[...]... energy 1 kWh 1 therm 6.895 kN/m2 101 .3 kN/m2 100 kN/m 2. 99 kN/m2 2. 49 N/m2 3. 39 kN/m2 133 N/m2 0.1 N s/m2 0.414 mN s/m2 10−4 m2 /s 0 .25 8 cm2 /s 0. 126 g/s 0 .28 2 kg/s 1 .35 6 g/s m2 3. 155 5.678 2. 32 6 4.187 1. 731 W/m2 W/m2 K kJ/kg kJ/kg K W/m K 3. 6 MJ 106.5 MJ calorific value 1 Btu/ft3 1 Btu/lb 37 .26 kJ/m3 2. 32 6 kJ/kg density 1 lb/ft3 16. 02 kg/m3 SECTION 2- 1 Particulate Solids PROBLEM 1.1 The size analysis... 4000 29 40 1 620 968 20 00 16 ,20 0 30 ,24 0 40,000 41,160 29 ,160 21 ,29 6 = 21 ,968 Thus: = 180,056 ds = (180,056 /21 ,968) = 8 .20 µm This is the size of a particle with the same specific surface as the mixture The volume of a particle 8 .20 µm in diameter = (π/6)8 .20 3 = 28 8.7 µm3 4 The surface area of a particle 8 .20 µm in diameter = (π × 8 .20 2 ) = 21 1 .2 µm2 and hence: the specific surface = (21 1 .2/ 288.7) = 0. 731 ... )/π] = 2. 76 × 10 3 m 2. 1 03 × 10−8 m3 6 × 10−9 m3 0 .28 5 6 × 10−4 m2 √ [(4 × 6 × 10−4 )/π] = 2. 76 × 10 2 m 2. 1 03 × 10−5 m3 6 × 10−7 m3 0. 028 5 3 (R0 /ρu2 )Re 02 = (4k / 2 π)(ρs − ρ)ρdp g (equation 3. 52) = [(4 × 0 .28 5)/(π × 0.0 12 )] (30 00 − 820 )( 820 × 2. 1 03 × 10−8 × 9.81) = 134 0 for smallest particle and 134 ,000 for largest particle Smallest particles log10 (R0 /ρu2 )Re 02 log10 Re0 Correction from Table 3. 6... to the First Edition of Volume 5 IN THE preface to the first edition of Chemical Engineering, Volume 4, we quoted the following paragraph written by Coulson and Richardson in their preface to the first edition of Chemical Engineering, Volume 1: ‘We have introduced into each chapter a number of worked examples which we believe are essential to a proper understanding of the methods of treatment given in. .. 0.598 0. 822 0.67 0.08 0.0 536 0. 035 9 0. 024 1 0.0161 0 .37 0.17 0.0 629 0. 0 23 3 0.0086 0.0 031 9 0.1875 0. 03 0.0056 0.00105 0.00 020 0.000 037 0. 125 0.05 0.00 625 0.00078 0.000098 0.0000 12 2 .28 4 8.616 4.00 14.58 2. 00 0.75 0.50 0 .25 0. 125 Totals: 37 .991 From equation 1.11, the mass mean diameter is: dv = 4 n1 d1 3 n1 d1 = (177. 92/ 37 .991) = 4.6 83 mm From equation 1.14, the surface mean diameter is: ds = 3 n1 d1 2 n1... is: 3 x1 = n1 k1 d1 ρs nkd 3 ρs In this case: d n kd 3 nρs x 1 3 6 10 14 18 22 20 0 600 140 40 15 5 2 5 ,30 0,000k 42, 930 ,000k 80, 136 ,000k 106,000,000k 109,074,000k 77 ,27 4,000k 56, 434 ,400k 0.011 0.090 0.168 0 .22 2 0 .22 9 0.1 62 0.118 = 477,148,400k = 1.0 The surface mean diameter is given by equation 1.14: ds = 3 (n1 d1 ) 2 (n1 d1 ) and hence: d n nd 2 nd 3 1 3 6 10 14 18 22 20 00 600 140 40 15 5 2 2000... mean size of 25 mm to a product with a mean size of 1 mm 11 Solution The mean size of the product may be obtained thus: n1 d1 3 n1 d1 4 n1 d1 0.5 7.5 45.0 19.0 16.0 8.0 3. 0 1.0 12. 5 7.5 5.0 2. 5 1.5 0.75 0.40 0 .20 39 06 31 64 5 625 29 6.9 54.0 3. 375 0.1 92 0.008 48, 828 23 , 731 28 , 125 7 42. 2 81.0 2. 531 0.0768 0.0016 Totals: 13, 049 101,510 and from equation 1.11, the mass mean diameter is: dv = 4 n1 d1 3 n1 d1 =... Btu/h 0.454 kg 1016 kg 25 .4 mm 0 .30 5 m 1.609 km 60 s 3. 6 ks 86.4 ks 31 .5 Ms 645 .2 mm2 0.0 93 m2 16 ,38 7.1 mm3 0. 028 3 m3 4546 cm3 37 86 cm3 0. 138 N 4.45 N 10−5 N 1 .36 J 4.187 J 10−7 J 1.055 kJ 745 W 0 .29 3 W pressure 1 lbf /in2 1 atm 1 bar 1 ft water 1 in water 1 in Hg 1 mm Hg viscosity 1 P 1 lb/ft h 1 stoke 1 ft2 /h mass flow 1 lb/h 1 ton/h 1 lb/h ft2 thermal 1 Btu/h ft2 1 Btu/h ft2 ◦ F 1 Btu/lb 1 Btu/lb... taken into account Solution If the total drag force is proportional to the square of the velocity, then when the terminal velocity u0 is attained: 2 F = k1 u2 dm 0 2 since the area is proportional to dp 3 and the accelerating force = (ρs − ρ)gk2 dp where k2 is a constant depending on the shape of the particle and dp is a mean projected area When the terminal velocity is reached, then: 2 3 k1 u2 dp =... which gives our solutions to the problems in the third edition of Chemical Engineering, Volume 2 The material is grouped in sections corresponding to the chapters in that volume and the present book is complementary in that extensive reference has been made to the equations and sources of data in Volume 2 at all stages The book has been written concurrently with the revision of Volume 2 and SI units . CHEMICAL ENGINEERING Solutions to the Problems in Chemical Engineering Volumes 2 and 3 Related Butterworth-Heinemann Titles in the Chemical Engineering Series by J. M. COULSON & J. F. RICHARDSON Chemical. mm 2 1 lb/h 0. 126 g/s 1ft 2 0.0 93 m 2 1 ton/h 0 .28 2 kg/s 1lb/hft 2 1 .35 6 g/s m 2 volume thermal 1in 3 16 ,38 7.1 mm 3 1 Btu/h ft 2 3. 155 W/m 2 1ft 3 0. 028 3 m 3 1 Btu/h ft 2 ◦ F 5.678 W/m 2 K 1 UK gal. 150 2- 13 Liquid–liquid extraction 171 2- 14 Evaporation 181 2- 15 Crystallisation 21 6 2- 16 Drying 22 2 2- 17 Adsorption 23 1 2- 18 Ion exchange 23 4 2- 19 Chromatographic separations 23 5 Solutions to Problems