Chemical Engineering Volume 4: Solutions to the Problems in Chemical Engineering Volume 1

Nội dung text: Chemical Engineering Volume 4: Solutions to the Problems in Chemical Engineering Volume 1

1. 308 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS PROBLEM 12.16 Explain the importance of the universal velocity profile and derive the relation between the dimensionless derivative of velocity uC, and the dimensionless derivative of distance C from the surface y , using the concept of Prandtl’s mixing length E. It may be assumed that the fully turbulent portion of the boundary layer starts at C y D 30, that the ratio of the mixing length E to distance y from the surface, E/y D 0.4, and that for a smooth surface uC D 14 at yC D 30. If the laminar sub-layer extends from yC D 0toyC D 5, obtain the equation for the relation between uC and yC in the buffer zone, and show that the ratio of the eddy viscosity to the molecular viscosity increases linearly from 0 to 5 through this buffer zone. Solution The importance of the universal velocity profile is discussed in Section 12.4. From equation 12.18, for isotropic turbulence, the eddy kinematic viscosity, E˛EuE where E is the mixing length and uE is some measure of the linear velocity of the fluid in the eddies. The momentum transfer rate per unit area in a direction perpendicular to the surface at position y is then: Ry DE dux/ dy and for constant density, Ry DE dux/ dy (equation 12.20) or: Ry/ D E dux/ dy
   Ł Ł2 where Ry/, the friction velocity, may be denoted by u and then u D Edux/dy Assuming E D EuE, that is a proportionality constant of unity, and uE D Ej dux/ dyj, then: Ł2 2 u D Edux/dyjdux/dyj and hence near the surface where dux/dy is positive: Ł u D Edux/dy Assuming E D Ky: Ł u dy/y D Kdux (equation 12.28) Ł Integrating: ux/u D 1/K ln y C B where B is a constant Ł Ł 0 or: ux/u D 1/K lnyu / C B (equation 12.29) Since uŁ/ is constant, B0 is also constant and, writing the dimensionless velocity Ł C Ł C term, ux/u  as u and the dimensionless derivative of yyu / as y , then: uC D 1/K ln yC C B0 (equation 12.30) Given that K D 0.4, then: uC D 2.5lnyC C B0
2. 310 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS Solution The derivation of the Taylor–Prandtl modification of the Reynolds analogy between heat and momentum transfer is presented in Section 12.8.3 and the result is summarised as: 2 St D h/Cpus D R/us /[1 C ˛Pr  1] (equation 12.117) 2 or: R/us  D [h/Cpus][1  ˛1  Pr] For turbulent pipe flow, us is approximately equal to umean/0.82 and: 2 0.82R/u  D [h/Cpu][1  ˛1  Pr] When Re D 10,000, then from Fig. 3.1, R/u2 D 0.0038, and for Pr D 10: 0.82 ð 0.0038 D St[1  ˛1  10] or: St1 C 9˛ D 0.0031 (i) But: Nu D 0.023 Re0.8 Pr0.33 and: St D Nu/Re ÐPr D 0.023 Re0.2 Pr0.67 D 0.02310,0000.2100.67 D 0.000777 Hence, substituting in equation (i): 0.0007771 C 9˛ D 0.0031 and ˛ D ub/us D 0.33 PROBLEM 12.18 Obtain a dimensionless relation for the velocity profile in the neighbourhood of a surface for the turbulent flow of a liquid, using Prandtl’s concept of a “Mixing Length” (Universal Velocity Profile). Neglect the existence of the buffer layer and assume that, outside the laminar sub-layer, eddy transport mechanisms dominate. Assume that in the turbulent fluid the mixing length E is equal to 0.4 times the distance y from the surface and that the dimensionless velocity uC is equal to 5.5 when the dimensionless distance yC is unity. Show that, if the Blasius relation is used for the shear stress R at the surface, the thickness of the laminar sub-layer υb is approximately 1.07 times that calculated on the assumption that the velocity profile in the turbulent fluid is given by Prandtl’s one seventh power law. Blasius Equation:   R u υ 0.25 D 0.0228 s 2 us  where ,  are the density and viscosity of the fluid, us is the stream velocity, and υ is the total boundary layer thickness.
