TEKNIK KIMIA: reaktorr

a. Jumlah Pengaduk

Jumlah Pengaduk =

Dimana,

WELH = water equivalent liguid high = ZL.sg

ID = diameter dalam reaktor

sg = specific gravity

Z_L= tinggi cairan dalam shell = meter

Sg =

WELH = ZL.sg = x = 2,87 m

Jumlah Pengaduk = ≈ 1 buah pengaduk

b. Kecepatan Pengaduk

Perhitungan Kecepatan Pengaduk Dipilih berdasarkan, Rase, H.F., dan J.R., Holmes, Chemical Reactor Design for Process Plants, Willey and Son, New York, 1977, vol.1, halaman 366. Kecepatan putar berkisar antara 500 ft/menit sampai 700 ft/menit, dengan tipe pengaduk propeller dan flat blade turbine impeller dengan 6 flate blade.

Dipilih: 500 ft/menit

N = 500 ft/menit x [0,3048 m/ft] = 152,4 m/menit

N = = 0,82 rps

Dari Wallas, halaman 288 untuk kecepatan pengaduk standar:

Sehingga, digunakan kecepatan pengadukan 56 rpm = 0,93rps

c. Menghitung bilangan Reynold

Dimana,

ρ = densitas campuran = kg/m³

N = kecepatan pengadukan (rpm) = 0,93 rps

Da = diameter impeller = m

µ = viskositas campuran = 18,01 Cp

= 0,018 kg/m.s

Maka:

Dari gambar 8.8 (Rase, 1957), untuk six blade turbine dengan NRe > 10⁵, maka nilai Np = 5

d. Menghitung Tenaga Pengaduk

Dihitung dengan persamaan:

Po = Np x ρ x N³ x Da⁵ (Mc Cabe hal 253 pers 9.20)

Keterangan:

Po = Daya penggerak (watt)

Np = Bilangan daya

ρ = Rapat massa fluida yang diaduk (kg/m³)

N = Kecepatan pengaduk (s^-1)

Di = Diameter pengaduk (m)

Maka:

Efisiensi motor pengaduk, diperoleh dari tabel 3.1 Towler, halaman 111:

Size (kW)	Efisiensi (%)
5	80
15	85
75	90
200	92
750	95
>4000	97

Diperoleh effisiensi = 80%, maka daya penggerak motor pengaduk yang diperlukan:

Daya = = 6,7HP

Digunakan motor pengaduk standar, diperoleh dari Ludwig, E.F., Applied Process design for Chemical and Petrochemical Plants, Gulf, Publishing, Co. Houston, Texas, 2001, edisi 3, halaman 628:

Sehingga, dipilih motor pengaduk standar yaitu 7 ½ HP.

NERACA PANAS REAKTOR

ΔH_Total

100 ^oC

25 ^oC

ΔH_f

ΔH_R

ΔH_P

T_input = 100 ^oC

Tabel 8. Harga H_f dan Cp masing-masing komponen

Komponen	H⁰f(kJ/mol)	Cp (J/mol.K)
C₆H₅CH₅CL	18,70	82,217 + 7,0948E-01 T – 1,7551E-03 T² + 1,8744E-06 T³
NaCN	-87,50	69,098 + 3,8057E-02 T – 4,1570E-05 T² + 1,3876E-08 T³
C₆H₅CH₅CN	117,28	87,722 + 7,6976E-01 T – 1,6788E-03 T² + 1,5257E-6 T³
NaCl	-411,20	95,016 – 3,1081E-02 T + 9,6789E-07 T²+ 5,5116E-09 T³
H₂O	–241,800	92,053 – 3,9953.10^-2 T – 2,1103.10^-4 T² + 5,3469.10^-7 T³

