Menghitung Panas Reaksi ∆Hr Menghitung Panas Masuk 1 Panas Masuk Alur 15 Menghitung Panas Keluar 1 Panas Keluar Alur 17

a. Menghitung Panas Reaksi ∆Hr

1 Panas Reaksi Reduksi Reaksi: SiO 2 + 3 C SiC + 2 CO ……. 1 ∆Hr 1 1600 o C = ∆H r o 1 +       1873 298 3 2 2 dT Cps Cps Cpg Cps C SiO CO SiC r 1 = 81,8841 kmoljam ∆H r o 1 = ∆H r o SiC + 2 ∆H r o CO –∆H r o SiO2 – 3 ∆H r o C = -117.230,4000 + 2 x -110.541,1580 - -851.385,7800 – 3 x 0 = 513.073,0640 Jmol ∆Hr 1 1600 o C = ∆H r o 1 +       1873 298 3 2 2 dT Cps Cps Cpg Cps C SiO CO SiC = 513.073,0640 + -43.900,5072 = 469.172,5568 Jmol 2 Panas Reaksi Pembakaran Karbon Monooksida Reaksi: CO + 2 1 O 2 CO 2 ……. 2 r 2 = 163,7683 kmoljam ∆H r o 2 = ∆H r o CO2 - ∆H r o CO – 2 1 ∆H r o O2 = -393504,7656 – -110541,1580 – 2 1 x 0 = -282.963,6076 Jmol ∆Hr 2 1600 o C = ∆H r o 2 +      1873 298 2 1 2 2 dT Cpg Cpg Cpg O CO CO = -282.963,6076 + 19.721,2229 = -263.242,3847 Jmol UNIVERSITAS SUMATERA UTARA Maka, panas reaksi total   T Hr r i i i    2 2 =   C Hr r i i i 2 2 1600    = r 1 ∆Hr 1 1600 o C + r 2 ∆Hr 2 1600 o C = 81,8841 x 469172,5568 + 163,7683 x -263242,3847 = -4.692.961,5647 kJjam

b. Menghitung Panas Masuk 1 Panas Masuk Alur 15

Panas masuk pada alur 15 Electric Furnace B-103 sama dengan panas keluar pada alur 15 Fluidized Bed Preheater B-102 Q SiO2 = 4.007.033,8625 kJjam Q C = 2.510.690,8311 kJjam Q FePO4 = 56.396,8703 kJjam Q Na2O = 88.032,5002 kJjam 2 Panas Masuk Alur 16 O 2 : Qi O2 = N 16 O2 .  303 298 2 dT Cpg O = 12.105,1121 kJjam N 2 : Qi N2 = N 16 N2  303 298 2 dT Cpg N = 57.577,1166 kJjam ∑ Qi = Q 15 + Q 16 = 6.731.836,2929 kJjam UNIVERSITAS SUMATERA UTARA

c. Menghitung Panas Keluar 1 Panas Keluar Alur 17

CO 2 : Qi CO2 = N 17 CO2 .  1673 298 2 dT Cpg CO = 11.774.571,9283 kJjam N 2 : Qi N2 = N 17 N2  1673 298 2 dT Cpg N = 37.057.546,7412 kJjam 2 Panas Keluar Alur 18 SiO 2 : Qo SiO2 = N 18 SiO2 .  1873 298 2 dT Cp SiO = 383.393,0281 kJjam C : Qo C = N 18 C .  1873 298 dT Cp C = 356.468,8006 kJjam FePO 4 : Qo FePO4 = N 18 FePO4 .  1873 298 4 dT Cp FePO = 214.863,0053 kJjam Na 2 O: Qo Na = N 18 Na2O .  1873 298 2 dT Cp O Na = 328.883,9479 kJjam SiC: Qo SiC = N 18 SiC .  1873 298 2 dT Cp SiO = 6.298.498,8766 kJjam UNIVERSITAS SUMATERA UTARA ∑ Qo = Q 17 + Q 18 = 56.414.226,3280 kJjam dQdt =   T Hr r i i i    2 2 + Qo – Qi = -4.692.961,5647 + 56.414.226,3280 – 6.731.836.2929 = 44.989.428,4704 kJjam Tabel B.9 Neraca Panas pada Electric Furnace B-103 Komponen Masuk kJjam Keluar kJjam H 15 H 16 H 17 H 18 SiO 2 4.007.033,8625 - - 383.393,0281 C 2.510.690,8311 - - 356.468,8006 FePO 4 56.396,8703 - - 214.863,0053 Na 2 O 88.032,5002 - - 328.883,9479 SiC - - - 6.298.498,8766 O 2 - 12.105,1121 - - N 2 - 57.577,1166 37.057.546,7412 - CO 2 - 11.774.571,9283 - Jumlah 6.662.154,0641 69.682,2287 48.832.118,6695 7.582.107,6585 Sub Total 6.731.836,2929 56.414.226,3280 ∆Hr 4.692.961,5647 - Q 44.989.428,4704 - Total 56.414.226,3280 56.414.226,3280 UNIVERSITAS SUMATERA UTARA B.2.6 Cooling Yard A-101 Fungsi: Menurunkan suhu produk SiC dari furnace menjadi 30 C. T: 30 o C SiC SiC C C T: 1600 o C SiO2 SiO 2 T: 30 o C FePO 4 FePO 4 Na 2 O Na 2 O T : 25 o C Untuk mendinginkan bahan keluaran furnace dari 1600 C menjadi 30 C, estimasi suhu udara pendingin 25 C, suhu udara panas keluar 30 o C. Panas Masuk Alur 18 Panas masuk pada alur 19 Cooling Yard A-101 sama dengan panas keluar pada alur 18 Electric Furnace B-103: ∑Q 19 : 7.582.107,6585 kJjam Panas Keluar Alur 20 SiO 2 : Qo SiO2 = N 20 SiO2 .  303 298 2 dT Cp SiO = 34,199.1793 kJjam C : Qo C = N

32 C

.  303 298 2 dT Cp SiO = 5,357.7440 FePO 4 : Qo FePO4 = N 32 FePO4 .  303 298 2 dT Cp SiO = 271.8152 kJjam UNIVERSITAS SUMATERA UTARA Na 2 O : Qo Na2O = N 32 Na2O .  303 298 2 dT Cp SiO = 585.6527 kJjam SiC : Qo SiC = N 32 SiC .  303 298 2 dT Cp SiO = 55,700.2387 kJjam ∑Q 20 = Qo SiO2 + Qo C + Qo FePO4 + Qo Na2O + Qo SiC = 96,114.6300 kJjam Beban Panas Udara Pendingin = ∑Q 19 - ∑Q 20 = 7,485,993.0285 kJjam Q udara = N O 2  303 298 2 dT Cp O + N N 2  303 298 2 dT Cp N 7,485,993.0285 kJjam = N O 2  303 298 2 dT Cp O + N N 2  303 298 2 dT Cp N 7,485,993.0285 kJjam = 0,21 N udara  303 298 2 dT Cp O + 0,79 N udara  303 298 2 dT Cp N N udara = 45,694.7376 kmoljam Diperoleh massa udara yang diperlukan = 1,314,444.4793 kgjam Tabel B.10 Neraca Panas pada Cooling Yard A-101 Komponen Q in kJjam Q out kJjam Umpan 7.582.107,6585 - Produk - 96.114,6300 Udara - 7.485.993,0285 Total 7.582.107,6585 7.582.107,6585 UNIVERSITAS SUMATERA UTARA B.2.7 Mixing Point M-102

a. Panas Masuk 1 Panas Masuk Alur 22 :