3. 312 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS may be written as: hD[1  ˛1  Sc] D h/Cp[1  ˛1  Pr] (equation 12.121) (i) The Schmidt group, Sc D /D D 1 ð 103/1000 ð 5 ð 109 D 200 3 3 The Prandtl group, Pr D Cp/k D 4 ð 10 ð 1 ð 10 /0.6 D 6.67 3 and: h/Cp D 4000/4 ð 10 ð 1000 D 0.001 Thus, in equation (i): hD[1  ˛1  200] D 0.001[1  ˛1  6.67] and: hD D 0.0011 C 5.67˛/1 C 199˛ m/s (ii) When ˛ D 0.2, then from equation (ii), 5 hD D 0.0011 C 1.134/1 C 39.8 D 5.2 ð 10 m/s When ˛ D 0.6, then from equation (ii), 5 hD D 0.0011 C 3.402/1 C 119.4 D 3.6 ð 10 m/s It is worth noting that even with a very large variation in ˛ (threefold in fact) the change in the mass transfer coefficient is less than 50%. PROBLEM 12.20 By using the simple Reynolds analogy, obtain the relation between the heat transfer coefficient and the mass transfer coefficient for the gas phase for the absorption of a soluble component from a mixture of gases. If the heat transfer coefficient is 100 W/m2 K, what will the mass transfer coefficient be for a gas of specific heat capacity 1.5 kJ/kg K and density 1.5 kg/m3? The concentration of the gas is sufficiently low for bulk flow effects to be negligible. Solution From Section 12.8.1, the heat transfer coefficient is given by: 2 R/u  D h/Cpus (equation 12.102) 2 and the mass transfer coefficient by: R/u  D hD/us (equation 12.103) Hence: hD D h/Cp (equation 12.105) 3 In this case: hD D 100/1.5 ð 10 ð 1.5 D 0.044 m/s PROBLEM 12.21 The velocity profile in the neighbourhood of a surface for a Newtonian fluid may be expressed in terms of a dimensionless velocity uC and a dimensionless distance yC from
4. 314 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS Evaluating: yC: yC RHS 10 11.26 15 12.27 20 13.00 25 13.5 8 10.7 7 10.4 12 11.71 11 11.5 11.5 11.6 11.6 11.62 and: yC D 11.6 The equation for buffer zone may be written as: uC D A ln yC C B When yC D 5,uC D 5 When yC D 30,uC D 2.5ln30C 5.5 Thus: 5 D A ln 5 C B 5.5 C 2.5ln30D A ln 30 C B Substracting: 0.5 C 2.5ln30D A ln 6 Thus: A D 0.5 C 2.5ln30/ln 6 D 5.02 and: B D 5  A ln 5 D3.08 The difference in the two values of uC is a maximum when yC D 11.6. From the two-layer theory: uC D 11.6 From the buffer-layer theory: uC D 5.02 ln C11.6  3.08 D 9.2 The maximum difference in the two values of uC is then: 2.4 PROBLEM 12.22 In the universal velocity profile a “dimensionless” velocity uC is plotted against ln yC, where yC is a “dimensionless” distance from the surface. For the region where eddy transport dominates (eddy kinematic viscosity × kinematic viscosity), the ratio of the mixing length E to the distance y from the surface may be taken as approximately constant and equal to 0.4. Obtain an expression for duC/dyC in terms of yC. In the buffer zone the ratio of duC/dyC to yC is twice the value calculated above. Obtain an expression for the eddy kinematic viscosity E in terms of the kinematic viscosity / C and y . On the assumption that the eddy thermal diffusivity EH and the eddy kinematic
5. 316 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS   C Ł 2  du u and: uŁ D C E  dyC /uŁ       duC  5 Hence: D C E D C E   dyC  yC When yC D 15, then: / D / C E/3 and: E D 2/ dT iii) For heat transfer in buffer zone: q Dk C C E  p H dy dT Writing E D E gives: q Dk C C E H p dy When yC D 15, then putting E D 2/ gives: dT dT q Dk C 2C  Dk1 C 2Pr p dy dy dT q Thus: D dy k1 C 2Pr Putting: k D 0.62 W/m K,PrD 7andq D 1000 W then: dT/dy D 108 deg K/m or 0.108 deg K/mm PROBLEM 12.23 Derive an expression relating the pressure drop for the turbulent flow of a fluid in a pipe to the heat transfer coefficient at the walls on the basis of the simple Reynolds analogy. Indicate the assumptions which are made and the conditions under which it would be expected to apply closely. Air at 320 K and atmospheric pressure is flowing through a smooth pipe of 50 mm internal diameter and the pressure drop over a 4 m length is found to be 