Reaktor 1

Tabel 9. Komponen Masuk dan Keluar Reaktor 1

Komponen	Masuk kmol/jam	Keluar kmol/jam
C₆H₅CH₅CL	67,390	14,852
NaCN	67,390	14,852
C₆H₅CH₅CN	0,646	53,183
NaCl	0,00	52,537
H₂O	373,530	373,530

a. Panas Reaktan (∆H_R)

T₁ = 100^oC = 373 K

T₂ = 25 ^oC = 298 K

∆H_R = ∑ n.Cp. dT

= (-978.200.581,4) + (-392.706.986) + (-10.401.996,87) + (-2.107.848.235)

= -3.494.504.516 Joule/jam

b. Panas Reaksi Standar

Reaksi yang terjadi :

C₆H₅CH₅CL_(l)+ NaCN_(l) C₆H₅CH₅CN_(l) + NaCl _(l)

∆H1 =

=(-14.515,4589)+(-5.827,3550)+(-16.091,977)+(-9.127,9300)+(-5.643,0403)

= -51.205,7335 J/mol

∆H2 =

=14.515,4589+5.827,3550+16.091,977+9.127,930+5.643,040

=-51.205,7335 J/mol

∆H⁰_R = H_f ⁰Produk – H_f ⁰Reaktan

= [(1 × C₆H₅CH₅CN) + (1 × NaCl )] − [(1 × C₆H₅CH₅CL) + (1 × NaCN)]

= ((117,28 kJ/mol) + (-411,2 kJ/mol)) – ((-87,5 kJ/mol) + (18,7 kJ/mol))

= -225,12 kJ/mol

= -225.120 J/mol

∆Hreaksi = ∆H1+∆H2+∆H⁰_R

₌-51.205,7335 + 51.205,7335 -225.120) J/mol

= -225.120 J/mol

Panas Reaksi = ∆Hreaksi. n_A0.X_A

₌ -225.120 J/mol×67.390,26mol/jam×77,96 %

= -11.827.230.000 J/jam

c. Panas Produk (∆H_P)

T₁ = 25 ^oC = 298 K

T₂ = 140 ^oC = 373 K

∆H_P = ∑ n.Cp. dT

=215.595.408,7+ 86.552.619,67+ 855.833.417,1+ 479.558.157,8+ 2.107.848.235

= 3.745.387.838 Joule/jam

d. Neraca Panas Di Reaktor 1

Q_S =∆H_R+ Panas Reaksi + ∆H_P

=( -3.494.504.516 -11.827.230.000 +3.745.387.838)Joule/jam

= Kjoule/jam

Reaktor 2

Tabel 10. Komponen Masuk dan Keluar Reaktor 2

Komponen	Masuk kmol/jam	Keluar kmol/jam
C₆H₅CH₅CL	14,852	6,073
NaCN	14,852	6,073
C₆H₅CH₅CN	53,183	61,317
NaCl	52,537	61,963
H₂O	373,530	373,5306

a. Panas Reaktan (∆H_R)

T₁ = 140^oC = 373 K

T₂ = 25 ^oC = 298 K

∆H_R = ∑ n.Cp. dT

=(-215.595.408,7)+(-86.552.619,67)+(-855.833.417,1)+(- 479.558.157,8)+( -2.107.848.235)

= -3.750.734.552Joule/jam

b. Panas Reaksi Standar

Reaksi yang terjadi :

C₆H₅CH₅CL_(l)+ NaCN_(l) C₆H₅CH₅CN_(l) + NaCl _(l)

∆H1 =

=(-14.515,4589)+(-5.827,3550)+(-16.091,977)+(-9.127,9300)+(-5.643,0403

= -51.205 J/mol

∆H2 =

=14.515,4589+5.827,3550+16.091,977+9.127,930+5.643,040

=51.205 J/mol

∆H⁰_R = H_f ⁰Produk – H_f ⁰Reaktan

= [(1 × C₆H₅CH₅CN) + (1 × NaCl )] − [(1 × C₆H₅CH₅CL) + (1 × NaCN)]

