Perhitungan Neraca MassaKapasitas produksi: 160.000 ton/tahunJumlah hari kerja dalam satu tahun : 330 hariJumlah hari kerja dalam satu hari: 24 jamkapasitas produksi dalam satu jam: 160.000 ton/ tahun * 1000 kg/ton * 1 tahun/330 hari *1 hari/ 24 jam: 20.202,02 kg/jamSpesifikasi produk Etilen Glikol 99,5% weight : 0,995 * 20.202,02 kg/jam: 20.101,01 kg/jam: 20.101,01 kg/ jam : Mr etilen glikol: 20.101,01kg/ jam : 62 kg/kmol: 324,2098403 kmol/jam

Berat MolekulKomponenBM,Kg/Kgmol

C2H4O44

H2O18

C2H6O262

C4H10O3106

C6H14O4150

Umpan masuk reaktor:C2H4O: 14.573,0764 kg/jam: 331,2062828 kmol/jamH2O (impuritas): kg/jam: 4,373 kg/jam (0,03% berat)H2O umpan: (8*EO 99,97 % berat) + impuritas H2O 0,03 % berat: (8 * 14.573,0764) kg/jam + 4,373 kg/jam: 116.588,985 kg/jam : 64.746,6799 kmol/jam

Sehingga kebutuhan bahan baku:C2H4O: ton/ tahun: 115.453,4017 ton/tahunH2O: ton/tahun: 923.350,1252 ton/tahun

NERACA MASSA REAKTOR

Perbandingan berat reaktan C2H4O : H2O = 1 : 8

(konversi C2H4O: 99 %)Reaksi 1: C2H4O yang bereaksi = mol C2H4O umpan * selektivitas C2H6O2 * konversi C2H4OC2H4O yang bereaksi = 331,2062828 * 0,989* 0,99 kmol= 327,53028 kmolC2H4O+ H2OC2H6O2(selektivitas: 98,90%)mula- mula:331,206282864.746,6799 -reaksi:327,53028327,53028327,53028sisa:3,6766.149,635575327,53028

Reaksi 2:C2H6O2 yang bereaksi = mol C2H4O umpan * selektivitas C4H10O3 * konversi C2H4O= 331,2062828 * 0,01* 0,99 kmol= 3,31173 kmolC2H6O2+ C2H4OC4H10O3 (selektivitas C4H10O3: 1%)mula- mula:327,5303,676 -reaksi:3,311733,311733,31173

sisa:324,21850,364293,31173

Reaksi 3:C2H6O2 yang bereaksi = mol C2H4O umpan * selektivitas C6H14O4 * konversi C2H4O= 331,2062828 * 0,001* 0,99 = 0,31173163 kmol C4H10O3+ C2H4OC6H14O4(selektivitas C6H14O4: 0,1%)mula- mula:3,311730,36429 -reaksi: 0,3311730,3311730,331173

sisa:2,980558460,0331206280,331173

NERACA MASSA REAKTORKomponenInputOutput

Kmol/jamkg/jamkmol/jamkg/jam

C2H4O331.206283414573.076470.0331206281.457307647

H2O6477.165833116588.9856149.635575110693.4404

C2H6O200324.218526420101.54864

C4H10O3002.980558465315.9391973

C6H14O4000.33117316349.67597442

TOTAL6808.372117131162.06156477.198954131162.0615

16

komponenFraksi beratFraksi mol

C2H4O0.00000.0000

H2O0.94940.8439

C2H6O20.05010.1533

C4H10O30.00050.0024

C6H14O40.00010.0004

Total11

NERACA MASSA EVAPORATOR 1 DAN 2

Neraca panas masing- masing efek:1) F. Hf + S . s1 = L1. h1 + V1. H1F. Hf + S . s1 =(F-V1). h1 + V1. H1

2) L1. h1 + V1. H1 = L2. h2 + V2. H2 + V1. hc2(F-V1). h1 + V1. H1 = (F-V1-V2). h2 + V2. H2 + V1. hc2

Keterangan:F = umpan masuk evaporator (kg)S = steam pemanas (kg)V = produk atas evaporator (kg)L = produk bawah evaporator (kg)H = entalphy (KJ/kg)s = panas laten steam (kJ/kg)

Langkah- langkah penyelesaian evaporator double effect:1. Fraksi mol evaporatorUntuk evaporator jenis Long Tube Vertical Natural Circulation, harga overall heat transfer coefficient(U) = 1100 - 4000 W/m2.K (Geankoplis p.496)Dipilih:U1 = 2500 (W/m2.K)U2 = 1900 (W/m2.K)

