STRUCTURED QUESTIONS: 100 MARKS SOALAN STRUKTUR: 100 MARKAH INSTRUCTION: This section consists of FOUR(4) structured questions. Answer ALL the questions ARAHAN : Bahagian ini mengandungi EMPAT (4) soalan struktur. Jawab SEMUA soalan. QUESTION 1 SOALAN 1 CLO1 C1 a) i) State FIVE (5) dimension in basic concepts measurement. Nyatakan LIMA (5) dimensi dalam pengukuran konsep asas. [5 marks] [5 markah] ii) If the density of nitrobenzene is 1200 kg/m 3,determine the specific gravity of nitrobenzene? Jika ketumpatan nitrobenzene ialah 1200 kg/m 3, berikan spesifik gravity nitrobenzene? [3 marks] [3 markah] CLO1 C1 b) Process is a set of tasks or operation that accomplish a chemical or material transformation to produce a product. Define process unit, process streams and process variables. Proses adalah satu set tugas atau operasi yang melakukan sesuatu bahan kimia atau bahan transformasi untuk menghasilkan produk. Takrifkan unit proses, aliran proses dan pembolehubah proses. [6 marks] [6 markah] 2 SULIT
CLO1 c) A 0.70 molar solution of sulfuric acid (H 2 SO 4 ) in water flows into a reactor at a rate of 2.35 m 3 /min. The specific gravity of the solution is 1.03 (relative to water at 4 0 C). The molecular weight of H 2 SO 4 is 98. 0.70 larutan molar of asid sulfurik (H 2 SO 4 ) di dalam air mengalir ke dalam reactor pada kadar 2.35 m 3 /min. spesifik gravity untuk larutan tersebut ialah 1.03 (relative kepada air pada 4 0 C). Berat molekul H 2 SO 4 ialah 98. i) Calculate the mass concentration of H 2 SO 4 in kg/m 3 Kira kepekatan jisim H 2 SO 4 dalam kg/m 3 [2 marks] [2 markah] ii) Calculate The mass flow rate of H 2 SO 4 in kg/s Kira kadar aliran jisim H 2 SO 4 dalam kg/s [2 marks] [2 markah] iii) Calculate The mass fraction of H 2 SO 4. Kira pecahan jisim H 2 SO 4. [7 marks] [7 markah] 3 SULIT
QUESTION 2 SOALAN 2 (a) In a distillation train a hydrocarbon liquid containing 20 mole% ethane (), 30 mole % propane (C3) and 50 mole % butane (C4) is to be fractionated into essentially pure components as shown in the Figure 2(a). Dalam penyulingan berperingkat cecair hidrokarbon yang mengandungi 20 mol % etana (), 30 mol % propane (C3) and 50 mol % butane (C4) akan dipecahkan kepada komponen tulen seperti di dalam Rajah 2(a) di bawah. E(mol/s) 86% 10 % C3 4% C4 P (mol/s) 96% C3 4% C4 200 mol/s 20% 30% C3 50% C4 A(mol/s) 1 2 X of C3 Y of C4 Diagram 2(a) / Rajah2(a) B(mol/s) 8.4% C3 91.6% C4 CLO1 Determine all the unknown flow rates and compositions Tentukan semua kadar alir dan komposisi yang tidak diketahui. [10 marks] [ 10markah] (b) Antimony is obtained by heating pulverized stibnite (Sb 2 S 3 ) with scrap iron and drawing-off the molten Antimony from the bottom of reaction vessel. 40 kmol of stibnite (Sb 2 S 3 ) and 60 kmol of iron (Fe) turning are heated together to produce Sb metal. The reaction proceeds to a point where the fractional conversion limiting reactant is 85 %. Antimony dihasilkan dengan memanaskan pulverized stibnite (Sb 2 S 3 ) dengan besi skrap dan leburan antimony dikeluarkan dari bawah tiub tindak balas. 40 4 SULIT
kmol stibnite (Sb 2 S 3 ) dan 60 kmol besi(fe) dipanaskan bersama untuk menghasilkan logam Sb. Tindak balas diteruskan sehingga mencapai fractional conversion limiting reactan sebanyak 85%. Sb 2 S 3 + 3 Fe 2 Sb + 3 FeS i). Determine the limiting reactant and excess reactant Tentukan bahan tindak balas terhad dan berlebihan. ii). Determine percentage of the excess reactant. Tentukan peratus bahan tindak balas berlebihan. iii). Calculate number of mole of all the products. Kirakan bilangan mol semua hasil. [ 4 marks] [4 markah] [2 mark] [2 markah] [ 9 marks] [9 markah] QUESTION 3 SOALAN 3 (a) Figure 3(a) describes a mixing tank of three streams F 1, F 2 and F 3. Rajah 3(a) menerangkan suatu tangki campuran bagi tiga aliran. F 2 = 85 kg/h of C F 1 =150 kg/h 55 wt% A 45 wt%b F 3 =190 kg/h 72 wt% A 28 wt% C F 4 Product Figure 3(a) / Rajah 3(a) 5 SULIT
