=×=

×=

io

tt

ioio

h

hh ϕ

ϕ



2

T

's

s ftP144

BCDa×××

= (Pers. (7.1), Kern, 1965)


Ds = Diameter dalam shell = 12 in




= 1 – 0,75 = 0,25 in

0,10421144

525,012=

×××

=sa ft 2


s

s awG = (Pers. (7.2), Kern, 1965)

18221.237,041042,0

23052.8998==sG lbm/jam.ft2

(5’) jamftlbLN

wGt

/5740,984020

23052.89983/23/2

" =×

==

(6’) Asumsi h-=ho = 150


FtThohio

hott cvcw01470,76)686,98(

150413,391715068)( =−

++=−

++=

FtT

t wvf

037,872

1470,766,982

=+

=+

=

Pada tf = 87,37 °F


sf = 0,60

µ = 0,1393 cP


Pressure drop


(1) Untuk Ret = 9.246,7662 f = 0,00028 ft2/in2 (Gambar 26, Kern, 1965)

s = 1 (Tabel 6, Kern, 1965)

φt = 1


(2) tφsID10105,22

nL2tGf

tΔP⋅⋅⋅⋅

⋅⋅⋅= (Pers. (7.53), Kern, 1965)

(1)(1)(0,0543))1010(5,22

4)()20(2) 67448.749,36((0,00028)tΔP

×××⋅

×××= = 1,59045 psi


2V = 0,027

psi 4320,0

.0,0271

(4).(4)2g'

2V.s

4nrΔP

=

=

=


= 1,59045 psi + 4320,0 psi



(1′) Untuk Res = 96.615,8414


φs =1

s = 0,9

(2′) BL x 12 1N =+

520 x 12 1N =+ = 48 (Pers. (7.43), Kern, 1965)

Ds = 12/12 = 1 ft

(3′)

sϕ .s. eD . 1010.22,5

1)(N .sD . 2sG f.

sP+

=∆ (Pers. (7.44), Kern, 1965)

(1)(0,9)(0,0608) 1010.22,5

(48) (1) 2418)(221.237,0 0,0013 sP×××

×××=∆ = 1,8520 psi



LC.42 Condensor III (E-113)

Fungsi : Mengkondensasikan bahan yang keluar dari hasil atas destilasi IV.


Jumlah : 1 unit



= 1.025,317371 Kg/jam x 2,2046 lb/kg

= 2.260,4495 lb/jam

T1 = 30 0C

= (300C x 1,8) + 32

= 86 0F

T2 = 300C

= (300C x 1,8) + 32

= 86 0F



= 510,787047 kg/jam x 2,2046 lb/kg

= 1.126,0985 lb/jam

t1 = 15 0C

= (15 0C x 1,8) + 32

= 59 0F

t2 = 25 0C

= (25 0C x 1,8) + 32

= 77 0F





Fluida Panas Fluida Dingin Selisih T1 = 86 0F Temperatur yang lebih tinggi t2 = 77 0F ∆t2 = 27 °F T2 =86 0F Temperatur yang lebih rendah t1 = 59 0F ∆t1 = 9 °F


16,403

9272,3log

18,00

ΔtΔt2,3log

ΔtΔtLMTD

1

2

12 =

=

−

= °F

(2) Tc dan tc

=+

=+

=2

86 862

TTT 21c 86 °F

=+

=+

=2

77592

ttt 21

c 68 °F






g. Dari Tabel 8, hal. 840, Kern, 1965, kondensor untuk fluida panas gas dan




2

oo2

D

ft 24,7338F403,16

FftjamBtu50


QA =×

⋅⋅

=×

=



ft 24,7338aL

AN 2

2

"t =×

=×

= buah


h. Dari Tabel 9, hal 842, Kern, 1965, nilai yang terdekat adalah 24 tube dengan

ID shell 8 in.

i. Koreksi UD

2

"t

ft 94,2240ft2/ft 0,196324ft 20

aNLA

=

××=××=

Fftjam

Btu125,31F16,403x ft2240,94


QU 22D °⋅⋅=

°=

⋅=



n144aN

a'tt

t ××

= (Pers. (7.48), Kern, 1965)

=×

×=

4144 0,33424

ta 0,0139 ft 2


t

t awG = (Pers. (7.2), Kern, 1965)

==0139,0

1.126,0985tG 80.917.,2545 lbm/jam.ft 2


Pada tc = 68 °F



ID = 0,652 in = 0,0543 ft

µ×

= tt


=×

=63682,2

4580.917.,250543,0Re t 1.667,3515


(11) Taksir jH dari Gambar 24 Kern (1965), diperoleh jH = 7 pada Ret

=1.667,3515

(12) Pada tc = 68 °F



=

×=

3

13

1

347,063682,299,0.

kc µ 1,9594

(13) 3

1.

××=

kc

IDkjH

h

t

i µϕ

=××= 1,95940,0543

347,07t

ihϕ

87,5971

76,151175,0

652,087,5971 =×=

×=

t

io

t

i

t

io

hODIDhh

ϕ

ϕϕ


F ft Btu/jam 76,1511176,1511 o2=×=

×=

io

tt

ioio

h

hh ϕ

ϕ



2

T

's

s ftP144

BCDa×××

= (Pers. (7.1), Kern, 1965)





= 1 – 0,75 = 0,25 in


0,10421144

525,012=

×××

=sa ft 2


s

s awG = (Pers. (7.2), Kern, 1965)

621.693,3731042,0

2.260,4495==sG lbm/jam.ft2

(5’) jamftlbLN

wGt

/13.58682420

2.260,44953/23/2

" =×

==

(6’) Asumsi h-=ho = 100

ioh = F ft /jam76,1511Btu o2

FtThohio

hott cvcw02184,78)6886(

10076,151110068)( =−

++=−

++=

FtT

t wvf

01092,822

2184,78862

=+

=+

=

Pada tf = 1092,82 °F


sf = 0,60

µ = 0,1327 cP


Pressure drop



s = 1 (Tabel 6, Kern, 1965)

φt = 1

(2) tφsID10105,22

nL2tGf

tΔP⋅⋅⋅⋅

⋅⋅⋅= (Pers. (7.53), Kern, 1965)


(1)(1)(0,0543))1010(5,22

4)()20(2)80917,2545((0,00043)tΔP

×××⋅

×××= = 0,07942 psi


2V = 0,016

psi 2560,0

.0,0161

(4).(4)2g'

2V.s

4nrΔP

=

=

=


= 0,07942 psi + 2560,0 psi

= 0,3354 psi

∆Pt yang diperbolehkan = 10 psi


(1′) Untuk Res = 9.473,6556


φs =1

s = 0,9

(2′) BL x 12 1N =+

520 x 12 1N =+ = 48 (Pers. (7.43), Kern, 1965)

Ds = 8/12 = 0.6667 ft

(3′)

sϕ .s. eD . 1010.22,5

1)(N .sD . 2sG f.

sP+

=∆ (Pers. (7.44), Kern, 1965)

(1)(0,9)(0,0608) 1010.22,5

(24) (0,6667) 26)(21693.373 0,0022 sP×××

×××=∆ = 0,0079 psi



LC.43 Condensor (E-109)

Fungsi : Mengkondensasikan bahan yang keluar dari hasil atas destilasi III.


Jumlah : 1 unit



= 3.136,9596 Kg/jam x 2,2046 lb/kg

= 6.915,8482 lb/jam T1 = 63,5 0C

= (63,50C x 1,8) + 32

= 146,3 0F

T2 = 63,50C

= (63,50C x 1,8) + 32

= 146,3 0F



= 2.129,945049 kg/jam x 2,2046 lb/kg

= 4.695,7495 lb/jam t1 = 15 0C

= (15 0C x 1,8) + 32

= 59 0F

t2 = 25 0C

= (25 0C x 1,8) + 32

= 77 0F


Panas yang diserap (Q) = 89.244,69755 kJ/jam = 84.587,3197 Btu/jam (1) ∆t = beda suhu sebenarnya

Fluida Panas Fluida Dingin Selisih T1 = 146,3 0F Temperatur yang lebih tinggi t2 = 77 0F ∆t2 = 87,3 °F T2 =146,3 0F Temperatur yang lebih rendah t1 = 59 0F ∆t1 = 69,3 °F



78.139

69,3 87,32,3log

18,00

ΔtΔt2,3log

ΔtΔtLMTD

1

2

12 =

=

−

= °F

(2) Tc dan tc

=+

=+

=2

146,3 146,32

TTT 21c 146,3 °F

=+

=+

=2

77592

ttt 21

c 68 °F






j. Dari Tabel 8, hal. 840, Kern, 1965, kondensor untuk fluida panas gas dan




2

oo2

D

ft 21,6505F042,78

FftjamBtu50


QA =×

⋅⋅

=×

=



ft21,6505aL

AN 2

2

"t =×

=×

= buah

k. Dari Tabel 9, hal 842, Kern, 1965, nilai yang terdekat adalah 24 tube dengan

ID shell 8 in.

l. Koreksi UD

2

"t

ft 94,2240ft2/ft 0,196324ft 20

aNLA

=

××=××=

Fftjam

Btu489,11F78,193x ft2240,94


QU 22D °⋅⋅=

°=

⋅=




n144aN

a'tt

t ××

= (Pers. (7.48), Kern, 1965)

=×

×=

4144 0,33424

ta 0,0139 ft 2


t

t awG = (Pers. (7.2), Kern, 1965)

==0139,0

4.695,7495tG 337.419,1233 lbm/jam.ft 2


Pada tc = 68 °F



ID = 0,652 in = 0,0543 ft

µ×

= tt


=×

=63682,2

33337.419,120543,0Re t 6.952,7357

Taksir jH dari Gambar 24 Kern (1965), diperoleh jH = 26pada Ret = 6.952,7357

(6) Pada tc = 68 °F



=

×=

3

13

1

347,063682,299,0.

kc µ 1,9594


(7)

31

.

××=

kc

IDkjH

h

t

i µϕ

=××= 1,95940,0543

347,026t

ihϕ

325,3607

282,846975,0

652,0325.3607 =×=

×=

t

io

t

i

t

io

hODIDhh

ϕ

ϕϕ


F ft u/jam282,8469Bt1282,8469 o2=×=

×=

io

tt

ioio

h

hh ϕϕ



2

T

's

s ftP144

BCDa×××

= (Pers. (7.1), Kern, 1965)





= 1 – 0,75 = 0,25 in

0,10421144

525,012=

×××

=sa ft 2


s

s awG = (Pers. (7.2), Kern, 1965)

366.370,9041042,0

6.915,8482==sG lbm/jam.ft2

(5’) jamftlbLN

wGt

/41,56872420

6.915,84823/23/2

" =×

==

(6’) Asumsi h-=ho = 300



FtThohio

hott cvcw03021,108)683,146(

300 282,846930068)( =−

++=−

++=

FtT

t wvf

03010,1272

3021,1083,1462

=+

=+

=

Pada tf = 3010,127 °F


sf = 0,69

µ = 0,18143 cP


Pressure drop



s = 1 (Tabel 6, Kern, 1965)

φt = 1

(2) tφsID10105,22

nL2tGf

tΔP⋅⋅⋅⋅

⋅⋅⋅= (Pers. (7.53), Kern, 1965)

(1)(1)(0,0543))1010(5,22

4)()20(2) 33337.419,12((0,0003)tΔP

×××⋅

×××= = 0,96342 psi


2V = 0,016

psi 0,2560

.0,0161

(4).(4)2g'

2V.s

4nrΔP

=

=

=


= 0,96342 psi + 0,2560 psi




(1′) Untuk Res = 28.984,6615


φs =1

s = 0,66

(2′) BL x 12 1N =+

520 x 12 1N =+ = 48 (Pers. (7.43), Kern, 1965)

Ds = 8/12 = 0.6667 ft

(3′)

sϕ .s. eD . 1010.22,5

1)(N .sD . 2sG f.

sP+

=∆ (Pers. (7.44), Kern, 1965)

(1)(0,9)(0,0608) 1010.22,5

(24) (0,6667) 215)(28.984,66 0,0013 sP×××

×××=∆ = 0,0740 psi

∆Ps yang diperbolehkan = 10 psi LC.44 Condensor V (E-115)

Fungsi : Mengkondensasikan bahan yang keluar dari hasil atas destilasi V.


Jumlah : 1 unit



= 5.636,1122 Kg/jam x 2,2046 lb/kg

= 12.425,5654 lb/jam


T1 = 30 0C

= (30 0C x 1,8) + 32

= 86 0F

T2 = 30 0C

= (30 0C x 1,8) + 32

= 86 0F



= 4.721,188002 kg/jam x 2,2046 lb/kg

= 10.408,4923 lb/jam

t1 = 15 0C

= (15 0C x 1,8) + 32

= 59 0F

t2 = 25 0C

= (25 0C x 1,8) + 32

= 77 0F




Fluida Panas Fluida Dingin Selisih T1 = 86 0F Temperatur yang lebih tinggi t2 = 77 0F ∆t2 = 27 °F T2 =86 0F Temperatur yang lebih rendah t1 = 59 0F ∆t1 = 9 °F


403,16

9 272,3log

18,00

ΔtΔt2,3log

ΔtΔtLMTD

1

2

12 =

=

−

= °F

(2) Tc dan tc

=+

=+

=2

86 862

TTT 21c 86 °F

=+

=+

=2

77592

ttt 21

c 68 °F







Dari Tabel 8, hal. 840, Kern, 1965, kondensor untuk fluida panas gas dan fluida




2

oo2

D

ft 228,6137F403,16

FftjamBtu50

Btu/jam 90187.494,33ΔtU

QA =×

⋅⋅

=×

=



ft228,6137aL

AN 2

2

"t =×

=×

= buah

Dari Tabel 9, hal 842, Kern, 1965, nilai yang terdekat adalah 76 tube dengan ID shell

12 in.

Koreksi UD

2

"t

ft 298,3760ft2/ft 0,196376ft 20

aNLA

=

××=××=

Fftjam

Btu38,310F16,403x ft298,3760

Btu/jam 90187.494,33ΔtA

QU 22D °⋅⋅=

°=

⋅=



n144aN

a'tt

t ××

= (Pers. (7.48), Kern, 1965)

=×

×=

4144 0,33476

ta 0,0441 ft 2



t

t awG = (Pers. (7.2), Kern, 1965)

==0441,0

310.408,492tG 236.183,8771 lbm/jam.ft 2


Pada tc = 68 °F



ID = 0,652 in = 0,0543 ft

µ×

= tt


=×

=63682,2

71236.183,870543,0Re t 4.866,7191

Taksir jH dari Gambar 24 Kern (1965), diperoleh jH = 17 pada Ret = 4.866,7191

(6) Pada tc = 68 °F



=

×=

3

13

1

347,063682,299,0.

kc µ 1,9594

(7)

31

.

××=

kc

IDkjH

h

t

i µϕ

=××= 1,95940,0543

347,017t

ihϕ

325,3607

282,846975,0

652,0325,3607 =×=

×=

t

io

t

i

t

io

hODIDhh

ϕ

ϕϕ



F ft Btu/jam 282,84691282,8469 o2=×=

×=

io

tt

ioio

h

hh ϕϕ



2

T

's

s ftP144

BCDa×××

= (Pers. (7.1), Kern, 1965)





= 1 – 0,75 = 0,25 in

0,10411144

525,012=

×××

=sa ft 2


s

s awG = (Pers. (7.2), Kern, 1965)

92119.361,811041,0

412.425,565==sG lbm/jam.ft2

(5’) jamftlbLN

wGt

/6366,437620

412.425,5653/23/2

" =×

==

(6’) Asumsi h-=ho = 400


FtThohio

hott cvcw05440,78)6886(

400282,846940068)( =−

++=−

++=

FtT

t wvf

02720,822

5440,78862

=+

=+

=

Pada tf = 2720,82 °F



sf = 0,68

µ = 0,1437 cP


Pressure drop



s = 1 (Tabel 6, Kern, 1965)

φt = 1

(2) tφsID10105,22

nL2tGf

tΔP⋅⋅⋅⋅

⋅⋅⋅= (Pers. (7.53), Kern, 1965)

(1)(1)(0,0543))1010(5,22

4)()20(2) 71236.183,87((0,0003)tΔP

×××⋅

×××= = 0,47204 psi


2V = 0,0056

psi 0,0896

.0,00561

(4).(4)2g'

2V.s

4nrΔP

=

=

=


= 0,47204 psi + 0,0896 psi



(1′) Untuk Res = 71.405,7306


φs =1


s = 0,68

(2′) BL x 12 1N =+

520 x 12 1N =+ = 48 (Pers. (7.43), Kern, 1965)

Ds = 12/12 = 1 ft

(3′)

sϕ .s. eD . 1010.22,5

1)(N .sD . 2sG f.

sP+

=∆ (Pers. (7.44), Kern, 1965)

(1)(0,68)(0,0608) 1010.22,5

(76) (1) 2192)(119.361,8 0,0014 sP×××

×××=∆ = 0,3237 psi



LAMPIRAN D

PERHITUNGAN SPESIFIKASI ALAT UTILITAS

LD.1 Screening (SC)

Fungsi : menyaring partikel-partikel padat yang besar

Jenis : bar screen

Jumlah : 1

Bahan konstruksi : stainless steel

Kondisi operasi :

Temperatur = 30 °C

Densitas air (ρ) = 995,68 kg/m3 (Geankoplis, 1997)


Laju alir volume (Q) = 3kg/m995,68

s jam/36001kg/jam 1.556,9329 × = 0,00043 m3/s

Dari tabel 5.1 Physical Chemical Treatment of Water and Wastewater

Ukuran bar :

Lebar bar = 5 mm; Tebal bar = 20 mm;

Bar clear spacing = 20 mm; Slope = 30°

Direncanakan ukuran screening:

Panjang screen = 2 m

Lebar screen = 2 m

Misalkan, jumlah bar = x

Maka, 20x + 20 (x + 1) = 2000

40x = 1980

x = 49,5 ≈ 50 buah

Luas bukaan (A2) = 20(50 + 1) (2000) = 2040000 mm2 = 2,04 m2

Untuk pemurnian air sungai menggunakan bar screen, diperkirakan Cd = 0,6 dan

30% screen tersumbat.


