KESIMPULAN

1. Ditinjau dari teknik yang meliputi pengadaan alat-alat produksi, penerapan teknologi, bahan baku, proses produksi, hasil produksi dan tenaga kerja maka pabrik asetanilida dari anilin dan asam asetat dengan kapasitas produksi 186.000 ton/tahun menarik untuk dikaji lebih lanjut.

2. Prarancangan pabrik asetanilida dari anilin dan asam asetat dengan kapasitas produksi 186.000 ton/tahun direncanakan akan didirikan pada tahun 2027 di Desa Brangsi, Kecamatan Laren, Kabupaten Lamongan, Jawa Timur, Indonesia.

3. Bentuk badan usaha perusahaan ini direncanakan adalah Perseroan Terbatas (PT) dengan jumlah tenaga kerja yang dibutuhkan sebanyak 95 orang.

4. Dari hasil analisa ekonomi diperoleh bahwa :

 ROI sebelum pajak adalah 23,27% dan sesudah pajak 18,61%

 POT sebelum pajak adalah 4,30 tahun dan sesudah pajak 5,37 tahun

 BEP adalah 49,75%

 SDP adalah 20,20%

 IRR adalah 30%

Berdasarkan hasil analisa di atas maka prarancangan pabrik asetanilida ini layak dan dapat dilanjutkan ke tahap perancangan sesuai dengan prosedur yang telah direncanakan.

92

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95

LAMPIRAN A

PERHITUNGAN NERACA MASSA

Prarancangan pabrik pembuatan Asetanilida dilaksanakan untuk kapasitas produksi dengan basis sebesar 186.000 ton/tahun dengan ketentuan sebagai berikut:

Basis perhitungan = 100 kg/jam Kapasitas produksi = 186.000 ton/tahun 1 tahun operasi = 330 hari/tahun 1 hari kerja = 24 jam

Maka kapasitas produk Asetanilida tiap jam:

= ^186.000

1 Tahun

x

^{1000 kg}_{1 Ton}

x

_{330 Hari}^{1 Tahun}

x

_{24 Jam}^{1 Hari}

= 23.484,848 kg/jam

Untuk menghasilkan Asetanilida 186.000 ton/tahun, maka diperkukan Asetanilida sebanyak 23.484.848 kg/jam kapasitas produksi.

Faktor pengali (FP) = Jumlah Asetanilida yang dibutuhkan Total Keluaran Produk

= 23.484,848 kg/jam 132,335 kg/jam

= 177,465 kg/jam Berat molekul tiap komponen :

- C₆H₅NHCOCH₃ : 135 kg/mol

- C₆H₅NH₂ : 93 kg/mol

- CH₃COOH : 60 kg/mol

- H₂O : 18 kg/mol

Basis perhitungan : 100 kg/jam

96

5

1. Reaktor

Fungsi : Tempat terjadinya reaksi pembentukan asetanilida C6H5NH2(l)

CH3COOH(l) Reaksi :

NM = A1 + A2 = A3 Umpan Masuk :

Arus 3 : Kemurnian Anilin = 99%

Massa C6H5NH2 = Massa Anilin erat Molekul

= ^{99 kg/jam}

93

= 1,065 kmol/jam Massa H2O = ^{Massa Air}

erat Molekul

= ^{1 kg/jam}

18

= 0,056 kmol/jam Arus 4 : Kemurnian Asam Asetat = 99%

Massa CH3COOH = Massa Asam Asetat erat Molekul

=

^{99 kg/jam}

60

= 1,650 kmol/jam Massa H₂O = ^{Massa Air}

erat Molekul

= ^{1 kg/jam}

18

= 0,056 kmol/jam

C6H5NHCOCH3(l) C6H5NH2(l) CH3COO(l) H2O(l)

97 Umpan Keluar

Konversi = 99,5%

Reaksi : C₆H₅NH₂ + CH₃COOH C₆H₅NHCOCH₃ + H₂O Umpan : 1,065 1,650 - - Reaksi : 1,059 1,059 1,059 1,059 Sisa : 0,005 0,591 1,059 1,170 Perhitungan Arus 3 :

Umpan C6H5NH2 = Massa Anilin / Berat Molekul

= 99 kg/jam / 93 kg/kmol

= 1,065 kmol/jam

Reaksi = Umpan Anilin x Konversi

= 1,065 kmol/jam x 99,5%

= 1,059 kmol/jam

Sisa = Umpan - Reaksi

= 1,065 – 1,059

= 0,005 kmol/jam x 93 kg/kmol

= 0,495 kg/jam

Umpan CH3COOH = Massa Asam Asetat / Berat Molekul

= 99 kg/jam / 60 kg/kmol

= 1,650 kmol/jam Reaksi = 1,059 kmol/jam

Sisa = Umpan - Reaksi

= 1,650 kmol/jam – 1,059 kmol/jam

= 0,591 kmol/jam x 60 kg/kmol

= 35,448 kg/jam Umpan C₆H₅NHCOCH₃ = -

Reaksi = 1,059 kmol/jam

Sisa = 1,059 kmol/jam x 135 kg/kmol

= 142,991 kg/jam Umpan H2O = -

Reaksi = 1,059 kmol/jam

98

Sisa = 1,059 + (0,056 + 0,056)

= 1,170 kmol/jam x 18 kg/kmol

= 21,065 kg/jam

Tabel A.1 Neraca Massa disekitar Reaktor

Komponen Masuk (kg/Jam) Keluar (kg/Jam)

Arus 3 Arus 4 Arus 5

C6H5NHCOCH3(l) - - 142,991

C6H5NH2(l) 99 - 0,495

CH₃COOH_(l) - 99 35,448

H₂O_(l) 1 1 21,065

TOTAL 100 100

200 200

Tabel A.2 Neraca Massa Sebenarnya di Reaktor

Komponen Masuk (Kg/Jam) Keluar (Kg/Jam)

Arus 3 Arus 4 Arus 5

C6H5NHCOCH3(l) - - 25.375,975

C6H5NH2(l) 17.569,072 - 87,845

CH3COOH(l) - 17.569,072 6.290,861

H₂O_(l) 177,465 177,465 3.738,394

TOTAL 17.746,538 17.746,538

35.493,076 35.493,076

2. Evaporator

Fungsi : Mengurangi kandungan anilin dan asam asetat dalam larutan asetanilida dengan cara memekatkan produk pada suhu 225^oC

