Calculate umf for particles of ZnS fluidized by air at 1200 K and 200 kPa. Assume d'p = 4 X lo^-4 m , mf = 0.5 , ρp = 3500 kg m-³, and μf = 4.6 X lo^-5 N s m-²; g = 9.81 m s-².

Question

Calculate [latex]u_{mf}[/latex] for particles of ZnS fluidized by air at 1200 K and 200 kPa. Assume [latex]d^\prime_p = 4 imes 10^{-4} m [/latex] , [latex]\bullet mf = 0.5 [/latex] , [latex] ho_p = 3500 kg m^{-3} [/latex] , and [latex]\mu_f = 4.6 imes 10^{-5} N s m^{-2} [/latex] ; [latex]g = 9.81 m s^{-2} [/latex]

Accepted Answer

At the (T,P) conditions given, the density of air ([latex] ho_f[/latex]), assumed to be an ideal gas (z = 1) with [latex]M_{a u} = 28.8 [/latex] , is

[latex]Pf = \frac{m}{V} = \frac{PM_{a u}}{RT} = \frac{200(1000)28.8}{8.314(1200)1000} = 0.577 kg m^{-3}[/latex]

Since the particles are relatively small, we compare the results obtained from equations 23.2-5 and -6. With given values of parameters inserted, the former becomes

[latex]u_{mf}^2 + \frac{150(1 - \epsilon _{mf})\mu _f}{1.75 ho_fd_p^\prime}u_{mf} - \frac{g( ho _p - ho _f)\epsilon ^3_{mf}d_p^\prime}{1.75 ho _f} = 0 [/latex] (23.2-5)

[latex] u_{mf} = \frac{g( ho _p - ho _f)(d^\prime_p)^2}{K\mu _f} (Re^\prime_{mf } small)[/latex] (23.2-6)

[latex]u_{mf}^2 + 8.542u_{mf} - 1.700 = 0 [/latex]

from which [latex]u_{mf} = 0.195 m s^{-1} [/latex] (the units of each term should be confirmed for consis-
tency). From equation 23.2-6

[latex]u_{mf} = \frac{9.81(3500)4^2(10^{-4})^2(1 - 0.5)^3}{150(1 - 0.5)4.6(10^{-5})} = 0.199 m s^{-1}[/latex]

which is within 2% of the more accurate value above. (See also file ex23-2.msp.)

Question 23.2: Calculate umf for particles of ZnS fluidized by air at 1200 ...

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