Centrifugation Calculation

TUTORIAL FST359 CHAPTER 2: CENTRIFUGE QUESTION 1 In a centrifuge operating at 70 L/minutes, 50% of the food particles required size was reduced to 15 µm (d). Given: Particle density (ps) Liquid density (pL) Liquid viscosity ()

= 1155 kg/m3 = 1053 kg/m3 = 1.5 Poise

i) Calculate the feasibility value of the centrifuge.

(5 marks) ii) By using the information shown in Table 1 below, decide if this operation is feasible to be carried out in a tubular, disc, or decanter centrifuge.

(1 mark) Table 1: List of centrifuge and its feasibility value Types of Centrifuge Tubular Disc Decanter Q liter/minute 10-100 2-2,000 15-2,300 -8 -8 Q/∑ m/s 5-35 x 10 8-50 x 10 500-150,000 x 10-8 ANSWER Step 1: Convert  into

kg value ms

u  liquid vis cos ity u  1.5Poise g 1kg 100cm u  1.5   cm  s 1000g 1m kg u  0.15 ms

Step 2: convert d into m value d  15m d  15  106 m

Step 3: use this formula u 

gd 2  s   l  18

gd 2  s   l  u 18

9.8115  106  1155  1053 18  0.15 m u  8.338  108 s 2

u

Step 4: use this formula Q  2 u

Q  2 u

Q  2u  Q m  28.338  108   s Q m  16.67  108  s Therefore, the operation is feasible for both tubular and disc bowl centrifuges

QUESTION 2 In a centrifuge operating at 120 L/minutes, 50% of the food particles required size was reduced to 20 µm (d). Given: Particle density (ps) Liquid density (pL) Liquid viscosity ()

= 1750 kg/m3 = 1250 kg/m3 = 2.5 Poise

i) Calculate the feasibility value of the centrifuge.

(5 marks) ii) By using the information shown in Table 1 below, decide if this operation is feasible to be carried out in a tubular, disc, or decanter centrifuge.

(1 mark) Table 1: List of centrifuge and its feasibility value Types of Centrifuge Tubular Disc Decanter Q liter/minute 10-100 2-2,000 15-2,300 Q/∑ m/s 5-35 x 10-8 8-50 x 10-8 500-150,000 x 10-8 ANSWER Step 1: Convert  into kg/m.s value  = 0.25 kg/m.s Step 2: convert d into m value d = 20 x 10-6 m Step 3: use this formula u = g d2 (s-l)/18µ u = 43.6 x 10-8 m/s Step 4: use this formula Q = 2 ∑ u Q/ = 87.2 x 10-8 m/s Therefore, the operation is not feasible for all tubular, decanter and disc bowl centrifuges.

QUESTION 3 In a centrifuge operating at 20 L/minutes, 50% of the food particles required size was reduced to 18 µm (d). Given: Particle density (ps) Liquid density (pL) Liquid viscosity ()

= 1050 kg/m3 = 930 kg/m3 = 0.5 Poise

i) Calculate the feasibility value of the centrifuge.

(5 marks) ii) By using the information shown in Table 1 below, decide if this operation is feasible to be carried out in a tubular, disc, or decanter centrifuge.

(1 mark) Table 1: List of centrifuge and its feasibility value Types of Centrifuge Tubular Disc Decanter Q liter/minute 10-100 2-2,000 15-2,300 Q/∑ m/s 5-35 x 10-8 8-50 x 10-8 500-150,000 x 10-8 ANSWER Step 1: Convert  into kg/m.s value  = 0.05 kg/m.s Step 2: convert d into m value d = 18 x 10-6 m Step 3: use this formula u = g d2 (s-l)/18µ u = 42.3 x 10-8 m/s Step 4: use this formula Q = 2 ∑ u Q/ = 84.6 x 10-8 m/s Therefore, the operation is not feasible for all tubular, decanter and disc bowl centrifuges.

QUESTION 4 A 1000 L tank containing 20% salt solution (density 1130 kg/m3 (l)) at 24oC was mixed with 17% fine sand (0.3 mm (d) diameter and density 2010 kg/m3 (s)). If the tank was 6.5 m high (h) and the viscosity of water was 1 x 10-3 kg/m.s (µ), estimate the time required for the sand to settle. u = g d2 (s-l)/18µ

g = 9.81 m/s2

t = h/uo (3 marks)

ANSWER Step 1: convert d into m value d  0.3mm

d  3  104 m Step 2: use this formula u 

u

gd 2  s   l  18

gd 2  s   l  18

9.813  104  2010  1130 u 181  103  m u  0.0043164 s 2

Step 3: use this formula t 

t t

h u0 6.5m

m s t  1505.8845s 0.0043164

h u0

QUESTION 5 A 1020 L tank containing 15% salt solution (density 1010 kg/m 3 (l)) at 14oC was mixed with 10% fine sand (0.7 mm (d) diameter and density 1910 kg/m3 (s)). If the tank was 10 m high (h) and the viscosity of water was 1 x 10-3 kg/m.s (µ), estimate the time required for the sand to settle. g = 9.81 m/s2

u = g d2 (s-l)/18µ

t = h/uo (3 marks)

ANSWER Step 1: convert d into m value d = 7 x 10-4 m Step 2: use this formula u = g d2 (s-l)/18µ u = 0.2403 m/s Step 3: use this formula t = h/uo t = 41.6146 s

QUESTION 6 A 5500 L tank containing 30% salt solution (density 3700 kg/m 3 (l)) at 27oC was mixed with 12% fine sand (1.3 mm (d) diameter and density 5910 kg/m3 (s)). If the tank was 16 m high (h) and the viscosity of water was 1 x 10-3 kg/m.s (µ), estimate the time required for the sand to settle. g = 9.81 m/s2

u = g d2 (s-l)/18µ

t = h/uo (3 marks)

ANSWER Step 1: convert d into m value d = 13 x 10-4 m Step 2: use this formula u = g d2 (s-l)/18µ u = 2.0355 m/s Step 3: use this formula t = h/uo t = 7.8604 s