EVAPORATOR

C6H5NHCOCH3(uap) C6H5NH2(uap)

CH3COOH(uap)

H2O(uap)

C6H5NHCOCH3(l) C6H5NH2(l)

CH3COOH(l)

H2O(l)

C6H5NHCOCH3(l)

C6H5NH2(l)

CH3COOH(l) H2O(l)

99 NM = A5 A6 + A7

Umpan masuk : Arus 5 :

Massa C₆H₅NHCOCH₃ = 142,991 kg/jam Massa C₆H₅NH₂ = 0,495 kg/jam Massa CH₃COOH = 35,488 kg/jam Massa H₂O = 21,065 kg/jam Umpan keluar :

Arus 6 :

Asumsi Efisiensi Evaporator = 95%

Massa C₆H₅NHCOCH_3(uap) = 142,991 kg/jam x 5%

= 7,510 kg/jam

Massa C₆H₅NH_2(uap) = 0,495 kg/jam x 95%

= 0,470 kg/jam

Massa CH₃COOH(uap) = 35,448 kg/jam x 95%

= 33,676 kg/jam

Massa H₂O(uap) = 21,065 kg/jam x 95%

= 20,012 kg/jam Arus 7 :

Massa C₆H₅NHCOCH_3(l) = 142,991 kg/jam – 7,150 kg/jam

= 135,842 kg/jam

Massa C₆H₅NH_2(l) = 0,470 kg/jam – 0,470 kg/jam = 0,025 kg/jam

Massa CH₃COOH(l) = 35,448 kg/jam – 33,676 kg/jam

= 1,772 kg/jam

Massa H₂O(l) = 21,065 kg/jam – 20,012 kg/jam

= 1,053 kg/jam

100 KRISTALIZER

Tabel A.3 Neraca Massa disekitar Evaporator Komponen Masuk (kg/Jam) Keluar (kg/Jam)

Arus 5 Arus 6 Arus 7

C6H5NHCOCH3(l) 142,991 7,150 135,842

C₆H₅NH_2(l) 0,495 0,470 0,025

CH₃COOH_(l) 35,448 33,676 1,772

H2O(l) 21,065 20,012 1,053

TOTAL 200 61,308 138,692

200 Tabel A.4 Neraca Massa Sebenarnya di Evaporator Komponen Masuk (Kg/Jam) Keluar (Kg/Jam)

Arus 5 Arus 6 Arus 7

C6H5NHCOCH3(l) 25.375,975 1.268,799 24.107,176

C₆H₅NH_2(l) 87,845 83,453 4,392

CH₃COOH_(l) 6.290,861 5.976,318 314,543

H2O(l) 3.738,394 3.551,474 186,920

TOTAL 35.493,076 10.880,045 24.613,031

35.493,076 3. Kriztalizer

Fungsi : Tempat terbentuknya kristal asetanilida

C6H5NHCOCH3(l)

C6H5NH2(l) CH3COOH(l) H2O(l)

C6H5NHCOCH3(S)

C6H5NHCOCH3(l)

C6H5NH2(l) CH3COOH(l) H2O (l)

101 NM = A7 A8

Umpan Masuk : Arus 7 :

Massa C₆H₅NHCOCH₃ = 135,842 kg/jam Massa C₆H₅NH₂ = 0,025 kg/jam Massa CH₃COOH = 1,772 kg/jam

Massa H₂O = 1,053 kg/jam

Umpan Keluar : Arus 8 :

Asumsi Efisiensi Kristalizer = 96%

Massa C₆H₅NHCOCH_3(s) = 135,842 kg/jam x 96%

= 130,408 kg/jam

Massa C₆H₅NHCOCH_3(l) = 135,842 kg/jam – 130,408 kg/jam

= 5,434 kg/jam Massa C₆H₅NH_2(l) = 0,025 kg/jam Massa CH₃COOH(l) = 1,772 kg/jam Massa H₂O(l) = 1,053 kg/jam

Tabel A.5 Neraca Massa disekitar Kristalizer

Komponen Masuk (kg/Jam) Keluar (kg/Jam)

Arus 7 Arus 8

C6H5NHCOCH3 (s) - 130,408

C6H5NH2(l) 0,025 0,025

CH₃COOH_(l) 1,772 1,772

H2O(l) 1,053 1,053

C6H5NHCOCH3 (l) 135,842 5,434

TOTAL 138,692 138,692

102 8

Tabel A.6 Neraca Massa Sebenarnya di Kristalizer

Komponen Masuk (Kg/Jam) Keluar (Kg/Jam)

Arus 7 Arus 8

C₆H₅NHCOCH_3(s) - 23.142,889

C₆H₅NH_2(l) 4,392 4,392

CH3COOH (l) 314,543 314,543

H2O (l) 186,920 186,920

C6H5NHCOCH3 (l) 24.107,176 964,287

TOTAL 24.613,031 24.613,031

4. Rotary Dryer

Fungsi : Mengurangi kadar air didalam asetanilida dengan cara pengeringan pada suhu 100^oC.

NM = A8 A9 + A10 Umpan Masuk :

Arus 8 :

Massa C6H5NHCOCH3(s) = 130,408 kg/jam Massa C6H5NH2(l) = 0,025 kg/jam Massa CH3COOH(l) = 1,772 kg/jam Massa H2O(l) = 1,053 kg/jam Massa C6H5NHCOCH3(l) = 5,434 kg/jam C6H5NHCOCH3(s)

C6H5NH2(l)

CH3COOH(l) H2O(l)

C6H5NHCOCH3(l)

C6H5NHCOCH3(s)

C6H5NH2(l)

CH3COOH(l)

H2O(l)

C6H5NHCOCH3(l)

C6H5NHCOCH3(l)

H2O(l)

103 Umpan Keluar :

Arus 9 :

Asumsi Efisiensi = 98%

Massa H₂O_(l) = 1,053 kg/jam x 98%

= 1,032 kg/jam

Massa C₆H₅NHCOCH_3(l) = 5,434 kg/jam x 98%

= 5,325 kg/jam Arus 10 :

Massa C6H5NHCOCH3(s) = 130,040 kg/jam Massa C6H5NH2(l) = 0,025 kg/jam Massa CH3COOH(l) = 1,772 kg/jam

Massa H2O(l) = 1,053 kg/jam - 1,032 kg/jam

= 0,021 kg/jam

Massa C6H5NHCOCH3(l) = 5,434 kg/jam - 5,325 kg/jam = 0,109 kg/jam

Tabel A.7 Neraca Massa disekitar Rotary Dryer Komponen Masuk (kg/Jam) Keluar (kg/Jam)

