Solutions Manual for
Numerical Methods for Engineers
(Indian Edition) 8e By Steven
Chapra, Raymond Canale (All
Chapters 1-32, 100% Original
Verified, A+ Grade)
All Chapters Arranged Reverse: 32-1
This is the Original Solutions Manual
for (Indian Edition) 8th Edition, All
Other Files in the
Market are Wrong/Old Questions.
, 1
CHAPTER 32
Chemical/Bio Engineering
32.1 Perform the same computation as in Sec. 32.1, but use Δx = 1.25.
First equation
6.075c0 - 3.2c1 = 262.5
Middle equations (i = 1 to 8)
-2.1ci -1 + 3.45ci - 1.1ci +1 = 0
Last equation
-3.2c8 + 3.45c9 = 0
The solution is
c0 = 76.53 c5 = 29.61
c1 = 63.25 c6 = 24.62
c2 = 52.28 c7 = 20.69
c3 = 43.22 c8 = 17.88
c4 = 35.75 c9 = 16.58
32.2 Develop a finite-element solution for the steady-state system of Sec. 32.1.
Element equation: (See solution for Prob. 31.4 for derivation of element equation.)
é a11
êë a21 {}{}
a12 ù c1
a22 úû c2
b
= 1
b2
where
D U k -D U -D U
a11 = - + ( x2 - x1 ) a12 = + a21 = -
x2 - x1 2 2 x2 - x1 2 x2 - x1 2
D U k
a22 = + + ( x2 - x1 )
x2 - x1 2 2
dc dc
b1 = - D ( x1 ) b2 = D ( x2 )
dx dx
Substituting the parameter values:
2 1 0.2 -2 1
a11 = - + (2.5) = 0.55 a12 = + = -0.3
2.5 2 2 2.5 2
-2 1 2 1 0.2
a21 = - = -1.3 a22 = + + (2.5) = 1.55
2.5 2 2.5 2 2
dc dc
b1 = -2 ( x1 ) b2 = 2 ( x2 )
dx dx
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written consent of McGraw-Hill Education.
, 2
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior
written consent of McGraw-Hill Education.
, 3
Assembly:
ì dc ü
-2 ( x )
é 0.55 -0.3 ù ìc0 ü ï dx 1 ï
ê -1.3 2.1 -0.3 ú ïï c1 ïï ïï 0 ï
ï
ê -1.3 2.1 -0.3 ú íc2 ý = í 0 ý
ê -1.3 2.1 -0.3ú ï c3 ï ï 0 ï
êë -1.3 1.55 úû ïîc4 ïþ ï dc ï
ïî 2 dx ( x2 ) ïþ
Boundary conditions:
A mass balance at the inlet can be written as:
dc
Ucin = Uc0 - D (0)
dx
which can be solved for
dc
-D (0) = Ucin - Uc0
dx
Substitute into first equation
0.55c0 - 0.3c1 = 100 - c0
1.55c0 - 0.3c1 = 100
Outlet:
dc
(10) = 0
dx
Therefore, the system of equations to be solved is
é1.55 -0.3 ù ìc0 ü ì100 ü
ê -1.3 2.1 -0.3 ú ïï c1 ïï ïï 0 ïï
ê -1.3 2.1 -0.3 ú íc2 ý = í 0 ý
ê -1.3 2.1 -0.3ú ï c3 ï ï 0 ï
êë -1.3 1.55 úû ïîc4 ïþ ïî 0 ïþ
Solution:
c0 = 74.4c1 = 51.08 c2 = 35.15 c3 = 24.72 c4 = 20.74
32.3 Compute mass fluxes for the steady-state solution of Sec. 32.1 using Fick’s first law.
According to Fick’s first law, the diffusive flux is
dc
J ( x) = - D ( x)
dx
where J(x) = flux at position x. If c has units of g/m3, D has units of m2/d and x is measured in m, flux
has units of g/m2/d. In addition, there will be an advective flux which can be calculated as
J ( x) = Uc ( x)
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior
written consent of McGraw-Hill Education.
