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Overall Shell Energy Balance. Forms of Energy Generation : Degradation of electrical energy to heat Heat from nuclear source (by fission) Heat from viscous dissipation. (S e ) (S n ) ( S v ). Energy Generation. Let S = rate of heat production per unit volume (W/m 3 ). - PowerPoint PPT Presentation
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Forms of Energy Generation:
1. Degradation of electrical energy to heat
2. Heat from nuclear source (by fission)
3. Heat from viscous dissipation
Overall Shell Energy Balance
Energy Generation
Let S = rate of heat production per unit volume (W/m3)
(Se)
(Sn)
(Sv)
Electrical Heat Source
Consider an electrical wire (solid cylinder):
Shell Heat Balance:
(2๐๐๐ฟ๐๐ ) |๐ โ (2๐๐๐ฟ๐๐ ) |๐+โ๐+ (2๐ ๐ โ๐๐ฟ )๐๐=0
Rate of Heat IN:
Rate of Heat OUT:
Generation:
(2๐๐๐ฟ๐๐ ) |๐(2๐๐๐ฟ๐๐ ) |๐+โ๐
(2๐๐ โ๐๐ฟ )๐๐
Electrical Heat Source
Rate of Heat IN:
Rate of Heat OUT:
Generation:
(2๐๐๐ฟ๐๐ ) |๐(2๐๐๐ฟ๐๐ ) |๐+โ๐
(2๐๐ โ๐๐ฟ )๐๐
The Shell:
(2๐๐๐ฟ๐๐ ) |๐=(2๐๐๐ฟ) โ (๐๐ |๐ )
Rate of Heat IN Area perpendicular to qr at r = r
Electrical Heat Source
Rate of Heat IN:
Rate of Heat OUT:
Generation:
(2๐๐๐ฟ๐๐ ) |๐(2๐๐๐ฟ๐๐ ) |๐+โ๐
(2๐๐ โ๐๐ฟ )๐๐
The Shell:
(2๐๐๐ฟ๐๐ ) |๐+โ๐=(2๐(๐ +โ๐ )๐ฟ) โ (๐๐ |๐ +โ๐ )
Rate of Heat OUT Area perpendicular to qr at r = r + dr
Electrical Heat Source
Rate of Heat IN:
Rate of Heat OUT:
Generation:
(2๐๐๐ฟ๐๐ ) |๐(2๐๐๐ฟ๐๐ ) |๐+โ๐
(2๐๐ โ๐๐ฟ )๐๐
The Shell:
Generation = Volume X Se
๐=๐ [ (๐+โ๐ )2โ๐2 ] ๐ฟToo small
โด๐=๐ [2๐ โ๐ ]๐ฟ โด๐บ๐๐=2๐ ๐ โ๐๐ฟ โ๐๐
ยฟ๐ [๐2+2๐ โ๐+ (โ๐ )2โ๐2 ] ๐ฟ
Electrical Heat Source
Consider an electrical wire (solid cylinder):
Shell Heat Balance:
(2๐๐๐ฟ๐๐ ) |๐ โ (2๐๐๐ฟ๐๐ ) |๐+โ๐+ (2๐ ๐ โ๐๐ฟ )๐๐=0
Dividing by
(๐ ๐๐ )|๐โ (๐ ๐๐ )|๐+โ๐
โ๐ =โ๐๐ ๐
Q: Why did we divide by and not by ?
(2๐๐๐ฟ๐๐ ) |๐ โ (2๐๐๐ฟ๐๐ ) |๐+โ๐=โ (2๐๐ โ๐๐ฟ)๐๐
Electrical Heat Source
Consider an electrical wire (solid cylinder):
We now have:(๐ ๐๐ )|๐โ (๐ ๐๐ )|๐+โ๐
โ ๐ =โ๐๐ ๐
Taking the limit as :
Q: Is this correct?
๐๐๐ (๐ ๐๐ )=โ๐๐ ๐
NO!
Electrical Heat Source
Consider an electrical wire (solid cylinder):
We now have:(๐ ๐๐ )|๐โ (๐ ๐๐ )|๐+โ๐
โ ๐ =โ๐๐ ๐
We must adhere to the definition of the derivative:
(๐ ๐๐ )|๐+โ ๐โ (๐ ๐๐ ) |๐โ๐ =+๐๐ ๐
limโ๐โ0
(๐ ๐๐ )|๐+โ ๐โ (๐ ๐๐ ) |๐โ ๐ = ๐
๐๐ (๐ ๐๐ )=๐๐๐
Electrical Heat Source
Consider an electrical wire (solid cylinder):
We now have:๐๐๐ (๐ ๐๐ )=๐๐๐
Boundary conditions:
๐๐ก ๐=0 ,๐๐= ๐๐๐๐๐ก๐
๐๐ก ๐=๐ ,๐=๐ 0
Note: The problem statement will tell you hints about what boundary conditions to use.
