A simple air purifier is designed to remove tiny particles in the air. In order not to scarify theair flow, a pressure drop of 0.25 atm is permissible. Determine the air velocity if the air purifier(1.5 m height, 0.2 m diameter) is fully packed with activated carbons (5 mm side, total mass of1200g, density of 1 g/cm3). Given that the R=0.082 atm.m3/kmol.K, and the viscosity of the airis 2.1 x10-5 Pa.s. VmSpd"v²-np4nReg =8n-1K1+ 3n.89²Ld5 = 4f -Reg = 390 d-2.05Heg = 5410.8 d²-(4)E = 1-6asodpas = aso(1 – €)d"v²-"p4nRec =8n-1K (1+ 3n)6464n(Rec)critical(2+n)/(1+n)(1 + 3n)² (G)61d,asoPV = nRTVmSp6rdpaso(1 - e)²n= 150(1– E)P „2(-ΔΡ)v+ 1.75e*d,?LA²(-AP)ApdtnG+ AR,)VrepVr(1– e)Pp1- MSM = 1+C =αρS Heg=0 10104105 106 107 108 10°01Realomcal0.010.001410102103104105106ReaFigure 1: Friction factor for Herschel-Bulkley fluids.1.00.200.90.50Laminar FlowV 0.8Vmax.0.751.001.502.00n=0.200.7Turbulent Flow0.6n-0.50n=0.75n1.000.5n-1.50A-2.000.40.310102103104105106ReaFigure 2: Variation (V/Vmax) as a function of the generalized Reynolds number for powerlaw fluids.

Permissible pressure drop, Δp = 0.25 atm = 25331.25 Pa Height of the purifier, L = 1.5 m Diameter of…

Answered: A simple air purifier is designed to…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

See similar textbooks

Related questions

Question

A simple air purifier is designed to remove tiny particles in the air. In order not to scarify the
air flow, a pressure drop of 0.25 atm is permissible. Determine the air velocity if the air purifier
(1.5 m height, 0.2 m diameter) is fully packed with activated carbons (5 mm side, total mass of
1200g, density of 1 g/cm3
). Given that the R=0.082 atm.m3
/kmol.K, and the viscosity of the air
is 2.1 x10-5 Pa.s.

Vm
Sp
d"v²-np
4n
Reg =
8n-1K
1+ 3n.
89²L
d5 = 4f -
Reg = 390 d-2.05
Heg = 5410.8 d²
-(4)
E = 1-
6
aso
dp
as = aso(1 – €)
d"v²-"p
4n
Rec =
8n-1K (1+ 3n)
6464n
(Rec)critical
(2+n)/(1+n)
(1 + 3n)² (G)
61
d,
aso
PV = nRT
Vm
Sp
6r
dp
aso
(1 - e)²n
= 150
(1– E)P „2
(-ΔΡ)
v+ 1.75
e*d,?
L
A²(-AP)
Ap
dt
nG+ AR,)
Vrep
Vr(1– e)Pp
1- MS
M = 1+
C =
αρS

Heg=0 10104105 106 107 108 10°
01
Realomcal
0.01
0.0014
10
102
103
104
105
106
Rea
Figure 1: Friction factor for Herschel-Bulkley fluids.
1.0
0.20
0.9
0.50
Laminar Flow
V 0.8
Vmax
.0.75
1.00
1.50
2.00
n=0.20
0.7
Turbulent Flow
0.6
n-0.50
n=0.75
n1.00
0.5
n-1.50
A-2.00
0.4
0.3
10
102
103
104
105
106
Rea
Figure 2: Variation (V/Vmax) as a function of the generalized Reynolds number for power
law fluids.

Expert Solution

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemical-engineering and related others by exploring similar questions and additional content below.

Recommended textbooks for you

Introduction to Chemical Engineering Thermodynami…

Chemical Engineering

ISBN:

9781259696527

Author:

J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:

McGraw-Hill Education