SOLUTIONS TO CHAPTER 9 EXERCISES: GAS CYCLONES

Transcription

1 SOLUTIONS TO CHAPTER 9 EXERCISES: GAS CYCLONES EXERCISE 9.1: A gas-particle separation device is tested and gives the results shown in the table below: Size range (μm) Range mean (μm) Feed mass (kg) Coarse product mass (kg) a) Find the total efficiency of the device. b) Produce a plot of the grade efficiency for this device and determine the equiprobable cut size. SOLUTION TO EXERCISE 9.1: (a) From the test results: Mass of feed, M = = 300 grams. Mass of coarse product, M c = = 185 grams. Therefore, from Text-Equation 9.5, total efficiency, E T = M c M = (or 61.7 %) (b) In this case, G(x) may be obtained directly from the results table as G(x) = m c m For example, for the size range 0-10 μm, G()= x = 0.3. For the remaining size 45 ranges: Size range (μm) G(x) Plotting this data gives x 50 = 19.4 μm, as may be seen from Solution-Manual-Figure SOLUTIONS TO CHAPTER 9 EXERCISES: GAS CYCLONES Page 9.1

2 EXERCISE 9.2: A gas-particle separation device is tested and gives the results shown in the table below: Size range (μm) feed size distribution Coarse product size distribution If the total mass of feed is 200 kg and the total mass of coarse product collected is kg, a) Find the total efficiency of the device b) Determine the size distribution of the fine product. c) Plot the grade efficiency curve for this device and determine the equiprobable size. d) If this same device was fed with a material with the size distribution below, what would be the resulting coarse product size distribution? Size range (μm) feed size distribution SOLUTION TO EXERCISE 9.2: (a) From the test results: Mass of feed, M = 200 kg and mass of coarse product, M c = kg. Therefore, from Text-Equation 9.5, total efficiency, E T = M c = (or %) M (b) Text-Equation 9.9 gives us the relationship between the size distributions of feed, coarse product and fine product: (df/dx) = E T (df c /dx) + (1-E T ) (df f /dx) Rearranging, df f dx = 1 1 E T df dx E T df c 1 E T dx = 5.97 df dx 4.97 df c dx Hence, we can calculate the fine product distribution from a knowledge of the feed distribution and the coarse product distribution: SOLUTIONS TO CHAPTER 9 EXERCISES: GAS CYCLONES Page 9.2

3 Size range (μm) Feed: df/dx Coarse: df c /dx Hence, Fine: df f /dx (c) Grade efficiency, G(x) = M c M df c dx df dx = E T df c dx df dx For example, for the size range μm, G()= x = And for the remaining size ranges: Size range (μm) Feed: df/dx Coarse: df c /dx Hence, G(x): Plotting this data gives x 50 = 10.5μm, as may be seen from Solution-Manual-Figure (d) To calculate the coarse product size distribution with the new feed to the same device, we rearrange Text-Equation 9.8 to give: df c dx = G(x) df E T dx hence, for the size range μm, df c dx = = And for the remaining size ranges: Size range (μm) New feed: df/dx G(x) G(x) / E T Hence, df c /dx SOLUTIONS TO CHAPTER 9 EXERCISES: GAS CYCLONES Page 9.3

4 EXERCISE 9.3: a) Explain what a "grade efficiency curve" is for a gas-solids separation device and sketch an example of such a curve for a gas cyclone separator. b) Determine the diameter and number of Stairmand HR gas cyclones to be operated in parallel to treat 3 m 3 /s of gas of density 0.5 kg/m 3 and viscosity 2 x 10-5 Pas carrying a dust of density 2000 kg/m 3. A x 50 cut size of at most 7 μm is to be achieved at a pressure drop of 1200 Pa. (For a Stairmand HR cyclone: Eu = 46 and Stk 50 = 6 x 10-3.) c) Give the actual cut size achieved by your design. d) A change in process conditions requirements necessitates a 50% drop in gas flowrate. What effect will this have on the cut size achieved by your design? SOLUTION TO EXERCISE 9.3: (b) From Text-Equation 9.1: Eu =Δp ( ρ f v 2 /2), with Δ p = 1200 Pa, ρ f = 0.5 kg/m 3 and Eu = 46: characteristic velocity, v = m/s Assuming that n cyclones in parallel are required and that the total flow is evenly split, then for each cyclone the flow rate will be q = Q/n = 3.0/n. Text-Equation 9.2 defines the characteristic velocity: v = 4q (πd 2 ) Substituting q and v, gives: D = n Substituting this expression for D together with the required cut size in Text-Equation 9.21, 2 ρp v Stk 50 = x 50 18μD, gives: ( ) = n hence, n = We will therefore need 2 cyclones. Now with n=2, we recalculate the cyclone diameter from D = /n 0.5 and the actual achieved cut size from Text-Equation SOLUTIONS TO CHAPTER 9 EXERCISES: GAS CYCLONES Page 9.4

