BAB III METODOLOGI PENELITIAN
3.1
Kerangka Pemecahan Masalah
Pada laporan ini dibuat flow chart yang menggambarkan bagaimana perancangan pemecahan masalah dari awal sampai akhir berdasarkan hasil dari dar i keseluruhan kerja praktik. Berikut Berikut ini Gambar pemecahan masalah.
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3.1 adalah adalah gambar flow chart
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Gambar 3.1 Flow chart pemecahan chart pemecahan masalah.
3.2.
Langkah-langkah Pemecahan Masalah
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yang tersusun dari material butiran (bulk (bulk material ) membutuhkan proses pemindahan yang harus diatur sedemikian rupa agar material dapat dipindahkan sekaligus dalam jumlah banyak dan tidak terganggu dengan gangguan dari luar, seperti temperatur lingkungan atau angin. Dalam masalah kali ini, alat pemindah berupa conveyor merupakan sarana yang cocok, karena conveyor dapat dapat bekerja secara terus-menerus, perawatannya mudah, dan dapat memindahkan material butiran dalam jumlah yang banyak. Desain conveyor harus disesuaikan dengan kondisi lapangan, material yang akan dibawa, dan kebutuhan proses produksi, agar menunjang kinerja dalam pembuatan produk tersebut.
3.2.2 Rumusan Masalah
Dari identifkasi masalah yang telah dijelaskan sebelumnya, maka rumusan masalah dalam penelitian kali ini adalah bagaimana merancang ulang sebuahconveyor sebuah conveyor menggunakan analisa dan perhitungan yang tepat, sehingga sesuai dengan kebutuhan produksi suatu perusahaan.
ower F low C onveyo onveyor r 3.2.3 Analisa P owe Power flow flo w atau aliran daya pada conveyor bertujuan bertujuan untuk memahami sistem kerja conveyor . Power flow flow adalah analisa aliran daya didalam sistem yang saling terhubung. Power terhubung. Power flow menunjukkan flow menunjukkan daya yang masuk kedalam sebuah sistem, proses konversi daya tersebut, dan beberapa variabel yang menyebabkan hilangnya daya tersebut yang akhirnya menjadi daya akhir atau output yang bekerja pada sistem tersebut. Power tersebut. Power flow digambarkan flow digambarkan menggunakan sebuah notasi yang disederhanakan, seperti diagram satu baris dan sistem per unit. Sistem kerja conveyor menggunakan prinsip kerja dari sabuk transmisi. Pada conveyor terdapat dua buah puli utama pada bagian kepala dan ekor. Motor
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Skema power flow conveyor . Gambar 3.2 Skema power
Diagram power flow flow conveyor . Gambar 3.3 Diagram power
Dari Gambar 3.3 diatas dapat dilihat jika conveyor bekerja bekerja menggunakan sumber daya dari motor listrik yang dipasangkan pada puli bagian atas. Motor listrik sebagai
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mekanik berupa torsi dan kecepatan sudut yang hasil akhirnya adalah berupa rotasi. Rotasi motor inilah yang akan memutar puli conveyor . Tetapi daya rotasi ini masih akan kehilangan dayanya (energy (energy loss) loss) karena beberapa penyebab, seperti gesekan pada bearing puli puli dan gesekan antara sabuk dan puli. Hasilnya adalah daya akhir ( P P out ) yang dapat menjalankan conveyor tersebut, tersebut, yang jauh lebih kecil daripada daya masuk ( P in). in).
3.2.4 Spesifikasi Umum Conveyor
Rencana spesifikasi umum conveyor akan yang dirancang adalah sebagai berikut : 1. Kapasitas
: 10000 kg/jam
2. Panjang
: 14 m
3. Lebar belt
: 900 mm
4. Sudut inklinasi : 20ᴼ 20ᴼ 5. Material angkut : Curah tembakau
3.2.5
Perhitungan Kebutuhan Daya Motor
Motor adalah bagian penting dari conveyor yang yang merupakan sumber energi agar conveyor dapat dapat bekerja. Motor yang ada di d i pasaran memiliki memiliki spesifikasi yang berbeda beda, yang dapat disesuikan sesuai dengan beban kerja. Perhitungan P erhitungan kebutuhan ke butuhan daya da ya penting agar motor yang akan digunakan pada conveyor tidak tidak terlalu kecil atau terlalu besar, agar conveyor dapat dapat bekerja seefektif mungkin.
