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:
2𝜋𝑟𝐿𝑞𝑟
𝑟
− 2𝜋𝑟𝐿𝑞𝑟
𝑟+∆𝑟
+ 2𝜋𝑟∆𝑟𝐿 𝑆𝑒 = 0
Rate of Heat IN:
2𝜋𝑟𝐿𝑞𝑟
Rate of Heat OUT:
2𝜋𝑟𝐿𝑞𝑟
Generation:
2𝜋𝑟∆𝑟𝐿 𝑆𝑒
𝑟
𝑟+∆𝑟
Electrical Heat Source
Rate of Heat IN
2𝜋𝑟𝐿𝑞𝑟
Area perpendicular
to qr at r = r
𝑟
= (2𝜋𝑟𝐿) ∙ 𝑞𝑟
𝑟
The Shell:
Rate of Heat IN:
2𝜋𝑟𝐿𝑞𝑟
Rate of Heat OUT:
2𝜋𝑟𝐿𝑞𝑟
Generation:
2𝜋𝑟∆𝑟𝐿 𝑆𝑒
𝑟
𝑟+∆𝑟
Electrical Heat Source
Rate of Heat OUT
2𝜋𝑟𝐿𝑞𝑟
𝑟+∆𝑟
Area perpendicular
to qr at r = r + dr
= (2𝜋(𝑟 + ∆𝑟)𝐿) ∙ 𝑞𝑟
𝑟+∆𝑟
The Shell:
Rate of Heat IN:
2𝜋𝑟𝐿𝑞𝑟
Rate of Heat OUT:
2𝜋𝑟𝐿𝑞𝑟
Generation:
2𝜋𝑟∆𝑟𝐿 𝑆𝑒
𝑟
𝑟+∆𝑟
Electrical Heat Source
Generation = Volume X Se
𝑉 = 𝜋 𝑟 + ∆𝑟
2
− 𝑟2 𝐿
∴ 𝑉 = 𝜋 2𝑟∆𝑟 𝐿
Too small
= 𝜋 𝑟 2 + 2𝑟∆𝑟 + ∆𝑟
2
− 𝑟2 𝐿
∴ 𝐺𝑒𝑛 = 2𝜋𝑟∆𝑟𝐿 ∙ 𝑆𝑒
The Shell:
Rate of Heat IN:
2𝜋𝑟𝐿𝑞𝑟
Rate of Heat OUT:
2𝜋𝑟𝐿𝑞𝑟
Generation:
2𝜋𝑟∆𝑟𝐿 𝑆𝑒
𝑟
𝑟+∆𝑟
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
Shell Heat Balance:
2𝜋𝑟𝐿𝑞𝑟
2𝜋𝑟𝐿𝑞𝑟
𝑟
− 2𝜋𝑟𝐿𝑞𝑟
𝑟
𝑟+∆𝑟
− 2𝜋𝑟𝐿𝑞𝑟
+ 2𝜋𝑟∆𝑟𝐿 𝑆𝑒 = 0
𝑟+∆𝑟
= − 2𝜋𝑟∆𝑟𝐿 𝑆𝑒
Dividing by 2𝜋∆𝑟𝐿:
𝑟𝑞𝑟
𝑟
− 𝑟𝑞𝑟
∆𝑟
𝑟+∆𝑟
= −𝑆𝑒 𝑟
Q: Why did we divide by
2𝜋∆𝑟𝐿 and not by 2𝜋𝒓∆𝑟𝐿?
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
We now have:
𝑟𝑞𝑟 𝑟 − 𝑟𝑞𝑟
∆𝑟
𝑟+∆𝑟
= −𝑆𝑒 𝑟
Taking the limit as ∆𝑟 → 0:
𝑑
𝑟𝑞𝑟 = −𝑆𝑒 𝑟
𝑑𝑟
Q: Is this correct?
NO!