Head loss (∆h) = 22

2

22

2d

2

(2,04) (0,6) (9,8) 2(0,00043)

A C g 2Q

=

= 0,6.10-8 m dari air

= 0,0006 mm dari air

1200

1200

20

Gambar LD.1 Sketsa sebagian bar screen , satuan mm (dilihat dari atas)

LD.2 Bak Sedimentasi (BS) Fungsi : untuk mengendapkan lumpur yang terikut dengan air. Jumlah : 1

Jenis : beton kedap air

Kondisi operasi :

Temperatur = 30 °C

Tekanan = 1 atm

Densitas air (ρ) = 995,68 kg/m3 = 62,1585 lbm/ft3 (Geankoplis, 1997)

Laju alir massa (F) = 1.556,9329 kg/jam = 1,3482 lbm/s

Laju air volumetrik, /sft 0,0153lbm/ft 62,1585lbm/s 1,3482

ρFQ 3

3 ===

= 0,92031 ft3/min

Desain Perancangan : Bak dibuat dua persegi panjang untuk desain efektif (Kawamura, 1991).

Perhitungan ukuran tiap bak : Kecepatan pengendapan 0,1 mm pasir adalah (Kawamura, 1991) :

0υ = 1,57 ft/min atau 8 mm/s

Desain diperkirakan menggunakan spesifikasi :


Kedalaman tangki 10 ft

Lebar tangki 2 ft

Kecepatan aliran ft/min 0,0460ft2ft x 10/minft 0,92031

AQv

3

t

===

Desain panjang ideal bak : L = K

0υh v (Kawamura, 1991)

dengan : K = faktor keamanan = 1,5 h = kedalaman air efektif ( 10 – 16 ft); diambil 10 ft.

Maka : L = 1,5 (14/1,57) . 0,0460 = 0,4396 ft

Diambil panjang bak = 3/4 ft = 0,2286 m

Uji desain :

Waktu retensi (t) : QVat =

min/ft 0,92031ft 1) x 2 x (10

3

3

= = 21,737 menit

Desain diterima ,dimana t diizinkan 6 – 16 menit (Kawamura, 1991).

Surface loading : =AQ

=

= 3,44246 gpm/ft2

Desain diterima, dimana surface loading diizinkan diantara 4 – 10 gpm/ft2 (Kawamura, 1991).

Headloss (∆h); bak menggunakan gate valve, full open (16 in) :

∆h = K v2

2 g

= 0,12 [0,04601 ft/min. (1min/60s) . (1 m/3,2808 ft) ]2 2 (9,8 m/s2)

= 3,3456. 10-10 m dari air

= panjang x lebar x tinggi laju alir volumetrik

laju alir volumetrik luas permukaan masukan air

0,92031 ft3/min (7,481 gal/ft3) 2 ft x 1 ft


LD.3 Tangki Pelarutan Alum [Al2(SO4)3] (TP-01)

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

Kondisi operasi :

Temperatur = 30 °C

Tekanan = 1 atm

Al2(SO4)3 yang digunakan = 50 ppm

Al2(SO4)3 yang digunakan berupa larutan 30 % (% berat)

Laju massa Al2(SO4)3 = 0,07784 kg/jam

Densitas Al2(SO4)3 30 % = 1363 kg/m3 = 85,0889 lbm/ft3 (Perry, 1999)

Kebutuhan perancangan = 1 hari

Faktor keamanan = 20 %

Perhitungan:

Ukuran Tangki

Volume larutan, 3l kg/m13630,3

hari30jam/hari24kg/jam 0,07784V

×××

=

= 0,1370 m3

Volume tangki, Vt = 1,2 × 0,1370 m3 = 0,16448 m3

Direncanakan perbandingan diameter dengan tinggi silinder tangki, D : H = 3 : 2

33

23

2

πD83m 0,16448

D23πD

41m 0,16448

HπD41V

=

=

=

Maka: D = 0,5117 m ; H = 0,7676 m

Tinggi cairan dalam tangki = silindervolume

silindertinggixcairanvolume


= )0,16448(

) 0,7676)(0,137074( = 0,6396 m = 2,0986 ft

Tebal Dinding Tangki

Tekanan hidrostatik, Phid = ρ x g x l

= 1363 kg/m3 x 9,8 m/det2 x 0,6396 m

= 8544,4483 kg/m.det2

= cm

mxkg

gx100

11

1000m.det

kg 8544,4483 2

= 85444,48343 g/cm.det2 = 1,239266 psia

Tekanan udara luar, Po = 1 atm = 14,696 psia

Poperasi = 14,696 psia + 1,239266 psia = 15,9352 psia


Maka, Pdesign = (1,05) (15,9352 psia)

= 16,7320 psia



Tebal shell tangki:

in 0,01667m 0,0004234psia) 01,2(16,732a)(0,8)2(12650psi

m) 0,5117 psia)( (16,73201,2P2SE

PDt

==−

=

−=

Faktor korosi = 0,125 in

Maka tebal shell yang dibutuhkan = 0,01667 in + 0,125 in = 0,141671 in

Daya Pengaduk

Jenis pengaduk : flat 6 blade turbin impeller

Jumlah baffle : 4 buah

Untuk turbin standar (McCabe, 1999), diperoleh:

Da/Dt = 1/3 ; Da = 1/3 x 0,5117 m = 0,170581 m

E/Da = 1 ; E = 0,170581 m

L/Da = ¼ ; L = ¼ x 0,170581 m = 0,0426 m

W/Da = 1/5 ; W = 1/5 x 0,170581 m = 0,03411 m


J/Dt = 1/12 ; J = 1/12 x 0,5117 m = 0,0426 m

dengan :

Dt = diameter tangki

Da = diameter impeller

E = tinggi turbin dari dasar tangki

L = panjang blade pada turbin

W = lebar blade pada turbin

J = lebar baffle

Kecepatan pengadukan, N = 1 putaran/det

Viskositas Al2(SO4)3 30 % = 6,72⋅10-4 lbm/ft⋅detik ( Othmer, 1967)

Bilangan Reynold,

( )μDNρ

N2

aRe = (Geankoplis, 1997)

( )( )( )39657,452

106,72 x3,28080,170581185,0889

N 4

2

Re =⋅

=−

NRe > 10.000, maka perhitungan dengan pengadukan menggunakan rumus:

c

5a

3T

gρ.D.nK

P = (McCabe,1999)

KT = 6,3 (McCabe,1999)

Hp 0,001663 ft.lbf/det 5501Hp x ft.lbf/det 0,914663

.detlbm.ft/lbf 32,174)lbm/ft (85,0889ft) 3,2808.(0,170581put/det) (1 6,3P 2

353

=

=

×=

Efisiensi motor penggerak = 80 %

Daya motor penggerak = 8,0

0,001663 = 0,002078781 hp

Maka daya motor yang dipilih 0,05 hp

LD.4 Tangki Pelarutan Soda Abu [Na2CO3] (TP-02)

Fungsi : Membuat larutan soda abu (Na2CO3)




Jumlah : 1

Kondisi operasi :

Temperatur = 30 °C

Tekanan = 1 atm

Na2CO3 yang digunakan = 27 ppm

Na2CO3 yang digunakan berupa larutan 30 % (% berat)

Laju massa Na2CO3 = 0,0420 kg/jam

Densitas Na2CO3 30 % = 1327 kg/m3 = 82,845 lbm/ft3 (Perry, 1999)



Perhitungan Ukuran Tangki


hari30jam/hari24kg/jam0420,0V

×××

=

= 0,0760 m3

Volume tangki, Vt = 1,2 × 0,060 m3

= 0,0912 m3


( )( )

33

23

2

πD83m 0,0912

3/2DπD41m 0,0912

HπD41V

=

=

=

Maka: D = 0,2838 m ; H = 0,4257 m



= ) 0,0912(

)4257,0)(0912,0( = 0,3547 m = 1,1640 ft




= 1327 kg/m3 x 9,8 m/det2 x 0,3547 m

= 4.614,002105 kg/m.det2

= cm

mxkg

gx100

11

1000m.det

kg 054.614,0021 2

= 46.140,02105 g/cm.det2 = 0,6692 psia





= 16,1334 psia



Tebal shell tangki:

in 0,0089m 0,000226psia) 41,2(16,133a)(0,8)2(12650psi

0,2838m) psia)( (16,13341,2P2SE

PDt

==−

=

−=



Daya Pengaduk




Da/Dt = 1/3 ; Da = 1/3 x 0,2838 m = 0,0946 m

E/Da = 1 ; E = 0,0946 m

L/Da = ¼ ; L = ¼ x 0,0946 m = 0,0236 m

W/Da = 1/5 ; W = 1/5 x 0,0946 m = 0,0189 m

J/Dt = 1/12 ; J = 1/12 x 0,2838 m = 0,03236 m

dengan :







J = lebar baffle


Viskositas Na2CO3 30 % = 3,69⋅10-4 lbm/ft⋅detik (Othmer, 1967)

Bilangan Reynold,

( )μDNρ

N2


( )( )( )320.157,175

103,69 x3,28080,0946182,845

N 4

2

Re =⋅

=−

NRe > 10.000, maka perhitungan dengan pengadukan menggunakan rumus :

c

5a

3T

gρ.D.nK

P = ( McCabe,1999)

KT = 6,3 (McCabe,1999)

Hp 0,000084ft.lbf/det 5501hp x ft.lbf/det 0,0467

.detlbm.ft/lbf 32,174)lbm/ft (82,845ft) 0,0946.(3,2808put/det) 6,3.(1P 2

353

=

=

×=



0,000084 = 0,000106Hp


LD.5 Tangki Pelarutan Asam Sulfat (H2SO4) (TP-03)

Fungsi : Membuat larutan asam sulfat



Bahan konstruksi : Low Alloy Steel SA–203 grade A

Kondisi operasi :

Temperatur = 30 °C

Tekanan = 1 atm

H2SO4 yang digunakan mempunyai konsentrasi 5 % (% berat)

Laju massa H2SO4 = 0,03630 kg/jam

Densitas H2SO4 = 1061,7 kg/m3 = 66,2801 lbm/ft3 (Perry, 1999)



Perhitungan :

Volume larutan, 3l kg/m1061,70,05

jam24hari30kg/jam 0,03630V

×××

= = 0,4923 m3



( )

33

23

2

πD83m 0,5908

3/2DπD41m 0,5908

HπD41V

=

=

=

Maka, D = 1,8381 m ; H = 2,7572 m

Tinggi larutan H2SO4 dalam tangki = silindervolume


= 0,5908752,24923,0 x

= 2,2977 m= 7,5384 ft




= 1061,7 kg/m3 x 9,8 m/det2 x 2,2977 m

= 23.907,20048 kg/m.det2

= cm

mxkg

gx100

11

1000m.det

kg 4823.907,200 2

= 239.072,0048 g/cm.det2 = 3,4674 psia





= 19,0716 psia



Tebal shell tangki:

in0,0682m 0,00173psia) 61,2(19,071a)(0,8)2(12650psi

m) 1,8381 psia)( (19,07161,2P2SE

PDt

==−

=

−=



Daya Pengaduk Jenis pengaduk : flat 6 blade turbin impeller



Da/Dt = 1/3 ; Da = 1/3 x 1,8381 m = 0,6127 m

E/Da = 1 ; E = 0,6127 m

L/Da = ¼ ; L = ¼ x 0,6127 = 0,1531 m

W/Da = 1/5 ; W = 1/5 x 0,6127 = 0,1225 m

J/Dt = 1/12 ; J = 1/12 x 1,8381 = 0,1531 m



Viskositas H2SO4 5 % = 0,012 lbm/ft⋅detik (Othmer, 1967)

Bilangan Reynold,

( )μDNρN

2a

Re = (Geankoplis, 1983)

( )( )722.320,326

0,012)3,2808(0,61271 66,2801

N2

Re =×

=


c

5a

3T

gρ.D.nK

P = (McCabe, 1999)

KT = 4,1 (McCabe, 1999)

Hp 0,5041 ft.lbf/det 5501Hpx

.detlbm.ft/lbf 32,174)lbm/ft (66,2801ft) 3,2808 .(0,6127put/det) (1 4,1P 2

353

=

×=



0,5041 = 0,6301 Hp

Maka daya motor yang dipilih 1 hp

LD.6 Tangki Pelarutan NaOH (TP-04)

Fungsi : Tempat membuat larutan NaOH


Bahan konstruksi : Carbon Steel SA-283 grade C

Jumlah : 1 unit

Kondisi operasi :

Temperatur = 30 °C

Tekanan = 1 atm

Laju alir massa NaOH = 0,01284 kg/jam

Waktu regenerasi = 24 jam

NaOH yang dipakai berupa larutan 4% (% berat)

Densitas larutan NaOH 4% = 1518 kg/m3 = 94,7689 lbm/ft3 (Perry, 1999)



Faktor keamanan = 20%

Perhitungan :

Volume larutan, (V1) =)kg/m8(0,04)(151

jam/hari) hari)(24kg/jam)(30 (0,012843

= 0,1523 m3

Volume tangki = 1,2 x 0,1523 m3 = 0,1828 m3

Volume silinder tangki (Vs) = 4

HsDiπ 2

(Brownell,1959)

Ditetapkan perbandingan tinggi tangki dengan diameter tangki Hs : Di = 3 : 2

Maka : Hs : Di = 3 : 2

Maka : Vs= ( )4

232 DiDiπ

0,1523 = ( )8

Di3Di2π

Di = 0,5688 m

Hs = 3/2 Di = 3/2 (0,5688)

= 0,8532 m



= 3

3

m31828,0m))(0,8532m(0,1523 = 0,7110 m



= 1518 kg/m3 x 9,8 m/det2 x 0,7110 m

= 10.577,8499 kg/m.det2

= cm

mxkg

gx100

11

1000m.det

kg 910.577,849 2

= 105.778,4994 g/cm.det2 = 1,5341 psia





= 17,0416 psia




Tebal shell tangki:

in 0,0188m 0,000479psia) 61,2(17,041a)(0,8)2(12650psi

m) 0,5688 psia)( (17,04161,2P2SE

PDt

==−

=

−=


Maka tebal shell yang dibutuhkan = 0,0188in + 0,125 in = 0,1438 in

Daya Pengaduk




Da/Dt = 1/3 ; Da = 1/3 x 0,5688 m = 0,1896 m

E/Da = 1 ; E = 0,1896 m

L/Da = ¼ ; L = ¼ x 0,1896 m = 0,0474 m

W/Da = 1/5 ; W = 1/5 x 0,1896 m = 0,0379 m

J/Dt = 1/12 ; J = 1/12 x 0,5688 m = 0,0474 m

dengan :






J = lebar baffle


Viskositas NaOH 4% = 4,302 . 10-4 lbm/ft.det (Othmer, 1967)

Bilangan Reynold,

( )μDNρ

N2



( )( )4371,249.58

10302,4)2808,3 0,1896(1 94,7689

N 4

2

Re =⋅

×=

−


c

5a

3T

gρ.D.nK

P = ( McCabe,1999)

KT = 4,1 (McCabe,1999)

Hp00000538,0ft.lbf/det 5501hpx

.detlbm.ft/lbf 32,174)lbm/ft (94,7662ft) .(0,1896put/det) 4,1.(1P 2

353

=

=



00000538,0 = 0,00000672 Hp


LD.7 Tangki Pelarutan Kaporit [Ca(ClO)2] (TP-05)

Fungsi : Membuat larutan kaporit [Ca(ClO)2]



Kondisi operasi:

Temperatur = 30 °C

Tekanan = 1 atm

Ca(ClO)2 yang digunakan = 2 ppm

Ca(ClO)2 yang digunakan berupa larutan 70 % (% berat)