Arus 8 Arus 9 Arus 10

C₆H₅NHCOCH_3(s) 131,659 - 130,659

C6H5NH2(l) 0,025 - 0,025

CH3COOH(l) 1,772 - 1,772

H2O(l) 1.053 1.032 0,021

C₆H₅NHCOCH_3(l) 5,434 5.325 0,109

TOTAL 138,692 0,955 132,335

138,692

104

Tabel A.8 Neraca Massa Sebenarnya di Rotary Dryer Komponen Masuk (Kg/Jam) Keluar (Kg/Jam)

Arus 8 Arus 9 Arus 10

C₆H₅NHCOCH_3(s) 23.142,889 - 23.142,889

C₆H₅NH_2(l) 4,392 - 4,392

CH3COOH(l) 314,543 - 314,543

H2O(l) 186,920 0, 183,181 3,738

C6H5NHCOCH3(l) 964,287 945,001 19,286

TOTAL 24.613,031 1.128,183 23.484,848

24.613,031

105 LAMPIRAN B

PERHITUNGAN NERACA PANAS Kapasitas produksi = 186.000 ton/tahun

Operasi pabrik = 330 hari/tahun Basis perhitungan = 1 jam operasi Temperatur referensi = 25^oC

Satuan perhitungan = kJ/jam Neraca Energi:

{(Energi masuk)-(Energi keluar)+(Generasi energy)-(Konsumsi energy)} = {Akumulasi energy}

(Himmelblau,ed.6,1996) Entalpi bahan pada temperatur dan tekanan tertentu adalah :

ΔH = ΔHT- ΔHf

(Himmelblau,ed.6,1996) Keterangan:

ΔH = Perubahan entalpi

ΔHT = Entalpi bahan pada suhu T

ΔHf = Entalpi bahan pada suhu referensi (25^oC)

Entalpi bahan untuk campuran dapat dirumuskan sebagai berikut : ΔH = ∑ n Cp t

(Himmelblau,ed.6,1996) Keterangan:

ΔH = Perubahan entalpi

N = mol

Cp = Kapasitas panas(J/mol.K) Dt = Perbedaan temperature (K) Kapasitas Panas

∫_Tref^T Cp dT =

∫ (

_Tref^T

A+ T+CT2+ T3+ T4) t

Cp dT = A (T-Tref) +

2(T²-T²ref) + ^C

3(T³-T³ref) +

4(T⁴-T⁴ref)+

5(T⁵-T⁵ref)

106 Keterangan :

Cp = Kapasitas panas (J/mol K) A,B,C,D,E = Konstanta

T_ref = Temperatur referensi = 298 K T = Temperatur operasi (K)

Tabel B.1 Data Kapasitas Panas Gas, Cp = A+BT+CT²+DT³+ET⁴ [J/mol K]

( Reklaitis, 1983 dan Carl.L Yaws "Chemical Properties Handbook")

Tabel B.2 Data Kapasitas Panas Liquid, Cp = A+BT+CT²+DT³ [J/mol K]

( Reklaitis, 1983 dan Carl.L Yaws "Chemical Properties Handbook") Tabel B.3 Data Kapasitas Panas Solid, Cp = A+BT+CT²

KOMPONEN A B C

C₆H₅NHCOCH₃ -115,731 0,78917 - (Carl.L Yaws "Chemical Properties Handbook")

KOMPONEN A B C D E

C₆H₅NHCOCH₃ -52,405 0,78304 -0,000592 2,33E-07 -4,1725E-11 C₆H5NH2 -2,26E+01 3,08E-01 2,42E-04 -5,38E-07 2,36E-10 CH₃COOH 6,90E+00 0,257 -1,92E-04 7,58E-08 -1,23E-11

H₂O 3,40E+01 -9.65E-03 3,30E-05 -2,04E-08 4,30E-12

N₂ 2,94E+01 -3,01E+03 5,45E-06 5,13E-09 -4,25E-12

O₂ 2,99E+01 -1,14E-02 4,34E-05 -3,70E-08 1,01E-11

KOMPONEN A B C D

C₆H5NHCOCH3 -54,663 1,7295 -3,33E-03 2,48E-06 C₆H₅NH₂ 63,288 0,9896 -2,36E-03 2,33E-06 CH₃COOH -3,61E+01 6,05E-01 -3,94E-04 -5,62E-07

H₂O 18,298 4,27E-01 -1,34E-03 1,31E-06

107

3

1. HEATER

Fungsi : Menaikkan temperatur anilin sebelum dimasukkan k e reaktor dari 30^oC menjadi 150^oC.

a. Entalpi Umpan Heater Anilin

Suhu umpan masuk = 30^oC = 303 K Suhu referensi = 25^oC = 298 K - C6H5NH2(l)

Cp dT = 63,288(303-298) + ^0,9896

2 (303² - 298²) + ^{-2,36 -03}

3 (303³ - 298³) + ^{-2,33 -06}₄ (303⁴ - 298⁴)

= 1.054,607 kJ/Kmol - H2O(l)

Cp dT = 18,298(303-298) + ^{4,27 -01}

2 (303² - 298²) + ^{-1,34 -03}

3 (303³ - 298³) + ^{1,31 -06}₄ (303⁴ - 298⁴)

= 305,780 kJ/Kmol Sehingga :

Q C₆H₅NH_2(l) = n x CpΔT

= (188,915 Kmol/Jam) x (1.054,607 kJ/Kmol) = 199.230,828 kJ/Jam

Q H₂O(l) = n x CpΔT

= (9,859 Kmol/Jam) x (305,780 kJ/Kmol) Kondensat

250^oC C6H5NH2(l)

C6H5NH2(l)

Saturated Steam 250^oC

1

108

= 3.014,741 kJ/Jam

Maka panas yang masuk Heater Anilin = 202.245,570 kJ/Jam b. Entalpi Keluar Heater Anilin

Suhu umpan keluar = 150^oC = 423 K Suhu referensi = 25^oC = 298 K - C₆H₅NH_2(l)

Cp dT = 63,288(423-298) + ^0,9896

2 (423² - 298²) + ^{-2,36 -03}

3 (423³ - 298³) + ^{-2,33 -06}₄ (423⁴ - 298⁴)

= 27.863,339 kJ/Kmol - H2O_(l)