Dirancang H2O yang teruapkan 98% berat dan etilen oksida teruapkan semua sebesar 0.00999702 kg/jam, sehingga diperoleh H2O sisa:H2O masuk - H2O teruapkan = H2O sisa110.693,4404 kg/jam -(0.98*110.693,4404) kg/jam = 2.213,868807 kg/jam

Produk bawah evaporator 2 (L2)KomponenBMkgfraksi beratkmolFraksi mol

C2H4O440000

H2O182213.8688070.097608819122.99271150.272999869

C2H6O26220101.548640.886271317324.21852640.719649271

C4H10O3106315.93919730.0139296652.9805584650.006615775

C6H14O415049.675974420.0021901990.3311731630.000735086

TOTAL38022681.032611450.52296951

Tebakan awal, uap air di V1 = V2Neraca Massa TotalF = L2 + (V1 + V2)131.162,0615 = 22.680,85764 + (V1+V2)V1 + V2 = 108.481,0289Sehingga:V2 = (108.481,0289- 1,4573)/2= 54.239,78577V1 = V2 + 1,4573= 54.241,24308L1 = F V1= 131.162,0615 54.241,24308= 76.920,8184

Produk bawah evaporator 1 (L-1)Komponen BMkgfraksi beratKmolFraksi mol

H2O1856453.65460.7339190583136.3141430.9054

C2H6O26220101.548640.261327805324.21852640.0936

C4H10O3106315.93919730.004107332.9805584650.0009

C6H14O415049.675974420.0006458070.3311731630.0001

L176920.818413463.8444011.00000

Uap keluar dari evaporator 1 (V1)KomponenBMkgFraksi beratKmolFraksi Mol

EO441.457307652.68672E-050.0331206281.09913E-05

H2O= (V1-EO)1854239.785774180.9999731333013.3214320.999989009

Total V154241.243083013.3545531

Uap keluar dari evaporator 2 (V2)KomponenBMkgFraksi beratKmolFraksi Mol

H2O ( L1-L2)1854239.785774213013.3214321

Total V254239.7857713013.3214321

2. Menentukan Kondisi Operasi Kedua Evaporatora. Efek 1Digunakan saturated steam pada Ts = Ts 1 = 171,11 oCDitetapkan P = 2,9 atm = 2204 mmHgSuhu bawah evaporator 1 dicari dengan trial sehingga y =1Trial T = 136.2472125 oC (409.3972125 K)

Komponenxi (fraksi mol)Psatki (Psat/ P)yi = ki * xi

H2O0.90542.4136.E+031.0951.E+009.9154.E-01

EG0.09368.4973.E+013.8554.E-023.6090.E-03

DEG0.00091.5818.E+017.1771.E-036.1762.E-06

TEG0.00013.1563.E+001.4321.E-031.3693.E-07

Total1--9.9515.E-01

Suhu atas keluar evaporator 1 dicari dengan trial sehingga y =1KomponenxiPsatki (Psat/P)yi = ki *xi

EO1.09922E-052.09426E+049.50210E+000.000104449

H2O0.9999890082.41359E+031.09509E+001.095081398

Total11.095185847

Dari trial didapat T1' = 132.85 oC (406 K)BPR 1 = Suhu bawah keluar evaporator 1 - Suhu atas keluar evaporator 1BPR 1 = 136oC 132.85 oC = 3.15 oCT effective 1 = Ts 1 - T1 = suhu steam suhu produk bawah evaporator 1= 171,11 136 oC= 35,11 oC

b. Efek 2Ditetapkan P = 0,2 atm (152mmHg)Suhu bawah keluar evaporator dicari dengan trial sehingga y = 1KomponenPok = Po/Pxy = k *x

H2O532.17301633.5011382650.2729841860.95575538

EG9.5171810940.0626130340.7196647950.045060396

DEG1.2330011710.008111850.0066159175.36673E-05

TEG0.2273975790.0014960370.0007351021.09974E-06

Total1.000870543

Dari trial didapat T2 = 90,31855 oC (363,46855 K)

Suhu atas keluar evaporator 2 dicari dengan trial sehingga y =1KomponenPok = Po/Pxy= k*x

H2O152.00451091.00002967711.000029677

Total1.000029677

Dari trial didapat T2' = 60,34819 oC (333,498 K)

BPR 2 = suhu bawah keluar evaporator 2 - suhu atas keluar evaporator 2T effective 2 = T1' - T2= suhu produk atas evaporator 1 - suhu produk bawah evaporator 2= 132.85 oC - 90.31855062 oC= 42,5314 oC