Laju massa Ca(ClO)2 = 0,002582 kg/jam

Densitas Ca(ClO)2 70 % = 1272 kg/m3 = 79,4088 lbm/ft3 (Perry, 1997)



Perhitungan :


hari9024jam/harikg/jam0,002582V

×××

= = 0,109 m3



Direncanakan perbandingan diameter dengan tinggi tangki, D : H = 3 : 2

33

23

2

πD83m0,130

D23πD

41m0,130

HπD41V

=

=

=

Maka, D = 0,407 m ; H = 0,611 m



= )130,0(

)611,0)(109,0( = 0,509 m

Tebal tangki :


= 1272 kg/m3 x 9,8 m/det2 x 0,509 m

= 6350,4 kg/m.det2

= cm

mxkg

gx100

11

1000m.det

kg6350,4 2

= 63504 kg/cm.det2 = 0,925 psia





= 16,402 psia



Tebal shell tangki:

in 0,013m 0,00033psia) 1,2(16,402a)(0,8)2(12650psi

m) 0,407 psia)( (16,4021,2P2SE

PDt

==−

=

−=




Daya Pengaduk :




Da/Dt = 1/3 ; Da = 1/3 x 0,407 m = 0,135 m

E/Da = 1 ; E = 0,135

L/Da = ¼ ; L = 1/4 x 0,135 m = 0,0339 m

W/Da = 1/5 ; W = 1/5 x 0,135 m = 0,0271 m

J/Dt = 1/12 ; J = 1/12 x 0,407 m = 0,0339 m

dengan :






J = lebar baffle


Viskositas kaporit = 6,7197⋅10-4 lbm/ft⋅detik (Othmer, 1967)

Bilangan Reynold,

( )µ

ρ 2

ReaDN

N = (Pers. 3.4-1, Geankoplis, 1983)

( )( )( )23683,3622

107194,63,28080,1351 79,4088

4

2

Re =⋅

×=

−N

NRe < 10.000, maka perhitungan dengan pengadukan menggunakan rumus:

cRe

5a

3T

gNρ.D.nK

P =

KT = 4,05


hp 10.7035,8

ft.lbf/det 5501hp x

).detlbm.ft/lbf )(32,17(4850,0323)lbm/ft (79,4088ft) .(0,135put/det) 4,05.(1P

7

2

353

−=

=


Daya motor penggerak = 8,010.7035,8 7−

= 1,0879.10-6 Hp

Maka daya motor yang dipilih 0,05 Hp

LD.8 Clarifier (CL)

Fungsi : Memisahkan endapan (flok-flok) yang terbentuk karena

penambahan alum dan soda abu

Tipe : External Solid Recirculation Clarifier

Bentuk : Circular desain

Jumlah : 1 unit

Bahan konstruksi : Carbon steel SA-283, Grade C

Data:

Laju massa air (F1) = 1.556,933007 kg/jam

Laju massa Al2(SO4)3 (F2) = 0,0778 kg/jam

Laju massa Na2CO3 (F3) = 0,0420 kg/jam

Laju massa total, m = 1.557,052891 kg/jam = 0,4325 kg/detik

Densitas Al2(SO4)3 = 2710 kg/m3 (Perry, 1999)

Densitas Na2CO3 = 2533 kg/m3 (Perry, 1999)

Densitas air = 996,2 kg/m3 (Perry, 1999)

Reaksi koagulasi:

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

Perhitungan:

Dari Metcalf & Eddy, 1984, diperoleh :

Untuk clarifier tipe upflow (radial):

Kedalaman air = 3-10 m

Settling time = 1-3 jam

Dipilih : kedalaman air (H) = 3 m, waktu pengendapan = 1 jam


Diameter dan Tinggi clarifier

Densitas larutan,

( )

2533 0,0420

2710 0,0778

2,996 071.556,9330

071.556,9330

++=ρ

ρ = 996,247 kg/m3 = 0,996 gr/cm3

Volume cairan, V = 3m1,5629 996,247

jam1kg/jam 071.556,9330=

×

V = 1/4πD2H

D = m3318,033,14

5629,14)

πH4V(

1/21/2 =

×

×=

Maka, diameter clarifier = 0,3318 m

Tinggi clarifier = 1,5 D = 0,4977 m



= 996,247 kg/m3 x 9,8 m/det2 x 0,4977 m

= 4.859,5918 kg/m.det2

= cm

mxkg

gx100

11

1000m.det

kg 4.859,5918 2

= 48.595,91825 kg/cm.det2 = 0,7048 psia





= 16,1708 psia




Tebal shell tangki:

in 0,01044m 0,000265psia) 81,2(16,170a)(0,8)2(12650psi

m) 0,3318 psia)( (16,17081,2P2SE

PDt

==−

=

−=

1Faktor korosi = 0,125 in


Desain torka yang diperlukan untuk operasi kontinu yang diperlukan untuk

pemutaran (turnable drive) : (Azad, 1976)

T, ft-lb = 0,25 D2 LF

Faktor beban (Load Factor) : 30 lb/ft arm (untuk reaksi koagulasi sedimentasi )

Sehingga : T = 0,25 [(0,3318 m).(3,2808 ft/m) ]2.30

T = 8,8889 ft-lb

Daya Clarifier

P = 0,006 D2 (Ulrich, 1984)

dimana: P = daya yang dibutuhkan, kW

Sehingga,

P = 0,006 × (0,3318)2 = 0,000660 kW = 8,85 x 10-9 Hp

Maka daya motor yang dipilih 0,05 Hp

LD.9 Sand Filter (SF)

Fungsi : Menyaring partikel – partikel yang masih terbawa dalam air

yang keluar dari clarifier

Bentuk : Silinder tegak dengan alas dan tutup ellipsoidal

Bahan konstruksi : Carbon steel SA-283 grade C

Jumlah : 1

Kondisi operasi :

Temperatur = 30 °C

Tekanan = 1 atm

Laju massa air = 1.556,933007 kg/jam

Densitas air = 995,68 kg/m3 = 62,1585 lbm/ft3 (Geankoplis, 1997)

Tangki filter dirancang untuk penampungan ¼ jam operasi.


Direncanakan volume bahan penyaring =1/3 volume tangki

Ukuran Tangki Filter

Volume air, 3a kg/m995,68

jam0,25 kg/jam 071.556,9330V

×= = 0,3909 m3

Volume total = 4/3 x 0,3909 m3 = 0,5212 m3

Faktor keamanan 20 %, volume tangki = 1,05 x 0,5212 = 0,6254 m3

Volume silinder tangki (Vs) = 4

HsDi. 2π

Direncanakan perbandingan tinggi tangki dengan diameter tangki Hs : Di = 3 : 1

Vs = 4.3 3Diπ

0,6254 m3 = 4.3 3Diπ

Di = 0,6462 m; H = 1,9278 m

Tinggi penyaring = ½ x 1,9278 m = 0,9639 m

Tinggi air = ½ x 1,9278 m = 0,9639 m

Perbandingan tinggi tutup tangki dengan diameter dalam adalah 1 : 4

Tinggi tutup tangki = ¼ (0,6426) = 0,1606 m

Tekanan hidrostatis, Phid = ρ x g x l

= 995,68 kg/m3 x 9,8 m/det2 x 0,9639 m

= 9.405,4272 kg/m.det2

= cm

mxkg

gx100

11

1000m.det

kg 9.405,4272 2

= 94.054,2724 kg/cm.det2 = 1,3641 psia





= 16,8631 psia




Tebal shell tangki:

in 0,0210m 0,000535psia) 11,2(16,863a)(0,8)2(12650psi

m) 0,6426 psia)( (16,86311,2P2SE

PDt

==−

=

−=



LD.10 Penukar Kation/Cation Exchanger (CE)

Fungsi : Mengurangi kesadahan air

Bentuk : Silinder tegak dengan alas dan tutup ellipsoidal


Kondisi operasi :

Temperatur = 30 °C

Tekanan = 1 atm

Data :

Laju massa air = 117,0011 kg/jam

Densitas air = 995,68 kg/m3 = 62,1985 lbm/ft3 (Geankoplis,1997)



Ukuran Cation Exchanger

Dari Tabel 12.4, The Nalco Water Handbook, diperoleh:

- Diameter penukar kation = 2 ft = 0,609 m

- Luas penampang penukar kation = 3,14 ft2

Tinggi resin dalam cation exchanger = 2,5 ft = 0,7620 m

Tinggi silinder = 1,2 × 2,5 ft

= 3,0 ft


Diameter tutup = diameter tangki = 2 ft

Rasio axis = 2 : 1

Tinggi tutup = ( ) ft 1 221

= = 0,3048

(Brownell,1959)

Sehingga, tinggi cation exchanger = 3,0 ft + 2(1) ft = 5 ft = 1,524 m



= 995,68 kg/m3 x 9,8 m/det2 x 1,524 m

= 14.870,860 kg/m.det2

= cm

mxkg

gx100

11

1000m.det

kg14.870,860 2

= 148.708,60 kg/cm.det2 = 2,156 psia





= 17,695 psia



Tebal shell tangki:

in 0,0209m 0,00053psia) 1,2(17,695a)(0,8)2(12650psi

m) 9psia)(0,60 (17,6951,2P2SE

PDt

==−

=

−=




LD.11 Tangki Penukar Anion (anion exchanger) (AE)

Fungsi : Mengikat anion yang terdapat dalam air umpan ketel

Bentuk : Silinder tegak dengan tutup atas dan bawah elipsoidal

Bahan konstruksi : Carbon Steel SA-283 grade C

Jumlah : 1

Kondisi operasi :

Temperatur = 30 oC

Tekanan = 1 atm

Laju massa air = 117,0011 kg/jam

Densitas air = 995,68 kg/m3 (Geankoplis, 1997)



Ukuran Anion Exchanger

Dari Tabel 12.3, The Nalco Water Handbook, diperoleh:

- Diameter penukar anion = 1 ft = 0,3048 m

- Luas penampang penukar anion = 3,14 ft2

Tinggi resin dalam anion exchanger = 2,5 ft

Tinggi silinder = 1,2 × 2,5 ft

= 3 ft = 0,9144 m


Rasio axis = 2 : 1

Tinggi tutup = ( ) m 0,15240,304821

= (Brownell,1959)

Sehingga, tinggi anion exchanger = 0,9144 + 2(0,1524) = 3,304 m


Tekanan hidrostatik Phid = ρ x g x l

= 996,24 kg/m3 x 9,8 m/det2 x 3,304 m

= 32.247,127 kg/m.det2


= cm

mxkg

gx100

11

1000m.det

kg32.247,127 2

= 322.471,27 kg/cm.det2 = 4,677 psia





= 20,341 psia



Tebal shell tangki:

in 0,0241m 0,00061psia) 1,2(20,341a)(0,8)2(12650psi

m) 48psia)(0,30 (20,3411,2P2SE

PDt

==−

=

−=



LD.12 Refrigerator (RF)

Fungsi : Mendingin air pendingin hingga 15 0C

Jenis : Refrigerator

Bahan konstruksi : Carbon Steel SA–53 Grade B

Jumlah : 1 buah

Kondisi operasi :

Suhu air masuk menara (TL2) = 25 °C = 77 °F

Suhu air keluar menara (TL1) = 15 °C = 59 °F

Suhu udara (TG1) = 25 °C = 77°F

Densitas air (15 °C) = 995,68 kg/m3 (Perry, 1999)

Laju massa air pendingin = 25.369,5565kg/jam

= 55.929,7242 lb/jam


H2 tetrafluoroethane, T= 10 0F- 10 0F = 0 0F = 103,015 btu/lbm

H4 tetrafluoroethane, T= 70 0F + 10 0F = 80 0F = 37,978 btu/lbm

( )sbtuQ

lbmbtuQ

HHQ

m

C

CC

/9677,859/978,37015,10342

=−

=−

=

LD.13 Tangki Utilitas-01 (TU-01)

Fungsi : Menampung air untuk didistribusikan



Kondisi penyimpanan : Temperatur 25°C dan tekanan 1 atm

Jumlah : 1 unit

Kondisi operasi :

Temperatur = 30 oC

Laju massa air = 1.556,9330 kg/jam

Densitas air = 995,68 kg/m3 = 62,1586 lbm/ft3 (Geankoplis, 1997)


Perhitungan :


jam3kg/jam 1.556,9330V

×= = 4,6910 m3


Direncanakan perbandingan diameter dengan tinggi silinder, D : H = 5 : 6

33

23

2

πD103m 5,6292

D56πD

41m 5,6292

HπD41V

=

=

=

D = 1,8141 m ; H = 2,1769 m




= ) 6292,5(

)1769,2)( 4,6910( = 1,8141 m = 5,9518 ft



= 995,68 kg/m3 x 9,8 m/det2 x 1,8141 m

= 17.701,6918 kg/m.det2

= cm

mxkg

gx100

11

1000m.det

kg 817.701,691 2

= 177.016,918 kg/cm.det2 = 2,5674 psia





= 18,1265 psia



Tebal shell tangki:

in 0,0640m 0,00162psia) 51,2(18,126a)(0,8)2(12650psi

m) 41psia)(1,81 (18,12651,2P2SE

PDt

==−

=

−=




LD.14 Tangki Utilitas - 02 (TU-02)

Fungsi : menampung air untuk didistribusikan ke domestik



Kondisi operasi :

Temperatur = 30 °C

Tekanan = 1 atm

Laju massa air = 904 kg/jam

Densitas air = 995,68 kg/m3 (Perry, 1997)



Perhitungan:

Volume air, 3a kg/m995,68jam24kg/jam904

V×

= = 21,7901 m3


Direncanakan perbandingan diameter dengan tinggi silinder, D : H = 2 : 3

33

23

2

πD83m 26,1481

D23πD

41m 26,1481

HπD41V

=

=

=

Maka, D = 2,8099 m

H = 4,2148 m

Tinggi air dalam tangki = ) 26,1481(

)2099,4)( 7901,21( = 3,5124 m



= 995,68 kg/m3 x 9,8 m/det2 x 3,5124 m

= 34272,90741 kg/m.det2


= cm

mxkg

gx100

11

1000m.det

kg134272,9074 2

= 342729,0741 kg/cm.det2 = 4,970psia





= 20,650 psia



Tebal shell tangki:

in0,113m 0,00287psia) 1,2(20,650a)(0,8)2(12650psi

m) 99psia)(2,80 (20,6501,2P2SE

PDt

==−

=

−=



LD.15 Deaerator (DE)

Fungsi : Menghilangkan gas-gas yang terlarut dalam air umpan ketel

Bentuk : Silinder horizontal dengan tutup atas dan bawah elipsoidal

Bahan konstruksi : Carbon steel SA-283 Grade C

Jumlah : 1

Kondisi operasi : Temperatur = 90 oC

Tekanan = 1 atm

Kebutuhan Perancangan = 24 jam

Laju alir massa air = 117,0011 kg/jam

Densitas air (ρ) = 995,68 kg/m3 = 62,1936 lbm/ft3 (Perry, 1999)



Perhitungan :


jam24kg/jam 117,0011V

×= = 2,8202 m3


Direncanakan perbandingan diameter dengan tinggi tangki, D : H = 2 : 3

33

23

2

πD83m 3,3842

D23πD

41m 3,3842

HπD41V

=

=

=

Maka: D = 1,4213 m ; H = 2,1320 m

Tinggi cairan dalam tangki = 1320,23,3842

2,8202 x = 1,7766 m


Direncanakan perbandingan diameter dengan tinggi tutup, D : H = 4 : 1

Tinggi tutup = m 0,3553 m 1,421341

=x (Brownell,1959)

Tinggi tangki total = 2,1320 x 2( 0,3553) = 1,5151 m

Tebal tangki


= 996,24 kg/m3 x 9,8 m/det2 x 1,5151 m

= 14.784,67154 kg/m.det2

= cm

mxkg

gx100

11

1000m.det

kg 5414.784,671 2

= 147.846,7154 kg/cm.det2 = 2,1443 psia





= 17,6823 psia




Tebal shell tangki:

in0,0489m 0,00124psia) 31,2(17,682a)(0,8)2(12650psi

m) 13psia)(1,42 (17,68231,2P2SE

PDt

==−

=

−=



LD.16 Ketel Uap (KU)

Fungsi : Menyediakan uap untuk keperluan proses

Jenis : Ketel pipa air

Jumlah : 1

Bahan konstruksi : Carbon steel

Data :

Uap jenuh yang digunakan bersuhu 150 °C

kalor laten steam 2745,0 kj/kg = 1180,6451 Btu/lbm

Panas entalpi steam = 2745,0 kj/kg = 1180,6451 Btu/lbm

Panas entalpi campuran = 549,4799 kj/kg = 236,3354 Btu/lbm

Total kebutuhan uap = 585,00563 kg/jam = 1.289,7034 lbm/jam Perhitungan:

