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calculate the Ergun parameter, ?. using the following parameters,  you will need to do some unit conversion, i think i did it correctly i have added the eqautions that need to be used,

calculate the Ergun parameter, ?. using the following parameters,  you will need to do some unit conversion, i think i did it correctly i have added the eqautions that need to be used,

Introduction to Chemical Engineering Thermodynamics
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
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calculate the Ergun parameter, ?. using the following parameters, 

you will need to do some unit conversion, i think i did it correctly i have added the eqautions that need to be used, 

α - m = 104.4 lbm/hour Air Operating conditions: Feed (air): Catalyst properties: Reactor dimensions: a = T = 260 °C (isothermal) Mo = 0.0673 lbm/ (ft * hr) Pcat = 72 lbm/ft³ p = 0.36 1.5 inch SCH 40 pipe (13) * 2* G * (1¹17) & b Po*Po*9c*Dp*Pcat* Ac Ac 2*G Po Po 9c*Dp*Pcat* Ac G = p*um * Ac P₁ = 21162.266 lbf ft² & gc = Catalyst particles 150*(1-0)*μ Dp 104.4 lbm hr Dpart = = 4.17 x 108. = + 1.75* Gor 150*(1-φ)*μ Dp 1 0.01414 ft² lbm*ft lbf*hr2 P = 10 atm Po = 0.413 lbm/ft³ Dpart = 0.64inches Ac = 0.01414 ft² 0.64 | 1 ft 0.641/7 12 in = +1.75 G → a= a*b G] = 7383.310 0.053 ft lbm ft² * hr
Transcribed Image Text:α - m = 104.4 lbm/hour Air Operating conditions: Feed (air): Catalyst properties: Reactor dimensions: a = T = 260 °C (isothermal) Mo = 0.0673 lbm/ (ft * hr) Pcat = 72 lbm/ft³ p = 0.36 1.5 inch SCH 40 pipe (13) * 2* G * (1¹17) & b Po*Po*9c*Dp*Pcat* Ac Ac 2*G Po Po 9c*Dp*Pcat* Ac G = p*um * Ac P₁ = 21162.266 lbf ft² & gc = Catalyst particles 150*(1-0)*μ Dp 104.4 lbm hr Dpart = = 4.17 x 108. = + 1.75* Gor 150*(1-φ)*μ Dp 1 0.01414 ft² lbm*ft lbf*hr2 P = 10 atm Po = 0.413 lbm/ft³ Dpart = 0.64inches Ac = 0.01414 ft² 0.64 | 1 ft 0.641/7 12 in = +1.75 G → a= a*b G] = 7383.310 0.053 ft lbm ft² * hr
Expert Solution
Step 1: given

g i v e n space equals space T space equals space 260 to the power of o c mu subscript o space equals space 0.0673 space l b subscript m divided by left parenthesis f t. h r right parenthesis P space equals space 10 space a t m rho subscript o space equals space 0.413 space l b m divided by f t cubed rho subscript c a t space end subscript equals space 72 space l b m divided by f t cubed empty set space equals space 0.36 space a n d space A subscript c space equals space 0.01414 space f t squared

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