### 16a

```Overall Shell Energy Balance
Forms of Energy Generation:
(Se) 1. Degradation of electrical energy to heat
(Sn) 2. Heat from nuclear source (by fission)
(Sv) 3. Heat from viscous dissipation
Let S = rate of heat production
per unit volume (W/m3)
Energy
Generation
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
Shell Heat Balance:
2

− 2
+∆
+ 2∆  = 0
Rate of Heat IN:
2
Rate of Heat OUT:
2
Generation:
2∆

+∆
Electrical Heat Source
Rate of Heat IN
2
Area perpendicular
to qr at r = r

= (2) ∙

The Shell:
Rate of Heat IN:
2
Rate of Heat OUT:
2
Generation:
2∆

+∆
Electrical Heat Source
Rate of Heat OUT
2
+∆
Area perpendicular
to qr at r = r + dr
= (2( + ∆)) ∙
+∆
The Shell:
Rate of Heat IN:
2
Rate of Heat OUT:
2
Generation:
2∆

+∆
Electrical Heat Source
Generation = Volume X Se
=   + ∆
2
− 2
∴  =  2∆
Too small
=   2 + 2∆ + ∆
2
− 2
∴  = 2∆ ∙
The Shell:
Rate of Heat IN:
2
Rate of Heat OUT:
2
Generation:
2∆

+∆
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
Shell Heat Balance:
2
2

− 2

+∆
− 2
+ 2∆  = 0
+∆
= − 2∆
Dividing by 2∆:

−
∆
+∆
= −
Q: Why did we divide by
2∆ and not by 2∆?
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
We now have:
−
∆
+∆
= −
Taking the limit as ∆ → 0:

= −

Q: Is this correct?
NO!
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
We now have:
−
∆
+∆
= −
definition of the derivative:

lim
∆→0

−
∆
+∆
−
∆
+∆

= +

=
=

Electrical Heat Source
Consider an electrical wire
(solid cylinder):

=

We now have:
Integrating:
1
=
+
2

Boundary conditions:
= 0,
= ,
=
= 0
Note: The problem statement will
tell you hints about what boundary
conditions to use.
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
We now have:
1
=
+
2

Applying B.C. 1:   = 0,  =
Because q has to be finite at r = 0, all
the terms with radius, r, below the
denominator must vanish. Therefore:
1 = 0

=
2
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
We now have:

=
2
Substituting Fourier’s Law:

−
=

2
−
=

2
−  2
=
+ 2
4
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
We now have:
−  2
=
+ 2
4
Applying B.C. 2:   = ,
− 2
0 =
+ 2
4
2
2 = 0 +
4
This is it! But,
we rewrite it
into a nicer
−  2  2
form…
=
+
+ 0
4
4
= 0
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
Temperature Profile:
2

− 0 =
1−
4

2
Important assumptions:
1. Temperature rise is not large so that
k and Se are constant & uniform.
2. The surface of the wire is
maintained at T0.
3. Heat flux is finite at the center.
Electrical Heat Source
Other important notes…
Let:
= electrical conductivity
= current density

2
1
Ω ∙
= voltage drop over a length
These imply the following :
2
=

=

2

− 0 =
1−
4

2
2  2 2
=
4
42
Electrical Heat Source
Heat flux profile:

=
2
The stress profile versus
the temperature profile:
Temperature Profile:
2

− 0 =
1−
4

2
Electrical Heat Source
Quantities that might be asked for:
1. Maximum Temperature
Substituting r = 0
to the profile T(r):

2
= 0 +
4
2. Average Temperature Rise
− 0 =
2
− 0
0
0
2

0
0

2 1
− 0 =
=
8
2
3. Heat Outflow Rate at the Surface

= =
=
2
2
=
= 2  ∙
Electrical Heat Source
Examples for Review:
Example 10.2-1 and Example 10.2-2
Bird, Stewart, and Lightfoot, Transport
Phenomena, 2nd Ed., p. 295
Nuclear Heat Source
Consider a spherical
nuclear fuel assembly
(solid sphere):
Before doing a balance, let:
= volumetric heat rate of
production within the
fissionable material only
0 = volumetric heat rate of
production at r = 0
= a dimensionless positive constant
Nuclear Heat Source
Consider a spherical
nuclear fuel assembly
(solid sphere):
Before doing a balance, let:

= temperature profile in the
fissionable sphere
()

= temperature profile in the
()
= heat flux in the fissionable
sphere
()
= heat flux in the Al cladding

Nuclear Heat Source
Consider a spherical
nuclear fuel assembly
(solid sphere):
For the fissionable material:
()
4 2
()

− 4 2
+∆
+ 4 2 ∆  = 0
Rate of Heat IN:
()
2
4
Rate of Heat OUT:
4 2
Generation:
4 2 ∆

()
+∆
Electrical Heat Source
Generation = Volume X Sn
4
=   + ∆
3
3
−
3
∴  = 4  2 ∆
Too small
4
=   3 + 3 2 ∆ + 3 ∆
3
2
+ ∆
3
− 3
∴  = 4 2 ∆
Rate of Heat IN:
()
2
4
Rate of Heat OUT:
4 2
Generation:
4 2 ∆

()
+∆
Nuclear Heat Source
For the fissionable material:
()
4 2
()

− 4 2
+∆
No
generation
here!