Menghitung Daya Ketel Uap

H,P,W 3970534 ××

=

dimana: P = daya ketel uap, Hp

W = kebutuhan uap, lbm/jam

H = kalor laten steam, Btu/lbm

Maka,

3,9705,34 ) 236,3354 - (1180,6451 1.289,7034

××

=P = 36,3813 Hp


Menghitung Jumlah Tube

Luas permukaan perpindahan panas, A = P × 10 ft2/Hp

= 36,3813 Hp × 10 ft2/Hp

= 363,3813 ft2

Direncanakan menggunakan tube dengan spesifikasi:

- Panjang tube, L = 30 ft

- Diameter tube 6 in

- Luas permukaan pipa, a′ = 1,734 ft2/ft (Kern, 1965)

Sehingga jumlah tube,

734,130 363,3813

' ×=

×=

aLANt = 6,9937 buah ≈ 7 buah

LD.17 Tangki Bahan Bakar (TB)

Fungsi : Menyimpan bahan bakar solar



Jumlah : 1 unit

Kondisi operasi :

Temperatur : 30 °C

Tekanan : 1 atm

Laju volume solar = 124,652 L/jam

Densitas solar = 0,89 kg/L (Perry, 1997)


Perhitungan :

a. Volume Tangki

Volume solar (Va) = 124,652 L/jam x 7 hari x 24 jam/hari

= 20.941,536 L = 20,9415 m3

Direncanakan membuat 1 tangki dan faktor kelonggaran 20%, maka :

Volume 1 tangki, Vl = 1,2 x 20,9415 m3 = 25,1298 m3



- Tinggi silinder (Hs) : Diameter (D) = 4 : 3

- Tinggi tutup (Hd) : Diameter (D) = 1 : 4

- Volume shell tangki (Vs) :

3

2s

2

D3πVs

34D

4πHπRVs

=

== D

- Volume tutup tangki (Ve) :

Vh = 32d

2 D24

D41D

6HR

32 π

=

π

=π (Brownell,1959)

- Volume tangki (V) :

Vt = Vs + Vh

= 3D83π

25,1298 m3 = 31,1775 D

D3 = 21,3416 m3

D = 2,7738 m

D = 109,2047 in

Hs = =D34 3,6984 m

c. Diameter tutup




diperoleh data :

- Allowable stress (S) = 12650 psia = 87218,714 kPa


- Corrosion allowance (C) = 0.125 in/tahun (Brownell,1959)

Volume cairan = 20,9415 m3


Tinggi cairan dalam tangki =3

3

m1298,25m 20,9415 × 3,6984 m = 3,0820 m

Tekanan Hidrostatik :


= 890,0712 kg/m3 × 9,8 m/det2 × 3,0820 m

= 26.883,4253 kg/m.det2

= cm

mxkg

gx100

11

1000m.det

kg 326.883,425 2

= 268.834,253 kg/cm.det2 = 3,8991 psia





= 19,5248 psia



Tebal shell tangki:

in0,1054m 0,00267psia) 81,2(19,524a)(0,8)2(12650psi

m) 38psia)(2,77 (19,52481,2P2SE

PDt

==−

=

−=



LD.18 Pompa Screening (PU-01)

Fungsi : Memompa air dari sungai ke bak sedimentasi Jenis : Pompa sentrifugal

Bahan konstruksi : Commercial Steel

Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30 oC



Densitas (ρ) = 995,68 kg/m3 = 62,1603 lbm/ft3

Viskositas (µ) = 0,8007 cP = 0,0005 lbm/ft.s

Laju alir volumetrik, === 3m

m

ft/lb 62,1258/seclb 0,9534

ρFQ 0,01533 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (0,01533)0,45 (62,1258)0,13

= 1,0181 in

Dari Appendiks A.5 Geankoplis (1997), dipilih pipa commercial steel :

Ukuran nominal : 1,25 in





Kecepatan linear, v = Q/A = 2

3

ft0,01040/sft0,01533

= 1,4748 ft/s

Bilangan Reynold : NRe = µ××ρ Dv

= lbm/ft.s0,0005

ft)49ft/s)(0,11 )(1,4748lbm/ft (62,1603 3

= 21.085,5271 (Turbulen)

Untuk pipa commercial steel, harga ε = 0,000046 (Geankoplis, 1997)

Pada NRe = 21.085,5271 dan ε/D = 0,0004

Dari Fig.2.10-3 Geankoplis (1997), diperoleh harga f = 0,0053


Friction loss :

1 Sharp edge entrance= hc = 0,55α2

12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

4748,1012

−



v.2

2

= 1(0,75))174,32(2

4748,1 2


1 check valve = hf = n.Kf.cg

v.2

2

= 1(2,0) ( )( )174,3212 4748,1 2

= 0,0676 ft.lbf/lbm

Pipa lurus 70 ft = Ff= 4fcgD

vL.2.. 2∆

= 4(0,006)( )( )

( ) ( )174,32.2.0,11494748,1.70 2

=0,4362 ft.lbf/lbm

1 Sharp edge exit = hex=cg

vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

4748,1012

− = 0,033 ft.lbf/lbm

Total friction loss : ∑ F = 0,5815 ft.lbf/lbm

Kerja yang diperlukan, -Wf ;

0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv

cc ρ (Geankoplis,1983)

Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 2,085 ft/s

P2 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

005815,016,62

92,2117174,32

)03(174,32174,322

04748,11

2

=++−

+−

+− P

x

0,0338 + 3 + 5815,0 +16,6292,2117 1P− = 0

P1 = 2.342,6527 lbf/ft2

= 2.342,6527 lbf/ft2 x 2

2

1441

inft = 16,2684 lbf/in2


Sehingga,

-Wf = 5815,0303380,016,62

92,2117 2342,6527+++

−

= 7,2307 ft.lbf/lbm

Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

01252,0./.550

/16,62/0,01533/. 7,2309550

..

33

==

−=

ρ

.

Untuk efisiensi alat 80%, maka :

Tenaga pompa yang dibutuhkan = hp01566,08,0

01252,0= (Geankoplis, 1983)

Maka dipilih pompa yang berdaya motor 0,05 hp

LD.19 Pompa Sedimentasi (PU-02)

Fungsi : Memompa air dari bak sedimentasi ke klarifier



Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30 oC





m

ft/lb 62,1258/seclb 0,9534

ρFQ 0,01533 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (0,01533)0,45 (62,1258)0,13

= 0,7830 in









3

ft0,01040/sft01533,0

= 1,4748 ft/s


= lbm/ft.s0,0005

ft)49ft/s)(0,11 )(1,4748lbm/ft (62,1585 3

= 21.085,52717 (Turbulen)


Pada NRe = 29.817,3989 dan ε/D = 0,0004


Friction loss :


12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

1,4748012

−

= 0,0185 ft.lbf/lbm


v.2

2

= 1(0,75))174,32(2

1,4748 = 0,0253 ft.lbf/lbm


v.2

2

= 1(2,0) ( )( )174,3212 1,4748 2

= 0,0676 ft.lbf/lbm

Pipa lurus 70 ft = Ff =4fcgD

vL.2.. 2∆

=4(0,006)( )( )

( ) ( )174,32.2.0,1149 1,4748.70 2

=0,0436 ft.lbf/lb

1 Sharp edge exit = hex=cg

vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

1,4748012





0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 2,085 ft/s

P2 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

005815,016,62

92,2117174,32

)03(174,32174,322

01,47481

2

=++−

+−

+− P

x

0,06755 + 3 + 0,5815 +16,6292,2117 1P− = 0

P1 = 2342,6527 lbf/ft2

= 2342,6527 lbf/ft2 x 2

2

1441


Sehingga,

-Wf = 65815,030338,016,62

92,2117 2342,6527+++

−

= 7,2307 ft.lbf/lbm

Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

01253,0./.550

/16,62/01533,0/. 7,2307550

..

33

==

−=

ρ

.



01253,0= (Geankoplis, 1983)


LD.20 Pompa Alum (PU-03)


Fungsi : Memompa alum dari tangki pelarutan alum ke klarifier



Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30 oC

Laju alir massa (F) = 0,07784 kg/jam = 4,7672.10-5 lbm/s

Densitas alum (ρ) = 1363 kg/m3 = 85,0920 lbm/ft3 (Othmer, 1967)

Viskositas alum (µ) = 6,72 10-4 cP = 4,5156.10-7 lbm/ft.s (Othmer, 1967)


m

ft/lb 85,0920/seclb 5-4,7672.10

ρFQ 5,6024.10-7 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (5,6024.10-7)0,45 (85,0920)0,13

= 0,0106 in








3

ft 0,0004/sft 7-5,6024.10

= 0,00140 ft/s


= lbm/ft.s4,5156.10

ft)24ft/s)(0,02 )(0,00140lbm/ft (85,09207-

3

= 5.916,3227 (Turbulen)



Pada NRe = 5.916,3227 dan ε/D = 0,0020


Friction loss :


12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

0,00140012

−

= 1,6767.10-8 ft.lbf/lbm

1 elbow 90° = hf = n.Kf.cg

v.2

2

= 1(0,75) ( )( )174,3212 0,00140 2

= 2,2864.10-8 ft.lbf/lbm


v.2

2

= 1(2,0) ( )( )174,3212 0,00140 2

= 6,0971.10-8 ft.lbf/lbm

Pipa lurus 30ft=Ff=4fcgD

vL.2.. 2∆

=4(0,008)( )( )( ) ( )174,32.2.0,0224

0,00140.30 2

=1,6319.10-6

ft.lbf/lbm

1Sharp edge exit=hex=cg

vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

0,00140012

− =3,048.10-8

ft.lbf/lbm

Total friction loss : ∑ F = 1,7630.10-6 ft.lbf/lbm


0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 0,0024 ft/s

P2 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :


00.107630,10898,8592,2117

174,32)03(174,32

174,322000140,0 6-1

2

=++−

+−

+− P

x

P1 = 2373,196 lbf/ft2

= 2373,196 lbf/ft2 x 2

2

1441


Sehingga,

-Wf = 6-8 .105675,13100962,60898,85

92,2117196,2374+++

− −x

= 6,00 ft.lbf/lbm

Daya pompa, Ws ;

hpxhpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

733

10200,5./.550

/0920,85/ 7-.106024,5/.00,6550

..

−==

−=

ρ


Tenaga pompa yang dibutuhkan = hpxx 77

10500,68,010200,5 −

−

= (Geankoplis, 1983)


LD.21 Pompa Soda Abu (PU-04)

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



Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30 oC



Densitas soda abu (ρ) = 1327 kg/m3 = 82,8446 lbm/ft3 (Othmer, 1967)

Viskositas soda abu (µ) = 3,69 10-4 cP = 2,4797.10-7 lbm/ft.s (Othmer, 1967)


m

ft/82,8446lb/seclb 5-2,5742.10

ρFQ 3,1073.10-7 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (3,1073.10-7)0,45 (82,8446)0,13

= 0,00816 in








3

ft 0,0004/s7ft-3,1073.10

= 0,00077 ft/s


= lbm/ft.s2,4797.10

ft)24ft/s)(0,02 )(0,00077lbm/ft (82,84467-

3

= 5.818,1983 (Turbulen)


Pada NRe = 5.818,1983 dan ε/D = 0,0020


Friction loss :


12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

0,00077012

−

= 5,1581.10-9 ft.lbf/lbm


v.2

2

= 1(0,75) ( )( )174,32120,001092

= 7,0338.10-9 ft.lbf/lbm



v.2

2

= 1(2,0) ( )( )174,3212 0,00077 2

= 1,8757.10-8 ft.lbf/lbm

Pipa lurus 30 ft=Ff=4fcgD

vL.2.. 2∆

=4(0,008)( )( )

( ) ( )174,32.2.0,0224 0,00077.30 2

=3,514.10-7 ft.lbf/lbm

1 Sharp edge exit = hex =cg

vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

0,00077012

− =9,37.10-9

ft.lbf/lbm



0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 0,00109 ft/s

P2 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

00109176,38446,8292,2117

174,32)03(174,32

174,322000077,0 71

2

=++−

+−

+−

−xP

x

P1 = 2366,4539 lbf/ft2

= 2366,4539 lbf/ft2 x 2

2

1441


Sehingga,

-Wf = 78 109788,8310875,18423,82

92,2117453,2366 −− +++− xx

= 6,00 ft.lbf/lb


Daya pompa, Ws ;

hpxhpslbfft

ftlbmxsftxxlbmlbfft

QWW f

S

7337

108083,2./.550

/8446,82/10 7-3,1073.10/.00,6550

..

−−

==

−=

ρ



105104,38,0

108083,2 −−

= (Geankoplis,

1983)


LD.22 Pompa Klarifier (PU-05)

Fungsi : Memompa air dari klarifier ke tangki filtrasi Jenis : Pompa sentrifugal


Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30 oC





m

ft/lb 62,1258/seclb 0,95344

ρFQ 0,01533 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (0,01533)0,45 (62,1258)0,13

= 1,0181 in









3

ft0,01040/sft 0,01533

= 1,4748 ft/s


= lbm/ft.s0,0005

ft)49ft/s)(0,11 )(1,4748lbm/ft (62,1585 3

= 19.594,5256 (Turbulen)


Pada NRe = 19.594,5256 dan ε/D = 0,0004


Friction loss :


12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

1,4748012

−

= 0,0185 ft.lbf/lbm


v.2

2

= 1(0,75) ( )( )174,3212 1,4748 2

= 0,0253 ft.lbf/lbm


v.2

2

= 1(2,0) ( )( )174,3212 1,4748 2

= 0,0676 ft.lbf/lbm

Pipa lurus 50 ft =Ff =4fcgD

vL.2.. 2∆

= 4(0,006)( )( )

( ) ( )174,32.2.0,1149 1,4748.50 2

= 0,5291 ft.lbf/lbm

1 Sharp edge exit = hex = cg

vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

1,4748012

− = 0,0338

ft.lbf/lbm




0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 2,0856 ft/s

P2 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

00 0,67441258,6292,2117

174,32)03(174,32

174,3220 1,4748

12

=++−

+−

+− P

x

P1 = 2.348,42661 lbf/ft2

= 2.348,42661 lbf/ft2 x 2

2

1441


Sehingga,

-Wf = 0,674430,033801258,62

92,2117 12.348,4266+++

−

= 7,4165 ft.lbf/lbm

Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

0128,0./.550

/1258,62/01533,0/. 7,4165550

..

33

==

−=

ρ



0128,0= (Geankoplis, 1983)



LD.23 Pompa Sand Filter (PU-06)

Fungsi : Memompa air dari tangki filtrasi ke tangki utilitas TU-01



Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30 oC





m

ft/lb 62,1258/seclb 071.556,9330

ρFQ 0,01533 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (0,01533)0,45 (62,1258)0,13

= 1,0181in








3

ft0,01040/sft 0,01533

= 1,4748 ft/s


= lbm/ft.s0,0005

ft)49ft/s)(0,11 )(1,4748lbm/ft (62,1585 3

= 19.594,5256 (Turbulen)



Pada NRe = 27.708,9486 dan ε/D = 0,0004


Friction loss :


12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

1,4748012

−

= 0,0185 ft.lbf/lbm


v.2

2

= 4(0,75) ( )( )174,32121,47482

= 0,1014 ft.lbf/lbm


v.2

2

= 1(2,0) ( )( )174,32121,47482

= 0,0676 ft.lbf/lbm

Pipa lurus 80 ft = Ff = 4fcgD

vL.2.. 2∆

= 4(0,0065)( )( )

( ) ( )174,32.2.0,11491,4748.80 2

= 0,84ft.lbf/lbm


vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

1,4748012

− = 0,0338 ft.lbf/lbm



0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 1,4748 ft/s

P2 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

00 1,06791258,6292,2117

174,32)03(174,32

174,3220 1,4748

12

=++−

+−

+− P

x

P1 = 2.372,8878 lbf/ft2

= 2.372,8878 lbf/ft2 x 2

2

1441



Sehingga,

-Wf = 1,067930,03381258,62

92,2117 2.372,8878+++

−

= 8,2035 ft.lbf/lbm

Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

0142,0./.550

/1258,62/0,01533/. 8,2035550

..

33

==

−=

ρ

.