Integrating: ๐๐=๐๐๐2
+๐ถ1
๐
Electrical Heat Source
Consider an electrical wire (solid cylinder):
We now have:
๐๐ก ๐=0 ,๐๐= ๐๐๐๐๐ก๐Applying B.C. 1:
๐๐=๐๐๐2
+๐ถ1
๐
Because q has to be finite at r = 0, all the terms with radius, r, below the denominator must vanish. Therefore:
๐ถ1=0
๐๐=๐๐๐2
Electrical Heat Source
Consider an electrical wire (solid cylinder):
We now have: ๐๐=๐๐๐2
Substituting Fourierโs Law:
โ๐ ๐๐๐๐ =
๐๐ ๐2
๐๐๐๐ =
โ๐๐๐2๐
๐=โ๐๐๐ 2
4๐ +๐ถ2
Electrical Heat Source
Consider an electrical wire (solid cylinder):
We now have: ๐=โ๐๐๐ 2
4๐ +๐ถ2
Applying B.C. 2: ๐๐ก ๐=๐ ,๐=๐ 0
๐ 0=โ๐๐๐ 2
4 ๐ +๐ถ2
๐ถ2=๐0+๐๐๐ 2
4 ๐
๐=โ๐๐๐ 2
4๐ +๐๐๐ 2
4๐ +๐ 0
This is it! But, we rewrite it into a nicer formโฆ
Electrical Heat Source
Consider an electrical wire (solid cylinder):
๐ โ๐0=๐๐๐ 2
4๐ [1โ( ๐๐ )2]
Temperature Profile:
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 notesโฆLet: electrical conductivity
current density
voltage drop over a length
[ 1ฮฉโ๐๐ ]
[ ๐๐๐๐๐2 ]
[๐ ]
๐๐=๐ผ 2๐๐
๐ผ=๐๐๐ธ๐ฟ
๐๐๐ 2
4๐ =๐ธ2๐ 2
4๐ฟ2 (๐๐
๐ )These imply the following : ๐ โ๐0=
๐๐๐ 2
4๐ [1โ( ๐๐ )2]
Electrical Heat Source
๐ โ๐0=๐๐๐ 2
4๐ [1โ( ๐๐ )2]
Temperature Profile:
The stress profile versus the temperature profile:
๐๐=๐๐๐2
Heat flux profile:
Electrical Heat Source
Quantities that might be asked for:
1. Maximum Temperature
2. Average Temperature Rise
3. Heat Outflow Rate at the Surface
Substituting r = 0 to the profile T(r): ๐๐๐๐ฅ=๐ 0+
๐๐๐ 2
4 ๐
โจ๐ โฉโ๐ 0=โซ0
2๐
โซ0
๐
(๐ โ๐0 )๐ ๐๐ ๐ ๐
โซ0
2 ๐
โซ0
๐
๐ ๐๐ ๐ ๐โจ๐ โฉโ๐ 0=
๐๐๐ 2
8๐ =12๐๐๐๐ฅ
๐๐=๐๐ด= ๐
2๐ ๐ ๐ฟ=๐๐๐2
|๐=๐ ๐=๐ ๐ 2๐ฟ โ๐๐
Electrical Heat Source
Examples for Review:
Example 10.2-1 and Example 10.2-2Bird, 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
volumetric heat rate of production at r = 0
Sn depends on radius parabolically:
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 Alcladding
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:
(4๐ ๐2๐๐(๐น ))|๐โ (4๐๐ 2๐๐
(๐น) )|๐+โ ๐+(4 ๐๐2โ๐ )๐๐=0
Rate of Heat IN:
Rate of Heat OUT:
Generation:
(4๐ ๐2๐๐(๐น ))|๐
(4๐ ๐2๐๐(๐น ))|๐+โ๐
(4๐๐ 2โ ๐ )๐๐
Electrical Heat Source
Rate of Heat IN:
Rate of Heat OUT:
Generation:
Generation = Volume X Sn
๐=43 ๐ [ (๐+โ๐ )3โ๐3 ]
Too small
โด๐=4๐ [๐2โ๐ ] โด๐บ๐๐=(4๐๐ 2โ๐ )๐๐
ยฟ43 ๐ [๐3+3๐ 2โ๐ +3๐ (โ ๐ )2+ (โ๐ )3โ๐ 3 ]
(4๐ ๐2๐๐(๐น ))|๐
(4๐ ๐2๐๐(๐น ))|๐+โ๐
(4๐๐ 2โ ๐ )๐๐
Nuclear Heat Source
For the fissionable material:
(4๐ ๐2๐๐(๐น ))|๐โ (4๐๐ 2๐๐
(๐น) )|๐+โ ๐+(4 ๐๐2โ๐ )๐๐=0
For the Al cladding:
(4๐ ๐2๐๐(๐ถ ))|๐โ ( 4๐๐ 2๐๐
(๐ถ))|๐ +โ๐=0
No generation
here!
