STRUCTURE : 100 MARKS STRUKTUR : 100 MARKAH INSTRUCTION: This section consists of FOUR (4) structured questions. Answer ALL questions. ARAHAN: Bahagian ini mengandungi EMPAT (4) soalan berstruktur. Jawab SEMUA soalan. QUESTION 1 SOALAN 1 C1 a) Unit is the means of expressing the dimensions, such as feet or centimeters for length, and hours or seconds for time. Dimensions and their respective units are classified as fundamental (basic) or derived. For the following dimensions, state the unit and identify whether it is base, multiple or derived unit: Unit adalah cara untuk menyatakan dimensi, seperti kaki atau sentimeter untuk panjang, dan jam atau saat untuk masa. Dimensi dan unit masingmasing dikelaskan sebagai asas atau derived. Untuk dimensi yang berikut, nyatakan unit dan kenal pasti sama ada ianya adalah unit asas, pelbagai atau terbitan: i. Volume. [2 Marks] Isipadu. ii. Density. [2 Marks] Ketumpatan. C2 b) A liquid has a specific gravity (SG) of 0.70 and flowing at a rate of 100 g/min which contains 0.30 mole fraction of benzene (MW = 78 g/mol) and the balance 0.70 mole fraction is toluene (MW = 92 g/mol). Calculate: 2 SULIT
Suatu cecair mempunyai gravity tentu (SG) 0.70 dan mengalir pada kadar 100 g/min mengandungi 0.30 pecahan mol benzene (MW = 78 g/mol) dan selebihnya ialah 0.70 pecahan mol toluene (MW = 92 g/mol). Kirakan: i. the mass of 30.0 L of this liquid. [5 Marks] Jisim bagi 30.0 L cecair ini. [5 Markah] ii. the molar flow rate of each component in the stream. [5 Marks] kadar aliran molar bagi setiap komponen dalam aliran [5 Markah] C3 c) If 100.00 g of air consists of 77.00% by weight of nitrogen (N 2, molecular weight = 28 g/mol) and 23.00% by weight of oxygen (O 2, molecular weight = 32 g/mol). The total pressure of air is 1.50 atm and the temperature is 25 C. Calculate: Jika 100.00 g udara terdiri daripada 77.00% mengikut berat nitrogen (N 2, berat molekul = 28 g/mol) dan 23.00% mengikut berat oksigen (O 2, berat molekul = 32 g/mol). Jumlah tekanan udara ialah 1.50 atm dan suhu ialah pada 25 C. Kirakan: i. the total of mixture. [5 Marks] jumlah mol campuran. [5 Markah] ii. the mole fraction of oxygen. [2 Marks] pecahan mol oksigen. iii. the concentration of oxygen in mol/m 3. [4 Marks] kepekatan oksigen dalam mol/m 3. [4 Markah] 3 SULIT
QUESTION 2 SOALAN 2 C1 a) Define the followings: Definisikan yang berikut: i. Fractional conversion. [2 Marks] Pecahan penukaran. ii. Selectivity. [2 Marks] Keterpilihan. C2 b) The feed to a continous ammonia formation reactor is 200 mol/s of nitrogen, 600 mol/s of hydrogen and 1.00 mol/s of argon (inert gas). The percentage conversion of the hydrogen in the reactor is 70.00%. Calculate: Suapan kepada reaktor pembentukan ammonia berterusan ialah 200 mol/s nitrogen, 600 mol/s hidrogen dan 1.00 mol/s argon (gas lengai). Peratus penukaran hidrogen di dalam reaktor ialah 70.00%. Kirakan: N 2 + 3H 2 2NH 3 i. the extent of the reaction. [8 Marks] takat tindak balas. [8 Markah] C3 ii. c) The reaction: the molar flow rate of each species as it exits the reactor. kadar aliran molar bagi setiap spesies yang keluar daripada reaktor [3 Marks] [3 Markah] CH 4 + O 2 CH 4 + 2O 2 HCHO + H 2 O CO 2 + 2H 2 O takes place in a plug flow reactor at steady state. The feed to a plug flow reactor contains equimolar amounts of methane and oxygen. Assuming a basis of 100 mol/s feed. The fractional conversion of methane is 0.90 and the fractional yield of formaldehyde is 0.855. Methane (CH 4 ) and oxygen are 4 SULIT