Cp dT = 18,298(423-298) + ^{4,27 -01}

2 (423² - 298²) + ^{-1,34 -03}

3 (423³ - 298³) + ^{1,31 -06}₄ (423⁴ - 298⁴)

= 7.444,887 kJ/Kmol - C₆H₅NH_2(uap)

Cp dT = -2,26E+01(423-298) + ^{3,08 -01}

2 (423² - 298²) + ^{2,42 -04}

3 (423³ - 298³) + ^{-5,38 -07}₄ (423⁴ - 298⁴)+ ^{2,36 -10}₅ (423⁵ – 298⁵)

= 12.307,813 kJ/Kmol - H₂O(uap)

Cp dT = 3,40E+01(423-298) + ^{-9,65 -03}

2 (423² - 298²) + ^{3,30 -05}

3 (423³ - 298³) + ^{-2,04 -08}₄ (423⁴ - 298⁴)+ ^{4,30 -12}₅ (423⁵ – 298⁵)

= 4.243,169 kJ/Kmol Sehingga :

Q C₆H₅NH_2(l) = n x CpliquidΔT + n x CpuapΔT

= (188,915 Kmol/Jam) x (27.863,339 kJ/Kmol) + (188,915 Kmol/Jam) x (12.307,813 kJ/Kmol)

= (5.263.795,891 kJ/Jam + 2.325.127,569 kJ/Jam)

109

= 7.588.923,460 kJ/Jam Q H₂O(l) = n x CpliquidΔT + n x CpuapΔT

= (9,859 Kmol/Jam) x (7.444,887 kJ/Kmol) + (9,859 Kmol/Jam) x (4.243,169 kJ/Kmol)

= (73.400,540 kJ/Jam + 41.834,204 kJ/Jam)

= 115.234,744 kJ/Jam

Maka panas keluar Heater Anilin = 7.704.158,204 kJ/Jam Jumlah panas yang dibutuhkan:

ΔQ = Qout – Qin

= (7.704.158,204 – 202.245,570) kJ/Jam

= 7.501.912,634 kJ/Jam

Sebagai pemanas pada heater digunakan saturated steam (uap jenuh) pada kondisi:

T = 250^oC

Hf = 1.085,8 kJ/kg Hg = 2.800,4 kJ/kg

(Appendix F steam table, Thermodynamics) λ steam = H_g – Hf

= 1.085,8 kJ/kg – 2.800,4 kJ/kg

= 1.714,6 kJ/kg Menghitung massa steam : Massa steam =

= 7.501.912,634 kJ/Jam 1 714,6 kJ /kg

= 4.375,314 kg/jam Panas steam masuk :

ΔHs IN = massa steam x hg

= 4.375,314 kg/Jam x 2.800,4 kJ/kg

= 12.252.628,100 kJ/Jam

110

4

Panas steam keluar :

ΔHs _OUT = massa steam x h_f

= 4.375,314 kg/Jam x 1.085,8 kJ/kg

= 4.750.715,466 kJ/Jam

Tabel B.4 Neraca Panas Heater Anilin

Komponen Masuk (kJ/jam) Keluar (kJ/jam) C6H5NH2(l) 199.230,828 7.588.923,460

H2O(l) 3.014,741 115.234,744

Q steam 12.252.628,100 4.750.715,466 Total 12.454.873,670 12.454.873,670

2. Heater

Fungsi : Menaikkan temperatur asam asetat sebelum dimasukkan ke reaktor dari 30^oC menjadi 150^oC.

a. Entalpi Umpan Heater Asam Asetat Suhu umpan masuk = 30^oC = 303 K Suhu referensi = 25^oC = 298 K

Kondensat 250^oC CH₃COOH(l)

CH₃COOH(l)

Saturred Steam 250^oC

111 - CH3COOH(l)

Cp dT = -3,61E+01(303-298) + ^{6,05 -01}

2 (303² - 298²) + ^{-3,94 -04}

3 (303³ -298³) + ^{-5,62 -07}₄ (303⁴ - 298⁴)

= 474,362 kJ/Kmol - H2O(l)

Cp dT = 18,298(303-298) + ^{4,27 -01}

2 (303² - 298²) + ^{-1,34 -03}

3 (303³ - 298³) + ^{1,31 -06}₄ (303⁴ - 298⁴)

= 305,780 kJ/Kmol Sehingga :

Q CH₃COOH_(l) = n x CpΔT

= (292,818 Kmol/Jam) x (474,362 kJ/Jam) = 138.901,568 kJ/Jam

Q H₂O(l) = n x CpΔT

= (9,859 Kmol/Jam) x (305,780 kJ/Jam) = 3.014,741 kJ/Jam

Maka panas yang masuk Heater Asam Asetat = 141.916,309 kJ/Jam b. Entalpi Keluar Heater Asam Asetat

Suhu umpan keluar = 150^oC = 423 K Suhu referensi = 25⁰C = 298 K - CH3COOH(l)

Cp dT = -3,61E+01(423-298) + ^{6,05 -01}

2 (423² - 298²) + ^{-3,94 -04}

3 (423³ -298³) + ^{-5,62 -07}₄ (423⁴ - 298⁴)

= 12.895,457 kJ/Kmol - H2O(l)

Cp dT = 18,298(423-298) + ^{4,27 -01}

2 (423² - 298²) + ^{-1,34 -03}

3 (423³ - 298³) + ^{1,31 -06}₄ (423⁴ - 298⁴)

112

= 7.444,887 kJ/Kmol - CH3COOH(uap)

Cp dT = 6,90E+00(423-298) + ^0,257

2 (423² - 298²) + ^{-1,92 -04}

3 (423³ -298³) + ^{7,58 -08}₄ (423⁴ - 298⁴) + ^{-1,23 -11}₅ (423⁵ – 298⁵)

= 9.722,986 kJ/Kmol - H2O_(uap)

Cp dT = 3,40E+01(423-298) + ^{-9,65 -03}

2 (423² - 298²) + ^{3,30 -05}

3 (423³ - 298³) + ^{-2,04 -08}₄ (423⁴ - 298⁴)+ ^{4,30 -12}₅ (423⁵ – 298⁵)

= 4.243,169 kJ/Kmol Sehingga :

Q CH₃COOH(l) = n x CpliquidΔT + n x CpuapΔT

= (292,818 Kmol/Jam) x (12.895,457 kJ/Jam) + (292,818 Kmol/Jam) x ( 9.722,986 kJ/Jam)