150 mm water gauge. By how much would the air temperature be expected to fall over the first metre if the wall temperature there is 290 K? Viscosity of air D 0.018 mN s/m2. Specific heat 3 capacity Cp D 1.05 kJ/kg K. Molecular volume D 22.4m/kmol at 1 bar and 273 K. Solution If a mass of fluid, m, situated at a distance from a surface, is moving parallel to the surface with a velocity of us, and it then moves to the surface, where the velocity is zero, it will give up its momentum mus in time t. If the temperature difference between the mass of fluid and the surface is Âs, then the heat transferred to the surface is mCpÂs and over a surface of area, A: mCpÂs/t DqA where q is the heat transferred from the surface per unit area per unit time.
6. SECTION 13 Humidification and Water Cooling PROBLEM 13.1 In a process in which benzene is used as a solvent, it is evaporated into dry nitrogen. The resulting mixture at a temperature of 297 K and a pressure of 101.3kN/m2 has a relative humidity of 60%. It is required to recover 80% of the benzene present by cooling to 283 K and compressing to a suitable pressure. What must this pressure be? Vapour pressures of benzene: at 297 K D 12.2kN/m2:at283KD 6.0kN/m2. Solution See Volume 1, Example 13.1 PROBLEM 13.2 0.6m3/s of gas is to be dried from a dew point of 294 K to a dew point of 277.5 K. How much water must be removed and what will be the volume of the gas after drying? Vapour pressure of water at 294 K D 2.5kN/m2. Vapour pressure of water at 277.5KD 0.85 kN/m2. Solution 2 When the gas is cooled to 294 K, it will be saturated and Pw0 D 2.5kN/m. From Section 13.2: 3 mass of vapour D Pw0Mw/RT D 2.5 ð 18/8.314 ð 294 D 0.0184 kg/m gas. When water has been removed, the gas will be saturated at 277.5 K, and 2 Pw D 0.85 kN/m . At this stage, mass of vapour D 0.85 ð 18/8.314 ð 277.5 D 0.0066 kg/m3 gas Hence, water to be removed D 0.0184  0.0066 D 0.0118 kg/m3 gas or: 0.0118 ð 0.6 D 0.00708 kg/s Assuming the gas flow, 0.6m3/s, is referred to 273 K and 101.3kN/m2, 0.00708 kg/s of water is equivalent to 0.00708/18 D 3.933 ð 104 kmol/s. 318
7. 320 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS Hence, the mass of moist air required at the inlet conditions D 3.76 C 0.0233 D 3.783 kg/s PROBLEM 13.4 30,000 m3 of cool gas (measured at 289 K and 101.3kN/m2 saturated with water vapour) is compressed to 340 kN/m2 pressure, cooled to 289 K and the condensed water is drained off. Subsequently the pressure is reduced to 170 kN/m3 and the gas is distributed at this pressure and 289 K. What is the percentage humidity after this treatment? The vapour pressure of water at 289 K is 1.8kN/m2. Solution 2 2 At 289 K and 101.3kN/m, the gas is saturated and Pw0 D 1.8kN/m. Thus from equation 13.2, H0 D [1.8/101.3  1.8]18/MA D 0.3256/MA kg/kg dry gas, where MA is the molecular mass of the gas. At 289 K and 340 kN/m2, the gas is in contact with condensed water and therefore 2 still saturated. Thus Pw0 D 1.8kN/m and: H0 D [1.8/340  1.8]18/MA D 0.0958/MA kg/kg dry gas At 289 K and 170 kN/m2, the humidity is the same, and in equation 13.2: 0.0958/MA D [Pw/170  Pw]18/MA 2 or: Pw D 0.90 kN/m The percentage humidity is then: D [P  Pw0/P  Pw]100Pw/Pw0 (equation 13.3) D [170  1.8/170  0.90]100 ð 0.90/1.8 D 49.73% PROBLEM 13.5 A rotary countercurrent dryer is fed with ammonium nitrate containing 5% moisture at the rate of 1.5 kg/s, and discharges the nitrate with 0.2% moisture. The air enters at 405 K and leaves at 355 K; the humidity of the entering air being 0.007 kg moisture/kg dry air. The nitrate enters at 294 K and leaves at 339 K. Neglecting radiation losses, calculate the mass of dry air passing through the dryer and the humidity of the air leaving the dryer. Latent heat of water at 294 K D 2450 kJ/kg. Specific heat capacity of ammonium nitrate D 1.88 kJ/kg K. Specific heat capacity of dry air D 0.99 kJ/kg K. Specific heat capacity of water vapour D 2.01 kJ/kg K.