3.2.5.1 Penentuan Kecepatan Conveyor
Perhitungan diawali dengan penentuan kecepatan. Kecepatan conveyor menentukan seberapa besar kapasitas material yang akan diangkut oleh conveyor .
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a. Massa material per meter yang diangkut. Diasumsikan per satuan panjang dar i conveyor mengangkut material sebanyak
kgm. Qm = F1 × γ F1 = Luas penampang material material (m ) kg m γ = Massa jenis material mater ial ( ) 5 kgm = F1 × 320 kgm kgm 5 F1= kg = 0,016 016 320 m 5
b. Kecepatan conveyor .
Qt =Qm ×v ×3600 10000 kgjam = 5 kgm ×v ×3600 kgjam 10000 v = kg 5 m ×3600 v=0,56 m⁄s =0,6 m⁄s 3.2.5.2 Perbandingan G aya T1 dan T2
Desain conveyor belt menyerupai kontruksi sabuk transmisi yang biasa digunakan pada mesin-mesin untuk mentrasmisikan daya dari motor menuju poros yang digerakkan. Didalam desain belt tersebut tersebut menghasilkan gaya yang dihasilkan dar i sisi kencang (T1) dan sisi kendur (T2) belt. Sisi kencang ini adalah bagian yang menerima tarikan dari putaran motor. Sedangkan sisi kendur adalah bagian yang
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Gambar 3.4 Distribusi gaya pada sabuk.
(Khurmi, 2005). Gambar 3.5 Diagram benda bebas gaya belt (Khurmi,
Sisi kencang T1 = (ΔT + T )
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R = ∆ ∆ ∆ (ii) R = (ΔT + T ) T ) Sin + T Sin R = (ΔT + T ) ∆ + T ∆ R = ΔT∆ + T∆ + T ∆ R = 2T ∆ R = T Δθ ∆ (iii) R = ∆ ∆ Δθ µ = dθ µ =
T Δθ =
Persamaan diselesaikan dengan mengintegralkan masing-masing variabel.
dT ∫ T =
μ ∫ dθ
log T1 T2 = μ θ T1 = e T2
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yang digunakan (Siegling, 2012). Gambar 3.6 Spesifikasi belt yang
Gaya sisi kencang T1
- Spesifikasi gaya dari belt =
9 Nmm
- Lebar belt yang akan digunakan = 900 mm
T1=Tb ×b T1=9 Nmm × 900 900 mm= mm = 8100 8100 N
Sudut kontak θ = 210ᴼ = 210ᴼ = 3,67 radian
Koefisien gesekan
42,6 μ=0,54− 152,6+v 42,6 =0,54−0,28=0,26 =0,54− 152,6+0,6
Rasio gaya
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Karena belt memiliki massa, maka gaya pada masing-masing sisi dipengaruhi oleh gaya sentrifugal. sentrifugal. Dimana gaya sentrifugal sentrifugal ini bekerja secara secara tangensial agar belt tetap tetap berada pada posisi yang seimbang.
(Khurmi, 2005). Gambar 3.7 Diagram benda bebas gaya sentr ifugal belt (Khurmi,
Gaya sentrifugal se ntrifugal belt.
Tc = m v Tc=2,7 kgm × 0,6 m⁄s Tc = 0,98 N (i) T1 = 8100 + 0,98 = 8100,98 N (ii) T2=3127,4+0,98=3128,38 N (iii) T =8100,98 −3128,38=4972,26 N
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d = T ×Cb ××180 β N ×25 ×180 = 4972,26 900 mm × 210 d = 119 mm Dibulatkan menjadi 120 mm atau 0,12 m.
3.2.5.4 Daya Motor Yang Dibutuhkan Conveyor
Perhitungan kebutuhan daya motor dalam conveyor diperlukan diperlukan supaya dapat dipilih spesifikasi motor motor yang tepat, dan tidak terlalu jauh dari daya teo ritis. ritis. a. Daya yang ditransmisikan oleh belt Daya yang ditransmisi d itransmisikan kan belt merupakan merupakan hasil perkalian torsi yang dihasilkan dan dikalikan dengan kecepatan sudut belt tersebut. tersebut. Perhitungan daya transmisi ini pent ing agar dalam perancangan conveyor ini dapat memilih daya motor yang tepat, tidak terlalu rendah atau terlalu tinggi, agar dapat menghemat biaya pembelian dari motor tersebut.