= (3.776.020,226 kJ/Jam + 2.847.064,009 kJ/Jam) = 6.623.084,235 kJ/Jam

Q H₂O(l) = n x CpliquidΔT + n x CpuapΔT

= (9,859 Kmol/Jam) x (7.444,887 kJ/Jam) + (9,859 Kmol/Jam) x (4.243,169 kJ/Jam) = (73.400,540 kJ/Jam + 41.834,204 kJ/Jam) = 115.234,744 kJ/Jam

Maka panas keluar Heater Asam Asetat = 6.738.318,979 kJ/Jam Jumlah panas yang dibutuhkan:

ΔQ = Q_out – Q_in

= (6.738.318,979 – 141.916,309) kJ/Jam

= 6.596.402,670 kJ/Jam

Sebagai pemanas pada heater digunakan saturated steam (uap jenuh) pada kondisi :

113

T = 250^oC

H_f = 1.085,8 kJ/kg

H_g = 2.800,4 kJ/kg

(Appendix F steam table, Thermodynamics) λ steam = H_g – Hf

= 2.800,4 kJ/kg – 1.085,8 kJ/kg

= 1.714,6 kJ/Kg Menghitung massa steam :

Massa steam =

= 6.596.402,670 kJ /Jam 1 714,6 kJ /kg

= 3.847,196 kg/Jam Panas steam masuk :

ΔHs _IN = massa steam x h_g

= 3.847,196 kg/Jam x 2.800,4 kJ/kg

= 10.773.688,345 kJ/Jam Panas steam keluar :

ΔHs OUT = massa steam x h_f

= 3.847,196 kg/Jam x 1.085,8 kJ/kg

= 4.177.285,675 kJ/ Jam

Tabel B.5 Neraca Panas Heater Asam Asetat Komponen Masuk (kJ/jam) Keluar (kJ/jam) CH₃COOH(l) 138.901,568 6.623.084,235 H₂O_(l) 3.014,741 115.234,744 Q steam 10.773.688,345 4.177.285,675 Total 10.915.604,654 10.915.604,654

114

5

3. Reaktor

Fungsi : Tempat terjadinya reaksi pembentukan asetanilida.

Panas masuk reaktor = Panas keluar heater anilin + Panas keluar heater

asam asetat

= (7.704.158,204 + 6.738.318,979) kJ/Jam

= 14.442.477,183 kJ/Jam

Reaksi yang berlangsung dalam reaktor :

C₆H₅NH_2(l) + CH₃COOH_(l) C₆H₅NHCOCH_3(s) + H₂O _(l) Tabel B.6 Panas Reaksi Pembentukan [kJ/mol]

(Reklaitis, 1983 dan Carl.L Yaws "Chemical Properties Handbook") a. Panas reaksi pada keaadaan standar:

ΔHro = (ΔHr^o Produk - ΔHr^o Bahan Baku)

ΔHro = (ΔHof Asetanilida + Δhof Air) – (ΔHof Anilin + ΔHof Asam Asetat)

KOMPONEN ∆Hf (kJ/mol) C6H5NHCOCH3 -78.4852 C6H5NH2 -20.76

CH3COOH -432.84

H2O -241.8

C6H5NHCOCH3(l) C6H5NH2(l) CH3COO(l) H2O(l)

C6H5NH2(l)

CH3COO(l)

Air Pendingin 30^oC

Air Pendingin 50^oC

115

= ((-78,4852) + (-241,8)) – ((-20,76) + (-432,84))

= 133.315 kJ/Kmol b. Entalpi Keluar Reaktor

Suhu umpan masuk = 150^oC = 423 K Suhu referensi = 25^oC = 298 K - C6H₅NHCOCH_3(l)

Cp dT = -54,663(423-298) + ^1.,2795

2 (423² - 298²) + ^{-3,33 -03}

3 (423³ - 298³) + ^{2,48 -06}₄ (423⁴ - 298⁴)

= 31.433,975 kJ/Kmol - C₆H₅NH_2(l)

Cp dT = 63,288(423-298) + ^0,9896

2 (423² - 298²) + ^{-2,36 -03}

3 (423³ - 298³) + ^{2,33 -06}₄ (423⁴ - 298⁴)

= 27.863,339 kJ/Kmol - CH₃COOH(l)

Cp dT = -3,61E+01(423-298) + ^{6,05 -01}

2 (423² - 298²) + ^{-3,94 -04}

3 (423³ -298³) + ^{-5,62 -07}₄ (423⁴ - 298⁴)

= 12.895,457 kJ/Kmol - H2O(l)

Cp dT = 18,298(423-298) + ^{4,27 -01}

2 (423² - 298²) + ^{-1,34 -03}

3 (423³ - 298³) + ^{1,31 -06}₄ (423⁴ - 298⁴)

= 7.444,887 kJ/Kmol Sehingga :

Q C6H5NHCOCH3(l) = n x CpΔT

= (187,970 Kmol/Jam) x (31.433,975 kJ/Kmol)

= 5.908.650,109 kJ/Jam

116

Q C6H5NH2(l) = n x CpΔT

= (0,047 Kmol/Jam) x (27.863,339 kJ/Kmol)

= 1.315,949 kJ/Jam Q CH3COOH(l) = n x CpΔT

= (104,848 Kmol/Jam) x (12.895,457 kJ/Kmol)

= 1.352.058,855 kJ/Jam

Q H2O(l) = n x CpΔT

= (207,689 Kmol/Jam) x (7.444,887 kJ/Kmol)

= 1.546.217,886 kJ/Jam Maka panas keluar Reaktor = 8.808.242,800 kJ/Jam c. Panas reaksi pada suhu operasi

Suhu operasi reaktor = 150^oC = 423 K

ΔHr _{(423 K)} = (Δhr^o - ∫ _{( )} dT - ∫ ( )

dT + ∫ ( )

dT + ∫ _{( )} dT)

= (133.314,8 – 27.863,3388 – 12.895,4567 + 31.433,9753 + 7.444,8872 )

= 131.434,867 kJ/Kmol Jumlah panas yang dibutuhkan:

ΔQ = Q_out – Qin

= (8.808.242,800 – 14.442.477,183) kJ/Jam = -5.634.234,383 kJ/Jam

Panas reaksi total = n C6H5NHCOCH3 x ΔHr

= (187,970 Kmol/Jam) x (131.434,867 kJ/Kmol)

