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: McGraw-Hill Education
Question
3.
A very viscous Newtonian fluid undergoes creeping flow in the
space between two stationary concentric spheres as shown to the right.
Simplify the equations of continuity and motion (see next page) to
derive a set of equations that describe the fluid velocity as a function of
pressure difference. Do not derive the velocity profile. Neglect the end
effects, assume steady state and creeping flow, assume that ve depends
only on r and 0, and assume that the other velocity components are zero.
Fluid in
Fluid out
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Transcribed Image Text:3. A very viscous Newtonian fluid undergoes creeping flow in the space between two stationary concentric spheres as shown to the right. Simplify the equations of continuity and motion (see next page) to derive a set of equations that describe the fluid velocity as a function of pressure difference. Do not derive the velocity profile. Neglect the end effects, assume steady state and creeping flow, assume that ve depends only on r and 0, and assume that the other velocity components are zero. Fluid in Fluid out
Problem 3 Equations
Equation of Continuity:
Spherical coordinates (r,0,0):
др
+
at
1 д
(pr²v₁) +
1
д
1
д
(pv sin 0) +
r² dr
r sin o de
(pv) = 0
(B.4-4)
r sin 0 дф
Equations of Motion:
Spherical coordinates (r,0,0):
до
dv
% до
P
+
+
-
ot
dr г де
r sin 0 do
a²
11+
ar2
1 д
2 sin 0 de
av,
sin 0.
de
ave
+
at
ave
dr r de
ve ave
r
=-
1
др
dr
r² sin² do²
V dve ve- cot e
+
+P&r
(B.6-7)
1 др
+
+
=-
r sin 0 дф
r
r de
дор
1 д
1 д
+1
+
(v sin 0)
2 dr
dr
12 до
sin 0 де
(0)) +
1
J²ve
2 dv,
+
r² sin² 0 do²
12 до
2 cote dv
2 sin 0 do
+ P&e
(B.6-8)
P
до
до
vedo
v
+
+
+
at
1 д
+μ
2 dr дт
V du v + cot
дг г де r sin 0 до
д до
1 д
0 де
e)
r
==
r sin 0 до
+()+((sin 0) + 12 sin' a des² + på sin o dig sin a do
"The quantity in the brackets in Eq. B.6-7 is not what one would expect from Eq. (M) for [V. Vv] in Table A.7-3, because we have added
to Eq. (M) the expression for (2/r)(V-v), which is zero for fluids with constant p. This gives a much simpler equation.
1
др
1
2 до
do
2 cot dve
+
+ P80
(B.6-9)
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Transcribed Image Text:Problem 3 Equations Equation of Continuity: Spherical coordinates (r,0,0): др + at 1 д (pr²v₁) + 1 д 1 д (pv sin 0) + r² dr r sin o de (pv) = 0 (B.4-4) r sin 0 дф Equations of Motion: Spherical coordinates (r,0,0): до dv % до P + + - ot dr г де r sin 0 do a² 11+ ar2 1 д 2 sin 0 de av, sin 0. de ave + at ave dr r de ve ave r =- 1 др dr r² sin² do² V dve ve- cot e + +P&r (B.6-7) 1 др + + =- r sin 0 дф r r de дор 1 д 1 д +1 + (v sin 0) 2 dr dr 12 до sin 0 де (0)) + 1 J²ve 2 dv, + r² sin² 0 do² 12 до 2 cote dv 2 sin 0 do + P&e (B.6-8) P до до vedo v + + + at 1 д +μ 2 dr дт V du v + cot дг г де r sin 0 до д до 1 д 0 де e) r == r sin 0 до +()+((sin 0) + 12 sin' a des² + på sin o dig sin a do "The quantity in the brackets in Eq. B.6-7 is not what one would expect from Eq. (M) for [V. Vv] in Table A.7-3, because we have added to Eq. (M) the expression for (2/r)(V-v), which is zero for fluids with constant p. This gives a much simpler equation. 1 др 1 2 до do 2 cot dve + + P80 (B.6-9)
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