262276972-Rhodes-Solutions-Ch6.pdf

262276972-Rhodes-Solutions-Ch6.pdf - SOLUTIONS TO CHAPTER 6...

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SOLUTIONS TO CHAPTER 6: FLOW THROUGH A PACKED BED OF PARTICLES EXERCISE 6.1: A packed bed of solid particles of density 2500 kg/m 3 , occupies a depth of 1 m in a vessel of cross-sectional area 0.04 m 2 . The mass of solids in the bed is 50 kg and the surface-volume mean diameter of the particles is 1 mm. A liquid of density 800 kg/m 3 and viscosity 0.002 Pas flows upwards through the bed, which is restrained at its upper surface. (a) Calculate the voidage (volume fraction occupied by voids) of the bed. (b) Calculate the pressure drop across the bed when the volume flow rate of liquid is 1.44 m 3 /h. SOLUTION TO EXERCISE 6.1: (a) Knowing the mass of particles in the bed, the density of the particles and the volume of the bed, the bed voidage may be calculated: mass of bed, M = M = AH 1 − ε ( ) ρ p giving voidage, ε = 1 50 2500 × 0.04 × 1 = 0.5 (b) With a liquid flow rate of 1.44 m 3 /h, the superficial liquid velocity through the bed, U is given by: U = 1.44 3600 × A = 0.01 m / s Use the Ergun equation (Text-Equation 6.15) to estimate the pressure drop across the bed at this flow rate: −Δ p ( ) H = 150 μ U x sv 2 (1 − ε ) 2 ε 3 + 1.75 ρ f U 2 x sv (1 − ε ) ε 3 With μ = 0.002 Pa.s, ρ f = 800 kg/m 3 , x sv = 1 mm and H = 1 m, −Δ p ( ) = 600 × 10 3 U + 5.6 × 10 6 U 2 = 6560 Pa Checking the Reynolds number, R e = U ρ f x sv μ 1 − ε ( ) = 8 (Text-Equation 6.12). Since the Reynolds number is less than 10, we might estimate the pressure drop using the Carman Kozeny equation (Text-Equation 6.16): −Δ p ( ) H = 180 μ U x sv 2 (1 − ε ) 2 ε 3 = 7200 Pa. Alternatively we could use the laminar part of the Ergun equation, which gives, (- Δ p) = 6000 Pa. SOLUTIONS TO CHAPTER 6: FLOW THROUGH A PACKED BED OF PARTICLES p. 6.1
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