2A.4 Loss of catalyst particles in stack gas.(a) Estimate the maximum diameter of microspherical catalyst particles that could be lost inthe stack gas of a fluid cracking unit under the following conditions:Gas velocity at axis of stackGas viscosityGas densityDensity of a catalyst particle1.0 ft/s (vertically upward)= 0.026 ср0.045 lb/ft³1.2 g/cm³=== 1µm).Express the result in microns (1 micron = 10-6 m =(b) Is it permissible to use Stokes' law in (a)?Answers: (a) 110 µm; Re = 0.93 6A.5 Sphere diameter for a given terminal velocity.(a) Explain how to find the sphere diameter D corresponding to given values of V∞, P, Psph, µ,and g by making a direct construction on Fig. 6.3-1.(b) Rework Problem 2A.4 by using Fig. 6.3-1.(c) Rework (b) when the gas velocity is 10 ft/s.

Answered: 2A.4 Loss of catalyst particles in…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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2A.4 Loss of catalyst particles in stack gas.
(a) Estimate the maximum diameter of microspherical catalyst particles that could be lost in
the stack gas of a fluid cracking unit under the following conditions:
Gas velocity at axis of stack
Gas viscosity
Gas density
Density of a catalyst particle
1.0 ft/s (vertically upward)
= 0.026 ср
0.045 lb/ft³
1.2 g/cm³
=
=
= 1µm).
Express the result in microns (1 micron = 10-6 m =
(b) Is it permissible to use Stokes' law in (a)?
Answers: (a) 110 µm; Re = 0.93

6A.5 Sphere diameter for a given terminal velocity.
(a) Explain how to find the sphere diameter D corresponding to given values of V∞, P, Psph, µ,
and g by making a direct construction on Fig. 6.3-1.
(b) Rework Problem 2A.4 by using Fig. 6.3-1.
(c) Rework (b) when the gas velocity is 10 ft/s.

Expert Solution

Step 1

2A.4:

Given;

$v = 1.0 f t / s μ = 0.026 c p ρ_{g} = 0.045 l b_{m} / f t^{3} ρ_{c} = 1.2 g / c m^{3}$

Solution:

$v = 1.0 f t / s \times \frac{30.48 c m}{1 f t} = 30.48 c m / s μ = 0.026 c p ρ_{g} = 0.045 l b_{m} / f t^{3} 1 l b m = 453.592 g m 1 f t = 2.54 c m 1 l b_{m} / f t^{3} = \frac{453.592 g m}{1 l b_{m}} \times {(\frac{1 f t}{30 c m})}^{3} = 0.017 g / c m^{3} 0.045 l b_{m} / f t^{3} = 0.045 \times 0.017 = 0.00076 g / c m^{3} ρ_{c} = 1.2 g / c m^{3}$

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