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بوالو CONSERVATION OF +V MOMENTUM (NAVIER- STOKES) r ər д 10 + Share √z (r) =√o = 0 1 ǝ² / 2 av ǝ²✓ ((ry.). Portrar д др 2(10 +V. pg, +μ dz dr ar Ve 1 др + +V. r ᎧᎾ др 10 dv. 1 + +V. H +μ r + Ir de rar dr r ᎧᎾ + Y₁ ar + Tr 20 ↑ +1 STEADY-STATE + P80+μ dr r dr 90² r² Do + + dz² 1 J² 2 dv/ d² v + 962 + до де IL N = 0 => f f(r) ар ≥ 31 = 412 (1317) 20170 az Ө 24 20 =0 => P+ f (o) BCS (1)@=a, √z=0 (2) @r=0, √z = FINITE

بوالو CONSERVATION OF +V MOMENTUM (NAVIER- STOKES) r ər д 10 + Share √z (r) =√o = 0 1 ǝ² / 2 av ǝ²✓ ((ry.). Portrar д др 2(10 +V. pg, +μ dz dr ar Ve 1 др + +V. r ᎧᎾ др 10 dv. 1 + +V. H +μ r + Ir de rar dr r ᎧᎾ + Y₁ ar + Tr 20 ↑ +1 STEADY-STATE + P80+μ dr r dr 90² r² Do + + dz² 1 J² 2 dv/ d² v + 962 + до де IL N = 0 => f f(r) ар ≥ 31 = 412 (1317) 20170 az Ө 24 20 =0 => P+ f (o) BCS (1)@=a, √z=0 (2) @r=0, √z = FINITE

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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How they find z value. Show that as well as explain Please explain r and teta too.
Applying the Navier-Stokes Equations, part 4 - Lecture 4.9-Chemical Engineering Flui chanics Share CONSERVATION OF MOMENTUM (NAVIER-STOKES) Uz (r) √=√0=0 др 10 DCA***CERE +V dr av + +v. dz ve + + ᎧᎾ VV dr pg, 1 др dr r dr a 1 д + 18² / 2 av 3² √√] r² 90² r² Do dz² y + 2 ǝv/ a²v -(rv + 1 a² ((()) ZP ǝt at + 三十 ǝr Trǝe STEADY-STATE ар r Z = 0 2P + +V₂ αν др Əz => f f(r) где + + + arrǝr r² 90² r² do dz² 10 dv. 1 ² ² + ] 32 = 41 31 (1317) az r Ər r ər 24 20 =0 => P + f (o) BCS (1) @ra, √z=0 (2) @r=0, √z = FINITE
Transcribed Image Text:Applying the Navier-Stokes Equations, part 4 - Lecture 4.9-Chemical Engineering Flui chanics Share CONSERVATION OF MOMENTUM (NAVIER-STOKES) Uz (r) √=√0=0 др 10 DCA***CERE +V dr av + +v. dz ve + + ᎧᎾ VV dr pg, 1 др dr r dr a 1 д + 18² / 2 av 3² √√] r² 90² r² Do dz² y + 2 ǝv/ a²v -(rv + 1 a² ((()) ZP ǝt at + 三十 ǝr Trǝe STEADY-STATE ар r Z = 0 2P + +V₂ αν др Əz => f f(r) где + + + arrǝr r² 90² r² do dz² 10 dv. 1 ² ² + ] 32 = 41 31 (1317) az r Ər r ər 24 20 =0 => P + f (o) BCS (1) @ra, √z=0 (2) @r=0, √z = FINITE
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