= 24.705.836,072 Kj/Jam ΔQ = (Q_out – Q_in) + (m C6H₅NHCOCH₃ x ΔHr

= (-5.634.234,383 + 24.705.836,072) kJ/Jam

= 19.071.601,689 kJ/Jam d. Air pendingin yang diperlukan

Media pendingin yang digunakan adalah air yang masuk pada suhu 30^oC dan keluar pada suhu 50^oC. Air pendingin yang diperlukan :

117

ΔH = H (50^oC) - H (30^oC) = [H (50^oC) - H (25^oC)] - [H (30^oC) - H (25^oC)]

=

∫

₂₉₈³²³

Cp

_{H2O (I)}

dT - ∫

₂₉₈³⁰³

Cp

_{H2O (I)}

dT

= ((18,298 x (323-298)) + ((4,27E-01/2) x (323²-298² )) + ((-1,34E 03/3) x (323³-298³)) + ((1,31E-06/4) x (323⁴-298⁴ ))) – ((18,298 x (303-298)) + ((4,27E-01/2) x (303²-298² )) + ((-1,34E-03/3) x (303³-298³)) + ((1,31E-06/4) x (303⁴-298⁴ )))

= 1.216,742 kJ/kg

Massa air pendingin yang diperlukan adalah:

m = ^Q

H (50^oC) - H (30^oC)

m = 19.071.601,689 kJ/Jam 1 216,742 kJ/kg m = 15.674,317 kg/jam

Tabel B.7 Neraca Panas Reaktor

Q Masuk (kJ/jam) Keluar (kJ/jam)

Umpan 14.442.477,183 -

Produk - 8.808.242,800

r x ∆Hr - 24.705.836,072

Air Pendingin 19.071.601,689 -

Total 33.514.078,872 33.514.078,872

118 4. Evaporator

Fungsi : Mengurangi kandungan anilin dan asam asetat dalam larutan asetanilida dengan cara memekatkan produk pada suhu 225^oC.

Panas masuk evaporator = Panas keluar reaktor = 8.808.242,800 kJ/Jam a. Entalpi Keluar Evaporator (Arus 7)

Suhu umpan keluar = 225^oC = 498 K Suhu referensi = 25^oC = 298 K - C6H₅NHCOCH_3(l)

Cp dT = -54,663(498-298) + ^1.,2795

2 (498² - 298²) + ^{-3,33 -03}

3 (498³ - 298³) + ^{2,48 -06}₄ (498⁴ - 298⁴)

= 52.283,960 kJ/Kmol - C6H₅NH_2(l)

Cp dT = 63,288(498-298) + ^0,9896

2 (498² - 298²) + ^{-2,36 -03}

3 (498³ - 298³) + ^{2,33 -06}₄ (498⁴ - 298⁴)

= 46.372,920 kJ/Kmol

EVAPORATOR

C6H5NHCOCH3(l) C6H5NH2(l)

CH3COOH(l)

H2O(l)

C6H5NHCOCH3(l)

C6H5NH2(l) CH3COOH(l) H2O(l)

C6H5NHCOCH3(uap) C6H5NH2(uap)

CH3COOH(uap)

H2O(uap)

Saturated Steam 250^oC

Kondensat 250^oC

119 - CH3COOH(l)

Cp dT = -3,61E+01(498-298) + ^{6,05 -01}

2 (498² - 298²) + ^{-3,94 -04}

3 (498³-298³) + ^{-5,62 -07}₄ (498⁴ - 298⁴)

= 20.659,512 kJ/Kmol - H2O(l)

Cp dT = 18,298(498-298) + ^{4,27 -01}

2 (498² - 298²) + ^{-1,34 -03}

3 (498³ - 298³) + ^{1,31 -06}₄ (498⁴ - 298⁴)

= 11.863,690 kJ/Kmol Sehingga :

Q C6H5NHCOCH3(l) = n x CpΔT

= (178,572 Kmol/Jam) x (52.283,960 kJ/Kmol)

= 9.336.434,284 kJ/Jam

Q C6H5NH2(l) = n x CpΔT

= (0,047 Kmol/Jam) x (46.372,920 kJ/Kmol)

= 2.190,132 kJ/Jam Q CH3COOH(l) = n x CpΔT

= (5,242 Kmol/Jam) x (20.659,512 kJ/Kmol)

= 108.305,103 kJ/Jam

Q H2O(l) = n x CpΔT

= (10,384 Kmol/Jam) x (11.863,690 kJ/Kmol)

= 123.197,636 kJ/Jam

Maka panas keluar Evaporator produk bawah = 9.570.127,156 kJ/Jam b. Entalpi Keluar Evaporator (Arus 6)

Suhu umpan keluar = 150^oC = 423 K Suhu referensi = 25^oC = 298 K

120 - C6H₅NHCOCH_3(uap)

Cp dT = -52,405(423-298) + ^0,78304

2 (423² - 298²) + -0,00059239

3 (423³ - 298³) + ^{2,33 -07}₄ (423⁴ - 298⁴) + -4,1725 -11

5 (423⁵– 298⁵)

= 20.328,648 kJ/Kmol - C6H₅NH_2(uap)

Cp dT = 0-2,26E+01(423-298) + ^{3,08 -01}

2 (423² - 298²) + ^{2,42 -04}

3 (423³ - 298³) + ^{-5,38 -07}₄ (423⁴ - 298⁴)+^{2,36 -10}

5 (423⁵ – 298⁵)

= 12.307,813 kJ/Kmol - CH3COOH(uap)

Cp dT = 6,90E+00(423-298) + ^0,257

2 (423² - 298²) + ^{-1,92 -04}

3 (423³-298³) + ^{7,58 -08}₄ (423⁴ - 298⁴) + ^{-1,23 -11}₅ (423⁵ – 298⁵)

= 9.722,986 kJ/Kmol - H₂O(uap)

Cp dT = 3,40E+01(423-298) + ^{-9,65 -03}

2 (423² - 298²) + ^{3,30 -05}

3 (423³ - 298³) + ^{-2,04 -08}₄ (423⁴ - 298⁴)+ ^{4,30 -12}₅ (423⁵ – 298⁵)

= 4.243,169 kJ/Kmol Sehingga :

Q C6H5NHCOCH3 (uap) = n x CpΔT

= (9,399 Kmol/Jam) x (20.328,648 kJ/Kmol)

= 191.058,984 kJ/Jam

Q C6H5NH2(uap) = n x CpΔT

= (0,897 Kmol/Jam) x (12.307,813 kJ/Kmol)