0142,0= (Geankoplis, 1983)


LD.24 Pompa Utilitas (PU-07)

Fungsi : Memompa air dari tangki utilitas TU-01 ke tangki kation



Jumlah : 1 unit

Kondisi operasi :

Tekanan = 1 atm

Temperatur = 25 oC

Laju alir massa (F) = 117,0011 kg/jam = 0,0716 lbm/detik




m

ft/lb 62,1258/seclb 0,0716

ρFQ 0,0011 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (0,0011)0,45 (62,1258)0,13

= 0,3176 in


Dari Appendiks A.5 Geankoplis (1997), dipilih pipa commercial steel : Ukuran nominal : 0,5 in






3

ft 0,00211/sft 0,0011

= 0,5462 ft/s


= lbm/ft.s0,0005

ft)18ft/s)(0,05 )(0,5462lbm/ft (62,1585 3

= 3.271,2749 (Turbulen)


Pada NRe = 3.271,2749 dan ε/D = 0,000887


Friction loss :


12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

0,5462012

−



v.2

2

= 3(0,75) ( )( )174,3212 0,5462 2

= 0,0104 ft.lbf/lbm


v.2

2

= 1(2,0) ( )( )174,3212 0,5462 2


Pipa lurus 50 ft= Ff = 4fcgD

vL.2.. 2∆

=4(0,0085)( )( )

( ) ( )174,32.2.0,0518 0,5462.50 2

= 0,053 ft.lbf/lbm


vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

0,5462012

− =0,004 ft.lbf/lbm




0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 0,5462 ft/s

P2 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

00 0,08051258,6292,2117

174,32)03(174,32

174,32200,5462

12

=++−

+−

+− P

x

P1 = 2.309,6982 lbf/ft2

= 2.309,6982 lbf/ft2 x 2

2

1441


Sehingga,

-Wf = 0,080530046,01258,62

92,2117 2.309,6982+++

−

= 6,1704 ft.lbf/lbm

Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

00080,0./.550

/1258,62/ 0,0011/. 6,1704550

..

33

==

−=

ρ

.



00080,0= (Geankoplis, 1983)




Fungsi : Memompa air dari tangki utilitas TU-01 ke cooling tower Jenis : Pompa sentrifugal


Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30 oC

Laju alir massa (F) = 535,9318 kg/jam = 0,3281 lbm/s




m

ft/lb 62,1258/seclb 0,3281

ρFQ 0,005 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (0,005)0,45 (62,1258)0,13

= 0,63 in

Dari Appendiks A.5 Geankoplis (1997), dipilih pipa commercial steel : Ukuran nominal : 1 in






3

ft 0,0060/sft 0,005

= 0,5462 ft/s


= lbm/ft.s0,0005

ft)1ft/s)(0,05 )(0,5462lbm/ft (62,1585 3

= 3.271,2749 (Turbulen)



Pada NRe = 3.271,2749 dan ε/D = 0,00052


Friction loss :


12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

0,5462012

−



v.2

2

= 2(0,75) ( )( )174,3212 0,5462 2

= 0,0069 ft.lbf/lbm


v.2

2

= 1(2,0) ( )( )174,3212 0,5462 2



vL.2.. 2∆

= 4(0,007)( )( )( ) ( )174,32.2.0,051

0,5462.130 2

= 0,139ft.lbf/lbm


vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

0,5462012

− =0,004 ft.lbf/lbm



0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 0,5462 ft/s

P2 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

00 0,27551258,6292,2117

174,32)03(174,32

174,3220 0,5462

12

=++−

+−

+− P

x

P1 = 2.322,2755 lbf/ft2

= 2.322,2755 lbf/ft2 x 2

2

1441



Sehingga,

-Wf = 0,27553 0,01201258,62

92,2117 2.322,2755+++

−

= 6,5751 ft.lbf/lbm

Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

0039,0./.550

/1258,62/ 0,005/. 6,5751550

..

33

==

−=

ρ

.



0039,0= (Geankoplis, 1983)



Fungsi : Memompa air dari tangki utilitas TU-01 ke tangki utilitas TU-02



Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30oC

Laju alir massa (F) = 904 kg/jam = 0,55359 lbm/s




m

ft/lb 62,1258/seclb 0,55359

ρFQ 0,0089 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (0,0089)0,45 (62,1258)0,13


= 0,7956 in




Diameter Luar (OD) : 1,315 in = 0,1096ft



3

ft 0,006/sft0,0089

= 1,4843 ft/s


= lbm/ft.s0,0005

ft)74ft/s)(0,08 )(1,4843lbm/ft (62,1585 3

= 14.990,3492 (Turbulen)


Pada NRe = 14990,3492 dan ε/D = 0,00046


Friction loss :


12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

1,4843012

−



v.2

2

= 1(0,75) ( )( )174,32121,48432

= 0,0256 ft.lbf/lbm


v.2

2

= 1(2,0) ( )( )174,32121,48432

= 0,0684 ft.lbf/lbm


vL.2.. 2∆

= 4(0,0055)( )( )

( ) ( )174,32.2.0,08741,4843.60 2

= 0,1292 ft.lbf/lbm


vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

1,4843012

− = 0,0342 ft.lbf/lbm




0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 1,4778 ft/s

P1 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

000,27641258,6292,2117

174,32)03(174,32

174,32204778,1

12

=++−

+−

+− P

x

P1 = 2323,7154 lbf/ft2

= 2323,7154 lbf/ft2 x 2

2

1441


Sehingga,

-Wf = 2764,030342,01258,62

92,2117 2323,7154+++

−

= 6,6214 ft.lbf/lbm

Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

00664,0./.550

/1258,62/ 0,0089/. 2323,7154550

..

33

==

−=

ρ

.



00664,0= (Geankoplis, 1983)



LD.27 Pompa H2SO4 (PU-10)

Fungsi : Memompa H2SO4 dari tangki H2SO4 ke tangki kation Jenis : Pompa sentrifugal


Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 25 oC


Densitas H2SO4 (ρ) = 1061,7 kg/m3 = 66,2801 lbm/ft3 (Othmer, 1967)

Viskositas H2SO4 (µ) = 5,2 cP = 0,012 lbm/ft.s (Othmer, 1967)


m

ft/lb 66,2801/seclb 5-2,2230..10

ρFQ 3,354.10-7 ft3/s

Desain pompa : Asumsi aliran laminar

Di,opt = 3,9 (Q)0,36 (µ)0,18 (Walas,1988)

= 3,9 (3,354.10-7)0,45 (5,2)0,18

= 0,008 in







37

ft 0,0004/sft 3,354.10 -

= 0,0008 ft/s


= lbm/ft.s0,0035

ft)24ft/s)(0,02 )(0,0008lbm/ft (66,2801 3


= 154,5017 (Laminar)

Untuk laminar, f = ReN

16 (Geankoplis, 1997)

= 154,5017

16 = 0,1035

Friction loss :


12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

0,0008012

−

= 6,009.10-9 ft.lbf/lbm


v.2

2

= 3(0,75) ( )( )174,3212 0,0008 2

= 2,458.10-8 ft.lbf/lbm


v.2

2

= 1(2,0) ( )( )174,3212 0,0008 2

= 2,185.10-8 ft.lbf/lbm


vL.2.. 2∆

= 4(0,1035)( )( )

( ) ( )174,32.2.0,0224 0,0008.20 2

= 2,0627.10-6

ft.lbf/lbm

1Sharpedge exit=hex=cg

vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

0,0008012

− =1,092.10-8 ft.lbf/lbm



0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 0,0008 ft/s

P2 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :


00 6-2,126.102801,6692,2117

174,32)03(174,32

174,3220 0,0008

12

=++−

+−

+− P

x

P1 = 2316,766 lbf/ft2

= 2316,766 lbf/ft2 x 2

2

1441


Sehingga,

-Wf = 6 -2,126.103 .101813,42801,66

92,2117766,2316 8- +++−

= 6,000 ft.lbf/lbm

Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

733

10.425,2./.550

/2801,66/ 7-3,354.10/.000,6550

..

−==

−=

ρ


Tenaga pompa yang dibutuhkan = hp77

10.031,38,010.425,2 −

−



LD.28 Pompa Kation (PU-11)

Fungsi : Memompa air dari tangki kation ke tangki anion Jenis : Pompa sentrifugal


Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30 oC






m

ft/lb 62,1258/seclb 0,071

ρFQ 0,0011 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (0,0011)0,45 (62,1258)0,13

= 0,3176 in




Diameter Dalam (ID) : 0.622 in = 0,0518 ft




3

ft 0,00211/sft 0,0011

= 0,5462 ft/s


= lbm/ft.s0,0005

ft)518ft/s)(0,0 )(0,5462lbm/ft (62,1585 3

= 3.271,2749 (Turbulen) Untuk pipa commercial steel, harga ε = 0,000046 (Geankoplis, 1997)

Pada NRe = 3.271,2749 dan ε/D = 0,00088


Friction loss :


12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

0,5462012

−



v.2

2

= 4(0,75) ( )( )174,3212 0,5462 2

= 0,0139 ft.lbf/lbm


v.2

2

= 1(2,0) ( )( )174,3212 0,5462 2

= 0,0092 ft.lbf/lbm


Pipa lurus 40 ft =Ff=4fcgD

vL.2.. 2∆

=4(0,003)( )

( ) ( )174,32.2.0,0518 0,5462.40 2

= 0,0429 ft.lbf/lbm


vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

0,5462012




0)(

.21212

21

22 =++

−+

−+

− ∑ WfFPP

gZZg

gvv

cc ρ

(Geankoplis,1983)

Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 0,5462 ft/s

P1 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

00 0,07331258,6292,2117

174,32)03(174,32

174,3220 0,5462

12

=++−

+−

+− P

x

P1 = 2.309,2470 lbf/ft2

= 2.309,2470 lbf/ft2 x 2

2

1441


Sehingga,

-Wf = 0,0733300463,01258,62

92,2117 2.309,2470+++

−

= 6,1559 ft.lbf/lbm

Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

0008,0./.550

/1258,62/ 0,0011/. 6,1559550

..

33

==

−=

ρ




0008,0= (Geankoplis, 1983)


LD.29 Pompa NaOH (PU-12)

Fungsi : Memompa NaOH dari tangki NaOH ke tangki anion



Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30 oC

Laju alir massa (F) = 0,0128 kg//jam = 7,8689.10-6 lbm/s

Densitas NaOH (ρ) = 1518 kg/m3 = 94,7662 lbm/ft (Othmer, 1967)

Viskositas NaOH (µ) = 0,00043 cP = 2,8909.10-7 lbm/ft.s (Othmer, 1967)


m-5

ft/lb 94,7662/seclb 7,8689.10

ρFQ 8,303 x 10-8 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (8,303 x 10-8)0,45 (94,7662)0,13

= 0,0045 in







3

ft 0,0004/sft 8-10 x 8,303

= 0,0002 ft/s



= lbm/ft.s10.8909,2

ft)24ft/s)(0,02 )(0,0002lbm/ft (94,76627

3

−

= 1.525,4670 (Turbulen)


Pada NRe = 1.525,4670 dan ε/D = 0,00205


Friction loss :


12

1

2 vAA

−

= 0,55 ( ) ( )( )174,3212

0,0002012

−

= 3,683 x 10-10 ft.lbf/lbm


v.2

2

= 3(0,75) ( )( )174,32120,00022

= 1,5067 x 10-9 ft.lbf/lbm


v.2

2

= 1(2,0) ( )( )174,32120,00022

= 1,339 x 10-9 ft.lbf/lbm


vL.2.. 2∆

= 4(0,004)( )( )

( ) ( )174,32.2.0,02240,0002.20 2

= 9,55.10-9

ft.lbf/lbm

1 Sharp edge exit hex= cg

vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

0,0002012

− = 6,696x10-10

ft.lbf/lbm



0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 0,0002 ft/s

P2 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2


Maka :

00 1,3443.107662,9492,2117

174,32)03(174,32

174,32200,0002 8-1

2

=++−

+−

+− P

x

P1 = 2402,226 lbf/ft2

= 2402,226 lbf/ft2 x 2

2

1441


Sehingga,

-Wf = 1,3443.10310 x 562,27662,94

92,2117226,2402 8-9- +++−

= 6,000 ft.lbf/lbm

Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

833

10.584,8./.550

/7662,94/ 8-10 x 8,303/.000,6550

..

−==

−=

ρ


Tenaga pompa yang dibutuhkan = hp78

10.0730,18,010.584,18 −

−



LD.30 Pompa Kaporit (PU-13)

Fungsi : Memompa kaporit dari tangki kaporit ke tangki utilitas TU-02 Jenis : Pompa sentrifugal


Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30 oC



Densitas kaporit (ρ) = 1272 kg/m3 = 79,4088 lbm/ft3 (Othmer, 1967)

Viskositas kaporit (µ) = 6,7197.10-4 cP = 4,5156.10-7 lbm/ft.s (Othmer, 1967)


m-6

ft/lb 79,4088/seclb 1,5817.10

ρFQ 1,991.10-8 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,36 (µ)0,18 (Walas,1988)

= 3,9 (1,991.10-8)0,36 (6,7197.10-4)0,18

= 0,0023 in







3-8

ft 0,0004/sft 1,991.10

= 4,9794.10-5 ft/s


= lbm/ft.s10.5156,4

ft)24ft/s)(0,02 0)(4,9794.1lbm/ft (79,40887

-53

−

= 196,3051 (Laminar)

Untuk laminar, f = ReN

16 (Geankoplis, 1997)

= 196,3051

16

= 0,0815

Friction loss :

1 Sharp edge entrance=hc=0,55α2

12

1

2 vAA

−

=0,55 ( ) ( )( )174,3212

)(4,9794.10012-5

−


= 2,1193.10-11 ft.lbf/lbm


v.2

2

= 1(0,75) ( )( )174,3212 )(4,9794.10 2-5

= 2,8899.10-11 ft.lbf/lbm


v.2

2

=1(2,0) ( )( )174,3212 )(4,9794.10 2-5

= 7,7066.10-11 ft.lbf/lbm

Pipalurus20ft=Ff=4fcgD

vL.2.. 2∆

=4(0,0819) ( )( )( ) ( )174,32.2.0,0224

4,9794.10.20 2-5

=1,120.10-8 ft.lbf/lbm

1 Sharp edge exit=hex=cg

vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

)(4,9794.10012-5

− =3,853.10-11

ft.lbf/lbm



0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 4,9794.10-5 ft/s

P1 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

00101373,14088,7992,2117

174,32)03(174,32

174,3220)104,9794( 81

25

=++−

+−

+−

−−

xP

xx

P1 = 2356,1464 lbf/ft2

= 2356,1464 lbf/ft2 x 2

2

1441


Sehingga,

-Wf = 811 101,13733108533,34088,79

92,21171464,2356 −− +++− xx

= 6,00 ft.lbf/lbm

Daya pompa, Ws ;


hpxhpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

8338-

107254,1./.550

/4088,79/1,991.10/.00,6550

..

−==

−=

ρ

.



101568,28,0107254,1 −

−




Fungsi : Memompa air dari tangki utilitas TU-01 ke distribusi domestik



Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30 oC

Laju alir massa (F) = 904 kg/jam = 0,55359 lbm/s




m

ft/lb 62,1258/seclb 0,55359

ρFQ 0,0089 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (0,0089)0,45 (62,1258)0,13


= 0,7956 in




Diameter Luar (OD) : 1,315 in = 0,1096ft



3

ft 0,006/sft0,0089

= 1,4843 ft/s


= lbm/ft.s0,0005

ft)74ft/s)(0,08 )(1,4843lbm/ft (62,1585 3

= 14990,3492 (Turbulen)


Pada NRe = 14990,3492 dan ε/D = 0,0005


Friction loss :


12

1

2 vAA

−

= 0,55 ( ) ( )( )174,3212

1,4843012

−

= 0,0188 ft.lbf/lbm


v.2

2

= 1(2,0) ( )( )174,3212 1,4843 2

= 0,0684 ft.lbf/lbm


vL.2.. 2∆

=4(0,0055)( )( )

( ) ( )174,32.2.0,08741,4843.120 2

= 13,1604

ft.lbf/lbm



vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

1,4843012

− = 0,0342

ft.lbf/lbm



0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 1,4843 ft/s

P2 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

002820,131258,6292,2117

174,32)03(174,32

174,32201,4843

12

=++−

+−

+− P

x

P1 = 3132.143 lbf/ft2

= 3132.143 lbf/ft2 x 2

2

1441


Sehingga,

-Wf = 2820,13303423,01258,62

92,2117 3132,143+++

−


Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

032,0./.550

/1258,62/0089,0/. 32,6325550

..

33

==

−=

ρ

.



032,0= (Geankoplis, 1983)



LD.32 Pompa Anion (PU-15)

Fungsi : Memompa air dari tangki anion ke deaerator Jenis : Pompa sentrifugal


Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30 oC





m

ft/lb 62,1258/seclb 0,0716

ρFQ 0,0011 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (0,0011)0,45 (62,1258)0,13

= 0,3176 in


Ukuran nominal : 1/2 in






3

ft 0,00211/sft0,0011

= 0,5462 ft/s



= lbm/ft.s0,0005

ft)18ft/s)(0,05 )(0,5462lbm/ft (62,1585 3

= 3.271,2749 (Turbulen)


Pada NRe = 3.271,2749 dan ε/D = 0,000887


Friction loss :


12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

0,5462012

−

= 0,00255ft.lbf/lbm


v.2

2

= 2(0,75) ( )( )174,3212 0,5462 2



v.2

2

= 1(2,0) ( )( )174,3212 0,5462 2

= 0,0092 ft.lbf/lbm

Pipa lurus 80 ft= Ff =4fcgD

vL.2.. 2∆

= 4(0,003)( )( )

( ) ( )174,32.2.0,0518 0,5462.80 2

= 0,0858 ft.lbf/lbm


vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

0,5462012

− = 0,0046

ft.lbf/lbm



0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 0,5462 ft/s

P1 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2


Maka :

00 0,109311258,6292,2117

174,32)03(174,32

174,3220 0,5462

12

=++−

+−

+− P

x

P1 = 2379,4842 lbf/ft2

= 2379,4842 2379,1851 lbf/ft2 x 2

2

1441

inft = 16,522lbf/in2

Sehingga,

-Wf = 0,10931300463,01258,62

92,2117 2379,1851+++

−

= 6,2279 ft.lbf/lbm

Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

000811,0./.550

/1258,62/ 0,0011/. 6,2279550

..