8. 322 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS Water entering with the air D 6.10 ð 0.007 D 0.0427 kg/s. Water evaporated D 0.07215 kg/s. Water leaving with the air D 0.0427 C 0.07215 D 0.1149 kg/s Humidity of outlet air D 0.1149/6.10 D 0.0188 kg/kg dry air. PROBLEM 13.6 Material is fed to a dryer at the rate of 0.3 kg/s and the moisture removed is 35% of the wet charge. The stock enters and leaves the dryer at 324 K. The air temperature falls from 341 K to 310 K, its humidity rising from 0.01 to 0.02 kg/kg. Calculate the heat loss to the surroundings. Latent heat of water at 324 K D 2430 kJ/kg. Specific heat capacity of dry air D 0.99 kJ/kg K. Specific heat capacity of water vapour D 2.01 kJ/kg K. Solution The wet feed is 0.3 kg/s and the water removed is 35%, or: 0.3 ð 35/100 D 0.105 kg/s If the flowrate of dry air is G kg/s, the increase in humidity D 0.02  0.01 D 0.01 kg/kg or: 0.01G D 0.105 and G D 10.5 kg/s This completes the mass balance, and the next step is to make an enthalpy balance along the lines of Problem 13.5. As the stock enters and leaves at 324 K, no heat is transferred from the air and the heat lost by the air must represent the heat used for evaporation plus the heat losses, say L kW. Thus heat lost by the inlet air and associated moisture is: [10.5 ð 0.99 C 0.01 ð 10.5 ð 2.01]341  310 D 328.8kW Heat leaving in the evaporated water D 0.105[2430 C 2.01310  324] D 252.2kW. Making a balance: 328.8 D 252.2 C L or L D 76.6kW PROBLEM 13.7 A rotary dryer is fed with sand at the rate of 1 kg/s. The feed is 50% wet and the sand is discharged with 3% moisture. The entering air is at 380 K and has an absolute humidity of 0.007 kg/kg. The wet sand enters at 294 K and leaves at 309 K and the air leaves at 310 K. Calculate the mass flowrate of air passing through the dryer and the humidity of the air leaving the dryer. Allow for a radiation loss of 25 kJ/kg dry air. Latent heat of water at 294 K D 2450 kJ/kg. Specific heat capacity of sand D 0.88 kJ/kg K. Specific heat capacity of dry air D 0.99 kJ/kg k. Specific heat capacity of vapour D 2.01 kg K.