Dividing by :
(๐ 2๐๐(๐น ))|๐+โ๐ โ (๐ 2๐๐
(๐น ) )|๐โ๐ =๐๐๐
2
Dividing by :
(๐ 2๐๐(๐ถ))|๐+โ๐โ (๐ 2๐๐
(๐ถ ))|๐โ๐ =0
Nuclear Heat Source
For the fissionable material:
(4๐ ๐2๐๐(๐น ))|๐โ (4๐๐ 2๐๐
(๐น) )|๐+โ ๐+(4 ๐๐2โ๐ )๐๐=0
For the Al cladding:
(4๐ ๐2๐๐(๐ถ ))|๐โ ( 4๐๐ 2๐๐
(๐ถ))|๐ +โ๐=0
No generation
here!
Taking :
๐๐๐ (๐ 2๐๐
(๐น ))=๐๐๐ 2
Taking :
๐๐๐ (๐ 2๐๐
(๐ถ))=0
Nuclear Heat Source
For the fissionable material:
(4๐ ๐2๐๐(๐น ))|๐โ (4๐๐ 2๐๐
(๐น) )|๐+โ ๐+(4 ๐๐2โ๐ )๐๐=0
For the Al cladding:
(4๐ ๐2๐๐(๐ถ ))|๐โ ( 4๐๐ 2๐๐
(๐ถ))|๐ +โ๐=0
No generation
here!
Taking :
๐๐๐ (๐ 2๐๐
(๐น ))=๐๐0[1+๐ ( ๐๐ (๐น ) )
2]๐ 2Taking :
๐๐๐ (๐ 2๐๐
(๐ถ))=0
Nuclear Heat Source
For the fissionable material:
(4๐ ๐2๐๐(๐น ))|๐โ (4๐๐ 2๐๐
(๐น) )|๐+โ ๐+(4 ๐๐2โ๐ )๐๐=0
For the Al cladding:
(4๐ ๐2๐๐(๐ถ ))|๐โ ( 4๐๐ 2๐๐
(๐ถ))|๐ +โ๐=0
No generation
here!
Integrating: Integrating:
Nuclear Heat Source
Integrating: Integrating:
Boundary Conditions: Boundary Conditions:
๐๐ก ๐=0 ,๐๐ ๐๐ ๐๐๐๐๐ก๐ ๐๐ก ๐=๐ (๐น ) ,๐๐(๐น )=๐๐
(๐ถ )
๐ถ1(๐น )=0 ๐ถ1
(๐ถ)=๐๐0 ( 13+๐5 )๐ (๐น )3
For the fissionable material For the Al cladding
Nuclear Heat Source
Inserting Fourierโs Law: Inserting Fourierโs Law:
For the fissionable material For the Al cladding
Nuclear Heat Source
For the fissionable material For the Al cladding
Boundary Conditions: Boundary Conditions:
At r = R(F),
T(F) = T(C) R(F)
R(C)
At r = R(C),
T(C) = T0
Nuclear Heat Source
For the fissionable material
For the Al cladding
Recall the Overall Shell Energy Balance:
Overall Shell Energy Balance
Q by Convective Transport Q by Molecular Transport
W by Molecular Transport W by External Forces
Energy Generation
Steady-State!
Overall Shell Energy Balance
Q by Convective Transport Q by Molecular Transport
W by Molecular Transport
How can we account for all these terms at once?