5 Thus, D = / 2 = m. Using this value for D in Text-Equation 9.21 together with required cut size and v = m/s (which is dictated by Δp and the cyclone geometry), we find that the actual cut size is 6.76 μm. Therefore, two m diameter Stairmand HR cyclones in parallel will give cut size of 6.76 μm using a pressure drop of 1200Pa. (d) To determine the influence of gas flowrate, we inspect Text-Equation 9.21, which shows that if all else is constant, x and so x v q hence, if x 50 represents the new cut size and q represents the new flow rate, x 50 = q 0.5 x 50 q and so, x 50 = 6.76 = 9.56 μm. 0.5 Thus if the flow rate drop by 50%, the cut size increases to 9.56 μm. EXERCISE 9.4: (a) Determine the diameter and number of Stairmand HE gas cyclones to be operated in parallel to treat 1 m 3 /s of gas of density 1.2 kg/m 3 and viscosity 18.5x10-6 Pas carrying a dust of density 1000 kg/m 3. An x 50 cut size of at most 5 μm is to be achieved at a pressure drop of 1200 Pa. (For a Stairmand HE cyclone: Eu = 320 and Stk 50 = 1.4 x 10-4.) b) Give the actual cut size achieved by your design. SOLUTION TO EXERCISE 9.4: (a) From Text-Equation 9.1: Eu =Δp ( ρ f v 2 /2), with Δ p = 1200 Pa, ρ f = 1.2 kg/m 3 and Eu = 320: characteristic velocity, v = 2.5 m/s Assuming that n cyclones in parallel are required and that the total flow is evenly split, then for each cyclone the flow rate will be q = Q/n = 1.0/n. Text-Equation 9.2 defines the characteristic velocity: v = 4q (πd 2 ) Substituting q and v, gives: D = n Substituting this expression for D together with the required cut size in Text-Equation 9.21, SOLUTIONS TO CHAPTER 9 EXERCISES: GAS CYCLONES Page 9.5

6 Stk 50 = x 50 2 ρ p v 18μD, gives: ( ) = n hence, n = We will therefore need only one cyclone. Now with n=1, we calculate the cyclone diameter from D = /n 0.5 and the actual achieved cut size from Text-Equation (b) Thus, D = /1= m. Using this value for D in Text-Equation 9.21 together with required cut size and v = 2.5 m/s, we find that the actual cut size is 3.65 μm. Therefore, one m diameter Stairmand HE cyclone will give cut size of 3.65 μm using a pressure drop of 1200Pa. EXERCISE 9.5: Stairmand HR cyclones are to be used to clean up 2.5 m 3 /s of ambient air (density 1.2 kg/m 3 and viscosity 18.5 x 10-6 Pas) laden with dust of particle density 2600 kg/m 3. The available pressure drop is 1200 Pa and the required cut size is to be not more than 6 μm. (a) What size of cyclones are required? (b) How many cyclones are needed and in what arrangement? (c) What is the actual cut size achieved? SOLUTION TO EXERCISE 9.5: From Text-Equation 9.1: Eu =Δp ( ρ f v 2 /2), with Δ p = 1200 Pa, ρ f = 1.2 kg/m 3 and Eu = 46 (see Text-Figure 9.5): characteristic velocity, v = m/s Assuming that n cyclones in parallel are required and that the total flow is evenly split, then for each cyclone the flow rate will be q = Q/n = 2.5/n. Text-Equation 9.2 defines the characteristic velocity: v = 4q (πd 2 ) Substituting q and v, gives: D = n SOLUTIONS TO CHAPTER 9 EXERCISES: GAS CYCLONES Page 9.6

7 Substituting this expression for D together with the required cut size in Text-Equation 9.21 (noting that Stk 50 = 6 x 10-3 ), Stk 50 = x 50 2 ρ p v 18μD, gives: = ( ) n hence, n = Since the required number of cyclones is only slightly greater than 5, we will try five cyclones in parallel. With n=5, we calculate the cyclone diameter from D = 0.695/n 0.5 and the actual achieved cut size from Text-Equation Thus, D = 0.695/ 5 = m. Using this value for D in Text-Equation 9.21 together with required cut size and v = m/s (which is dictated by Δp and the cyclone geometry), we find that the actual cut size is 6.1 μm. Therefore, five m diameter Stairmand HR cyclones in parallel will give cut size of 6.1 μm using a pressure drop of 1200Pa. (Six cyclones may be used if felt necessary, in which case, D = m and x 50 = 5.75 μm). SOLUTIONS TO CHAPTER 9 EXERCISES: GAS CYCLONES Page 9.7

8 Figure 9.1.1: Grade efficiency curve for Exercise 9.1. SOLUTIONS TO CHAPTER 9 EXERCISES: GAS CYCLONES Page 9.8

9 Figure 9.2.1: Grade efficiency curve for Exercise 9.2. SOLUTIONS TO CHAPTER 9 EXERCISES: GAS CYCLONES Page 9.9