Preparasi dan Karakterisasi Liquid Natural Rubber (LNR) Sebagai Kompatibiliser Untuk Meningkatkan Sifat Mekanik dan Sifat Termal Kompon Karet Alam

  LAMPIRAN

  Lampiran 1. Proses Pengenceran Lateks Pekat menjadi Lateks 30 % dan Penentuan Kadar Karet Kering

  Lateks Pekat Proses Pengenceran Lateks 30 % Plat tipis lapisan karet untuk menentukan kadar karet kering 30 %

  Lampiran 2. Perhitungan Penentuan Kadar Karet Kering (KKK) dan Persen Ralat (PR)

  Berat Lateks = 100 gram KKK teori = 30 % Berat Karet lapis tipis = 30,173 gram KKK praktik = 30,173 %

  −

  KKK = × 100% PR = × 100%

  30,173 30 %−30,173 %

  KKK = × 100% PR = × 100%

  100 30 %

  KKK = 30,173 % PR = 0,577 %

  Lampiran 3. Proses Pembuatan Liquid Natural Rubber (LNR) beserta Uji

  Hasil LNR tanpa destilasi Proses Pembuatan LNR dengan mengalirkan gas O ² Proses destilasi LNR

  LNR sebelum dikeringkan LNR sesudah dikeringkan Proses penimbangan LNR Penentuan BM LNR Penentuan BM Karet

  Penentuan sifat kelarutan dari LNR Analisa FTIR dari LNR

  Lampiran 4. Perhitungan Kadar Air (KA) dan Total Padatan (TP) dalam Liquid Natural Rubber (LNR)

  Berat sebelum kering (m ) = 124,02 gram _bk Berat sesudah kering (m k ) = 120 gram

  

m m ℎ

bk − k

  KA = × 100% TP = × 1

  124,02 −120 120

  KA = × 100% TP = × 100%

  124,02 124,02

  KA = 3,241 % TP = 96,759 %

  Lampiran 5. Spektrum FTIR Liquid Natural Liquid (LNR)

  Lampiran 6. Spektrum FTIR Karet SIR-10 Lampiran 7. Perhitungan Penentuan Bobot Molekul Liquid Natural Rubber _{toluena = 45,4466 s} (LNR) _{3 = 126,2633 s = 5,58.10 kg/ms 4}

  t̅ t̅ η _{1 = 73,8633 s 4 = 170, 7667 s} toluena t̅ t̅ _{2 = 95,6500 s 5 = 204, 3500 s} t̅ t̅

  Perhitungan Penentuan Viskositas Relatif ( ) t̅

  η = ( ) × η

  toluena t̅ toluena 73,8633 ⁴

  a. kg/ms η = ( ) × 5,58.10

  1 45,4466 s ₄

  kg/ms η = 9,0690.10

  1

  4 95,6500

  b. kg/ms η = ( ) × 5,58.10

  2 45,4466 s ₄

  kg/ms η = 11,7440.10

  2 ₄ 126,2633

  c. kg/ms η = ( ) × 5,58.10

  3 45,4466 s ₄

  kg/ms η = 15,5028.10

  3 170,7667 ⁴

  d. kg/ms η = ( ) × 5,58.10

  4 45,4466 s ₄

  kg/ms η = 20,9669.10

  4 ₄ 204,3500

  e. kg/ms η = ( ) × 5,58.10

  5 45,4466 s ₄

  kg/ms η = 25,0904.10

5 Perhitungan Penentuan Viskositas Spesifik ( )

  η − η

  toluena

  η = ( ) η

  toluena

  4

  4 9,0690.10 kg/ms−5,58.10 kg/ms a.

  η = ( )

  1

  4 5,58.10 kg/ms

  4

  3,489. 10 kg/ms η = ( )

  1

  4

  5,58. 10 kg/ms η = 0,6253

  1

  4

  4 11,7440.10 kg/ms−5,58.10 kg/ms b.

