TUGAS 1PEMODELAN MATEMATIS DAN KOMPUTASI C

DISUSUN OLEH :

HANI PRANOWO JATI (11/319084/TK/38217)

JURUSAN TEKNIK KIMIAUNIVERSITAS GADJAH MADA

2014

Diketahui : Sebuah Menara Absorber

Ditanya : a. Yaout? b. Jika diinginkan 90% CO2 hilang, berapakah tinggi menara?

c. Sama dengan kasus b, jika tinggi menara tetap, berapakah perubahan dimensi menara?

Jawab : Persamaan yang dipakai :

dYadz

=−kya×sG

× (Y−Y * )

dXadz

= kya×sL

× (Ya−Y * )− ra×sL

dXbdz

= ra×sL

Dengan ra=k ×Ca×Cb

Y * = H × Xa

Ca= Xa×ρMr

Cb= Xb× ρMr

Xain = ?Xbin =?L=1350 kmol/jamP=20 atmT= 320C

Yaout =?

Yain=0,15G=3280 kg/jam

Xaout = ?Xbout =0,25Xcout = ?

Z + ∆Z

Z

k = 2,07x109 exp(-5912,7/T) (m3/kmol.detik) Kya = 47,1144 kmol A / m3.detik Tetapan Henry, H = 0,1048 Diameter menara = 1,68 m Tinggi menara = 8 mDensitas larutan penyerap = 60 kmol/m3

Simulasi :clear clc// DataG = 3280 YGas = 0.85YCO2awal = 0.15L = 1350 XSolvent = 0.50XMDEAakhir = 0.25

// data reaksi didalam absorberT = 32+273.15 k = 2.07*10^9* exp(-5912.7/T)rhoH2O = 1000mrH2O = 18Ct = rhoH2O/mrH2Okya = 47.1144H = 0.1048

// data data fisik absorberd = 1.68z =8S = %pi/4*(d^2)

mprintf(' Tugas 1 PEMATKOM C\n')mprintf(' Kasus :\t Absorbsi Reaktif CO2 \n\t\t menggunakan solven MDEA\n')mprintf(' Nama :\t Hani Pranowo Jati\n')mprintf(' NIM :\t 11/319084/TK/38217\n')mprintf('=====================================================\n')

function dx=f(z, m) YCO2 = m(1) XCO2 = m(2) XMDEA =m(3) Ys = H*m(2) ra = k*m(2)*m(3)*Ct^2 dx(1)=-kya*S*(m(1)-Ys)/G dx(2)=(-ra+kya*(m(1)-Ys))*S/L dx(3)=S/L*raendfunctionm0=[YCO2awal 0 XMDEAakhir]'Z=0:0.5:100Z0=0m=ode(m0,Z0,Z,f)mprintf ("Z absorber\t YCO2\t\t XCO2\t\t XMDEA\n")mprintf ("-------\t\t -------\t -------\t -------\n")m1=m(1,:)m2=m(2,:)m3=m(3,:)mprintf("%4.4f\t\t %4.4f\t\t %4.4f\t\t %4.4f\t\t \n",[Z' m1' m2' m3'])clf()

plot(Z,m2)//soal Amprintf ("jadi konsentrasi gas CO2 keluar dari menara sebesar %4.4f \n",[m1(17)])

// soal B// Jika ingin dihilangkan 90% gas CO2, makaYO = YCO2awal*(0.1)// Dilakukan trial ketinggian absorber sampai 100 meter dengan delta z sebesar 0.5 dan di cari di ketinggian berapa yang menghasilkan nilai YOmprintf ("jadi ketinggian arbsorber harus %4.4f \n",[Z(146)])

//soal Cclear clc// DataG = 3280 YGas = 0.85YCO2awal = 0.15L = 1350 XSolvent = 0.50XMDEAakhir = 0.25

// data reaksi didalam absorberT = 32+273.15 k = 2.07*10^9* exp(-5912.7/T)rhoH2O = 1000mrH2O = 18Ct = rhoH2O/mrH2Okya = 47.1144H = 0.1048

// data data fisik absorber

mprintf(' Tugas 1 PEMATKOM C\n')mprintf(' Kasus :\t Absorbsi Reaktif CO2 \n\t\t menggunakan solven MDEA\n')mprintf(' Nama :\t Hani Pranowo Jati\n')mprintf(' NIM :\t 11/319084/TK/38217\n')mprintf('=====================================================\n')

d =input("Masukkan nilai diameter = ")z = 8S = %pi/4*(d^2)function dx=f(z, m) YCO2 = m(1) XCO2 = m(2) XMDEA =m(3) Ys = H*m(2) ra = k*m(2)*m(3)*Ct^2 dx(1)=-kya*S*(m(1)-Ys)/G dx(2)=(-ra+kya*(m(1)-Ys))*S/L

dx(3)=S/L*raendfunctionm0=[YCO2awal 0 XMDEAakhir]'Z=0:1:8Z0=0m=ode(m0,Z0,Z,f)m1=m(1,:)m2=m(2,:)m3=m(3,:)

mprintf('Z\t\t YCO2\t\t\XCO2\t\t\XMDEA\n')for i=1:length(Z) Ynew=0.1*YCO2awal K=m'mprintf(" %2.4f \t %2.4f \t %2.4f \t %2.4f\n",[,Z(i)',K(i,1),K(i,2),K(i,3)])if K(i,1)<=Ynew then breakendendmprintf('=============================================================\n')