A suspension made of aspirin particles and ethanol is flowing in a pipe of 50 mm diameter. Ethanol and aspirin properties are reported in the table below. Property Ethanol density (kg/m³) Ethanol dynamic viscosity (Pas) Particle density (g/cm³) Particle mean diameter (µm) Region Re single particle range CD Value 789 0.00109 Assume the single particle Reynolds number is equal to 0.1. The empirical parameters a and b are 0.7 and 0.2 respectively. Stokes < 0.3 24 Rep 1.4 20 Intermediate 0.3< Rep < 500 24 Rep (1 + 0.15 Rep 0.687) Newton's law 500 Rep < 2 x 105 ~0.44 (a) Calculate the critical velocity above which particles will remain suspended during the suspension flow in the pipe. Consider the suspension is flowing under Stokes flow regime. (b) How does the critical velocity of the suspension travelling in the pipe change if the solvent is ethylene glycol rather than ethanol? Ethylene glycol density is 1110 kg/m³ and dynamic viscosity is 0.0162 Pa.s. (c) How does the critical velocity of the suspension travelling in the pipe change if the mean diameter of the suspended particles is 300 μm? (d) How does the critical velocity of the suspension travelling in the pipe change if the diameter of the pipe is 10 cm?

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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W6/T1 (a) 0.23 m/s (b) 0.13 m/s (c) 1.16 m/s (d) 0.22 m/s

 

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TASK 1:
A suspension made of aspirin particles and ethanol is flowing in a pipe of 50 mm diameter.
Ethanol and aspirin properties are reported in the table below.
Property
Ethanol density (kg/m³)
Ethanol dynamic viscosity (Pas)
Particle density (g/cm³)
Particle mean diameter (um)
Region
Re single particle
range
Assume the single particle Reynolds number is equal to 0.1. The empirical parameters a and
b are 0.7 and 0.2 respectively.
CD
Value
789
0.00109
Stokes
< 0.3
24
Rep
1.4
20
Intermediate
0.3 < Rep < 500
24
Rep
(1 + 0.15 Rep 0.687)
Newton's law
500 <
Rep < 2 x 105
~0.44
(a) Calculate the critical velocity above which particles will remain suspended during the
suspension flow in the pipe. Consider the suspension is flowing under Stokes flow regime.
(b) How does the critical velocity of the suspension travelling in the pipe change if the solvent
is ethylene glycol rather than ethanol? Ethylene glycol density is 1110 kg/m³ and dynamic
viscosity is 0.0162 Pa.s.
(c) How does the critical velocity of the suspension travelling in the pipe change if the mean
diameter of the suspended particles is 300 μm?
(d) How does the critical velocity of the suspension travelling in the pipe change if the
diameter of the pipe is 10 cm?
Transcribed Image Text:TASK 1: A suspension made of aspirin particles and ethanol is flowing in a pipe of 50 mm diameter. Ethanol and aspirin properties are reported in the table below. Property Ethanol density (kg/m³) Ethanol dynamic viscosity (Pas) Particle density (g/cm³) Particle mean diameter (um) Region Re single particle range Assume the single particle Reynolds number is equal to 0.1. The empirical parameters a and b are 0.7 and 0.2 respectively. CD Value 789 0.00109 Stokes < 0.3 24 Rep 1.4 20 Intermediate 0.3 < Rep < 500 24 Rep (1 + 0.15 Rep 0.687) Newton's law 500 < Rep < 2 x 105 ~0.44 (a) Calculate the critical velocity above which particles will remain suspended during the suspension flow in the pipe. Consider the suspension is flowing under Stokes flow regime. (b) How does the critical velocity of the suspension travelling in the pipe change if the solvent is ethylene glycol rather than ethanol? Ethylene glycol density is 1110 kg/m³ and dynamic viscosity is 0.0162 Pa.s. (c) How does the critical velocity of the suspension travelling in the pipe change if the mean diameter of the suspended particles is 300 μm? (d) How does the critical velocity of the suspension travelling in the pipe change if the diameter of the pipe is 10 cm?
Critical velocity:
Archimedes number:
F (when Ar > 80):
F (when Ar < 80):
F√ √GD (2-1)
Vc = FgD
4 3
Ar = 1 x ³ pf (Ps - Pf) - /2/2
F = a Arb
F = √² [2 + 0.3 log₁0 (x)]
Unhindered settling, single particle velocity: vs
=
g(Ps-Pf)x²
18μ
Transcribed Image Text:Critical velocity: Archimedes number: F (when Ar > 80): F (when Ar < 80): F√ √GD (2-1) Vc = FgD 4 3 Ar = 1 x ³ pf (Ps - Pf) - /2/2 F = a Arb F = √² [2 + 0.3 log₁0 (x)] Unhindered settling, single particle velocity: vs = g(Ps-Pf)x² 18μ
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