  η = ( )

  2

  4 5,58.10 kg/ms

  4

  6,164. 10 kg/ms η = ( )

  2

  4

  5,58. 10 kg/ms η = 1,1047

  2

  4

  4 15,5028.10 kg/ms−5,58.10 kg/ms c.

  η = ( )

  3

  4 5,58.10 kg/ms

  4

  9,9228. 10 kg/ms η = ( )

  3

  4

  5,58. 10 kg/ms η = 1,7783

  3

  4

  4 20,9669.10 kg/ms−5,58.10 kg/ms d.

  η = ( )

  4

  4 5,58.10 kg/ms

  4

  15,3869. 10 kg/ms η = ( )

  4

  4

  5,58. 10 kg/ms η = 2,7575

  4

  4

  4 25,0904.10 kg/ms−5,58.10 kg/ms e.

  η = ( )

  5

  4 5,58.10 kg/ms

  4

  19,5104. 10 kg/ms η = ( )

  5

  4

  5,58. 10 kg/ms η = 3,4965

  5

  Perhitungan Penentuan Viskositas Reduksi ( ) η

  =( η ) dimana, C = Konsentrasi (g/ml)

  C

  = 0,01 = 0,04 C C

  1

  4

  = 0,02 = 0,05 C C

  2

  5

  = 0,03 C

  3 0,6253 a.

  η = ( )

  1 0,01

  η = 62,5300

  1 1,1047

  b.

  η = ( )

  2 0,02

  η = 55,2350

  2 1,7783

  c.

  η = ( )

  3 0,03

  η = 59,2767

  3 2,7575

  d.

  η = ( )

  4 0,04

  η = 68,9375

  4 3,4965

  e.

  η = ( )

  5 0,05

  η = 69,9300

5 Perhitungan Penentuan Viskositas Intrinsik (

  [ ]) Viskositas intrinsik = nilai intersept dari hasil plot grafik linier

  Konsentrasi (%) viskositas reduksi (

  ) 1 62,5300 2 55,2350 3 59,2767 4 68,9375 5 69,9300

  Viskositas Reduksi terhadap Konsentrasi

  )

  80 ed ηr (

  60 si k y = 2.8503x + 54.631 u d

40 R² = 0.5158

  re s ta

  20 si o k is

  V

  2

  4

  6 Konsentrasi (C)

  Maka nilai [η] = 2,850 ml/g

  Perhitungan Penentuan Bobot molekul Rata-Rata Viskositas ( ̅̅̅̅)

  2

  toluena a.

  η

  1

  = (

  141,3400 45,4466 s

  ) × 5,58.10 ⁴ kg/ms η

  1

  = 17,354.10 ⁴ kg/ms b.

  η

  2

  = (

  141,6150 45,4466 s

  ) × 5,58.10 ⁴ kg/ms η

  = 17,388.10 ⁴ kg/ms c.

  t̅ t̅ toluena

  η

  3

  = (

  255,4300 45,4466 s

  ) × 5,58.10 ⁴ kg/ms η

  3

  = 31,362.10 ⁴ kg/ms d.

  η

  4

  = (

  690,0825 45,4466 s

  ) × 5,58.10 ⁴ kg/ms η

  4

  ) × η

  η = (

  [η] = kM

  v

  v

  ̅̅̅̅ ^a [η] = Viskositas intrinsik (2,850 ml/g) M

  v

  ̅̅̅̅ = bobot molekul rata-rata viskositas k = Tetapan 50,2 × 10

• ^-3 ml/g a = nilai tetapan viskositas karet 0,67

  M

  v

  ̅̅̅̅ ^a =

  [η] k

  M

  v

  ̅̅̅̅ ^0,67 =

  2,850 ml/g 50,2 × 10 −3 ml/g

  log M

  ̅̅̅̅ ^0,67 = log 0,06 × 10 ³

  Perhitungan Penentuan Viskositas Relatif ( )