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
We now have:
𝑟𝑞𝑟 𝑟 − 𝑟𝑞𝑟
∆𝑟
𝑟+∆𝑟
= −𝑆𝑒 𝑟
We must adhere to the
definition of the derivative:
𝑟𝑞𝑟
lim
∆𝑟→0
𝑟𝑞𝑟
− 𝑟𝑞𝑟
∆𝑟
𝑟+∆𝑟
− 𝑟𝑞𝑟
∆𝑟
𝑟+∆𝑟
𝑟
𝑟
= +𝑆𝑒 𝑟
𝑑
=
𝑟𝑞𝑟 = 𝑆𝑒 𝑟
𝑑𝑟
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
𝑑
𝑟𝑞𝑟 = 𝑆𝑒 𝑟
𝑑𝑟
We now have:
Integrating:
𝑆𝑒 𝑟 𝐶1
𝑞𝑟 =
+
2
𝑟
Boundary conditions:
𝑎𝑡 𝑟 = 0,
𝑎𝑡 𝑟 = 𝑅,
𝑞𝑟 = 𝑓𝑖𝑛𝑖𝑡𝑒
𝑇 = 𝑇0
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:
𝑆𝑒 𝑟 𝐶1
𝑞𝑟 =
+
2
𝑟
Applying B.C. 1: 𝑎𝑡 𝑟 = 0, 𝑞𝑟 = 𝑓𝑖𝑛𝑖𝑡𝑒
Because q has to be finite at r = 0, all
the terms with radius, r, below the
denominator must vanish. Therefore:
𝐶1 = 0
𝑆𝑒 𝑟
𝑞𝑟 =
2
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
We now have:
𝑆𝑒 𝑟
𝑞𝑟 =
2
Substituting Fourier’s Law:
𝑑𝑇 𝑆𝑒 𝑟
−𝑘
=
𝑑𝑟
2
𝑑𝑇 −𝑆𝑒 𝑟
=
𝑑𝑟
2𝑘
−𝑆𝑒 𝑟 2
𝑇=
+ 𝐶2
4𝑘
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
We now have:
−𝑆𝑒 𝑟 2
𝑇=
+ 𝐶2
4𝑘
Applying B.C. 2: 𝑎𝑡 𝑟 = 𝑅,
−𝑆𝑒 𝑅2
𝑇0 =
+ 𝐶2
4𝑘
𝑆𝑒 𝑅2
𝐶2 = 𝑇0 +
4𝑘
This is it! But,
we rewrite it
into a nicer
−𝑆𝑒 𝑟 2 𝑆𝑒 𝑅2
form…
𝑇=
+
+ 𝑇0
4𝑘
4𝑘
𝑇 = 𝑇0
Electrical Heat Source
Consider an electrical wire
(solid cylinder):
Temperature Profile:
𝑆𝑒 𝑅2
𝑟
𝑇 − 𝑇0 =
1−
4𝑘
𝑅
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:
𝑘𝑒 = electrical conductivity
𝐼 = current density
𝑎𝑚𝑝
𝑐𝑚2
1
Ω ∙ 𝑐𝑚
𝐸 = voltage drop over a length
These imply the following :
𝐼2
𝑆𝑒 =
𝑘𝑒
𝐸
𝐼 = 𝑘𝑒
𝐿
𝑉
𝑆𝑒 𝑅2
𝑟
𝑇 − 𝑇0 =
1−
4𝑘
𝑅
2
𝑆𝑒 𝑅2 𝐸 2 𝑅2 𝑘𝑒
=
4𝑘
4𝐿2 𝑘
Electrical Heat Source
Heat flux profile:
𝑆𝑒 𝑟
𝑞𝑟 =
2
The stress profile versus
the temperature profile:
Temperature Profile:
𝑆𝑒 𝑅2
𝑟
𝑇 − 𝑇0 =
1−
4𝑘
𝑅
2
Electrical Heat Source
Quantities that might be asked for:
1. Maximum Temperature
Substituting r = 0