33

==

−=

ρ



000811,0= (Geankoplis, 1983)


LD.33 Pompa Cooling Tower (PU-16)

Fungsi : Memompa air dari cooling tower ke air pendingin



Jumlah : 1 unit

Kondisi operasi :

Tekanan = 1 atm

Temperatur = 25 oC






m

ft/lb 62,1258/seclb 15,5359

ρFQ 0,2499 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (0,2499)0,45 (62,1258)0,13

= 3,5747 in



Diameter Dalam (ID) : 5 in = 0,4205 ft




3

ft 0,1390/sft 0,2499

= 1,7980 ft/s


= lbm/ft.s0,0005

ft)05ft/s)(0,42 )(1,7980lbm/ft (62,1585 3

= 87.367,6112 (Turbulen)


Pada NRe = 87.367,6112 dan ε/D = 0,0001


Friction loss :


12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

1,7980012

−

= 0,0276 ft.lbf/lbm


v.2

2

= 2(0,75) ( )( )174,3212 1,7980 2

= 0,0753 ft.lbf/lbm



v.2

2

= 1(2,0) ( )( )174,3212 1,7980 2

= 0,1004ft.lbf/lbm


vL.2.. 2∆

= 4(0,005)( )( )( ) ( )174,32.2.0,4205

1,7980.150 2

= 0,358

ft.lbf/lbm


vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

3,5858012

− = 0,0502

ft.lbf/lbm



0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 1,7980 ft/s

P2 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

00 0,61211258,6292,2117

174,32)03(174,32

174,3220 1,7980

12

=++−

+−

+− P

x

P1 = 2.345,5738 lbf/ft2

= 2.345,5738 lbf/ft2 x 2

2

1441


Sehingga,

-Wf = 0,61213 0,05021258,62

92,2117 2.345,5738+++

−

= 7,3247 ft.lbf/lbm

Daya pompa, Ws ;


hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

206,0./.550

/1258,62/ 0,2499/. 7,3247550

..

33

==

−=

ρ



206,0= (Geankoplis, 1983)


LD.34 Pompa Deaerator (PU-17)

Fungsi : Memompa air dari deaerator ke ketel uap



Jumlah : 1 unit

Kondisi operasi :

P = 1 atm

Temperatur = 30 oC





m

ft/lb 62,1258/seclb 0,0716

ρFQ 0,0011 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (0,0011)0,45 (62,1258)0,13

= 0,3176 in

Dari Appendiks A.5 Geankoplis (1997), dipilih pipa commercial steel : Ukuran nominal : 1/2 in







3

ft 0,00211/sft0,0011

= 0,5462 ft/s


= lbm/ft.s0,0005

ft)18ft/s)(0,05 )(0,5462lbm/ft (62,1585 3

= 3.271,2749 (Turbulen)


Pada NRe = 3.271,2749 dan ε/D = 0,000887


Friction loss :


12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

0,5462012

−

= 0,00255ft.lbf/lbm


v.2

2

= 2(0,75) ( )( )174,3212 0,5462 2



v.2

2

= 1(2,0) ( )( )174,3212 0,5462 2

= 0,0092 ft.lbf/lbm

Pipa lurus 80 ft= Ff =4fcgD

vL.2.. 2∆

= 4(0,003)( )( )

( ) ( )174,32.2.0,0518 0,5462.80 2

= 0,0858 ft.lbf/lbm


vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

0,5462012

− = 0,0046

ft.lbf/lbm



0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv



Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 0,5462 ft/s

P1 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

00 0,109311258,6292,2117

174,32)03(174,32

174,3220 0,5462

12

=++−

+−

+− P

x

P1 = 2379,4842 lbf/ft2

= 2379,4842 2379,1851 lbf/ft2 x 2

2

1441

inft = 16,522lbf/in2

Sehingga,

-Wf = 0,10931300463,01258,62

92,2117 2379,1851+++

−

= 6,2279 ft.lbf/lbm

Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

000811,0./.550

/1258,62/ 0,0011/. 6,2279550

..

33

==

−=

ρ



000811,0= (Geankoplis, 1983)


LD.35 Pompa Bahan Bakar 1 (PU-18)

Fungsi : Memompa bahan bakar solar dari TB-01 ke ketel uap KU-01



Jumlah : 1 unit


Kondisi operasi :

Tekanan = 1 atm

Temperatur = 25 oC

Laju alir massa (F) = 110,9399 kg/jam = 0,0679 lbm/s




m

ft/lb 55,5656/seclb 0,0679

ρFQ 0,001222 ft3/s

Desain pompa :

Di,opt = 3,9 (Q)0,45 (ρ)0,13 (Walas,1988)

= 3,9 (0,001222)0,45 (55,5656)0,13

= 0,3214 in


Ukuran nominal : 1/2 in






3

ft 0,00211/sft0,00139

= 0,6592 ft/s


= lbm/ft.s0,0007

ft)18ft/s)(0,05 )(0,6592lbm/ft (55,5656 3

= 2.354,7083 (Turbulen)


Pada NRe = 2.354,7083 dan ε/D = 0,00088


Friction loss :



12

1

2 vAA

− = 0,55 ( ) ( )( )174,3212

0,6592012

−



v.2

2

= 1(0,75) ( )( )174,32121,068662



v.2

2

= 2(2,0) ( )( )174,32120,65922

= 0,0270 ft.lbf/lbm


vL.2.. 2∆

= 4(0,013)( )( )

( ) ( )174,32.2.0,08740,6592.20 2

= 0,1355

ft.lbf/lbm


vAA

..21

22

2

1

α

− = ( ) ( )( )174,3212

0,6592012

− = 0,0067

ft.lbf/lbm



0)(.2

121221

22 =++

−+

−+

− ∑ WfFPPg

ZZggvv


Bila :

Wf = 0

Z1 = 0 ; Z2 = 3 ft

v1 = 0 ; v2 = 0,6592 ft/s

P2 = 1 atm = 14,696 lbf/in2 = 2117,92 lbf/ft2

Maka :

00 0,178065656,5592,2117

174,32)03(174,32

174,32200,6592

12

=++−

+−

+− P

x

P1 = 2294,8912 lbf/ft2


= 2294,8912 lbf/ft2 x 2

2

1441


Sehingga,

-Wf = 0,17806306750,05656,55

92,2117 2294,8912+++

−

= 6,369 ft.lbf/lbm

Daya pompa, Ws ;

hphpslbfft

ftlbmxsftxlbmlbfft

QWW f

S

0007764,0./.550

/5656,55/ 0,001222/. 6,369550

..

33

==

−=

ρ

.



0007764,0= (Geankoplis, 1983)



LAMPIRAN E

PERHITUNGAN ASPEK EKONOMI

Dalam pra rancangan pabrik Heptenadigunakan asumsi sebagai berikut:

Pabrik beroperasi selama 330 hari dalam setahun.

Kapasitas maksimum adalah 40.000 ton/tahun.

Perhitungan didasarkan pada harga peralatan tiba di pabrik atau purchased-

equipment delivered (Peters et.al., 2004).

Harga alat disesuaikan dengan nilai tukar dolar terhadap rupiah adalah :

US$ 1 = Rp.9.025 ,- (Bank Indonesia, 3 Desember 2010)

LE.1 Modal Investasi Tetap (Fixed Capital Investment)

LE.1.1 Modal Investasi Tetap Langsung (MITL)

LE.1.1.1 Biaya Tanah Lokasi Pabrik

Menurut keterangan masyarakat setempat. biaya tanah pada lokasi pabrik berkisar

Rp 3.000.000/m2.

Luas tanah seluruhnya = 12.320 m2

Harga tanah seluruhnya = 12.320 m2 × Rp 3.000.000/m2 = Rp.36.960.000.000,-

Biaya perataan tanah diperkirakan 5% dari harga tanah seluruhnya

(Timmerhaus,2004).

Biaya perataan tanah = 0,05 × Rp. 36.960.000.000,- = Rp. 1,848,000,000 ,-

Biaya Administrasi Surat tanah = 0,05 x Rp. 36.960.000.000,- = Rp. 1,848,000,000,-

Total biaya tanah (A) = Rp. 36.960.000.000,- + Rp. 1,848,000,000,- +

Rp.1,848,000,000,-

= Rp. 40.656.000.000

,-


LE.1.1.2 Harga Bangunan Tabel LE.1 Perincian Harga Bangunan. dan Sarana Lainnya

No Nama Bangunan Luas (m2) Harga (Rp/m2) Jumlah (Rp)

1 Pos Keamanan 30 1.000.000 30.000.000 2 Tempat Parkir 250 800.000 200.000.000 3 Taman 200 1.000.000 200.000.000 4 Areal Bahan Baku 200 1.500.000 300.000.000 5 Ruang Kontrol 150 1.500.000 225.000.000 6 Areal Proses 4.000 3.000.000 12.000.000.000 7 Areal Produk 500 1.500.000 750.000.000 8 Perkantoran 250 1,500,000 375.000.000 9 Laboratorium 150 1,500,000 225.000.000 10 Poliklinik 50 1.500.000 75.000.000 11 Kantin 80 1,000,000 80.000.000 12 Ruang Ibadah 80 1.500.000 120.000.000 13 Gudang Peralatan 100 750.000 75.000.000 14 Bengkel 100 750.000 75.000.000 15 Perpustakaan 80 1,000,000 80.000.000 16 Unit Pemadam Kebakaran 100 1,000,000 100.000.000 17 Unit Pengolahan Air 1.000 1.500.000 1.500.000.000 18 Pembangkit Listrik 300 1,500,000 450.000.000 19 Pengolahan Limbah 400 1,500,000 600.000.000 20 Area Perluasan 2000 750.000 1.500.000.000

21 Perumahan Karyawan 1500 1,500,000 2.250.000.000

22 Sarana Olahraga 400 350.000 140.000.000

23 Jalan 400 350.000 140.000.000

Jumlah 12320 Rp. 21,490,000,000

Total harga bangunan saja (B) = Rp. 21,490,000,000,-


LE.1.1.3 Perincian Harga Peralatan Harga peralatan yang diimpor dapat ditentukan dengan menggunakan

persamaan berikut (Timmerhaus et al, 2004):

=

y

x

m

1

2yx I

IXXCC

dimana: Cx = harga alat pada tahun 2010

Cy = harga alat pada tahun dan kapasitas yang tersedia

X1 = kapasitas alat yang tersedia

X2 = kapasitas alat yang diinginkan

Ix = indeks harga pada tahun 2010

Iy = indeks harga pada tahun yang tersedia

m = faktor eksponensial untuk kapasitas (tergantung jenis alat)

Untuk menentukan indeks harga pada tahun 2010 digunakan metode regresi koefisien korelasi:

[ ]( )( ) ( )( )2

i2

i2

i2

i

iiii

ΣYΣYnΣXΣXn

ΣYΣXYΣXnr

−⋅×−⋅

⋅−⋅⋅= (Montgomery, 1992)

Tabel LE.2 Harga Indeks Marshall dan Swift


No.

Tahun

(Xi)

Indeks

(Yi)

Xi.Yi

Xi²

Yi²

1 1989 895 1780155 3956121 801025

2 1990 915 1820850 3960100 837225

3 1991 931 1853621 3964081 866761

4 1992 943 1878456 3968064 889249

5 1993 967 1927231 3972049 935089

6 1994 993 1980042 3976036 986049

7 1995 1028 2050860 3980025 1056784

8 1996 1039 2073844 3984016 1079521

9 1997 1057 2110829 3988009 1117249

10 1998 1062 2121876 3992004 1127844

11 1999 1068 2134932 3996001 1140624

12 2000 1089 2178000 4000000 1185921

13 2001 1094 2189094 4004001 1196836

14 2002 1103 2208206 4008004 1216609

Total 27937 14184 28307996 55748511 14436786

(Sumber: Tabel 6-2 Timmerhaus et al, 2004)

Data : n = 14 ∑Xi = 27937 ∑Yi = 14184

∑XiYi = 28307996 ∑Xi² = 55748511 ∑Yi² = 14436786

Dengan memasukkan harga-harga pada Tabel LE.2, maka diperoleh harga

koefisien korelasi:

r ])14184()14436786)(14[(])27937()55748511)(14[(

)14184)(27937()28307996)(14(22 −×−

−=

= 0.98 ≈ 1

Harga koefisien yang mendekati 1 menyatakan bahwa terdapat hubungan linier antar variabel X dan Y. sehingga persamaan regresi yang mendekati adalah persamaan regresi linier.

Persamaan umum regresi linier: Y = a + b ⋅ X

dengan: Y = indeks harga pada tahun yang dicari (2010)


X = variabel tahun ke n – 1

a, b = tetapan persamaan regresi

Tetapan regresi ditentukan oleh :

( ) ( )( ) ( )2

i2

i

iiii

ΣXΣXnΣYΣXYΣXnb

−⋅⋅−⋅

=

a 22

2

Xi)(Xin.Xi.YiXi.XiYi.

Σ−ΣΣΣ−ΣΣ

=

Maka :

a = 8,325283185

103604228)27937()55748511)(14(

)28307996)(27937()55748511)(14184(2 −=

−=

−−

b = 809,163185

53536)27937()55748511)(14(

)14184)(27937()28307996)(14(2 ==

−−

Sehingga persamaan regresi liniernya adalah:

Y = a + b ⋅ X

Y = 16,8088X – 32528,8

Dengan demikian harga indeks pada tahun 2010 adalah:

Y = 16,8088(2010) – 32528,8

Y = 1256,8703

Perhitungan harga peralatan menggunakan harga faktor eksponsial (m)

Marshall & Swift. Harga faktor eksponen ini beracuan pada Tabel 6-4 Timmerhaus et

al, 2004. Untuk alat yang tidak tersedia, faktor eksponensialnya dianggap 0,6

(Timmerhaus et al, 2004).

Contoh perhitungan harga peralatan:

a. Tangki Propena

Kapasitas tangki. X2 = 400,6808 m3. Dari Gambar LE.1. diperoleh untuk harga

kapasitas tangki (X1) 1 m³ pada tahun 2002 adalah (Cy) US$ 5700. Dari tabel 6-4.

Peters et.al., 2004. faktor eksponen untuk tangki adalah (m) 0,57. Indeks harga pada

tahun 2002 (Iy) 1103.


Capacity, m3

Purc

hase

d co

st, d

olla

r

106

105

104

103

102 103 104 105Capacity, gal

10-1 1 10 102 103

P-82Jan,2002

310 kPa (30 psig) Carbon-steel tank (spherical)

Carbon steel304 Stainless stellMixing tank with agitator

Gambar LE.1 Harga Peralatan untuk Tangki Penyimpanan (Storage)

(Peters et.al., 2004)

Indeks harga tahun 2010 (Ix) adalah 1256,8703 Maka estimasi harga

tangki untuk (X2) 400,6808 m3adalah:

Cx = US$ 5700 × 57,0

1 400,6808 ×

11031256,8703

Cx = US$ 395.562,38 × (Rp 9025,-)/(US$ 1)

Cx = Rp.3.569.950.440 ,-/unit

Dengan cara yang sama diperoleh perkiraan harga alat lainnya yang dapat dilihat

pada Tabel LE.2 untuk perkiraan peralatan proses dan Tabel LE.3 untuk perkiraan

peralatan utilitas.