  0,67 log M

  v

  ̅̅̅̅ = 1,7782 log M

  v

  ̅̅̅̅ =

  1,7782 0,67

  M

  v

  ̅̅̅̅ =

  1 log

  (2,6540) = 4,5082

  × 10 ² Lampiran 8. Perhitungan Penentuan Bobot Molekul Karet SIR-10 t̅ ^{toluena = 45,4466 s} t̅ ^{3 = 255,4300 s} t̅ ^{1 = 141,3400 s} t̅ ^{4 = 690,0825 s} t̅ ^{2 = 141,6150 s} η

  toluena

  = 5,58.10 ⁴ kg/ms

  = 84,729.10 ⁴ kg/ms

  Perhitungan Penentuan Viskositas Spesifik ( )

  η − η

  toluena

  η = ( ) η

  toluena

  4

  4 17,354.10 kg/ms−5,58.10 kg/ms a.

  η = ( )

  1

  4 5,58.10 kg/ms

  4

  11,774. 10 kg/ms η = ( )

  1

  4

  5,58. 10 kg/ms η = 2,110035

  1

  4

  4 17,388.10 kg/ms−5,58.10 kg/ms b.

  η = ( )

  2

  4 5,58.10 kg/ms

  4

  11,808. 10 kg/ms η = ( )

  2

  4

  5,58. 10 kg/ms η = 2,116129

  2

  4

  4 31,362.10 kg/ms−5,58.10 kg/ms c.

  η = ( )

  3

  4 5,58.10 kg/ms

  4

  25,782. 10 kg/ms η = ( )

  3

  4

  5,58. 10 kg/ms η = 4,620430

  3

  4

  4 84,729.10 kg/ms−5,58.10 kg/ms d.

  η = ( )

  4

  4 5,58.10 kg/ms

  4

  79,149. 10 kg/ms η = ( )

  4

  4

  5,58. 10 kg/ms η = 14,184409

  4 Perhitungan Penentuan Viskositas Reduksi ( ) η

  =( η ) dimana, C = Konsentrasi (g/ml)

  C

  = 0,01 = 0,04 C C

  1

  4

  = 0,02 = 0,05 C C

  2

  5

  = 0,03 C

  3 2,110035 a.

  η = ( )

  1 0,01

  η = 211,0035

  1 2,116129

  b.

  η = ( )

  2 0,02

  η = 105,8065

  2 4,620430

  c.

  η = ( )

  3 0,03

  η = 154,0143

  3 14,184409

  d.

  η = ( )

  4 0,04

  η = 354,6102

  4

  Perhitungan Penentuan Viskositas Intrinsik (

  [ ]) Viskositas intrinsik = nilai intersept dari hasil plot grafik linier

  Konsentrasi (%) viskositas reduksi (

  ) 1 211,0035 2 105,8065 3 154,0143 4 354,6102

  Viskositas reduksi terhadap Konsentrasi

  ) 400 d re y = 47.903x + 86.602

  (η 300 R² = 0.3292 si k u

  200 ed r as 100 sit o k Vis

  1

  2

  3

  

4

  5 Konsentrasi (C) Perhitungan Penentuan Bobot molekul Rata-Rata Viskositas ( ̅̅̅̅) _a

  ̅̅̅̅ [η] = kM

  v

  [η] = Viskositas intrinsik (47,90 ml/g) ̅̅̅̅ = bobot molekul rata-rata viskositas M

• _-3 v

  k = Tetapan 50,2 × 10 ml/g a = nilai tetapan viskositas karet 0,67 _a [η] ̅̅̅̅ = M

  v k _{0,67 47,90 ml/g}

  ̅̅̅̅ =

  M

  v −3 _0,67 50,2 × 10 ml/g

  log ̅̅̅̅ = log 954,1832 M

  v

  0,67 log ̅̅̅̅ = 2,979632 M

  v 2,979632

  log = ̅̅̅̅ M

  v 0,67

  1

  = (4,447212) ̅̅̅̅ M

  v log ₄

  = 2,8003 × 10

  Lampiran 9. Proses Pembuatan Kompon NR/CB/LNR

  Penimbangan bahan Proses pembuatan kompon Proses persiapan bahan

  Proses hotpress Analisa reologi Analisa viskositas mooney

  Bentuk uji viskositas Uji tarik dan uji sobek Spesimen uji tarik

  Uji morfologi SEM Spesimen uji sobek Uji termal TGA dan DSC

  Lampiran 10. Data Karakterisasi Sifat Pematangan Kompon NR/CB/LNR dari Hasil Rheometer