to the profile T(r):
𝑇𝑚𝑎𝑥
𝑆𝑒 𝑅2
= 𝑇0 +
4𝑘
2. Average Temperature Rise
𝑇 − 𝑇0 =
2𝜋 𝑅
𝑇 − 𝑇0 𝑟 𝑑𝑟
0
0
2𝜋 𝑅
𝑟 𝑑𝑟 𝑑𝜃
0
0
𝑑𝜃
𝑆𝑒 𝑅2 1
𝑇 − 𝑇0 =
= 𝑇𝑚𝑎𝑥
8𝑘
2
3. Heat Outflow Rate at the Surface
𝑄
𝑄
𝑆𝑒 𝑟
𝑞𝑟 = =
=
𝐴 2𝜋𝑅𝐿
2
𝑟=𝑅
𝑄 = 𝜋𝑅2 𝐿 ∙ 𝑆𝑒
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:
𝑆𝑛 = volumetric heat rate of
production within the
fissionable material only
𝑆𝑛0 = volumetric heat rate of
production at r = 0
Sn depends on radius parabolically:
𝑏 = a dimensionless positive constant
Nuclear Heat Source
Consider a spherical
nuclear fuel assembly
(solid sphere):
Before doing a balance, let:
𝑇𝑟
𝐹
= temperature profile in the
fissionable sphere
(𝐶)
𝑇𝑟
= temperature profile in the
Alcladding
(𝐹)
= heat flux in the fissionable
sphere
(𝐶)
= heat flux in the Al cladding
𝑞𝑟
𝑞𝑟
Nuclear Heat Source
Consider a spherical
nuclear fuel assembly
(solid sphere):
For the fissionable material:
(𝐹)
4𝜋𝑟 2 𝑞𝑟
(𝐹)
𝑟
− 4𝜋𝑟 2 𝑞𝑟
𝑟+∆𝑟
+ 4𝜋𝑟 2 ∆𝑟 𝑆𝑛 = 0
Rate of Heat IN:
(𝐹)
2
4𝜋𝑟 𝑞𝑟
Rate of Heat OUT:
4𝜋𝑟 2 𝑞𝑟
Generation:
4𝜋𝑟 2 ∆𝑟 𝑆𝑛
𝑟
(𝐹)
𝑟+∆𝑟
Electrical Heat Source
Generation = Volume X Sn
4
𝑉 = 𝜋 𝑟 + ∆𝑟
3
3
−
𝑟3
∴ 𝑉 = 4𝜋 𝑟 2 ∆𝑟
Too small
4
= 𝜋 𝑟 3 + 3𝑟 2 ∆𝑟 + 3𝑟 ∆𝑟
3
2
+ ∆𝑟
3
− 𝑟3
∴ 𝐺𝑒𝑛 = 4𝜋𝑟 2 ∆𝑟 𝑆𝑛
Rate of Heat IN:
(𝐹)
2
4𝜋𝑟 𝑞𝑟
Rate of Heat OUT:
4𝜋𝑟 2 𝑞𝑟
Generation:
4𝜋𝑟 2 ∆𝑟 𝑆𝑛
𝑟
(𝐹)
𝑟+∆𝑟
Nuclear Heat Source
For the fissionable material:
(𝐹)
4𝜋𝑟 2 𝑞𝑟
(𝐹)
𝑟
− 4𝜋𝑟 2 𝑞𝑟
𝑟+∆𝑟
No
generation
here!
+ 4𝜋𝑟 2 ∆𝑟 𝑆𝑛 = 0
For the Al cladding:
(𝐶)
4𝜋𝑟 2 𝑞𝑟
Dividing by 4𝜋∆𝑟:
(𝐹)
𝑟 2 𝑞𝑟
𝐹
2
𝑟+∆𝑟 − 𝑟 𝑞𝑟
∆𝑟
(𝐶)
𝑟
− 4𝜋𝑟 2 𝑞𝑟
𝑟+∆𝑟
=0
Dividing by 4𝜋∆𝑟:
𝑟
(𝐶)
= 𝑆𝑛 𝑟
2
𝑟 2 𝑞𝑟
𝐶
2𝑞
−
𝑟
𝑟+∆𝑟
𝑟
∆𝑟
𝑟
=0
Nuclear Heat Source
For the fissionable material:
(𝐹)
4𝜋𝑟 2 𝑞𝑟
(𝐹)
𝑟
− 4𝜋𝑟 2 𝑞𝑟
𝑟+∆𝑟
No
generation
here!