C1 (i) Write all the balance equations involved. Tuliskan semua persamaan seimbang yang terlibat. [ 4 marks] [4 markah] (ii) Calculate the mass flow rate of product F 4 and its composition (%). Kirakan jisim produk F 4 dan komposisi(%) nya. [ 4 marks] [4 markah] (iii). Calculate the mass ratio of A in product to mass of A in feed. Kirakan nisbah jisim A dalam produk kepada jisim A di dalam suapan. [ 2 marks] [ 2 markah] (b) 100 kmol/h of butane, C 4 H 8 and 4500 kmol/h of air are fed into a combustion reactor. Calculate the percentage of excess air. 100 kmol/jam butana C 4 H 8 dan 4500kmol/jam udara telah dimasukkan ke dalam reaktor pembakaran. Kirakan peratus udara berlebihan. [ 5 marks] [5 markah] C3 c) A stream at 1600 kpa exiting a fuel cell contains 5.7 mol% of H 2, 8.6 mol% of O 2, 17.1 mol% of H 2 O, and the balance being N 2. Molecular weight of each of the gas is 2.0 g/mol, 32.0 g/mol, 18.0 mol/g, and 28.0 g/mol respectively. Satu aliran gas keluar dari fuel cell pada tekanan 1600kPa mengandungi 5.7 mol% H 2, 8.6 mol% O 2, 17.1 mol% H 2 O, dan selebihya adalah gas nitrogen, N 2. i) Calculate the mole of all gases Tentukan bilangan mol semua gas. [ 2 marks] [2 markah] 6 SULIT
ii) Calculate the mass fraction of all gases in the exit stream. Tentukan pecahan jisim semua gas yang keluar. [ 4 marks] [4 markah] iii) Calculate the partial pressure of all gases. Tentukan tekanan separa semua gas. [ 4 marks] [4 markah] QUESTION 4 SOALAN 4 CLO1 (a) Calculate the gas constant in (unit m 3.Pa/mole.K) occupied by 72 liters, 100 g of nitrogen gas at 23C and 4 psi, assuming ideal gas behaviour. Molecular weight of nitrogen is 28. Kirakan nilai pemalar gas dalam (unit m 3.Pa/mol.K) yang menempati 72 liters, 100 g nitrogen pada suhu 23 C dan 4 psi dengan menganggap keadaan bersifat gas ungul. Berat molekul nitrogen ialah 28. [ 10 marks] [10 markah] C3 (b) (i) Describe the Hess s law. Terangkan Hukum Hess. [ 2 marks] [ 2 markah] (ii) Define the Potential energy. Takrifkan Tenaga Keupayaan. [ 2 marks] [ 2 markah] 7 SULIT
(iii) Give the S.I unit of potential energy. Berikan unit S.I untuk tenaga keupayaan. [ 2 marks] [ 2 markah] C3 (c ) Calculate the heat required to bring 150 mole/h of stream containing 70% of component A and 30% of component B by mole from 0C to 100C. Determine the heat capacity for the mixture as part of the problem solution. Given the heat polynomial heat capacity formula for component A and B are: Heat capacity of component A : C p ( J / mol. C) 22.64 6.2810 3 T Heat capacity of component B : C p ( J / mol. C) 20.67 12.3810 3 T Kirakan haba yang diperlukan untuk menaikkan 150mol/hr aliran yang mengandungi 60% C 2 H 6 and 40% C 3 H 8 mengikut mol daripada 0C to 400C. Tentukan muatan haba untuk campuran adalah ebahagian daripada penyelesaian masalah ini. Diberi fomula polinomial muatan haba bagi etana dan propana ialah: Mua tan Haba komponen A : C p ( J / mol. C) 22.64 6.2810 3 T Mua tan haba komponen B : C p ( J / mol. C) 20.67 12.3810 3 T [ 9 marks] [ 9 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) + 32 9 5. Standard Condition for gases system T s P s V s n s SI 273 K 1 atm 0.02245 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 T1 4. Heat of reaction ΔH = n ΔH (products) - nδh (reactants) 12 SULIT