= 11.044,356 kJ/Jam

Q CH3COOH(uap) = n x CpΔT

= (59,191 Kmol/Jam) x (9.722,986 kJ/Kmol)

121

= 575.515,575 kJ/Jam

Q H2O(uap) = n x CpΔT

= (181,052 Kmol/Jam) x (4.243,169 kJ/Kmol)

= 768.235,233 kJ/Jam

Maka panas keluar Evaporator (uap) = 1.545.854,147 kJ/Jam Jumlah panas yang dibutuhkan:

ΔQ = Q_out – Qin

= ((1.545.854,147 + 9.570.127,156) - (8.808.242,800)) kJ/Jam

= 2.307.738,503 kJ/Jam

Sebagai pemanas pada evaporator digunakan saturated steam (uap jenuh) pada kondisi:

T = 250^oC

Hf = 1.085,8 kJ/kg Hg = 2.800,4 kJ/kg λ steam = Hg – Hf

= 1.085,8 kJ/kg – 2.800,4 kJ/kg

= 1.714,6 kJ/kg Menghitung massa steam : Massa steam =

= 2.307.738,503 kj /Jam 1 714,6 kj /Kg

= 1.345,934 Kg/jam Panas steam masuk :

ΔHs IN = massa steam x h_g

= 1.345,934 Kg/Jam x 2.800,4 kJ/Kg

= 3.769.153,682 kJ/Jam Panas steam keluar :

ΔHs OUT = massa steam x hf

= 1.345,934 Kg/Jam x 1.085,8 kJ/Kg

= 1.461.415,179 kJ/Jam

122 KRISTALIZER

Tabel B.8 Neraca Panas Evaporator

Q Masuk (kJ/jam) Keluar (kJ/jam)

Umpan 8.808.242,800 -

Produk - 1.545.854,147

Uap - 9.570.127,156

Steam 3.769.153,682 1.461.415,179 Total 12.577.396,482 12.577.396,482

5. Kristalizer

Fungsi : Tempat terbentuknya kristal asetanilida.

Panas masuk kristalizer = Panas keluar evaporator = 9.570.127,156 kJ/jam a. Panas Kristalisasi

Panas kelarutan C₆H₅NHCOCH₃ = -2.732 kkal/kmol

Panas Kristalisasi C₆H₅NHCOCH₃ = - (Panas kelarutan C₆H₅NHCOCH₃) (Perry’s, 1999)

Panas kristalisasi C₆H₅NHCOCH₃ = -(-2732) kkal/kmol

= 2.732 kkal/mol C₆H₅NHCOCH₃ yang terkristal = 23.142,889 kg/jam

= 171,429 kmol/jam C6H5NHCOCH3(l)

C6H5NH2(l) CH3COOH(l) H2O(l)

C6H5NHCOCH3(S)

C6H5NHCOCH3(l)

C6H5NH2(l) CH3COOH(l) H2O (l)

Air Pendingin 50^oC

Air Pendingin 30^oC

123 Maka panas kristalisasi C₆H₅NHCOCH₃ = (2.732 kkal/mol) x (171,429 kmol/jam)

= 468.343,502 kkal/jam

= 11.960.954,243 kJ/jam b. Entalpi Keluar Kristalizer

Suhu umpan keluar = 60^oC = 333 K Suhu referensi = 25^oC = 298 K - C6H₅NHCOCH_3(s)

Cp dT = -115,731(333-298) + ^0,78917

2 (333² - 298²) + ⁰

3(333³ - 298³)

= 4.663,825 kJ/Kmol - C₆H₅NH_2(l)

Cp dT = 63,288(333-298) + ^0,9896

2 (333² - 298²) + ^{-2,36 -03}

3 (333³ - 298³) + ^{2,33 -06}₄ (333⁴ - 298⁴)

= 7.486,722 kJ/Kmol - CH3COOH(l)

Cp dT = -3,61E+-01(333-298) + ^{6,05 -01}

2 (333² - 298²) + ^{-3,94 -04}

3 (333³-298³) + ^{-5,62 -07}₄ (333⁴ - 298⁴)

= 3.423,509 kJ/Kmol - H2O_(l)

Cp dT = 18,298(333-298) + ^{4,27 -01}

2 (333² - 298²) + ^{-1,34 -03}

3 (333³ - 298³) + ^{1,31 -06}₄ (333⁴ - 298⁴)

= 2.126,699 kJ/Kmol - C6H₅NHCOCH_3(l)

Cp dT = -54,633(333-298) + ^1,7295

2 (333² - 298²) + ^{-3,33 -03}

3 (333³ - 298³) + ^{2,48 -06}₄ (333⁴ - 298⁴)

124

= 8.307,311 kJ/Kmol Sehingga :

Q C₆H₅NHCOCH_3(s) = n x CpΔT

= (171,429 Kmol/Jam) x (4.663,825 kJ/Kmol) = 799.513,911 kJ/Jam

Q C₆H₅NH_2(l) = n x CpΔT

= (0,047 Kmol/Jam) x (7.486,722 kJ/Kmol) = 353,588 kJ/Jam

Q CH₃COOH(l) = n x CpΔT

= (5,242 Kmol/Jam) x (3.423,509 kJ/Kmol) = 17.947,350 kJ/Jam

Q H₂O_(l) = n x CpΔT

= (10,384 Kmol/Jam) x (2.126,699 kJ/Kmol) = 86.266,670 kJ/Jam

Q C₆H₅NHCOCH_3(l) = n x CpΔT

= (7,143 Kmol/Jam) x (8.307,311 kJ/Kmol) = 15.190,729 kJ/Jam

Maka panas keluar Kristalizer = 919.272,248 kJ/jam Jumlah panas yang dibutuhkan:

ΔQ = (Q_out – Q_in) - Panas Kristalisasi

= (919.272,248 – 1.959.549,213) – 1.960.954,243

= -10.611.809,151 kJ/jam c. Air pendingin yang diperlukan

Media pendingin yang digunakan adalah air yang masuk pada suhu 30^oC dan keluar pada suhu 50^oC. Air pendingin yang diperlukan :

Air = H (50^oC) – H (30^oC) = [H (50^oC) – H (25^oC)] – [H (30^oC) – H (25^oC)]