9. 324 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS PROBLEM 13.8 Water is to be cooled in a packed tower from 330 to 295 K by means of air flowing countercurrently. The liquid flows at the rate of 275 cm3/m2 s and the air at 0.7m3/m2 s. The entering air has a temperature of 295 K and a humidity of 20%. Calculate the required height of tower and the condition of the air leaving at the top. The whole of the resistance to heat and mass transfer can be considered as being within the gas phase and the product of the mass transfer coefficient and the transfer surface per 1 unit volume of column hDa may be taken as 0.2s . Solution Assuming, the latent heat of water at 273 K D 2495 kJ/kg specific heat capacity of dry air D 1.003 kJ/kg K specific heat capacity of water vapour D 2.006 kJ/kg K then the enthalpy of the inlet air stream is: HG1 D 1.003295  273 C H 2495 C 2.006295  273 From Fig. 13.4, when  D 295 K, at 20% humidity, H D 0.003 kg/kg, and: HG1 D 1.003 ð 22 C 0.0032495 C 2.006 ð 22 D 29.68 kJ/kg In the inlet air, the humidity is 0.003 kg/kg dry air or 0.003/18/1/29 D 0.005 kmol/kmol dry air. Hence the flow of dry air D 1  0.0050.70 D 0.697 m3/m2 s. Density of air at 295 K D 29/22.4273/295 D 1.198 kg/m3. and hence the mass flow of dry air D 0.697 ð 1.198 D 0.835 kg/m2 s and the mass flow of water D 275 ð 106 m3/m2 sor275 ð 106 ð 1000 D 0.275 kg/m2 s. The slope of the operating line, given by equation 13.37 is: LCL/G D 0.275 ð 4.18/0.835 D 1.38 The coordinates of the bottom of the operating line are: ÂL1 D 295 K and HG1 D 29.7kJ/kg Hence, on an enthalpy–temperature diagram (Fig. 13a), the operating line of slope 1.38 is drawn through the point (29.7, 295). The top point of the operating line is given by ÂL2 D 330 K, and from Fig. 13a, HG2 D 78.5kJ/kg. From Figs 13.4 and 13.5 the curve representing the enthalpy of saturated air as a function of temperature is obtained and drawn in. This plot may also be obtained by calculation using equation 13.60. The integral:
   dHG/Hf  HG
10. 326 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS to HG2 D 78.5 kJ/kg is 300 K. From Fig. 13.5 the exit air therefore has a humidity of 0.02 kg/kg which from Fig. 13.4 corresponds to a percentage humidity of 90%. PROBLEM 13.9 Water is to be cooled in a small packed column from 330 to 285 K by means of air flowing countercurrently. The rate of flow of liquid is 1400 cm3/m2s and the flowrate of the air, which enters at 295 K with a humidity of 60% is 3.0m3/m2s. Calculate the required height of tower if the whole of the resistance to heat and mass transfer can be considered as being in the gas phase and the product of the mass transfer coefficient and the transfer surface per unit volume of column is 2 s1. What is the condition of the air which leaves at the top? Solution As in Problem 13.8, assuming the relevant latent and specific heat capacities: HG1 D 1.003295  273 C H 2495 C 2.006295  273 From Fig. 13.4, at  D 295 and 60% humidity, H D 0.010 kg/kg and hence: HG1 D 1.003 ð 22 C 0.0102495 C 44.13 D 47.46 kJ/kg In the inlet air, water vapour D 0.010 kg/kg dry air or 0.010/18/1/29 D 0.016 kmol/kmol dry air. Thus the flow of dry air D 1  0.0163.0 D 2.952 m3/m2s. Density of air at 295 K D 29/22.4273/293 D 1.198 kg/m3. and mass flow of dry air D 1.198 ð 2.952 D 3.537 kg/m2s. Liquid flow D 1.4 ð 103 m3/m2s and mass flow of liquid D 1.4 ð 103 ð 1000 D 1.4 kg/m2s. The slope of the operating line is thus: LCL/G D 1.40 ð 4.18/3.537 D 1.66 and the coordinates of the bottom of the line are: ÂL1 D 285 K,HG1 D 47.46 kJ/kg From these data, the operating line may be drawn in as shown in Fig. 13b and the top point of the operating line is: ÂL2 D 330 K,HG2 D 122 kJ/kg Again as in Problem 13.8, the relation between enthalpy and temperature at the interface Hf vs. Âf is drawn in Fig. 13b. It is seen that the operating line cuts the saturation curve, which is clearly an impossible situation and, indeed, it is not possible to cool the water to 285 K under these conditions. As discussed in Section 13.6.1, with mechanical draught towers, it is possible, at the best, to cool the water to within, say, 1 deg K of the wet