Untuk harga alat impor sampai di lokasi pabrik ditambahkan biaya sebagai berikut:

- Biaya transportasi = 5 %

- Biaya asuransi = 1 %

- Bea masuk = 15 % (Rusjdi, 2004)

- PPn = 10 % (Rusjdi, 2004)

- PPh = 10 % (Rusjdi, 2004)

- Biaya gudang di pelabuhan = 0,5 %

- Biaya administrasi pelabuhan = 0,5 %

- Transportasi lokal = 0,5 %

- Biaya tak terduga = 0,5 %

Total = 43 %

Untuk harga alat non impor sampai di lokasi pabrik ditambahkan biaya sebagai

berikut:

- PPn = 10 % (Rusjdi, 2004)

- PPh = 10 % (Rusjdi, 2004)

- Transportasi lokal = 0,5 %

- Biaya tak terduga = 0,5 %

- Total = 21 %

Tabel LE.3 Estimasi Harga Peralatan Proses

No Nama Alat Jumlah Ket*) Harga/unit

(US$)

Total Harga

(Rp)

1 Tangki Propena 3 NI 2.001.982,98 18.067.896.424

2 Tangki Butena

3

NI

1.119.991,31 10.107.921.556

3 Tangki Propana

2

NI

395.562,38 3.569.950.440

4 Tangki Butana

4

NI

1.043.704,17 9.419.430.130

5 Tangki Heksena

4

NI

955.038,61 8.619.223.470


6 Tangki Heptena

6

NI

1.644.204,70 14.838.947.439

7 Tangki Oktena 2 NI 2,001,982.98 18,067,896,424

8 Reaktor-01 1 NI 5,599.61 140,786,495

9 Pompa 01 1 NI 2,008.20 18,124,050

10 Pompa 02 1 NI 3,411.02 30,784,428

11 Pompa 03 1 NI 1,253.45 11,312,403

12 Pompa 04 1 NI 990.28 8,937,265

13 Pompa 05 1 NI 990.28 8,937,265

14 Pompa 06 1 NI 990.28 8,937,265

15 Pompa 07 1 NI 990.28 8,937,265

16 Pompa 08 1 NI 990.28 8,937,265

17 Pompa 09 1 NI 990.28 8,937,265

18 Pompa 10 1 NI 990.28 8,937,265

19 Pompa 11 1 NI 990.28 8,937,265

20 Pompa 12 1 NI 990.28 8,937,265

21 Pompa 13 1 NI 990.28 8,937,265

22 MD-101 1 I 1,271,075.53 11,471,456,691

23 MD-102 1 I 883,479.81 7,973,405,274

24 MD-103 1 I 732,801.67 6,613,535,030

25 MD-104 1 I 732,801.74 6,613,535,709

26 MD-105 1 I 581,165.97 5,245,022,861

27 Heater 01 (E-101) 1 I 7,055.91 63,679,621

28 Heater 02 (E-104) 1 I 7,055.91 63,679,621

29 Heater (E-108) 1 I 10,694.77 96,520,256

30 Heater (E-108) 1 I 10,694.77 96,520,256

31 Cooler (E-105) 1 I 4,655.17 42,012,882

32 Cooler (E-110) 1 I 4,655.17 42,012,882

33 Cooler (E-111) 1 I 4,655.17 42,012,882

34 Cooler (E-116) 1 I 6,324.77 57,081,020

35 Cooler (E-118) 1 I 8,066.76 72,802,531


36 Reboiler I 1 I 47,926.60 432,537,565

37 Reboiler II 1 I 46,447.34 419,187,201

38 Reboiler III 1 I 44,011.90 397,207,424

39 Reboiler IV 1 I 40,296.99 363,680,325

40 Reboiler V 1 I 37,777.78 340,944,465

41 Accumulator I 1 I 15,836.46 142,924,078

42 Accumulator II 1 I 13,000.49 117,329,465

43 Accumulator III 1 I 5,602.13 50,559,209

44 Accumulator IV 1 I 2,807.69 25,339,426

45 Accumulator V 1 I 10,194.27 92,003,269

46 Kondensor I 1 I 99,550.30 898,441,485

47 Kondensor II 1 I 69,599.05 628,131,395

48 Kondensor III 1 I 34,853.14 314,549,566

49 Kondensor IV 1 I 34,853.14 314,549,566

50 Kondensor V 1 I 92,455.13 834,407,585

Harga total 97,235,656,890

Impor 28.137.535,958

Non impor 69.098.120,932

*)Keterangan : I: untuk peralatan impor. NI: untuk peralatan non impor.

Tabel LE.4 Estimasi Harga Peralatan Utilitas

No Nama Alat Jumlah Ket*) Harga/unit

(US$)

Total Harga

(Rp)

1 PU-01 1 NI 39,392.77 355,519,741

2 PU -02 1 NI 452.64 4,085,118

3 PU -03 1 NI 452.64 4,085,118

4 PU -04 1 NI 452.64 4,085,118

5 PU -05 1 NI 452.64 4,085,118

6 PU -06 1 NI 452.64 4,085,118

7 PU -07 1 NI 452.64 4,085,118

8 PU -08 1 NI 452.64 4,085,118


9 PU -09 1 NI 452.64 4,085,118

10 PU -10 1 NI 572.94 5,170,766

11 PU -11 1 NI 452.64 4,085,118

12 PU -12 1 NI 452.64 4,085,118

13 PU -13 1 NI 452.64 4,085,118

14 PU -14 1 NI 452.64 4,085,118

15 PU -15 1 NI 452.64 4,085,118

16 PU -16 1 NI 452.64 4,085,118

17 PU -17 1 NI 452.64 4,085,118

18 PU -18 1 NI 452.64 4,085,118

19 PU -19 1 NI 452.64 4,085,118

20 SC 1 NI 39,392.77 355,519,741

21 Tangki Utilitas 1 1 NI 24,961.51 225,277,660

22 Tangki Utilitas 2 1 NI 41,632.63 375,734,503

23 Tangki Pelarut 1 1 NI 3,331.70 30,068,599





28 Clarifier I 205,025.9812 1,850,359,480

Tangki Filtrasi 1 I 37,057.6590 334,445,373

Cation Exchanger 1 I 51,009.9487 460,364,787

Refrigerator 1 I 13,349.8421 120,482,325

Dearator 1 I 72,975.8887 658,607,396

Anion Exchanger 1 I 51,009.9487 460,364,787

Ketel Uap 1 I 43,938.1215 396,541,547

Bak Sedementasi 2 NI 5,540.1662 50,000,000

Generator 1 I 4,986.1495 45,000,000

Harga total 5.361.992,107

Impor 3.509.153,766

Non impor 1.852.838,341


*)Keterangan : I: untuk peralatan impor. NI: untuk peralatan non impor.

Total harga peralatan tiba di lokasi pabrik (purchased-equipment delivered):

Total = 1,43 x (Rp 28,137,535,958 ,- + Rp 3,509,153,766 ,-) +

1,21 (Rp. 1.852.838,341,- + Rp. 97,235,656,890 3 ,-)= Rp. 131,105,427,025.69 ,-

Biaya pemasangan diperkirakan 20 % dari total harga peralatan (Timmerhaus. 2004).

= 0,2 x Rp. 131,105,427,025.69,-

= Rp 26.221.085.405,14

Sehingga total harga peralatan ditambah biaya pemasangan adalah:

Total Harga Peralatan (C) = Rp 157.326.512.430,83

LE.1.1.4 Instrumentasi dan Alat Kontrol

Diperkirakan biaya instrumentasi dan alat kontrol 40% dari total harga peralatan

(Timmerhaus et al. 2004).

Biaya instrumentasi dan alat kontrol (D) = 0,4 × Rp 157.326.512.430,83

= Rp 62.930,604.972.33 ,-

LE.1.1.5 Biaya Perpipaan

Diperkirakan biaya perpipaan 60% dari total harga peralatan


Biaya perpipaan (E) = 0,6 × Rp 157.326.512.430,83

= Rp 94.395.907,458.50


LE.1.1.6 Biaya Instalasi Listrik

Diperkirakan biaya instalasi listrik 35% dari total harga peralatan.


Biaya instalasi listrik (F) = 0,35 × Rp 157.326.512.430,83

= Rp 55.064.279.350,79

,-

LE.1.1.7 Biaya Insulasi

Diperkirakan biaya insulasi 60 % dari total harga peralatan.


Biaya insulasi (G) = 0,6 × Rp 157.326.512.430,83

= Rp 94.395.907.458,50

LE.1.1.8 Biaya Inventaris Kantor

Diperkirakan biaya inventaris kantor 15% dari total harga peralatan dan

pemasangan. (Timmerhaus et al. 2004)

Biaya inventaris kantor (H) = 0,15 × Rp 157.326.512.430,83

= Rp 23.598.976.864,62

LE.1.1.9 Biaya Perlengkapan Kebakaran dan Keamanan

Diperkirakan biaya perlengkapan kebakaran dan keamanan 30% dari total harga

peralatan dan pemasangan. (Timmerhaus et al. 2004)

Biaya perlengkapan kebakaran dan keamanan ( I ) = 0,3 × Rp 157.326.512.430,83

= Rp 47.197.953.729,25

LE.1.1.10 Sarana Transportasi

Tabel LE.5 Biaya Sarana Transportasi

No. Jenis Kendaraan Unit Tipe Harga/ Unit Harga Total (Rp)


(Rp)

1 Direktur 1 Fortuner 457,000,000.00 457,000,000.00

2 Sekretaris 1 Avanza 165,000,000.00 165,000,000.00

3 Manager 5 Kijang Inova 290,000,000.00 1,450,000,000.00

4 Bus Karyawan 2 Bus 410,000,000.00 820,000,000.00

5 Truk 10 Truk 350,000,000.00 3,500,000,000.00

6 Mobil Pemasaran 12 Avanza 165,000,000.00 1,650,000,000.00

7 Ambulance 1 Minibus 450,000,000.00 900,000,000.00

8 Mobil Pemadam

Kebakaran 2 Truk Tangki

110,000,000.00 110,000,000.00

Total

9.052.000.000,00

(Sumber: www.autocarprices.com)

Total MITL = A + B + C + D + E + F + G + H + I + J

= Rp 606.108.142.264,83

LE.1.2 Modal Investasi Tetap Tak Langsung (MITTL)

LE.1.2.1 Pra Investasi

Diperkirakan 20% dari total MITL. (Timmerhaus et al. 2004).

Pra Investasi (A) = 0,2 x Rp 606.108.142.264,83 = Rp 121.221.628.452,97

LE.1.2.2 Biaya Engineering dan Supervisi


Biaya Engineering dan Supervisi (B) = 0,30 × Rp 606.108.142.264,83

= Rp 181.832.442.679,45

LE.1.2.3 Biaya Legalitas


Biaya Legalitas (C) = 0,1 × Rp 606.108.142.264,83 = Rp 60.610.814.226


LE.1.2.4 Biaya Kontraktor


Biaya Kontraktor (D) =0,2 × Rp 606.108.142.264,83 = Rp 121.221.628.452,97

LE.1.2.5 Biaya Tak Terduga


Biaya Tak Terduga (E) = 0,2 x Rp 606.108.142.264,83 = Rp 121.221.628.452,97

Total MITTL = A + B + C + D + E = Rp 606.108.142.264,83

Total MIT = MITL + MITTL

= Rp 606.108.142.264,83 + Rp 606.108.142.264,83

= Rp 1.212.216.284.529,66

LE.2 Modal Kerja

Modal kerja dihitung untuk pengoperasian pabrik selama 3 bulan (90 hari).

LE.2.1 Persediaan Bahan Baku

LE.2.1.1 Bahan Baku Proses

Propena

Kebutuhan = 1.221,37 kg/jam

Harga = Rp 78600,-/kg (PT.PGN, 2010)

Harga total = 90 hari × 24 jam/hari × 1.221,37 kg/jam × Rp 78600,-

= Rp. 207.359.856.403,20

Butena

Kebutuhan = 3.384 kg/jam

Harga = Rp 61500,-/kg (PT.PGN, 2010)

Harga total = 90hari × 24 jam/hari × 3.384 kg/jam × Rp 61500,-

= Rp. 449.483.361.948,00


Katalis

Kebutuhan = 1,4518 Kg/jam

Harga = Rp. 25.000

Harga total = 90 hari x 24 jam/hari x 1,4518 kg/jam x Rp. 25.000

= Rp. 3.266.550

LE.2.1.2 Persediaan Bahan Baku Utilitas

1. Alum (Al2(SO4)3

Kebutuhan = 0,11008 kg/jam

Harga = Rp. 2.500 (PT. Bratachem, 2010)

Harga Total = 90 hari × 24 jam/hari × 0,11008 kg/jam × Rp 2.500,-/kg

= Rp. 594.432

2. Soda Abu Na2CO3


Harga = Rp. 3.500 (PT. Bratachem, 2010)

Harga Total = 90 hari × 24 jam/hari × 0,05944 kg/jam × Rp 3.500,-/kg

= Rp. 449.366,40

Kaporit


Harga = Rp 11.000,-/kg (PT. Bratachem, 2010)

Harga total = 90 hari × 24 jam/hari 0,002582 kg/jam × Rp 11.000,-/kg

= Rp.61.348,32,-

NaOH


Harga = Rp.7.350,-/kg (PT. Bratachem, 2010)

Harga total = 90 hari × 24 jam/hari × 0,3079 kg/jam × Rp 7.350,-/kg

= Rp.203.675,85,-

Solar

Kebutuhan = 47,9915 ltr/jam

Harga solar untuk industri = Rp.7.680,-/liter (PT.Pertamina, 2009)

Harga total = 90 hari × 24 jam/hari × 47,9915 ltr/jam × Rp. 7.680,-/liter


= Rp796.121.395,20,-

H2SO4


Harga = Rp.35.000,-/kg (PT. Bratachem, 2010)

Harga total = 90 hari × 24 jam/hari × 0,8712 kg/jam × Rp 35.000,-/kg

= Rp.2.744.280,00

Total biaya persediaan bahan baku proses dan utilitas selama 3 bulan (90 hari) adalah

Rp. 657.643.847.798,25 ,-

LE.2.2 Kas

LE.2.2.1 Gaji Pegawai

Tabel LE.6 Perincian Gaji Pegawai

Jabatan Jumlah Gaji/bulan

(Rp)

Jumlah gaji/bulan

(Rp)

Direktur 1 25,000,000 25,000,000

Sekretaris 1 3,000,000 3,000,000

Manajer Pemasaran 1 7,000,000 7,000,000

Manajer Keuangan 1 7,000,000 7,000,000

Manajer Personalia 1 7,000,000 7,000,000

Manajer Teknik 1 7,000,000 7,000,000

Manajer Produksi 1 7,000,000 7,000,000

Kepala Bagian Penjualan dan Pembelian 1 5,000,000 5,000,000

Kepala Bagian Pembukuan dan

Perpajakan 1

5,000,000 5,000,000

Kepala Bagian Kepegawaian dan Humas 1 5,000,000 5,000,000

Kepala Bagian Mesin 1 5,000,000 5,000,000

Kepala Bagian Proses 1 5,000,000 5,000,000

Kepalas Bagian Listrik dan Instrumentasi 1 5,000,000 5,000,000

Kepala Bagian Sipil 1 5,000,000 5,000,000

Kepala Seksi 13 4,000,000 52,000,000


Karyawan Produksi 56 2,500,000 140,000,000

Karyawan Teknik 30 2,500,000 75,000,000

Karyawan Keuangan dan Personalia 6 2,500,000 15,000,000

Karyawan Pemasaran dan penjualan 6 2,500,000 17,000,000

Dokter 2 4.000,000 8,000,000

Perawat 2 1,500,000 3,000,000

Petugas Keamanan 12 2,500,000 30,000,000

Petugas Kebersihan 6 1,200,000 7,200,000

Supir 4 1,500,000 6,000,000

Jumlah 151 Rp 449.200.000

Total gaji pegawai selama 1 bulan = Rp 449.200.000

Total gaji pegawai selama 3 bulan = Rp 1.347.600.000,00

LE.2.2.2 Biaya Administrasi Umum

Diperkirakan 10% dari gaji pegawai = 0,1 × Rp. 1.347.600,000. (Peters et.al., 2004)

= Rp 134,760.000,00

LE.2.2.3 Biaya Pemasaran

Diperkirakan 10% dari gaji pegawai = 0,1 × Rp. 1.347.600,000. (Peters et.al., 2004)

= Rp 134,760.000,00

LE.2.2.4 Pajak Bumi dan Bangunan

Dasar perhitungan Pajak Bumi dan Bangunan (PBB) mengacu kepada

Undang-Undang RI No. 20 Tahun 2000 Jo UU No. 21 Tahun 1997 tentang Bea

Perolehan Hak atas Tanah dan Bangunan sebagai berikut:

Yang menjadi objek pajak adalah perolehan hak atas tanah dan atas bangunan

(Pasal 2 ayat 1 UU No.20/00).

Dasar pengenaan pajak adalah Nilai Perolehan Objek Pajak (Pasal 6 ayat 1 UU

No.20/00).

Tarif pajak ditetapkan sebesar 5% (Pasal 5 UU No.21/97).


Nilai Perolehan Objek Pajak Tidak Kena Pajak ditetapkan sebesar Rp.

30.000.000.- (Pasal 7 ayat 1 UU No.21/97).

Besarnya pajak yang terutang dihitung dengan cara mengalikkan tarif pajak

dengan Nilai Perolehan Objek Kena Pajak (Pasal 8 ayat 2 UU No.21/97).

Maka berdasarkan penjelasan di atas. perhitungan PBB ditetapkan sebagai berikut :

Wajib Pajak Pabrik Pembuatan Heptena

Nilai Perolehan Objek Pajak

Tanah Rp. 1.232.000.000,00

Bangunan Rp. 3,696,000,000.00

Total NPOP Rp 4.928.000.000,-

Nilai Perolehan Objek Pajak Tidak Kena Pajak (Rp. 8.000.000,-)

Nilai Perolehan Objek Pajak Kena Pajak Rp 4.920.000.000,-

Pajak yang Terutang (5% x NPOPKP) Rp. 12,300,000.00 ,-

Tabel LE.7 Perincian Biaya Kas

No. Jenis Biaya Jumlah (Rp) 1. Gaji Pegawai Rp. 1,649,100,000.00

2. Administrasi Umum Rp. 164,910,000.00

3. Pemasaran Rp. 164,910,000.00 4. Pajak Bumi dan Bangunan Rp. 12,300,000.00

Total

Rp.1,629,420,000.00

LE.2.3 Biaya Start-Up

Diperkirakan 12% dari Modal Investasi Tetap. (Timmerhaus et al. 2004).

