# MSG Engineering Computation II [Pengiraan Kejuruteraan II]

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1 UNIVERSITI SAINS MALAYSIA Second Semester Examination 013/014 Academic Session June 014 MSG Engineering Computation II [Pengiraan Kejuruteraan II] Duration : 3 ours [Masa : 3 jam] Please ceck tat tis examination paper consists of SIX pages of printed materials before you begin te examination. [Sila pastikan baawa kertas peperiksaan ini mengandungi ENAM muka surat yang bercetak sebelum anda memulakan peperiksaan ini.] Instructions: [Araan: Answer all four [4] questions. Jawab semua empat [4] soalan.] In te event of any discrepancies, te Englis version sall be used. [Sekiranya terdapat sebarang percanggaan pada soalan peperiksaan, versi Baasa Inggeris endakla diguna pakai]. /-

2 -- [MSG 389] 1. (a) Consider a large tank olding 1000 liter of pure water into wic a salt solution begins to flow at a constant rate of 6 liter/min. Te solution inside te tank is kept well stirred, and is flowing out of te tank at a rate of 6 liter/min. If te concentration of salt in te solution entering te tank is 0.1 kg/liter, determine wen te concentration of salt will reac 0.05 kg/liter? [40 marks] Consider te initial value problem ' y x + 3xy + y = 0 y 1 = 0.5, = 0.1. Solve te initial value problem to determine y (1.3) using: (i) Euler metod (ii) Second order Taylor series metod (iii) Second order Runge Kutta metod (iv) Fourt order Runge Kutta metod (v) Euler metod as predictor metod and Heun metod as corrector metod (vi) Midpoint metod [60 marks] 1. (a) Pertimbangkan sebua tangki besar mengandungi 1000 liter air tulen di mana larutan garam mula mengalir pada kadar yang tetap 6 liter / min. Campuran di dalam tangki itu dikacau seragam, dan mengalir keluar dari tangki pada kadar 6 liter / min. Jika kepekatan garam dalam campuran yang memasuki tangki itu adala 0.1 kg / liter, tentukan masa yang diperlukan untuk kepekatan garam mencapai 0.05 kg / liter? [40 marka] Pertimbangkan masala nilai awalan ' y x xy y y = 0 1 = 0.5, = 0.1. Selesaikan masala nilai awalan ini bagi menentukan y (1.3) menggunakan: (i) Kaeda Euler (ii) Kaeda siri Taylor tertib kedua (iii) Kaeda Runge Kutta tertib kedua (iv) Kaeda Runge Kutta tertib keempat (v) Kaeda Euler metod sebagai penganggar dan kaeda Heun sebagai pembaiki (vi) Kaeda titik tenga [60 marka] 3/-

3 -3- [MSG 389]. (a) Find te approximate value of y(0.5) for te initial value problem ' y = x+ y, y 0 = 1, using te multistep metod yi+ 1 = yi 1+ ( fi fi + fi 3 wit =0.1. Compute te starting value using Runge-Kutta fourt order metod wit te same step lengt =0.1. [50 marks] Solve te initial value problem ' y = x + y 3, y( = 0, on te interval [1, 1.6] using te predictor-corrector metod, Predictor : yi+ 1 = yi + ( 3fi fi Corrector : yi+ 1 = yi + ( 5 fi fi fi 1 wit te step lengt =0.. Perform tree corrector iterations per step. Compute te starting value using Taylor series order metod wit te same step lengt. [50 marks]. (a) Dapatkan nilai anggaran y(0.5) bagi masala nilai awalan ' y = x+ y, y( 0) = 1, dengan menggunakan kaeda multi langka: yi+ 1 = yi 1+ ( fi fi + fi 3 dengan =0.1. Dapatkan nilai awalan menggunakan kaeda Runge-Kutta tertib keempat dengan saiz langka yang sama =0.1. [50 marka] Selesaikan masala nilai awalan ' y = x + y 3, y( = 0, dalam selang [1, 1.6] menggunakan kaeda pengangar dan pembaiki, Penganggar : yi+ 1 = yi + ( 3fi fi Pembaiki : yi+ 1 = yi + ( 5 fi fi fi 1 dengan saiz langka =0.. Lakukan tiga langka pembaiki bagi setiap lelaran. Dapatkan nilai permulaan dengan mengunakan kaeda siri Taylor dengan saiz langka, yang sama. [50 marka] 4/-