We need all these terms for viscous dissipation:
Combined Energy Flux Vector
Convective Energy FluxHeat Rate from Molecular Motion
Work Rate from Molecular Motion
Combined Energy Flux Vector:
๐=( 12 ๐๐ฃ2+๐ ๏ฟฝฬ๏ฟฝ)๐+ [๐ โ๐ ]+๐
We introduce something new to replace q:
Combined Energy Flux Vector
Combined Energy Flux Vector:
We introduce something new to replace q:
๐ =๐๐น+๐Recall the molecular stress tensor:When dotted with v: [๐ โ๐ ]=๐ ๐+[๐ โ๐ ]
Substituting into e:
๐=( 12 ๐๐ฃ2+๐ ๏ฟฝฬ๏ฟฝ)๐+๐๐+[๐ โ๐ ]+๐
Combined Energy Flux Vector
Combined Energy Flux Vector:
We introduce something new to replace q:
๐=( 12 ๐๐ฃ2+๐ ๏ฟฝฬ๏ฟฝ)๐+๐๐+[๐ โ๐ ]+๐
Simplifying the boxed expression:
๐ ๏ฟฝฬ๏ฟฝ ๐+๐ ๐=๐ (๏ฟฝฬ๏ฟฝ+ ๐๐ )๐=๐ (๐+๐ ๏ฟฝฬ๏ฟฝ ) ๐=๐ ๏ฟฝฬ๏ฟฝ ๐
Finally: ๐=( 12 ๐๐ฃ2+๐ ๏ฟฝฬ๏ฟฝ )๐+[๐ โ๐ ]+๐
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
๐ฃ ๐ง (๐ฅ )=๐ฃ๐( ๐ฅ๐ )
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:
๐๐ฟ๐๐ |๐ฅ
๐๐ฟ๐๐ |๐ฅ+โ ๐ฅ
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:
We now make a shell balance shown in red on the left.
Rate of Energy IN:
Rate of Energy OUT:
๐๐ฟ๐๐ |๐ฅ
๐๐ฟ๐๐ |๐ฅ+โ ๐ฅ
When the combined energy flux vector is used, the generation term will automatically appear from e.
๐๐ฟ๐๐ |๐ฅ+โ ๐ฅโ๐๐ฟ๐๐ |๐ฅ=0๐๐๐
๐๐ฅ =0
๐๐=๐ถ1
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
When the combined energy flux vector is used, the generation term will automatically appear from e.
๐๐=๐ถ1 ๐=( 12 ๐๐ฃ2+๐ ๏ฟฝฬ๏ฟฝ )๐+[๐ โ๐ ]+๐
Fourierโs Law:
Newtonโs Law:
๐๐ฅ=โ๐๐๐๐๐ฅ
๐๐ฅ๐ง=โ๐๐๐ฃ ๐ง
๐๐ฅ
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
When the combined energy flux vector is used, the generation term will automatically appear from e.
๐๐=๐ถ1 โ๐ ๐๐๐๐ฅ โ๐๐ฃ๐ง
๐๐ฃ๐ง
๐๐ฅ =๐ถ1
Substituting the velocity profile:
Integrating:
โ๐ ๐๐๐๐ฅ โ๐๐ฅ (๐ฃ๐
๐ )2
=๐ถ1
๐=โ ๐๐ ( ๐ฃ๐
๐ )2 ๐ฅ22โ๐ถ1
๐ ๐ฅ+๐ถ2
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
When the combined energy flux vector is used, the generation term will automatically appear from e.
Boundary Conditions:
After applying the B.C.:
๐ โ๐ 0
๐๐โ๐0=12 [ ๐๐ฃ๐
2
๐ (๐ ๐โ๐ 0 ) ]( ๐ฅ๐ )(1โ ๐ฅ๐ )+๐ฅ๐
๐=โ ๐๐ ( ๐ฃ๐
๐ )2 ๐ฅ22โ๐ถ1
๐ ๐ฅ+๐ถ2
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
When the combined energy flux vector is used, the generation term will automatically appear from e.
๐ โ๐ 0
๐๐โ๐0=12 [ ๐๐ฃ๐
2
๐ (๐ ๐โ๐ 0 ) ]( ๐ฅ๐ )(1โ ๐ฅ๐ )+๐ฅ๐
Q: So where is Sv?
๐๐ฃ=๐(๐ฃ๐
๐ )2
๐=โ ๐๐ ( ๐ฃ๐
๐ )2 ๐ฅ22โ๐ถ1
๐ ๐ฅ+๐ถ2
After applying the B.C.:
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
๐ โ๐ 0
๐๐โ๐0=12 [ ๐๐ฃ๐
2
๐ (๐ ๐โ๐ 0 ) ]( ๐ฅ๐ )(1โ ๐ฅ๐ )+๐ฅ๐Temperature
Profile:
New Dimensionless Number:Dim. Group Ratio Equation
Brinkman, Br viscous heat dissipation/ molecular heat transport
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 high-
speed 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 earthโs atmosphere.