  Lampiran 11. Perhitungan Selisih Torsi Maksimum dengan Torsi Minimum (S’) dalam Karakterisasi Sifat Pematangan

  90

  100 1,12 menit

  ) =

  100 1,82 menit−0,70 menit

  ILP ^{2 =} (

  = 85,47 menit ^-1 b.

  100 1,17 menit

  ) =

  100 1,93 menit−0,76 menit

  ILP ^{1 =} (

  ) a.

  100 TC 90−TS2

  0,76 0,70 0,68 0,65 0,68 Indeks Laju Pematangan (ILP) = (

  2

  1,93 1,82 1,79 1,60 1,56 TS

  = Waktu pematangan dini (menit) TC

  S _max = Torsi maksimum (dNm) S min = Torsi minimum (dNm)

  2

  = Waktu pematangan optimum (menit) TS

  90

  TC

  Lampiran 12. Perhitungan Indeks Laju Pematangan (ILP) Kompon

  S’ ^{5 = 6,71 dNm}

  = 7,17 dNm e. S’ ^{5 = 6,82 dNm - 0,11 dNm}

  S’ ^{4 = 7,37 dNm - 0,20 dNm} S’ ⁴

  = 8,18 dNm - 0,27 dNm S’ ^{3 = 7,91 dNm} d.

  S’ ^{2 = 7,88 dNm} c. S’ ³

  = 7,05 dNm b. S’ ^{2 = 8,18 dNm - 0,30 dNm}

  S’ ^{1 = 7,28 dNm - 0,23 dNm} S’ ¹

  S’ = S ^{max - S min} a.

  S max 7,28 8,18 8,18 7,37 6,82 S min 0,23 0,30 0,27 0,20 0,11

  = 89,29 menit ^-1

  100 c. ^{3 =}

  ILP ( )

  1,79 menit−0,68 menit 100

  =

  1,11 menit _-1

  = 90,09 menit

  100 d. ^{4 =}

  ILP ( )

  1,60 menit−0,65 menit 100

  =

  0,95 menit _-1

  = 105,26 menit

  100 e.

  ILP ^{5 =} ( )

  1,56 menit−0,68 menit 100

  =

  0,88 menit _-1

  = 113,64 menit

  Lampiran 13. Analisa dan Data Pengukuran Viskositas Mooney

  Lampiran 14. Data Hasil Analisa Sifat Mekanik Uji Tarik, Perpanjangan

Putus dan Modulus pada Kompon NR/CB/LNR

  Lampiran 15. Perhitungan Luas Penampang (A) Kompon pada Uji Tarik

  σ = kekuatan tarik (MPa) F = beban tarik (N) A = luas penampang (m ² )

2 Spesimen Kompon Perbandingan 100:50:2,5 a.

  = 9,36 m

  1

  ) A

  199 N 19,3 MPa

  = (

  1

  A

  3

  2 b.

  ) A

  204 N 21,8 MPa

  = (

  3

  A

  2 c.

  = 10,31 m

  2

  A

  = (

  

  A F 

  = 10,46 m

  3

  ) A

  205 N 19,6 MPa

  3

  2

  A

  2 c.

  = 10,32 m

  2

  ) A

  192 N 18,6 MPa

  = (

  = 9,31 m

  ) A

  σ ^{15,8 17,9 15.9 19,5 18,8 21,8 19,3 18,6 19,6 21,0 22,1 19,9 20,7 19,1 20,7} F ^{165 179 173 183 175 204 199 192 205 209 216 201 198 182 198}

  2 b.

  2

  ) A

  179 N 17,9 MPa

  = (

  2

  A

  = 10,44 m

  2 c.