+ 4𝜋𝑟 2 ∆𝑟 𝑆𝑛 = 0
For the Al cladding:
(𝐶)
4𝜋𝑟 2 𝑞𝑟
Taking ∆𝑟 → 0:
𝑑 2 (𝐹)
𝑟 𝑞𝑟
= 𝑆𝑛 𝑟 2
𝑑𝑟
(𝐶)
𝑟
− 4𝜋𝑟 2 𝑞𝑟
Taking ∆𝑟 → 0:
𝑑 2 (𝐶)
𝑟 𝑞𝑟
=0
𝑑𝑟
𝑟+∆𝑟
=0
Nuclear Heat Source
For the fissionable material:
(𝐹)
4𝜋𝑟 2 𝑞𝑟
(𝐹)
𝑟
− 4𝜋𝑟 2 𝑞𝑟
𝑟+∆𝑟
No
generation
here!
+ 4𝜋𝑟 2 ∆𝑟 𝑆𝑛 = 0
For the Al cladding:
(𝐶)
4𝜋𝑟 2 𝑞𝑟
Taking ∆𝑟 → 0:
𝑑 2 (𝐹)
𝑟
𝑟 𝑞𝑟
= 𝑆𝑛0 1 + 𝑏
𝑑𝑟
𝑅𝐹
(𝐶)
𝑟
− 4𝜋𝑟 2 𝑞𝑟
Taking ∆𝑟 → 0:
2
𝑟2
𝑑 2 (𝐶)
𝑟 𝑞𝑟
=0
𝑑𝑟
𝑟+∆𝑟
=0
Nuclear Heat Source
For the fissionable material:
(𝐹)
4𝜋𝑟 2 𝑞𝑟
(𝐹)
𝑟
− 4𝜋𝑟 2 𝑞𝑟
𝑟+∆𝑟
No
generation
here!
+ 4𝜋𝑟 2 ∆𝑟 𝑆𝑛 = 0
For the Al cladding:
(𝐶)
4𝜋𝑟 2 𝑞𝑟
Integrating:
(𝐶)
𝑟
− 4𝜋𝑟 2 𝑞𝑟
Integrating:
𝑟+∆𝑟
=0
Nuclear Heat Source
Boundary Conditions:
𝑎𝑡 𝑟 = 0,
(𝐹)
𝐶1
𝑞𝑟 𝑖𝑠 𝑓𝑖𝑛𝑖𝑡𝑒
=0
Integrating:
For the fissionable material
Boundary Conditions:
(𝐹)
𝑎𝑡 𝑟 = 𝑅(𝐹) ,
(𝐶)
𝐶1
= 𝑆𝑛0
𝑞𝑟
1 𝑏
+
𝑅
3 5
Integrating:
For the Al cladding
(𝐶)
= 𝑞𝑟
𝐹 3
Nuclear Heat Source
For the fissionable material
Inserting Fourier’s Law:
For the Al cladding
Inserting Fourier’s Law:
Nuclear Heat Source
For the fissionable material
For the Al cladding
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
For the Al cladding
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
SteadyState!
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
𝒆=
𝜌𝑣 + 𝜌𝑈 𝒗 + 𝝅 ∙ 𝒗 + 𝒒
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:
𝝅 = 𝑝𝜹 + 𝝉
𝝅 ∙ 𝒗 = 𝑝𝒗 + [𝝉 ∙ 𝒗]
Substituting into e:
1 2
𝒆=
𝜌𝑣 + 𝜌𝑈 𝒗 + 𝑝𝒗 + [𝝉 ∙ 𝒗] + 𝒒
2
Combined Energy Flux Vector
We introduce something new to replace q:
Combined Energy Flux Vector:
1 2
𝒆=
𝜌𝑣 + 𝜌𝑈 𝒗 + 𝑝𝒗 + [𝝉 ∙ 𝒗] + 𝒒
2
Simplifying the boxed expression:
𝑝
𝜌𝑈𝒗 + 𝑝𝒗 = 𝜌 𝑈 +
𝒗 = 𝜌 𝑈 + 𝑝𝑉 𝒗 = 𝜌𝐻𝒗
𝜌
Finally:
1 2
𝒆=
𝜌𝑣 + 𝜌𝐻 𝒗 + [𝝉 ∙ 𝒗] + 𝒒
2
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid
between 2 coaxial cylinders:
𝑥
𝑣𝑧 𝑥 = 𝑣𝑏
𝑏
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:
𝑊𝐿𝒆𝒙
Rate of
Energy OUT:
𝑊𝐿𝒆𝒙
When the combined energy flux vector is used, the
generation term will automatically appear from e.
