Oil, of density 900 kg/m3 and viscosity 3 mns/m2, is passed vertically upwards through a bed of catalyst consisting of approximately spherical particles of diameter 0.1 mm and density 2600 kg/m3. At approximately what mass rate of flow per unit area of bed will (a) fluidisation, and (b) transport of particles occur? Use e=0.48
Answers
Answer:
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Answer:
(a) Equations 4.9 and 6.1 may be used to determine the fluidising velocity, umf .
u = (1/K′′)(e3/(S2
(1 − e)2
)(1/µ)(− P/l) (equation 4.9)
− P = (1 − e)(ρs − ρ)lg (equation 6.1)
where S = surface area/volume, which, for a sphere, = πd2
/(πd3
/6) = 6/d.
Substituting K′′ = 5, S = 6/d and − P/l from equation 6.1 into equation 4.9 gives:
umf = 0.0055(e3
/(1 − e))(d2
(ρs − ρ)g)/µ
Hence : G
′
mf = ρu = (0.0055e
3
/(1 − e))(d2
(ρs − ρ)g)/µ
In this problem, ρs = 2600 kg/m3
, ρ = 900 kg/m3
, µ = 3.0 × 10−3 Ns/m2
and d = 0.1 mm = 1.0 × 10−4 m.
As no value of the voidage is available, e will be estimated by considering eight closely packed
spheres of diameter d in a cube of side 2d. Thus:
volume of spheres = 8(π/6)d3
volume of the enclosure = (2d)3 = 8d
3
and hence: voidage, e = [8d
3 − 8(π/6)d3
]/8d
3 = 0.478, say, 0.48.
Thus : G
′
mf = 0.0055(0.48)
3
(10−4
)
2
((900 × 1700) × 9.81)/((1 − 0.48) × 3 × 10−3
= 0.059 kg/m2
s
Explanation:
(b) Transport of the particles will occur when the fluid velocity is equal to the terminal falling
velocity of the particle.
Using Stokes’ law : u0 = d
2
g(ρs − ρ)/18µ (equation 3.24)
= ((10−4
)
2 × 9.81 × 1700)/(18 × 3 × 10−3
)
= 0.0031 m/s
The Reynolds number = ((10−4 × 0.0031 × 900)/(3 × 10−3
) = 0.093 and hence Stokes’ law
applies.
The required mass flow = (0.0031 × 900) = 2.78 kg/m2
s
An alternative approach is to make use of Figure 3.6 and equation 3.35,
(R/ρu2
)Re2 = 2d
3
ρg(ρs − ρ)/3µ
2
= (2 × (10−4
)
3 × (900 × 9.81) × 1700)/(3(3 × 10−3
)
2
) = 1.11
From Figure 3.6, Re = 0.09
Hence: u0 = Re(µ/ρd) = (0.09 × 3 × 10−3
)/(900 × 10−4
) = 0.003 m/s
and: G
′ = (0.003 × 900) = 2.7 kg/m2
s