Forms of Energy Generation : Degradation of electrical energy to heat

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:

(2𝜋𝑟𝐿𝑞𝑟 ) |𝑟 − (2𝜋𝑟𝐿𝑞𝑟 ) |𝑟+∆𝑟+ (2𝜋 𝑟 ∆𝑟𝐿 )𝑆𝑒=0

Rate of Heat IN:

Rate of Heat OUT:

Generation:

(2𝜋𝑟𝐿𝑞𝑟 ) |𝑟(2𝜋𝑟𝐿𝑞𝑟 ) |𝑟+∆𝑟

(2𝜋𝑟 ∆𝑟𝐿 )𝑆𝑒


Rate of Heat IN:

Rate of Heat OUT:

Generation:



The Shell:

(2𝜋𝑟𝐿𝑞𝑟 ) |𝑟=(2𝜋𝑟𝐿) ∙ (𝑞𝑟 |𝑟 )

Rate of Heat IN Area perpendicular to qr at r = r


Rate of Heat IN:

Rate of Heat OUT:

Generation:



The Shell:

(2𝜋𝑟𝐿𝑞𝑟 ) |𝑟+∆𝑟=(2𝜋(𝑟 +∆𝑟 )𝐿) ∙ (𝑞𝑟 |𝑟 +∆𝑟 )

Rate of Heat OUT Area perpendicular to qr at r = r + dr


Rate of Heat IN:

Rate of Heat OUT:

Generation:



The Shell:

Generation = Volume X Se

𝑉=𝜋 [ (𝑟+∆𝑟 )2−𝑟2 ] 𝐿Too small

∴𝑉=𝜋 [2𝑟 ∆𝑟 ]𝐿 ∴𝐺𝑒𝑛=2𝜋 𝑟 ∆𝑟𝐿 ∙𝑆𝑒

¿𝜋 [𝑟2+2𝑟 ∆𝑟+ (∆𝑟 )2−𝑟2 ] 𝐿



We now have:(𝑟 𝑞𝑟 )|𝑟− (𝑟 𝑞𝑟 )|𝑟+∆𝑟

∆ 𝑟 =−𝑆𝑒 𝑟

Taking the limit as :

Q: Is this correct?

𝑑𝑑𝑟 (𝑟 𝑞𝑟 )=−𝑆𝑒 𝑟

NO!



We now have:𝑑𝑑𝑟 (𝑟 𝑞𝑟 )=𝑆𝑒𝑟

Boundary conditions:

𝑎𝑡 𝑟=0 ,𝑞𝑟= 𝑓𝑖𝑛𝑖𝑡𝑒

𝑎𝑡 𝑟=𝑅 ,𝑇=𝑇 0

Note: The problem statement will tell you hints about what boundary conditions to use.

Integrating: 𝑞𝑟=𝑆𝑒𝑟2

+𝐶1

𝑟



We now have:

𝑎𝑡 𝑟=0 ,𝑞𝑟= 𝑓𝑖𝑛𝑖𝑡𝑒Applying B.C. 1:

𝑞𝑟=𝑆𝑒𝑟2

+𝐶1

𝑟

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

𝐶1=0




We now have: 𝑞𝑟=𝑆𝑒𝑟2

Substituting Fourier’s Law:

−𝑘 𝑑𝑇𝑑𝑟 =

𝑆𝑒 𝑟2

𝑑𝑇𝑑𝑟 =

−𝑆𝑒𝑟2𝑘

𝑇=−𝑆𝑒𝑟 2

4𝑘 +𝐶2



We now have: 𝑇=−𝑆𝑒𝑟 2

4𝑘 +𝐶2

Applying B.C. 2: 𝑎𝑡 𝑟=𝑅 ,𝑇=𝑇 0

𝑇 0=−𝑆𝑒𝑅2

4 𝑘 +𝐶2

𝐶2=𝑇0+𝑆𝑒𝑅2

4 𝑘

𝑇=−𝑆𝑒𝑟 2

4𝑘 +𝑆𝑒𝑅2

4𝑘 +𝑇 0

This is it! But, we rewrite it into a nicer form…



𝑇 −𝑇0=𝑆𝑒𝑅2

4𝑘 [1−( 𝑟𝑅 )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.