reacting in the presence of catalyst to form formaldehyde (HCHO). In a parallel reaction, methane is oxidized to carbon dioxide and water. Calculate: Tindak balas: CH 4 + O 2 HCHO + H 2 O CH 4 + 2O 2 CO 2 + 2H 2 O berlaku dalam sebuah reactor aliran palam pada keadaan seragam. Suapan kepada reactor aliran palam mengandungi jumlah molar yang sama bagi metana dan oksigen. Andaikan suapan ialah 100 mol/s. Pecahan penukaran metana adalah 0.90 dan hasil pecahan formaldehid adalah 0.855. Metana (CH 4 ) dan oksigen bertindak balas dalam kehadiran pemangkin untuk membentuk formaldehid (HCHO). Dalam satu tindak balas yang sama, metana dioksidakan kepada karbon dioksida dan air. Kirakan: i. the extent of the first and second reactions. [7 Marks] takat tindak balas pertama dan kedua. [7 Markah] ii. the molar of the product gas. [3 Marks] molar bagi produk gas. [3 Markah] QUESTION 3 SOALAN 3 C1 a) 100 mol/hr of butane (C 4 H 10 ) and 5000 mol/hr of air are fed into a combustion reactor. Calculate the percentage of excess air. 100 mol/hr butane (C 4 H 10 ) dan 5000 mol/hr udara dimasukkan ke dalam reaktor pembakaran. Kirakan peratus udara berlebihan. [4 Marks] [4 Markah] 5 SULIT
C2 b) The reaction: 2C 2 H 4 + O 2 2C 2 H 4 O takes place in a reactor. 100 kmol C 2 H 4 and 100 kmol O 2 are fed to a reactor where they combine to produce ethylene oxide. Determine: Tindak balas: 2C 2 H 4 + O 2 2C 2 H 4 O berlaku dalam sebuah reaktor. 100 kmol C 2 H 4 dan 100 kmol O 2 disuap ke dalam reactor untuk menghasilkan etena oksida. Tentukan: i. the limiting reactant. [2 Marks] reaktan penghad. ii. the percentage excess of the excess reactant. [2 Marks] komposisi produk berdasarkan asas mol. iii. the excess reactant will be left out and C 2 H 4 O will be formed if the reaction proceeds to completion. reaktan berlebihan yang tinggal dan C 2 H 4 O yang terhasil sekiranya tindakbalas telah lengkap. [6 Marks] [6 Markah] C3 c) Hexane (C 6 H 14 ) at 320 o C and 4.61 bar flows into a reactor at a rate of 1200 kg/hr. Given the molecular weight of hexane is 86. Calculate the volumetric flow rate of this stream in m 3 /hr by using: Heksana (C 6 H 14 ) pada 320 o C dan 4.61 bar mengalir ke dalam reaktor pada kadar 1200 kg/jam. Diberi berat molekul heksana ialah 86. Kirakan kadar alir isipadu dalam m 3 /jam dengan menggunakan: i. the ideal gas equation of state. [8 Marks] persamaan ideal gas. [8 Markah] 6 SULIT
ii. the conversion from standard conditions. [3 Marks] penukaran daripada keadaan piawai. [3 Markah] QUESTION 4 SOALAN 4 C1 a) Define the followings: Definasikan yang berikut: i. Hess s Law. [2 Marks] Hukum Hess. ii. Heat of formation. [2 Marks] Haba pembentukan. C2 b) Calculate the mole required if 25000 kj of heat is used to raise nitrous oxide temperature from 20 C to 25 C. The constant volume heat capacity of N 2 O in this temperature range is given by the equation: Kirakan mol yang diperlukan sekiranya 25000 kj haba diperlukan untuk menaikkan suhu daripada 20 C kepada 25 C. Isipadu muatan haba tentu N 2 O untuk julat suhu terbabit diberi dalam persamaan: C v (kj/kg C) = 2T + 3T 2 where T is in C [10 Marks] [10 Markah] C3 c) Calculate the heat rate required to bring 200 mol/h of a stream containing 70.00% of pentane, C 5 H 12 and 30.00% of hexane, C 6 H 14 by mole from 0 C to 500 C. Given the polynomial heat capacity formula for pentane and hexane are: 7 SULIT
