Kỹ Thuật phân tán pha: Bài tập chương 3

Kỹ Thuật phân tán pha: Bài tập chương – Trần Minh Mẫn (MSSV:1511952) 1.(i) A powder is contained in a vessel to form a cylindrical plug 0,8cm in diameter and 3cm long The powder density is 2,5gcm-3 and 2,2g of powder was used to form the plug The porosity inside the plug The porosity inside the plug of powder is (-): a.0,58 b.0,42 c.0,75 d 0,25 1.(ii) Air was drawn through the plug at a rate of 6.6 cm3 per minute A mercury manometer was used to measure the pressure drop during this process: a pressure drop of 60 mm Hg was recorded The specific gravity of mercury is 13.6, thus the pressure drop across the plug was (Pa): a.80 b.800 c.8000 d.80000 1.(iii) The superficial gas velocity in (ii) was (ms-1): a.0,0022 b.3,65.10-5 c.2,2.10-6 d.0,00365 1.(iv) The viscocity of the air was 1,8.10-5 Pas, using the Kozeny-Carman equation, the specific surface area per unit volume of the powder was (m-1): a.2310 b.3,0.1011 c 5,5.105 d.1,2.106 1.(v) The Sauter mean diameter of the powder was (  m): a.2600 b.22 c.11 d.5 1.(vi) The air density was 1,2kgm-3, the Modified Reynolds Number of the system was (-): a.0,1 b.8,4.10-10 c.4,6.10-4 d.0,00021 Solution 1(i): The volume of the cylindrical plug:  D2 3,14  0,8cm  H  Solution 1(iii):  106 m3  6,   Q  60 s   0, 0022 ms-1 Uo    D 3,14  0,8.102 m 2 4 Solution 1(iv): Sv   P      L  KU o  1   2   0, 423     3.102 cm 1,8.105 Pa.s   0, 0022ms 1   1  0, 42 2  8000 Pa  5,5.105 m 1 Solution 1(v): 6 xSv    11 m Sv 5,5.105 m1 Solution 1(vi): Re1  dU            Sv  Uo  U o         Sv   1, 2kgm  0, 0022ms  1  0, 42   5,5.10 m  1,8.10 3  1.(vii) Comment on whether your use of the KozenyCarman equation was valid or not:  3cm   1,5072cm3 4 The total volume of the powders: m 2, g Vs  s   0,88cm3 s 2,5 gcm3 The total volume of the emties: V  Vo  Vs  1,5072cm3  0,88cm3  0, 6272cm3 The porosity inside the plug of powder: V 0, 6272cm3    0, 42 Vo 1,5072cm3 Solution 1(ii): The pressure drop: P  60mmHg  8000Pa Vo  Solution 1(vii): 1 1 5 Pa.s   4, 6.104 2.(i) A cylindrical ion exchange bed composed of spherical particles 2mm in diameter packed at abed voidage of 0,45 is to be used to deionise aliquid of density and viscosity 1100kgm-3 and 0,0075Pas respectively The design flow rate is 5m3h-1 and the bed height and diameter are and 0,2m respectively, using the KozenyCarman equation the pressure drop is (Pa): a.99000 b.1,32.107 c.4400 d.44000 Solution 2(i): Kozeny-Carman equation: P1 K U o Sv 1     L 3 Tính:  m3  5  Q  3600 s   0, 0442ms 1 Uo    D 3,14  0, 2m 2 4 6 Sv    3000m1 d 2.103 m  2m  0, 0075Pas   0, 0442ms 1  3000m 1  1  0, 45   P1  0, 453  99000 Pa 2.(ii) The Modified Reynolds Number is: a.3,91 b.4,78 c.1290 d.0,478 Solution 2(ii): 2.(iii) Comment on your use of the KozenyCarman equation: 2.(iv) The interstitial liquid velocity inside the bed is (ms-1): a.0,02 b.0,098 c.0.08 d.0,044 Solution 2(iii): 2.(v) Using the Carman correlation the shear stress on the ion exchange beads is (Pa): a.17,2 b.3,5 c.13,5 d.27 Solution 2(v): 2.(vi) Hence the dynamic pressure drop over the bed is (kPa): a.84 b.130 c.99 d.150 Solution 2(vi): 2.(vii) Why is the answer to (vi) different to that in (i)? Solution 2(vii): (i) Không kể ma sát (ii) Kể đến ma sát ??? Solution 2(viii): 2.(viii) If the liquid has a datum height equal to the position at the base of the ion exchange vessel and, therefore, needs raising to the top of the column before it enters the ion exchange bed the additional pressure drop to effect this, i.e the static pressure drop over the bed, is (kPa): a.2,16 b.21,6 c.216 d.0,22 1100kgm  0, 0442ms   3,91 U o   1    Sv  0, 0075Pas 1  0, 45  3000m1  3 Re1  1 Solution 2(iv): U Uo   0,0442ms 1  0,098ms 1 0, 45  0,  0,   R  U   0,1   1100kgm 3  0, 098 ms 1    0,1   3,91 3,91   Re1 Re1   17, Pa P2 L L L   P2  R  R R d d       1    Sv  2m  17, Pa   130kPa   0, 45   1  1  0, 45   3000m   Pstatic   gL  1100kgm3  9,81m.s 1   2m  21600Pa  21,6kPa ...  L 3 Tính:  m3  5  Q  36 00 s   0, 0442ms 1 Uo    D 3, 14  0, 2m 2 4 6 Sv    30 00m1 d 2.10 3 m  2m  0, 0075Pas   0, 0442ms 1  30 00m 1  1  0, 45   P1  0, 4 53 ...  3, 91 3, 91   Re1 Re1   17, Pa P2 L L L   P2  R  R R d d       1    Sv  2m  17, Pa   130 kPa   0, 45   1  1  0, 45   30 00m   Pstatic   gL  1100kgm 3. .. 0442ms   3, 91 U o   1    Sv  0, 0075Pas 1  0, 45  30 00m1  3 Re1  1 Solution 2(iv): U Uo   0,0442ms 1  0,098ms 1 0, 45  0,  0,   R  U   0,1   1100kgm 3  0,