Torsi yang dihasilkan
τ = T1−T2R τ = 8100, 8100,98− 98 − 3128, 3128,38 38 N 0,06 m τ = 298, 298,35 3 5 Nm
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Tabel 3.2 Katalog motor (SEW Eurodrive, 2013).
Motor yang dipilih harus memiliki daya yang lebih besar daripada daya teoritis dari perhitungan, agar selisihnya selisihnya tidak terlalu berdekatan. Dari katalog dipilih dipilih motor motor dengan tipe DRS100M2.
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= 60210 = 95,5 = 96
Jadi putaran pada pulley pada pulley head conveyor adalah sebesar 96 rpm. Sedangkan motor yang digunakan memiliki kecepatan 2820 rpm. Jadi dibutuhkan rasio gearbox rasio gearbox sebesar 1 : 30. Diasumsikan sudut tekan roda gigi adalah sebesar 20ᴼ 20ᴼ. A. Jumlah minimum gigi roda gigi agar tidak terjadi interference. interference. Interferance yaitu suatu gangguan dimana sudut tekan roda gigi penggerak menabrak ro da gigi yang digerakkan sebelum mengenai sudut tekan roda gigi yang digerakkan.
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15 √ 30=82,2=82 3 0=82,2=82 82 ) = 29,8 = 30 e=( 82 )( ) ( 15 15 Asumsi rasio antar roda gigi yang berpasangan.
e=65=30 = 6 dan = 5 Diasumsikan jumlah gigi roda gigi N2 adalah sama dengan jumlah minimum NP yaitu 15 gigi.
N1 = 6 N2 = 615 = 90 N3 = 5 N4 = 515 = 75 75 ) = 30 e=( 90 )( ) ( 15 15 Spesifikasi kemungkinan jumlah gigi N1 6 N3 5
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Persamaan diatas dapat disederhanakan menggunakan perkalian dengan angka 6. Karena jumlah minimal gigi roda gigi adalah 15, maka diharuskan N2 adalah nilai kelipatan 6 yang melebihi 15, yaitu N2 = 18.
N2=18 N1=6N2=618 =108 N1+N2=108+18=126=N3+N4 126=5N4+N4=6N5 N4 = 126 6 = 21 N3=5N4=521 =105 Pengecekan
N3 ) e=( N1 ) ( N2 N4 105 ) = 65 = 30 e=( 108 )( ) ( 18 21
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d = NP = 184 = 4,5 inc incii b. Modul Rasio diameter pitch diameter pitch dengan dengan jumlah gigi roda gigi. m = modul (mm)
m = Nd = 4,518inci =0,25 c. Circular pitch (p) pitch (p) Jarak antara dua titik pitch titik pitch pada pada sisi luar gigi roda gigi yang bersebelahan.
5 = 0,785 p = πNd = π 4, inci 18 785 inci d. Lebar gigi
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2. Roda gigi yang digerakkan pinion (N1). a. Diameter pitch Diameter pitch P = diametral pitch = 4 N = Jumlah gigi = 108 d = Diameter pitch Diameter pitch (inci) (inci)
P = Nd d N 108 27 inci inci
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g. Working depth (hk) depth (hk)
hk= 2P = 24 = 0,5 inci inci
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35
3. Roda gigi N3 a. Diameter pitch Diameter pitch P = diametral pitch = 4 N = Jumlah gigi = 105 d = Diameter pitch Diameter pitch (inci) (inci)
P = Nd d N 105 26,25 25 i
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36
g. Working depth (hk) depth (hk)
hk= 2P = 24 = 0,5 inci inci
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37
4. Roda gigi N4
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g. Working depth (hk) depth (hk)
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39
Berikut adalah susunan roda gigi pada gearbox reducer setelah dilakukan
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3.2.7
Pembuatan Desain Conveyor
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41
b.
Sidewall Pengencang. Pengencang.
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