Heat capacity of pentane: C p (kj/mol C) = 114.8 x 10-3 + 34.09 x 10-5 T 18.99 x 10-8 T 2 + 42.26 x 10-12 T 3 Heat capacity of hexane: C p (kj/mol C) = 137.44 x 10-3 + 40.85 x 10-5 T 23.92 x 10-8 T 2 + 57.66 x 10-12 T 3 Kirakan kadar haba yang diperlukan untuk membawa 200 mol/h aliran yang mengandungi 70.00% pentana, C 5 H 12 dan 30.00% heksana, C 6 H 14 mengikut mol daripada 0 C kepada 500 C. Diberi formula polinomial muatan haba bagi pentana dan heksana adalah: Muatan haba pentana: C p (kj/mol C) = 114.8 x 10-3 + 34.09 x 10-5 T 18.99 x 10-8 T 2 + 42.26 x 10-12 T 3 Muatan haba heksana: C p (kj/mol C) = 137.44 x 10-3 + 40.85 x 10-5 T 23.92 x 10-8 T 2 + 57.66 x 10-12 T 3 [11 Marks] [11 Markah] SOALAN TAMAT 8 SULIT
Appendix 1 Table of Unit Conversions Quantity Mass Equivalent Values 1 kg = 1000 g = 0.001 metric ton = 2.20462 Ib m = 35.27392 oz 1 Ib m = 16 oz = 5 X 10-4 ton = 453.593 g = 0.453593 kg Length 1 m = 100 cm = 1000 mm = 10 6 microns ( m ) = 10 10 angstroms (A) = 39.37 in. = 3.2808 ft = 1.0936 yd = 0.0006214 mile 1 m 3 = 1000 liters = 10 6 cm 3 = 10 6 ml Volume = 35.3145 ft 3 = 220.83 imperial gallons = 264.17 gal = 1056.68 qt 1 ft 3 = 1728 in 3 = 7.4805 gal = 0.028317 m 3 = 28.317 liters = 28 317 cm 3 1 N = 1 kg.m/s 2 = 10 5 dynes = 10 5 g.cm/s 2 = 0.22481 Ib f Force 1 Ib f = 32.174 Ibm.ft/s 2 = 4.4482 N = 4.4482 X 10 4 dynes Pressure 1 atm = 1.01325 x 10 5 N/m 2 (Pa) = 101.325 kpa = 1.01325 bars = 1.01325 x 10 6 dynes/cm 2 = 760 mm Hg at 0 C (torr) = 10.333 m H 2 O at 4 C = 14.696lb f /in 2 (psi) = 33.9 ft H 2 O at 4 C = 29.921 in Hg at 0 C 1 J = 1 N.m = 10 7 ergs = 10 7 dyne.cm Energy = 2.778 x 10-7 kw.h = 0.23901 cal = 0.7376 ft-lb f = 9.486 x 10-4 Btu Power 1 W = 1J/s = 0.23901 cal/s = 0.7376 ft.lb f /s = 9.468 x 10-4 Btu/s = 1.341 x 10-3 hp 9 SULIT
Appendix ll FORMULAS & EQUATIONS CHAPTER 1 1. W = mg 2. g = 9.8066 m/s 2 = 980.66 cm/s 2 = 32.174 ft/s 2 3. Kinetic Energy = ½ mv 2 4. Potential Energy = mgh CHAPTER 2 1. SG = ρ / ρ ref 2. ρ ref ( H 2O, 4 C) = 1.000 g/cm 3 = 1000 kg/m 3 = 62.43 Ib m/ft 3 3. ρ = m/v 4. Avogadro s Number = 6.02 X 10 23 m 5. Mass Fraction, x and m Total Mole Fraction, y n n total CHAPTER 3 1. General Balance Equation for steady state process: input + generation = output + consumption 2. Frational ( fed ) excess ( fed ) 3. percentage excess 100% 4. fractional conversion, f ( Fed ) 5. % fractional conversion 100% ( Fed ) 6. Yield ( desired product) ( LR) stoichiometry coefficien t stoichiometry coefficien t ( LR) ( DP) 100% 10 SULIT
7 Selectivit y ( desired product) ( undesired product) 8. Overall conversion = reactant input to the process - reactant output from process reactant input to process 9. Single-pass conversion = reactant input to reactor - reactant output from reactor reactant input to reactor 10. Percentage excess air = ( air) fed ( air) theoretical X 100 % ( air) theoretical 11. 100 mol air 79 mol nitrogen 21 mol oxygen CHAPTER 4 1. Ideal gas law : PV = nrt : PV nt PV : 1 1 T1 PV n T P2V 2 T2 s s s s 2. P absolute = P atmospheric + P gauge 3. Gas constant, R = 8.314 m 3.Pa / mol.k = 0.08314 liter.bar / mol. K =0.08206 liter.atm/mol.k = 63.36 liter.mm Hg/mol.K = 0.7302 ft 3.atm/Ib-mole. R = 10.73 ft 3.psia / Ib-mole. R = 8.314 J/mol.K = 1.987 cal/mol.k = 1.987 Btu / Ib-mole. R 4. T (K) = T ( C ) + 273 T ( R) = T ( F) + 460 T( F) = 5 T( C) 9 + 32 5. Standard Condition for gases system T s P s V s n s SI 273 K 1 atm 0.022415 m 3 1 mol 11 SULIT
6. Vs/ns = 0.0224 m 3 (STP)/mol = 22.4 liters(stp)/mol = 359 ft 3 (STP)/Ib-mole CHAPTER 5 1. First Law of Thermodynamics for closed system: ΔU + ΔE kinetic + ΔE potential = Q + W 2. Energy balance for closed system: Q = ΔU = m ΔŨ T2 3 Specific internal energy, U Cv( T) dt T 1 4 Heat of reaction ΔH = n ΔH (products) - nδh (reactants 12 SULIT