3. Menentukan entalpi cairan masing-masing evaporatora. Aliran FeedDari perhitungan, didapat hf = 330.3299324 kJ/kg

b. Aliran L-1Dari perhitungan, didapat h1 = 445.4505718 kJ/kg

c. Aliran L-2Dari perhitungan didapat h2 = 265.422997 kJ/kg

4. Menghitung entalpi uap pada masing-masing evaporatorDari tabel steam

a. Efek 1

BPR 13.15oC

Ts = Ts1 =171.11oC

Hs2808,64kJ/kg

s2073.97kJ/kg

T1'132.85oC

H1'2763.09kJ/kg

hc2561.92kJ/kg

H1 = H1' + 1,884.BPR 1

2769.0246kJ/kg

b. Efek 2BPR 229.97035955oC

T2'60.34819107oC

H2'2609.9kJ/kg

H2 = H2' + 1,884.BPR 2

2666.364157kJ/kg

Dari persamaan neraca panas: F. Hf + S . s1 =(F-V1). h1 + V1. H1..1131.162,0615 * 330.3299324 + S * 2073.97 = (131.162,0615-V1) * 445.4505718 + V1 * 2769.490348-15099460.4 + 2073.97S = 2323.574028 V1

(F-V1). h1 + V1. H1 = (F-V1-V2). h2 + V2. H2 + V1. hc2(131.162,0615 - V1)* 445.4505718 + V1 * 2769.490348 = (131.162,0615 - V1 - (744.1407467 - V1)) * 265.422997 + (744.1407467 - V1)-236843779 = -4428,018186 V1..(2)

DimanaV1 + V2 = 108.481,0289V2 = 108.481,0289 V1(3)

Penyelesaian persamaan 1, 2, dan 3 secara simultan diperoleh:V1 = 53.487,5354V2 = 54.993,49345S = 59.974,74678

Q1 = S * sQ1 = 124.385.825,6 kJ/jam (34.551,61822 kW)

Q2 = V1 * (H1 - hc2)Q2 = 118.052.585,4 kJ/jam (32.792,38484 kW)

A1 = Q1 / (U1* T effective 1)= 410,0400909 m2A2 = Q2/ (U2* T effective 2)= 405,797361 m2

Neraca massa baruEvaporator 1KomponenBMinput

arus 3

kmol/jamkg/jam

C2H4O443.31.E-021.4573

H2O186.15.E+03110693.4404

C2H6O2623.24.E+0220101.5486

C2H10O31062.98.E+00315.9392

C2H14O41503.31.E-0149.6760

6477.199

Total131162.06

output

arus 4 , Varus 5, L

kmol/jam kg/jamkmol/jam kg/jam

3.3121E-021.45730760.0000.E+000

2.9714E+0353486.07813.1782.E+0357207.3623

3.2422.E+0220101.5486

2.9806.E+00315.9392

3.3117.E-0149.6760

2971.481953487.53543505.717177674.5261

6477.199

131162.06

Evaporator 2KomponenBMinput

arus 5

kmol/jamkg/jam

H2O183.18.E+0357207.3623

C2H6O2623.24.E+0220101.5486

C2H10O31062.98.E+00315.9392

C2H14O41503.31.E-0149.6760

3505.71777674.5261

Total77674.53

output

arus 6 , Varus 7, L

kmol/jam kg/jamkmol/jam kg/jam

3.0552E+0354993.49351.2299.E+022213.8688

3.2422.E+0220101.54864

2.9806.E+00315.9391973

3.3117.E-0149.67597442

3055.194154993.4935450.523022681.0326

3505.717

77674.53

NERACA MASSA STRIPPER

Dirancang H2O yang dibuang sebanyak 96,3 % wt dari H2O yang masuk.

Komposisi fluida masuk Stripper- 01 ( Arus 7)KomponenBMkmol/jam kg/jam

H2O181.2299.E+022213.868807

C2H6O2623.2422.E+0220101.54864

C2H10O31062.9806.E+00315.9391973

C2H14O41503.3117.E-0149.67597442

Total450.523022681.0326

Neraca Massa Total :M7 = M8 + M9Hasil Bawah Stripper :M H2O(9) = M H2O(7) - M H2O(8) = 2.213,86807 (0,963 * 2.213,86807)= 81,91314586 kgM DEG(7) = M DEG(9) = 315,9392 kgM TEG(7) = M TEG(9) = 49,6760 kgM EG(9) = M EG(7) - M EG(8) = 20101.5486 (20101.5486 * 0.001/100)= 20101,3476 kgHasil atas Stripper:M EG(8) = M EG yang hilang di stripper = 0,1 % * M EG(7)Maka M EG(9) = 99,999 % * M EG (7)M EG(8) = 0,001/100 * MEG7 = (0.001/100) * 20101.5486= 0.20102 kgM H2O(8) = 96,3 % * M H2O(7)= 0.963 * 2.213,86807= 20.101,3476 kg