4 -4- [MSG 389] 3. (a) Set up te equations for a value of u tat satisfies te Laplace equation of a regular plate wit 0 x 4 and 0 y. Use te conditions: x= y = 1 T( 0, y) = 5 T( x ) T x,0 = 0 and, = 10 T = 3 at x = 4 [40 marks] Solve te boundary value problem " ' y 4y + 3y = 0, y 0 = 1, y 1 = 0. using second order finite difference metod wit (i) =1/ (ii) =1/3. " 1 ' 1 [Hint: yi = ( yi 1 yi + yi+, yi = ( yi+ 1 yi ] [60 marks] 3. (a) Bangunkan persamaan untuk nilai u yang memenui persamaan Laplace bagi sata biasa dengan 0 x 4 dan 0 y. Gunakan syarat : x= y = 1 T( 0, y) = 5 T x,0 = 0 dan T x, = 10 T = 3 pada x = 4 [40 marka] Selesaikan masala nilai sempadan " ' y 4y + 3y = 0, y 0 = 1, y 1 = 0. mengunakan kaeda beza teringga peringkat kedua dengan (i) =1/ (ii) =1/3. " 1 ' 1 [Petunjuk: y = ( y 1, ( 1 i i y + y y y y i i + = i i + i ] [60 marka] 5/-

5 -5- [MSG 389] 4. (a) Te partial differential equation of te temperature in a long tin rod is given by T T = α. t If α = 0.8 cm / s, te initial temperature of rod is 40 C, and te rod is divided into tree equal segments, find te temperature at node 1 (using t = 0.1s) for t=0. sec by using: (i) Crank-Nicolson metod (ii) Forward Time Centered Space (FTCS) sceme node = T = 80 o C T = 0 o C 9 cm [70 marks] A typical steady-state eat flow problem is te following: A tin steel plate is a unit square in m. If two of its edges are eld at 0 C and te oter two are eld at te temperatures sown below: u u u = + = 0, 0 < x< 1, 0 < y< 1. y u 0, x = 100x u y,0 = 100y u x,1 = 100x u 1, y = 100y Write down te five point formula for te following coordinates: (1,, (,, (3,, (1,), (,), (3,), (1,3), (,3) and (3,3). [30 marks] 6/-

6 -6- [MSG 389] 4. (a) Masala pembezaan separa bagi suu suatu rod nipis yang panjang diberikan ole T T = α. t Sekiranya α = 0.8 cm / s, suu awalan rod adala 40 C, dan rod dibaagikan kepada tiga baagian yang seragam, dapatkan suu di nod 1 (mengunakan t = 0.1s) untuk t=0.s dengan menggunakan: (i) (ii) Kaeda Crank- Nicolson Skema Beza Ke Depan Teradap Masa Beza Pusat Teradap Ruang nod = T = 80 o C T = 0 o C 9 cm [70 marka] Aliran aba kaeda mantap biasa diberikan sebagai : Satu plat besi yang nipis adala satu unit persegi dalam m. Sekiranya dua pengujuna diletakkan pada suu 0 C dan dua pengujung lagi diletakkan dalam suu berikut : u u u = + = 0, 0 < x< 1, 0 < y< 1. y u( 0, x) = 100x u( y,0) = 100y u( x, = 100x u 1, y = 100y Tuliskan formula lima titik untuk koordinat berikut: (1,, (,, (3,, (1,), (,), (3,), (1,3), (,3) dan (3,3). [30 marka] -ooo000ooo-

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