=

∫

₂₉₈³²³

Cp

_{H2O (I)}

dT – ∫

₂₉₈³⁰³

Cp

_{H2O (I)}

dT

= ((18,298 x (323-298)) + ((4,27E-01/2) x (323²-298² )) + ((-1,34E 03/3) x (323³-298³)) + ((1,31E-06/4) x (323⁴-298⁴ ))) – ((18,298 x

125 8

(303-298)) + ((4,27E-01/2) x (303²-298² )) + ((-1,34E-03/3) x (303³-298³)) + ((1,31E-06/4) x (303⁴-298⁴ )))

= 1.216,742 kJ/kg

Massa air pendingin yang diperlukan adalah:

m = ^Q

H (50^oC)-H (30^oC)

m = -10.611.809,151 kJ/Jam 1 216,742 kJ/kg m = 8.721,494 kg/jam

Tabel B.9 Neraca Panas Kristalizer

Q Masuk (kJ/jam) Keluar (kJ/jam)

Umpan 9.570.127,156 -

Produk - 919.272,248

Panas kristalizer 1.959.549,213 -

Air Pendingin - 10.611.809,151

Total 11.531.081,399 11.531.081,399

6. Rotary Dryer

Fungsi : Mengurangi kadar air didalam asetanilida dengan cara pengeringan pada suhu 100^oC.

C6H5NHCOCH3(s)

C6H5NH2(l)

CH3COOH(l) H2O(l)

C6H5NHCOCH3(l) C6H5NHCOCH3(s)

C6H5NH2(l)

CH3COOH(l)

H2O(l)

C6H5NHCOCH3(l)

C6H5NHCOCH3(l)

H2O(l)

Saturated Steam 250^oC

126

Panas masuk rotary dryer = Panas keluar kristalizer = 919.272,248 kj/jam a. Entalpi Keluar (Arus 9)

Suhu umpan keluar = 60^oC = 333 K Suhu referensi = 25^oC = 298 K - H2O(l)

Cp dT = 18,298(333-298) + ^{4,27 -01}

2 (333² - 298²) + ^{-1,34 -03}

3 (333³ - 298³) + ^{1,31 -06}₄ (333⁴ - 298⁴)

= 2.126,699 kJ/Kmol - C₆H₅NHCOCH_3(l)

Cp dT = -54,633(333-298) + ^1,7295

2 (333² - 298²) + ^{-3,33 -03}

3 (333³ - 298³) + ^{2,48 -06}₄ (333⁴ - 298⁴)

= 8.307,311 kJ/Kmol Sehingga :

Q H2O(l) = n x CpΔT

= (10,177 Kmol/Jam) x (2.126,699 kJ/Kmol)

= 21.642,862 kJ/Jam Q C6H5NHCOCH3(l) =n x CpΔT

= (7,000 Kmol/Jam) x (8.307,311 kJ/Kmol)

= 58.151,255 kJ/Jam Maka panas keluar = 79.794,118 kJ/Jam b. Entalpi Keluar Rotary Dryer (Arus 10)

Suhu umpan keluar = 100^oC = 373 K Suhu referensi = 25^oC = 298 K - C6H₅NHCOCH_3(s)

Cp dT = -115,731(373-298) + ^0,78917

2 (373² - 298²) + ⁰

3(373³ - 298³)

= 11.177,665 kJ/Kmol

127 - C6H₅NH_2(l)

Cp dT = 63,288(373-298) + ^0,9896

2 (373² - 298²) + ^{-2,36 -03}

3 (373³ - 298³) + ^{2,33 -06}₄ (373⁴ - 298⁴)

= 16.336,240 kJ/Kmol - CH3COOH(l)

Cp dT = -3,61E+-01(373-298) + ^{6,05 -01}

2 (373² - 298²) + ^{-3,94 -04}

3 (373³-298³) + ^{-5,62 -07}₄ (373⁴ - 298⁴)

= 7.564,168 kJ/Kmol - H₂O(l)

Cp dT = 18,298(373-298) + ^{4,27 -01}

2 (373² - 298²) + ^{-1,34 -03}

3 (373³ - 298³) + ^{1,31 -06}₄ (373⁴ - 298⁴)

= 4.513,989 kJ/Kmol - C₆H₅NHCOCH_3(l)

Cp dT = -54,663(373-298) + ^1,7295

2 (373² - 298²) + ^{-3,33 -03}

3 (373³ - 298³) + ^{2,48 -06}₄ (373⁴ - 298⁴)

= 18.305,730 kJ/Kmol Sehingga :

Q C6H5NHCOCH3(s) = n x CpΔT

= (171,429 Kmol/Jam) x (11.177,665 kJ/Kmol) = 1.916.173,803 kJ/Jam

Q C6H5NH2(l) = n x CpΔT

= (0,047 Kmol/Jam) x (16.336,2403 kJ/Kmol) = 771,539 kJ/Jam

Q CH3COOH(l) = n x CpΔT

= (5,242 Kmol/Jam) x (7.564,168 kJ/Kmol) = 39.654,278 kJ/Jam

128 Q H2O(l) = n x CpΔT

= (0,208 Kmol/Jam) x (4.513,986 kJ/Kmol) = 937,503 kJ/Jam

Q C6H5NHCOCH3(l) = n x CpΔT

= (0,143 Kmol/Jam) x (18.305,730 kJ/Kmol) = 2.615,108 kJ/Jam

Maka panas keluar Rotary Dryer = 1.960.152,232 kJ/Jam Jumlah panas yang dibutuhkan:

ΔQ = (Qout+ Quap) – Qin

= (1.960.152,232 + 79.794,118) – 919.272,248

= 1.120.674,102 kJ/Jam

Tabel B.11 Neraca Panas Rotary Dryer

Q Masuk (kJ/jam) Keluar (kJ/jam)

Umpan 919.272,248 -

Produk - 1.960.152,232

Uap Air - 79.794,118

Steam 1.120.674,102 -

Total 2.039.946,350 2.039.946,350 7. Heater Udara

Fungsi : Menaikkan temperatur udara sebelum masuk ke rotary dryer dari 30^oC menjadi 100^oC.

Panas yang dibutuhkan rotary dryer = Panas keluar heater udara

= 1.020.674,102 kJ/Jam a. Entalpi Masuk Heater udara

Suhu umpan = 30^oC = 303 K Suhu referensi = 25^oC = 298 K

Q Udara yang dibutuhkan rotary dryer = 1.020.674,102 kJ/Jam Cp dT Udara = 292,813 kJ/Kg

Maka, massa udara (m) = ^Q

CpdT Udara