  1

  ) A

  165 N 15,8 MPa

  = (

  1

  A

  Spesimen Kompon Perbandingan 100:50:0 a.

  = 10 m

  A

  175 N 18,8 MPa

  3

  = (

  2

  A

  2 b.

  = 9,38 m

  1

  ) A

  183 N 19,5 MPa

  = (

  1

  A

  = 10,88 m

  3

  ) A

  173 N 15,9 MPa

  = (

2 Spesimen Kompon Perbandingan 100:50:5 a.

2 Spesimen Kompon Perbandingan 100:50:10

  a.

2 Spesimen Kompon Perbandingan 100:50:15 a.

  EB 520 540 540 530 510 550 580 530 550 580 600 570 600 570 610 Spesimen Kompon Perbandingan 100:50:0 a.

  ) A

  2

  = 9,53 m

  2 c.

  A

  3

  = (

  198 N 20,7 MPa

  ) A

  3

  = 9,57 m

  Perpanjangan putus (EB) = −

  × 100 % EB = % Elongation at Break d = Panjang saat putus (mm) a = Panjang mula-mula (mm), dimana a = 97 mm

  1

  = (

  = (

  97 mm × 520 % 100 %

  − 97 mm)

  1

  = ( 504,4 mm − 97 mm )

  1

  = 407,40 mm b.

  2

  = (

  97 mm × 540 % 100 %

  − 97 mm)

  2

  = ( 523,8 mm − 97 mm )

  182 N 19,1 MPa

  2

  A

  2 c.

  1

  = (

  209 N 21,0 MPa

  ) A

  1

  = 9,95 m

  2 b.

  A

  2

  = (

  216 N 22,1 MPa

  ) A

  2

  = 9,77 m

  A

  A

  3

  = (

  201 N 19,9 MPa

  ) A

  3

  = 10,10 m

  A

  1

  = (

  198 N 20,7 MPa

  ) A

  1

  = 9,57 m

  2 b.

2 Lampiran 16. Perhitungan Perpanjangan saat Putus (d) pada Kompon

  2

  97 mm × 550 % 100 %

  1

  Spesimen Kompon Perbandingan 100:50:10 a.

  = 436,50 mm

  3

  = ( 533,5 mm − 97 mm )

  3

  − 97 mm)

  = (

  97 mm × 580 % 100 %

  3

  = 417,10 mm c.

  2

  = ( 514,1 mm − 97 mm )

  2

  − 97 mm)

  97 mm × 530 % 100 %

  = (

  = (

  − 97 mm)

  = 465,60 mm b.

  2

  3

  = ( 552,9 mm − 97 mm )

  3

  − 97 mm)

  97 mm × 570 % 100 %

  = (

  3

  = 485 mm c.

  = ( 582 mm − 97 mm )

  1

  2

  − 97 mm)

  97 mm × 600 % 100 %

  = (

  2

  = 465,60 mm b.

  1

  = ( 562,6 mm − 97 mm )

  2

  1

  = 426,80 mm c.

  1

  = 417,10 mm b.

  1

  = ( 514,1 mm − 97 mm )

  1

  − 97 mm)

  97 mm × 530 % 100 %

  = (

  Spesimen Kompon Perbandingan 100:50:2,5 a.

  = (

  = 426,80 mm

  3

  = ( 523,8 mm − 97 mm )

  3

  − 97 mm)

  97 mm × 540 % 100 %

  = (

  3

  2

  97 mm × 510 % 100 %

  = ( 562,6 mm − 97 mm )

  = ( 533,5 mm − 97 mm )

  1

  − 97 mm)

  97 mm × 580 % 100 %

  = (

  1

  Spesimen Kompon Perbandingan 100:50:5 a.

  = 436,50 mm

  3

  3

  − 97 mm)

  − 97 mm)

  97 mm × 550 % 100 %

  = (

  3

  = 397,70 mm c.

  2

  = ( 494,7 mm − 97 mm )

  2

  = 455,90 mm

  Spesimen Kompon Perbandingan 100:50:15 a.