= ((117,28 kJ/mol) + (-411,2 kJ/mol)) – ((-87,5 kJ/mol) + (18,7 kJ/mol))

= -225,12 kJ/mol

= -225.120 J/mol

∆Hreaksi = ∆H1+∆H2+∆H⁰_R

₌(-51.205 + 51.205 -225.120) J/mol

= -225.120 J/mol

Panas Reaksi = ∆Hreaksi. n_A0.X_A

₌ −225.120J/mol×14,852,81 mol/jam×90,98 %

= -3.042.066.826 J/jam

c. Panas Produk (∆H_P)

T₁ = 25 ^oC = 298 K

T₂ = 100 ^oC = 373 K

∆H_P = ∑ n.Cp. dT

= 86.238.163,49+ 34.621.047,83+ 999.240.101,3+ 560.903.425,6+ 2.107.848.235

= 3.794.197.687Joule/jam

d. Neraca Panas Di Reaktor 2

Q_S =∆H_R+ Panas Reaksi + ∆H_P

=(-3.750.734.552)+ (-3.042.066.826) + 3.794.197.687) = −2.998.603.691Joule/jam

= −2.998.603,691 kJ/jam

Reaktor 3

Tabel 11. Komponen Masuk dan Keluar Reaktor 2

Komponen	Masuk kmol/jam	Keluar kmol/jam
C₆H₅CH₅CL	6,073	3,3694
NaCN	6,073	3,3694
C₆H₅CH₅CN	61,963	64,6671
NaCl	61,317	64,0207
H₂O	373,5306	373,5306

a. Panas Reaktan (∆H_R)

T₁ = 100^oC = 373 K

T₂ = 25 ^oC = 298 K

∆H_R = ∑ n.Cp. dT

=(-86.238.163,49)+(-34.621.047,83)+(-999.240.101,3)+(- 560.903.425,6)+( -2.107.848.235)

= -3.794.197.687 Joule/jam

b. Panas Reaksi Standar

Reaksi yang terjadi :

C₆H₅CH₅CL_(l)+ NaCN_(l) C₆H₅CH₅CN_(l) + NaCl _(l)

∆H1 =

=(-14.515,4589)+(-5.827,3550)+(-16.091,977)+(-9.127,9300)+(-5.643,0403)

= -51.205 J/mol

∆H2 =

=14.515,4589+5.827,3550+16.091,977+9.127,930+5.643,040

=51.205 J/mol

∆H⁰_R = H_f ⁰Produk – H_f ⁰Reaktan

= [(1 × C₆H₅CH₅CN) + (1 × NaCl )] − [(1 × C₆H₅CH₅CL) + (1 × NaCN)]

= ((117,28 kJ/mol) + (-411,2 kJ/mol)) – ((-87,5 kJ/mol) + (18,7 kJ/mol))

= -225,12 kJ/mol

= -225.120 J/mol

∆Hreaksi = ∆H1+∆H2+∆H⁰_R

₌(-74756,6636 + 74.756,6636 -225.120) J/mol

= -225.120 J/mol

Panas Reaksi = ∆Hreaksi. n_A0.X_A

= −225.120J/mol×5,9411 mol/jam×95 %

= -1.270.592.871 J/jam

c. Panas Produk (∆H_P)

T₁ = 25 ^oC = 298 K

T₂ = 140 ^oC = 373 K

∆H_P = ∑ n.Cp. dT

= 48.909.543,23+ 19.635.154,15+ 1.040.622.968+ 560.903.425,6+ 584.377.229,5

= 3.806.739.844 Joule/jam

d. Neraca Panas Di Reaktor 3

Q_S =∆H_R+ Panas Reaksi + ∆H_P

= (-3.794.197.687)+ (-1.270.592.871) + 3.806.739.844 = −1.258.050.714 Joule/jam

= −1.258.050,714 kJ/jam

TEKNIK KIMIA

Selasa, 23 Maret 2021

reaktorr

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