+ 4 2 ∆  = 0
()
4 2
Dividing by 4∆:
()
2

2
+∆ −
∆
()

− 4 2
+∆
=0
Dividing by 4∆:

()
=
2
2

2
−

+∆

∆

=0
Nuclear Heat Source
For the fissionable material:
()
4 2
()

− 4 2
+∆
No
generation
here!
+ 4 2 ∆  = 0
()
4 2
Taking ∆ → 0:
2 ()

=   2

()

− 4 2
Taking ∆ → 0:
2 ()

=0

+∆
=0
Nuclear Heat Source
For the fissionable material:
()
4 2
()

− 4 2
+∆
No
generation
here!
+ 4 2 ∆  = 0
()
4 2
Taking ∆ → 0:
2 ()

= 0 1 +

()

− 4 2
Taking ∆ → 0:
2
2
2 ()

=0

+∆
=0
Nuclear Heat Source
For the fissionable material:
()
4 2
()

− 4 2
+∆
No
generation
here!
+ 4 2 ∆  = 0
()
4 2
Integrating:
()

− 4 2
Integrating:
+∆
=0
Nuclear Heat Source
Boundary Conditions:
= 0,
()
1

=0
Integrating:
For the fissionable material
Boundary Conditions:
()
= () ,
()
1
= 0

1
+

3 5
Integrating:
()
=
3
Nuclear Heat Source
For the fissionable material
Inserting Fourier’s Law:
Inserting Fourier’s Law:
Nuclear Heat Source
For the fissionable material
Boundary Conditions:
Boundary Conditions:
R(C)
At r = R(F),
T(F) = T(C)
R(F)
At r = R(C),
T(C) = T0
Nuclear Heat Source
For the fissionable material
Overall Shell Energy Balance
Recall the Overall Shell Energy Balance:
Q by Convective Transport
W by Molecular Transport
Q by Molecular Transport
W by
External
Forces
Energy
Generation
Overall Shell Energy Balance
We need all these terms for viscous dissipation:
Q by Convective Transport
Q by Molecular Transport
How can we account for
all these terms at once?
W by Molecular Transport
Combined Energy Flux Vector
We introduce something new to replace q:
Combined Energy Flux Vector:
Heat Rate from Molecular Motion
Convective Energy Flux
1 2
=
+   +  ∙  +
2
Work Rate from Molecular Motion
Combined Energy Flux Vector
We introduce something new to replace q:
Combined Energy Flux Vector:
Recall the molecular stress tensor:
When dotted with v:
=  +
∙  =  + [ ∙ ]
Substituting into e:
1 2
=
+   +  + [ ∙ ] +
2
Combined Energy Flux Vector
We introduce something new to replace q:
Combined Energy Flux Vector:
1 2
=
+   +  + [ ∙ ] +
2
Simplifying the boxed expression:

+  =   +
=   +   =

Finally:
1 2
=
+   + [ ∙ ] +
2
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid
between 2 coaxial cylinders:

=

Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid
between 2 coaxial cylinders:
We now make a shell
balance shown in red
on the left.
Rate of
Energy IN:

Rate of
Energy OUT:

When the combined energy flux vector is used, the
generation term will automatically appear from e.

+∆
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid
between 2 coaxial cylinders:

+∆
−

=0

= 1

=0
We now make a shell
balance shown in red
on the left.
Rate of
Energy IN:

Rate of
Energy OUT:

When the combined energy flux vector is used, the
generation term will automatically appear from e.

+∆
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid
between 2 coaxial cylinders:
1 2
=
+   + [ ∙ ] +
2
= 1
Fourier’s Law:
Newton’s Law:

= −

= −

When the combined energy flux vector is used, the
generation term will automatically appear from e.
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid
between 2 coaxial cylinders:
= 1
Substituting the
velocity profile:
Integrating:

−
−
= 1

−
−

=−

2
2
= 1
2 1
−  + 2
2

When the combined energy flux vector is used, the
generation term will automatically appear from e.
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid
between 2 coaxial cylinders:
Boundary Conditions:

=−

2
2 1
−  + 2
2

After applying the B.C.:
− 0
1
2
=
− 0 2   − 0

1− +

When the combined energy flux vector is used, the
generation term will automatically appear from e.
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid
between 2 coaxial cylinders:

=−

2
2
1
−  + 2
2

Q: So where is Sv?

=

After applying the B.C.:
− 0
1
2
=
− 0 2   − 0

1− +

When the combined energy flux vector is used, the
generation term will automatically appear from e.
2
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid
between 2 coaxial cylinders:
Temperature
Profile:
− 0
1
2
=
− 0 2   − 0

1− +

New Dimensionless Number:
Dim. Group
Brinkman, Br
Ratio
viscous heat dissipation/
molecular heat transport
Equation
2
− 0
Viscous Dissipation Source
Scenarios when viscous heating is significant:
1. Flow of lubricant between rapidly moving parts.
2. Flow of molten polymers through dies in highspeed extrusion.
3. Flow of highly viscous fluids in high-speed
viscometers.
4. Flow of air in the boundary layer near an earth
satellite or rocket during reentry into the earth’s
atmosphere.
```