UNIVERSITAS KATHOLIK PARAHYANGAN

ICE 202 Komputasi Teknik Kimia

Soal Ujian Praktikum Komputasi Teknik Kimia ke 2

Waktu : 60 menit

Sifat : Open Book, catatan, dan file milik sendiri

Petunjuk umum:

1. Waktu pengerjaan selama 1 jam dan soal berikutnya hanya boleh dikerjakan bila soal sebelumnya telah

selesai dikerjakan. TIDAK DIPERKANANKAN UNTUK BEKERJASAMA ATAU COPY-PASTE

JAWABAN PESERTA LAIN.

2. Jawaban dan soal ditulis kembali dalam Microsoft Word yang berisi scipt MATLAB, dan apa yang ditanyakan

dalam soal (Grafik, jawaban angka, dsb). Sertakan script MATLAB di akhir jawaban. Sertakan hanya hasil akhir

(hasil akhir hitungan MATLAB, grafik dsb). Buatlah dokumen Microsoft Word yang rapi dan mudah dibaca.

3. Sertakan script matlab. Kumpulkan jawaban dalam sebuah folder. Beri nama folder dengan format KUIS 2

NAMA NRP. Contoh: TUGAS KUIS 2 Budi 6213200. Kemudian simpanlah folder dan isinya dalam bentuk ZIP

ataupun rar.

4. Semua file jawaban dikirim dalam bentuk Rar ataupun Zip yang didapat dari folder (lihat petunjuk nomor (4).

Kirimkan jawaban melalui email ke leonardoarief@yahoo.com dengan format subjek email KUIS 2 KOMPUT

UNPAR 1415 NAMA NRP. Contoh: KUIS 2 KOMPUT UNPAR 1415 Budi 6213200.

5. Selamat mengerjakan. Semoga berhasil.

1. Solve the following problem with the fourth-order RK method:

+ 0.5

+ 7 = 0

where y(0)=4 and y(4)=3.5030. Solve from x=0 to 5 with h=0.5. Plot your results.

2. Use the fourth-order RK method to solve

= 2 + 4

=

over the range x=0 to 1 using a step size of 0.2 with y(0)=2 and z(0.8)=1.3924. Plot your results.

3. Solve the following set of differential equations using RK method, assuming that at x=0, y1=4, and at

x= 0.5, y2=6.8577. Integrate to x=2 with a step size of 0.5. Plot your results.

= 0.5

= 4 0.3 0.1

UNIVERSITAS KATHOLIK PARAHYANGAN

ICE 202 Komputasi Teknik Kimia

Soal Ujian Praktikum Komputasi Teknik Kimia ke 2

Waktu : 60 menit

Sifat : Open Book, catatan, dan file milik sendiri

4. The following equations define the concentrations of three reactants:

= 10 +

= 10

= 10 + 2

If the initial condition is ca=50, cb=0, and cc=40, find the concentrations for the times from 0 to 3 s.

Plot your results.

5. Suppose that the position of a falling object is governed by the following differential equation,

+

= 0

where c= a first-order drag coefficient=12.5 kg/s, m=mass=70 kg, and g=gravitational

acceleration=9.81 m/s2. Use the shooting method to solve this equation for position and velocity

given the boundary conditions, x(0)=0 and x(12)=500. Plot your results.

6. The motion of a damped spring-mass system (Fig. P25.16) is described by the following ordinary

differential equation:

+

+ = 0

where x=displacement from equilibrium position (m), t=time (s), m=20-kg mass, and c=the damping

coefficient (N.s/m). The damping coefficient c takes on three values of 5 (under-damped), 40

(critically damped), and 200 (over-damped). The spring constant k=20 N/m. The initial velocity is zero,

and the initial displacement x=1 m. Solve this equation using a numerical method over the time

period 0t15 s. Plot the displacement versus time for each of the three values of the damping

coefficient on the same curve.