𝑥
𝑥+∆𝑥
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid
between 2 coaxial cylinders:
𝑊𝐿𝒆𝒙
𝑥+∆𝑥
− 𝑊𝐿𝒆𝒙
𝑑𝒆𝒙
=0
𝑑𝑥
𝒆𝒙 = 𝐶1
𝑥
=0
We now make a shell
balance shown in red
on the left.
Rate of
Energy IN:
𝑊𝐿𝒆𝒙
Rate of
Energy OUT:
𝑊𝐿𝒆𝒙
When the combined energy flux vector is used, the
generation term will automatically appear from e.
𝑥
𝑥+∆𝑥
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid
between 2 coaxial cylinders:
1 2
𝒆=
𝜌𝑣 + 𝜌𝐻 𝒗 + [𝝉 ∙ 𝒗] + 𝒒
2
𝒆𝒙 = 𝐶1
Fourier’s Law:
Newton’s Law:
𝑑𝑇
𝑞𝑥 = −𝑘
𝑑𝑥
𝜏𝑥𝑧
𝑑𝑣𝑧
= −𝜇
𝑑𝑥
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:
𝒆𝒙 = 𝐶1
Substituting the
velocity profile:
Integrating:
𝑑𝑇
𝑑𝑣𝑧
−𝑘
− 𝜇𝑣𝑧
= 𝐶1
𝑑𝑥
𝑑𝑥
𝑑𝑇
𝑣𝑏
−𝑘
− 𝜇𝑥
𝑑𝑥
𝑏
𝜇 𝑣𝑏
𝑇=−
𝑘 𝑏
2
2
= 𝐶1
𝑥 2 𝐶1
− 𝑥 + 𝐶2
2
𝑘
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:
𝜇 𝑣𝑏
𝑇=−
𝑘 𝑏
2
𝑥 2 𝐶1
− 𝑥 + 𝐶2
2
𝑘
After applying the B.C.:
𝑇 − 𝑇0
1
𝜇𝑣𝑏2
=
𝑇𝑏 − 𝑇0 2 𝑘 𝑇𝑏 − 𝑇0
𝑥
𝑏
𝑥
𝑥
1− +
𝑏
𝑏
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:
𝜇 𝑣𝑏
𝑇=−
𝑘 𝑏
2
𝑥2
𝐶1
− 𝑥 + 𝐶2
2
𝑘
Q: So where is Sv?
𝑣𝑏
𝑆𝑣 = 𝜇
𝑏
After applying the B.C.:
𝑇 − 𝑇0
1
𝜇𝑣𝑏2
=
𝑇𝑏 − 𝑇0 2 𝑘 𝑇𝑏 − 𝑇0
𝑥
𝑏
𝑥
𝑥
1− +
𝑏
𝑏
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:
𝑇 − 𝑇0
1
𝜇𝑣𝑏2
=
𝑇𝑏 − 𝑇0 2 𝑘 𝑇𝑏 − 𝑇0
𝑥
𝑏
𝑥
𝑥
1− +
𝑏
𝑏
New Dimensionless Number:
Dim. Group
Brinkman, Br
Ratio
viscous heat dissipation/
molecular heat transport
Equation
𝜇𝑣𝑏2
𝑘 𝑇𝑏 − 𝑇0
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.