11. 328 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS 350 300 250 kJ/kg) G 200 H 150 Enthaolpy ( Enthaolpy 100 50 0 280 290 300 310 320 330 Temperature (q k) Figure 13c. Thus:
    HG2 height of packing,z D [dHG/Hf  HG]G/hDa (equation 13.53) HG1 D 0.35 ð 3.537/2.0 ð 1.198 D 0.52 m Due to the close proximity of the operating line to the line of saturation, the gas will be saturated on leaving the column and will therefore be at 100% humidity. From Fig. 13c the exit gas will be at 306 K. PROBLEM 13.10 Air containing 0.005 kg water vapour/kg dry air is heated to 325 K in a dryer and passed to the lower shelves. It leaves these shelves at 60% humidity and is reheated to 325 K and passed over another set of shelves, again leaving with 60% humidity. This is again reheated for the third and fourth sets of shelves after which the air leaves the dryer. On the assumption that the material in each shelf has reached the wet bulb temperature and that heat losses from the dryer can be neglected, determine: (a) the temperature of the material on each tray, (b) the rate of water removal if 5 m3/s of moist air leaves the dryer,
12. 330 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS At 295 K: Enthalpy of moist air D 0.0884 ð 0.995295  273 C 0.017 ð 0.0884 ð [4.18295  273 C 2445] D 5.75 kW At 285 K: Enthalpy of moist air D 0.0884 ð 0.995285  273 C 0.009 ð 0.0884 ð [4.18285  273 C 2468] D 3.06 kW At 275 K: Enthalpy of moist air D 0.0884 ð 0.995275  273 C 0.0045 ð 0.0884 ð [4.18275  273 C 2491] D 1.17 kW and hence in cooling from 305 to 295 K, heat to be removed D 6.89  5.75 D 1.14 kW in cooling from 295 to 285 K, heat to be removed D 5.75  3.06 D 2.69 kW in cooling from 285 to 275 K, heat to be removed D 3.06  1.17 D 1.89 kW The mass of water condensed D 0.08840.018  0.0045 D 0.0012 kg/s. The humid heats at the beginning and end of the process are: 1.082 and 1.001 kJ/kg K respectively. PROBLEM 13.12 A hydrogen stream at 300 K and atmospheric pressure has a dew point of 275 K. It is to be further humidified by adding to it (through a nozzle) saturated steam at 240 kN/m2 at the rate of 1 kg steam: 30 kg of hydrogen feed. What will be the temperature and humidity of the resultant stream? Solution At 275 K, the vapour pressure of water D 0.72 kN/m2 (from Tables) and the hydrogen is saturated. 3 The mass of water vapour: Pw0Mw/RT D 0.72 ð 18/8.314 ð 275 D 0.00567kg/m and the mass of hydrogen: P  Pw0MA/RT D 101.3  0.722/8.314 ð 275 D 0.0880 kg/m3 Therefore the humidity at saturation, H0 D 0.00567/0.0880 D 0.0644 kg/kg dry hydrogen and at 300 K, the humidity will be the same, H1 D 0.0644 kg/kg. At 240 kN/m2 pressure, steam is saturated at 400 K at which temperature the latent heat is 2185 kJ/kg. The enthalpy of the steam is therefore: H2 D 4.18400  273 C 2185 D 2715.9kJ/kg Taking the mean specific heat capacity of hydrogen as 14.6 kJ/kg K, the enthalpy in 30 kg moist hydrogen or 30/1 C 0.0644 D 28.18 kg dry hydrogen is: 28.18 ð 14.6300  273 D 11,110 kJ
13. 332 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS Temperature (K) Vapour pressure of n-butanol kN/m2 330 5.99 335 7.89 340 10.36 345 14.97 350 17.50 Solution See Volume 1, Example 13.10 PROBLEM 13.14 Estimate the height and base diameter of a natural draught hyperbolic cooling tower which will handle 5000 kg/s water entering at 300 K and leaving at 294 K. The dry-bulb air temperature is 287 K and the ambient wet-bulb temperature is 284 K. Solution See Volume 1, Example 13.8