= 0,15 × Rp 1.212.216.284.529,66

= Rp 181.832.442.679,455


LE.2.4 Piutang Dagang

HPT12IPPD ×=

dimana: PD = piutang dagang

IP = jangka waktu kredit yang diberikan (1 bulan)

HPT = hasil penjualan tahunan

Penjualan:

1. Harga jual Propana = Rp 20.000kg (www.alibaba.com)

Produksi Propana = 1.025,3137 kg/jam

Hasil penjualan Propana tahunan yaitu:

= 1.025,3137 kg/jam × 24 jam/hari × 330 hari/tahun × Rp 20.000,-/kg

= 162.410.260.320,00,-

2. Harga Jual Butana = Rp 17.000,-/kg

Produksi Butana = 3.485,47 kg/jam

Hasil penjualan Butana tahunan yaitu:


= 469,284,259,752.00,-

3. Harga Jual Heksena = Rp 25.000,-/kg

Produksi Heksena = 3.136,96 kg/jam

Hasil penjualan Heksena tahunan yaitu:


= 621.118.000.800,00

4. Harga Jual Heptena = Rp 53.000,-/kg

Produksi Heptena = 5.050,5050 kg/jam

Hasil penjualan Heptena tahunan yaitu:



= Rp. 2,119,999,999,995,-

5. Harga Jual Oktena = Rp 21.000,-/kg

Produksi Oktena = 1.683,46 kg/jam

Hasil penjualan Oktena tahunan yaitu:


= 279.993.117.096,00

Hasil penjualan total tahunan = Rp 3.898.620.095.040.00

Piutang Dagang = 123× Rp 3.898.620.095.040.00

= Rp 974.655.023.760,00

Perincian modal kerja dapat dilihat pada tabel di bawah ini.

Tabel LE.8 Perincian Modal Kerja

No. Jenis Biaya Jumlah (Rp)

1. Bahan Baku Proses dan Utilitas Rp. 657.643.847.798,25

2. Kas Rp. 1,629,420,000,00

3. Start Up Rp 181.832.442.679,455

4. Piutang Dagang Rp 974.655.023.760,00

Total Rp. 1.802.589.606.369,70

Total Modal Investasi = Modal Investasi Tetap + Modal Kerja

= Rp 1.212.216.284.529,66 + Rp 1.815.760.734.237,70

= Rp 3.027.977.01.767,36

Modal ini berasal dari:

- Modal sendiri = 60% dari total modal investasi

= 0,6 × Rp 3.027.977.01.767,36


= Rp 1.816.786.211.260,42

- Pinjaman dari Bank = 40% dari total modal investasi

= 0,4 × Rp 3.027.977.01.767,36

= Rp 1.211.190.807.506,94

LE.3 Biaya Produksi Total

LE.3.1 Biaya Tetap (Fixed Cost = FC)

LE.3.1.1 Gaji Tetap Karyawan

Gaji tetap karyawan terdiri dari gaji tetap tiap bulan ditambah 2 bulan gaji

yang diberikan sebagai tunjangan. Sehingga :

Gaji total = (12 + 3) × Rp. 449.200.000,- = Rp 403.574.037.764,03 ,-

LE.3.1.2 Bunga Pinjaman Bank

Bunga pinjaman bank adalah 15% dari total pinjaman. ` (Bank Mandiri. 2010)

= 0,15 × Rp. 1.204.380.356.359,74,-

= Rp 181.678.621.126,04

LE.3.1.3 Depresiasi dan Amortisasi

Pengeluaran untuk memperoleh harta berwujud yang mempunyai masa

manfaat lebih dari 1 (satu) tahun harus dibebankan sebagai biaya untuk

mendapatkan,menagih dan memelihara penghasilan melalui penyusutan (Rusdji.

2004). Pada perancangan pabrik ini. dipakai metode garis lurus atau straight line

method. Dasar penyusutan menggunakan masa manfaat dan tarif penyusutan sesuai

dengan Undang-undang Republik Indonesia No.17 Tahun 2000 Pasal 11 ayat 6 dapat

dilihat pada tabel di bawah ini.

Tabel LE.9 Aturan depresiasi sesuai UU Republik Indonesia No. 17 Tahun 2000

Kelompok Harta

Berwujud

Masa

(tahun)

Tarif

(%) Beberapa Jenis Harta

I.Bukan Bangunan

1.Kelompok 1

4

25

Mesin kantor. perlengkapan. alat perangkat/

tools industri.


2. Kelompok 2

3. Kelompok 3

8

16

12.5

6.25

Mobil. truk kerja

Mesin industri kimia. mesin industri mesin

II. Bangunan

Permanen

20

5

Bangunan sarana dan penunjang

Sumber: Waluyo. 2000 dan Rusdji.2004

Depresiasi dihitung dengan metode garis lurus dengan harga akhir nol.

n

LPD −=

dimana: D = depresiasi per tahun

P = harga awal peralatan

L = harga akhir peralatan

n = umur peralatan (tahun)

Tabel LE.10 Perhitungan Biaya Depresiasi sesuai UU RI No. 17 Tahun 2000

Komponen Biaya (Rp) Umur

(tahun) Depresiasi (Rp)

Bangunan 21,490,000,000.00 20 1,074,500,000.00

Peralatan proses dan utilitas

157,326,512,430.83 10 15,732,651,243.08

Instrumentrasi dan pengendalian proses

62,930,604,972.33

10

6,293,060,497.23

Perpipaan

94,395,907,458.50

10

9,439,590,745.85

Instalasi listrik 55,064,279,350.79

10

5,506,427,935.08

Insulasi 94,395,907,458.50

10

9,439,590,745.85

Inventaris kantor 23,598,976,864.62 10 2,359,897,686.46

Perlengkapan keamanan dan kebakaran

47,197,953,729.25

10

4,719,795,372.92

Sarana transportasi

8,972,500,000.00

10

897,250,000.00


Total

55,462,764,226.48

Semua modal investasi tetap langsung (MITL) kecuali tanah mengalami

penyusutan yang disebut depresiasi. sedangkan modal investasi tetap tidak langsung

(MITTL) juga mengalami penyusutan yang disebut amortisasi.

Pengeluaran untuk memperoleh harta tak berwujud dan pengeluaran lainnya

yang mempunyai masa manfaat lebih dari 1 (satu) tahun untuk mendapatkan.

menagih. dan memelihara penghasilan dapat dihitung dengan amortisasi dengan

menerapkan taat azas (UURI Pasal 11 ayat 1 No. Tahun 2000). Para Wajib Pajak

menggunakan tarif amortisasi untuk harta tidak berwujud dengan menggunakan masa

manfaat kelompok masa 4 (empat) tahun sesuai pendekatan perkiraan harta tak

berwujud yang dimaksud (Rusdji. 2004).

Untuk masa 4 tahun. maka biaya amortisasi adalah 3% dari MITTL sehingga:

Biaya amortisasi = 0,03 × Rp. 604.180.642.264,83

= Rp 18.125.419.267,94

Total Biaya Depresiasi dan Amortisasi

= Rp. 55.462.764.226,48 + Rp 18.125.419.267,94

= Rp 73.588.183.494,43

LE.3.1.4 Biaya Perawatan Tetap

Biaya tetap perawatan terbagi menjadi:

1. Perawatan mesin dan alat-alat proses

Perawatan mesin dan peralatan dalam industri proses berkisar 2 sampai 20%.

diambil 10% dari harga peralatan terpasang di pabrik (Peters et.al., 2004).

Biaya perawatan mesin = 0,1 × Rp 60.418.064.226,48 ,-

= Rp 6.041.806.422,65

2. Perawatan bangunan

Diperkirakan 10% dari harga bangunan. (Peters et.al., 2004)

= 0,1 × Rp. 21,490,000,000.00

= Rp. 2,149,000,000.00

3. Perawatan kendaraan

Diperkirakan 10% dari harga kendaraan. (Peters et.al., 2004)


= 0,1 × Rp 9.052.000.000,00

= Rp. 905.2000.000,-

4. Perawatan instrumentasi dan alat kontrol

Diperkirakan 10% dari harga instrumentasi dan alat kontrol. (Peters et.al., 2004)

= 0,1 × Rp 62.930.604.972,33 ,-

= Rp. 6.293.060.497,233

5. Perawatan perpipaan

Diperkirakan 10 % dari harga perpipaan. (Peters et.al., 2004)

= 0,1 × Rp 94.395.907.458,50 ,-

= Rp. 9.439.590.745,850

6. Perawatan instalasi listrik

Diperkirakan 10% dari harga instalasi listrik. (Peters et.al., 2004)

= 0,1 × Rp 55.064.279.350,79

= Rp 5.506.427.935,079

7. Perawatan insulasi

Diperkirakan 10% dari harga insulasi. (Peters et.al., 2004)

= 0,1 × Rp 94.395.907.458,50

= Rp 9.439.590.745,850

8. Perawatan inventaris kantor

Diperkirakan 10% dari harga inventaris kantor. (Peters et.al., 2004)

= 0,1 × Rp 23.598.976.864,62

= Rp. 2.359.897.686,462

9. Perawatan perlengkapan kebakaran

Diperkirakan 10% dari harga perlengkapan kebakaran. (Peters et.al., 2004

= 0,1 × Rp 47.197.953.729,25

= Rp. 4.719.795.372,925

Total Biaya Perawatan = Rp 56.545.214.226,48

LE.3.1.5 Biaya Tambahan Industri (Plant Overhead Cost)

Biaya tambahan industri ini diperkirakan 5% dari modal investasi tetap. (Peters et.al.,

2004)


Plant Overhead Cost = 0,05 x Rp 1.208.361.284.529,66

= Rp 60.610.814.226,48

LE.3.1.6 Biaya Administrasi Umum

Diperkirakan 10% dari biaya tambahan

= 0,1 × Rp 60.610.814.226,48,- = Rp 6.041.806.422,648

LE.3.1.7 Biaya Pemasaran dan Distribusi

Diperkirakan 5% dari biaya tambahan

= 0,05 × Rp 60.610.814.226,48,- = Rp 3.030.540.711,32

LE.3.1.8 Biaya Laboratorium. Penelitian dan Pengembangan

Diperkirakan 5% dari biaya tambahan industri. (Peters et.al., 2004)

= 0,05 × Rp 60.610.814.226,48,- = Rp 3.030.540.711,32

LE.3.1.9 Biaya Asuransi

1. Biaya asuransi pabrik adalah 1 % dari modal investasi tetap langsung

(Jamsostek, 2010)

= 0,01 × Rp 1.208.361.284.529,66

= Rp 12.083.612.845,29

2. Biaya asuransi karyawan

Premi asuransi 1% ditanggung oleh perusahaan dan dibayar 2 % langsung

dari gaji karyawan. (Jamsostek, 2010)

Maka biaya asuransi karyawan = 0,01 x Rp. 6,596,400,000.00

= Rp. 65.964.000,00

Total Asuransi = Rp 12.149.576.845,30

LE.3.1.10 Pajak Bumi dan Bangunan

Pajak Bumi dan Bangunan adalah Rp. 49,200,000.00

Total Biaya Tetap (Fixed Cost) = Rp 403.574.037.764,03

LE.3.2 Biaya Variabel


LE.3.2.1 Biaya Variabel Bahan Baku Proses dan Utilitas per tahun

Total biaya persediaan bahan baku proses dan utilitas selama 1 tahun adalah:

= Rp 2.630.575.391.193,00

LE.3.2.2 Biaya Variabel Tambahan

Biaya variabel tambahan terbagi menjadi:

1. Biaya Perawatan dan Penanganan Lingkungan

Diperkirakan 5% dari biaya variabel bahan baku

= 0,05 × Rp 56.537.264.226,48

= Rp 2.826.863.211,32

2. Biaya Variabel Pemasaran dan Distribusi

Diperkirakan 5% dari biaya variabel bahan baku

= 0,05 × Rp 3.020.903.211,32

= Rp 151.045.160,57

LE.3.2.3 Biaya Variabel Lainnya

Diperkirakan 5% dari biaya variabel tambahan

= 0.05 × Rp 60.418.064.226,48

= Rp 3.020.903.211,32

Total Biaya Variabel = Rp 2.636.584.719.651,21

Total Biaya Produksi = Biaya Tetap + Biaya Variabel

= Rp 403.574.037.764,03 ,- + Rp 2.636.584.719.651,21

= Rp 3.040.158.757.415,24


LE.4 Perkiraan Laba/Rugi Perusahaan

LE.4.1 Laba Sebelum Pajak (Bruto)

Laba atas penjualan = Total penjualan – Total biaya produksi

= Rp 3.898.620.095.040,00 . – Rp 3.040.158.757.415,24

= Rp 858.461.337.624,76

Bonus perusahaan untuk karyawan 0,5% dari keuntungan perusahaan

= 0,005 × Rp 858.461.337.624,76

= Rp 3.002.067,94

Pengurangan bonus atas penghasilan bruto sesuai dengan UU RI No. 17/00 Pasal 6

ayat 1 sehingga:

Laba sebelum pajak (bruto) = Rp 858.461.337.624,76

LE.4.2 Pajak Penghasilan

Berdasarkan UURI Nomor 17 ayat 1 Tahun 2000. Tentang Perubahan Ketiga

atas Undang-undang Nomor 7 Tahun 1983 Tentang Pajak Penghasilan adalah

(Rusjdi. 2004):

Penghasilan sampai dengan Rp 50.000.000,- dikenakan pajak sebesar 10 %.

Penghasilan Rp 50.000.000,- sampai dengan Rp 100.000.000,- dikenakan pajak

sebesar 15 %.

Penghasilan di atas Rp 100.000.000,- dikenakan pajak sebesar 30 %.

Maka pajak penghasilan yang harus dibayar adalah:

- 10 % × Rp 50.000.000 = Rp 5.000.000 ,-

- 15 % × (Rp 100.000.000 - Rp 50.000.000) = Rp 7.500.000,-

- 30 % × (Rp 841.876.025.691.63 - Rp 100.000.000) = Rp 257,282,168,077.73

Total PPh =Rp 257.294.668.077,73

LE.4.3 Laba setelah pajak

Laba setelah pajak = laba sebelum pajak – PPh

= Rp 858.461.337.624,76- Rp 257.294.668.077,73

= Rp 600.412.558.848,03


LE.5 Analisa Aspek Ekonomi

LE.5.1 Profit Margin (PM)

PM = penjualanTotal

pajaksebelumLaba× 100 %

PM = %100 095.040,003.898.620. Rp

7.624,76858.461.33 Rp ×

PM = 22,0 %

LE.5.2 Break Even Point (BEP)

BEP = VariabelBiayaPenjualanTotal

TetapBiaya−

× 100 %

BEP = 719.651,212.636.584. Rp ,- 095.040,003.898.620. Rp

,- 7.764,03403.574.03 Rp −

×100 %

BEP = 31,97 %

Kapasitas produksi pada titik BEP = 31,97 %× 40.000.000 kg/tahun

= 12.788.000 kg/tahun

Nilai penjualan heptena pada titik BEP = 31,97 % × Rp 3.898.620.095.040,00 .

= Rp 756.540.832.103,80

LE.5.3 Return on Investment (ROI)

ROI = InvestasiModalTotal

pajaksetelahLaba× 100 %

ROI = 01.767,363.027.977. Rp 8.848,03600.412.55 Rp

× 100 %

ROI = 19,82 %

LE.5.4 Pay Out Time (POT)

POT = tahun10,1982

1×

POT = 5,04 tahun = 5 tahun


LE.5.5 Return on Network (RON)

RON = sendiriModal

pajaksetelahLaba× 100 %

RON = ,- 211.260,421.816.786. Rp

8.848,03600.412.55 Rp × 100 %

RON = 33,04 %

LE.5.6 Internal Rate of Return (IRR)

Untuk menentukan nilai IRR harus digambarkan jumlah pendapatan dan

pengeluaran dari tahun ke tahun yang disebut “Cash Flow”. Untuk memperoleh cash

flow diambil ketentuan sebagai berikut:

- Laba kotor diasumsikan mengalami kenaikan 10 % tiap tahun.

- Masa pembangunan disebut tahun ke nol.

- Jangka waktu cash flow dipilih 10 tahun.

- Perhitungan dilakukan dengan menggunakan nilai pada tahun ke – 10.

- Cash flow adalah laba sesudah pajak ditambah penyusutan.

Dari Tabel LE.11. diperoleh nilai IRR = 27,78 %,


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