  2

  Modulus Young (MY) = Stress = Tegangan tarik (MPa) Strain = Perpanjangan putus (%) _{stress 15,8 17,9 15.9 19,5 18,8 21,8 19,3 18,6 19,6 21,0 22,1 19,9 20,7 19,1 20,7 strain 520 540 540 530 510 550 580 530 550 580 600 570 600 570 610}

  Lampiran 17. Perhitungan Modulus 100 % pada Kompon NR/CB/LNR

  = 494,70 mm

  3

  = ( 591,7 mm − 97 mm )

  3

  − 97 mm)

  97 mm × 610 % 100 %

  = (

  3

  = 455,90 mm c.

  = ( 552,9 mm − 97 mm )

  1

  2

  − 97 mm)

  97 mm × 570 % 100 %

  = (

  2

  = 485 mm b.

  1

  = ( 582 mm − 97 mm )

  1

  − 97 mm)

  97 mm × 600 % 100 %

  = (

  MY 1,05 1,06 1,04 1,27 1,39 1,40 1,18 1,26 1,27 1,23 1,21 1,22 1,23 1,21 1,16 Lampiran 18. Data Hasil Analisa Sifat Mekanik Uji Ketahanan Sobek

  Lampiran 19. Perhitungan Kekuatan Maksimum (L) dari Kekuatan Sobek

  L x t

1 Kekuatan Sobek (N/mm) =

  t

  2 L = Kekuatan maksimum (N/mm) t ₁ = Ketebalan standard potongan uji (2,5 mm) t _{Kekuatan sobek 35,0 37,9 38,9 38,6 39,7 44,3 42,1 33,1 34,8 39,1} ^{2 = Ketebalan spesimen uji (1 mm)}

  Spesimen Kompon Perbandingan 100:50:0 a.

  1

  = 15,88 N/mm b. L

  2

  = (

  44,3 N/mm × 1 mm 2,5 mm

  ) L

  2

  = 17,72 N/mm

  Spesimen Kompon Perbandingan 100:50:10 a.

  L

  1

  = (

  42,1 N/mm × 1 mm 2,5 mm

  ) L

  = 16,84 N/mm b. L

  ) L

  2

  = (

  33,1 N/mm × 1 mm 2,5 mm

  ) L

  2

  = 13,24 N/mm

  Spesimen Kompon Perbandingan 100:50:15 a.

  L

  1

  = (

  34,8 N/mm × 1 mm 2,5 mm

  ) L

  1

  1

  39,7 N/mm × 1 mm 2,5 mm

  L

  L

  1

  = (

  35,0 N/mm × 1 mm 2,5 mm

  ) L

  1

  = 14 N/mm b. L

  2

  = (

  37,9 N/mm × 1 mm 2,5 mm

  ) L

  2

  = 15,16 N/mm

  Spesimen Kompon Perbandingan 100:50:2,5 a.

  1

  = (

  = (

  38,9 N/mm × 1 mm 2,5 mm

  ) L

  1

  = 15,56 N/mm b. L

  2

  = (

  38,6 N/mm × 1 mm 2,5 mm

  ) L

  2

  = 15,44 N/mm

  Spesimen Kompon Perbandingan 100:50:5 a.

  L

  1

  = 13,92 N/mm b.

  L

  2

  = (

  39,1 N/mm × 1 mm 2,5 mm

  ) L

  2

  = 15,64 N/mm

  Lampiran 20. Termogram TGA Kompon NR/CB/LNR Perbandingan 100:50:0 (A) dan 100:50:10 (B)

  Lampiran 21. Kurva DSC Kompon NR/CB/LNR Perbandingan 100:50:0 (A) dan 100:50:10 (B) A B

  A B Lampiran 22. Morfologi Hasil Foto SEM Kompon Perbandingan 100:50:0 (A) dan 100:50:10 (B)

  A B

  Lampiran 23. Data Uji Sifat Mekanik Kompon Ban Karet Skala Industri Menurut Mohd Bijamiri, et al (2010) dalam “ Mechanical

Properties of Industrial Tyre Rubber Compounds

  ”