Neraca Massa StripperMasuk

Arus 7

KomponenBMkmol/jam kg/jam

H2O181.2299.E+022213.868807

C2H6O2623.2422.E+0220101.54864

C2H10O31062.9806.E+00315.9391973

C2H14O41503.3117.E-0149.67597442

Total4.5052.E+0222681.03261

Keluar

Arus 8Arus 9

kmol/jam kg/jamkmol/jam kg/jam

1.1843.E+022.1318E+034.5504.E+008.1907E+01

3.2422.E-032.0102E-013.2422.E+022.0101E+04

0.0000.E+000.0000E+002.9806.E+003.1594E+02

0.0000.E+000.0000E+003.3117.E-014.9676E+01

118.43592131.9882332.077420548.8694

450.513322680.8576

NERACA MASSA MENARA DESTILASI 1

Fungsi: memisahkan produk utama (C2H6O2) dengan produk samping (C4H10O3 dan C6H14O4)F = produk bawah stripper = 332.0777 kmol/jamKomp LK =C2H6O2

Komp HK=C2H10O3

Diinginkan C2H6O2 sesuai spesifikasi sebagai hasil atas : 98,6 %Diinginkan C2H6O2 sesuai spesifikasi sebagai hasil bawah : 0,17 %Diinginkan C2H4O sebagai hasil atas (Distilat)

NM Total:F = B + D

NM Komponen:F.Xf = Xd. D + Xb. B

F = B + DF . Xf = D . Xd + B . Xb

D = F- BF. Xf = (F-B). Xd + B.XbF. Xf = F. Xd - B.Xd + B.XbF. Xf F.Xd = B . (Xb - Xd)B = F (Xf Xd)/ (Xb- Xd)..x -1B= (F. Xd) (F.Xf) / (Xd Xb)

Dari perhitungan NM Total dan komponenB :

= 3,2646 kmol/jam

D= F B= 332,0777 3,2646 kmol/jam= 328,8131 kmol/jam

Neraca Massa Menara Destilasi- 01KomponenBMInput

arus 7

kmol/jam kg/jam

H2O184.5507.E+008.1913.E+01

C2H6O2623.2422.E+022.0101.E+04

C2H10O31062.9806.E+003.1594.E+02

C2H14O41503.3117.E-014.9676.E+01

332.077720548.8759

Total

output

arus 8 (D)arus 9 (B)

kmol/jam kg/jamkmol/jam kg/jam

4.4893E+0080.80816.1393.E-021.105.E+00

3.2421E+0220101.01015.4439.E-033.375.E-01

1.5247E-0116.16162.8281.E+002.998.E+02

2.6936E-024.04043.0424.E-014.564.E+01

328.878620202.02023.1992346.8557

332.0777

20548.8759

Spesifikasi Etilen Glikol

C2H4O% berat

H2O0.41%

C2H6O299.56%

C2H10O30.01%

C2H14O40.02%

Total100.00%

NERACA MASSA MENARA DESTILASI- 02

F = produk bawah keluaran menara destilasi-01F = 3,2646 kmol/jam

Komp LK =C4H10O3

Komp HK=C6H14O4

Diinginkan C4H10O3 sesuai spesifikasi sebagai hasil atas : 99,8 %Diinginkan C4H10O3 sesuai spesifikasi sebagai hasil bawah : 0,71 %

NM Total:F = B + D

NM Komponen:F.Xf = Xd. D + Xb. B

F = B + DF . Xf = D . Xd + B . Xb

D = F- BF. Xf = (F-B). Xd + B.XbF. Xf = F. Xd - B.Xd + B.XbF. Xf F.Xd = B . (Xb - Xd)B = F (Xf Xd)/ (Xb- Xd)..x -1B= (F. Xd) (F.Xf) / (Xd Xb)

Dari perhitungan neraca massa total dan komponen:B: B = 0,3060 kmol/jam

D= F BD= 2,9586 kmol/jam

Neraca Massa Menara Destilasi-02KomponenBMInput

arus 9

kmol/jam kg/jam

C2H6O2625.5553.E-033.4443E-01

C4H10O31062.9548.E+003.1321E+02

C6H14O41503.0424.E-014.5636E+01

3.2646359.1934

Total

output

arus 10 (D)arus 11 (B)

kmol/jam kg/jamkmol/jam kg/jam

5.5553E-030.34440.0000.E+000.000.E+00

2.9527E+00312.98352.1609.E-032.291.E-01

3.6193E-040.05433.0388.E-014.558.E+01

2.9586313.38230.306045.8111

3.2646

359.1934

Spesifikasi Dietilen Glikol

Komposisi% berat

C2H6O20.11%

C4H10O399.87%

C6H14O40.02%

Total100.00%

Spesifikasi Trietilen GlikolKomposisi% berat

C2H6O20.00%

C4H10O30.50%

C6H14O499.50%

Total100.00%