130 kg of crushed glass particles with a diameter of 60 mu m and particle densit
ID: 521582 • Letter: 1
Question
130 kg of crushed glass particles with a diameter of 60 mu m and particle density of 2500 kg/m^3 are fluidized by water in a circular bed of cross-sectional area 0.2 m^2. The voltage at minimum fluidization is known to be 0.47. Calculate: a) the bed height at minimum fluidization. b) the pressure drop at minimum conditions c) the superficial and interstitial velocities at minimum fluidization conditions. d) the fluidized bed voltage and length when the liquid flow rate is 2 times 10^-5 m^3/s. e) the maximum allowable velocity.Explanation / Answer
mass of solids in the bed, M = 1 mf ( )p AHmf
Therefore, with M = 130 kg,
mf = 0.47,
Pp = 1500 kg/m3 and
A = 0.2 m2,
Hmf=130 /0.2 × (1 0.47) ×1500
=0.818 m
Bed height at incipient fluidization, Hmf = 0.818 m.
Bed height when liquid flow rate = 2 x 10^-5 m3/s:
Use Richardson-Zaki equation (
U = UT ^n
To determine exponent n, calculate single particle Reynolds number, Rep at U=UT:
To determine exponent n, calculate single particle Reynolds number,
Rep at U=UT:
Rep = UTfx/
=(0.68 ×10^3) ×1000 ×(50 ×10^6 /0.001
= 0.034, which is less than 0.3. Hence, n = 4.65
Hence, applying the Richardson-Zaki equation
= 1 ×10^4 = (0.68 ×10^3 )^ 4.65
which gives, = 0.6622 hence,
bed voidage at a liquid flow rate = 2 x 10^-5 m3/s is = 0.6622
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