Numerical Methods for Engineers
(Indian Edition) 8e By Steven
Chapra, Raymond Canale (All
Chapters 1-32, 100% Original
Verified, A+ Grade)
All Chapters Arranged Reverse: 32-1
This is the Original Solutions Manual
for (Indian Edition) 8th Edition, All
Other Files in the
Market are Wrong/Old Questions.
, 1
CHAPTER 32
Chemical/Bio Engineering
32.1 Perform the same computation as in Sec. 32.1, but use Δx = 1.25.
First equation
6.075c0 - 3.2c1 = 262.5
Middle equations (i = 1 to 8)
-2.1ci -1 + 3.45ci - 1.1ci +1 = 0
Last equation
-3.2c8 + 3.45c9 = 0
The solution is
c0 = 76.53 c5 = 29.61
c1 = 63.25 c6 = 24.62
c2 = 52.28 c7 = 20.69
c3 = 43.22 c8 = 17.88
c4 = 35.75 c9 = 16.58
32.2 Develop a finite-element solution for the steady-state system of Sec. 32.1.
Element equation: (See solution for Prob. 31.4 for derivation of element equation.)
é a11
êë a21 {}{}
a12 ù c1
a22 úû c2
b
= 1
b2
where
D U k -D U -D U
a11 = - + ( x2 - x1 ) a12 = + a21 = -
x2 - x1 2 2 x2 - x1 2 x2 - x1 2
D U k
a22 = + + ( x2 - x1 )
x2 - x1 2 2
dc dc
b1 = - D ( x1 ) b2 = D ( x2 )
dx dx
Substituting the parameter values:
2 1 0.2 -2 1
a11 = - + (2.5) = 0.55 a12 = + = -0.3
2.5 2 2 2.5 2
-2 1 2 1 0.2
a21 = - = -1.3 a22 = + + (2.5) = 1.55
2.5 2 2.5 2 2
dc dc
b1 = -2 ( x1 ) b2 = 2 ( x2 )
dx dx
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior
written consent of McGraw-Hill Education.
, 2
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior
written consent of McGraw-Hill Education.
, 3
Assembly:
ì dc ü
-2 ( x )
é 0.55 -0.3 ù ìc0 ü ï dx 1 ï
ê -1.3 2.1 -0.3 ú ïï c1 ïï ïï 0 ï
ï
ê -1.3 2.1 -0.3 ú íc2 ý = í 0 ý
ê -1.3 2.1 -0.3ú ï c3 ï ï 0 ï
êë -1.3 1.55 úû ïîc4 ïþ ï dc ï
ïî 2 dx ( x2 ) ïþ
Boundary conditions:
A mass balance at the inlet can be written as:
dc
Ucin = Uc0 - D (0)
dx
which can be solved for
dc
-D (0) = Ucin - Uc0
dx
Substitute into first equation
0.55c0 - 0.3c1 = 100 - c0
1.55c0 - 0.3c1 = 100
Outlet:
dc
(10) = 0
dx
Therefore, the system of equations to be solved is
é1.55 -0.3 ù ìc0 ü ì100 ü
ê -1.3 2.1 -0.3 ú ïï c1 ïï ïï 0 ïï
ê -1.3 2.1 -0.3 ú íc2 ý = í 0 ý
ê -1.3 2.1 -0.3ú ï c3 ï ï 0 ï
êë -1.3 1.55 úû ïîc4 ïþ ïî 0 ïþ
Solution:
c0 = 74.4c1 = 51.08 c2 = 35.15 c3 = 24.72 c4 = 20.74
32.3 Compute mass fluxes for the steady-state solution of Sec. 32.1 using Fick’s first law.
According to Fick’s first law, the diffusive flux is
dc
J ( x) = - D ( x)
dx
where J(x) = flux at position x. If c has units of g/m3, D has units of m2/d and x is measured in m, flux
has units of g/m2/d. In addition, there will be an advective flux which can be calculated as
J ( x) = Uc ( x)
Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior
written consent of McGraw-Hill Education.