Other important notes…Let: electrical conductivity

current density

voltage drop over a length

[ 1Ω∙𝑐𝑚 ]

[ 𝑎𝑚𝑝𝑐𝑚2 ]

[𝑉 ]

𝑆𝑒=𝐼 2𝑘𝑒

𝐼=𝑘𝑒𝐸𝐿

𝑆𝑒𝑅2

4𝑘 =𝐸2𝑅2

4𝐿2 (𝑘𝑒

𝑘 )These imply the following : 𝑇 −𝑇0=

𝑆𝑒𝑅2

4𝑘 [1−( 𝑟𝑅 )2]


𝑇 −𝑇0=𝑆𝑒𝑅2

4𝑘 [1−( 𝑟𝑅 )2]

Temperature Profile:

The stress profile versus the temperature profile:


Heat flux profile:


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): 𝑇𝑚𝑎𝑥=𝑇 0+

𝑆𝑒𝑅2

4 𝑘

⟨𝑇 ⟩−𝑇 0=∫0

2𝜋

∫0

𝑅

(𝑇 −𝑇0 )𝑟 𝑑𝑟 𝑑 𝜃

∫0

2 𝜋

∫0

𝑅

𝑟 𝑑𝑟 𝑑 𝜃⟨𝑇 ⟩−𝑇 0=

𝑆𝑒𝑅2

8𝑘 =12𝑇𝑚𝑎𝑥

𝑞𝑟=𝑄𝐴= 𝑄

2𝜋 𝑅𝐿=𝑆𝑒𝑟2

|𝑟=𝑅 𝑄=𝜋 𝑅2𝐿 ∙𝑆𝑒


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


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


For the fissionable material:

(4𝜋 𝑟2𝑞𝑟(𝐹 ))|𝑟− (4𝜋𝑟 2𝑞𝑟

(𝐹) )|𝑟+∆ 𝑟+(4 𝜋𝑟2∆𝑟 )𝑆𝑛=0

Rate of Heat IN:

Rate of Heat OUT:

Generation:

(4𝜋 𝑟2𝑞𝑟(𝐹 ))|𝑟

(4𝜋 𝑟2𝑞𝑟(𝐹 ))|𝑟+∆𝑟

(4𝜋𝑟 2∆ 𝑟 )𝑆𝑛


Rate of Heat IN:

Rate of Heat OUT:

Generation:

Generation = Volume X Sn

𝑉=43 𝜋 [ (𝑟+∆𝑟 )3−𝑟3 ]

Too small

∴𝑉=4𝜋 [𝑟2∆𝑟 ] ∴𝐺𝑒𝑛=(4𝜋𝑟 2∆𝑟 )𝑆𝑛

¿43 𝜋 [𝑟3+3𝑟 2∆𝑟 +3𝑟 (∆ 𝑟 )2+ (∆𝑟 )3−𝑟 3 ]

(4𝜋 𝑟2𝑞𝑟(𝐹 ))|𝑟

(4𝜋 𝑟2𝑞𝑟(𝐹 ))|𝑟+∆𝑟

(4𝜋𝑟 2∆ 𝑟 )𝑆𝑛

Nuclear Heat Source



(𝐹) )|𝑟+∆ 𝑟+(4 𝜋𝑟2∆𝑟 )𝑆𝑛=0


(4𝜋 𝑟2𝑞𝑟(𝐶 ))|𝑟− ( 4𝜋𝑟 2𝑞𝑟

(𝐶))|𝑟 +∆𝑟=0

No generation

here!

Taking :

𝑑𝑑𝑟 (𝑟 2𝑞𝑟

(𝐹 ))=𝑆𝑛𝑟 2

Taking :


(𝐶))=0

Nuclear Heat Source



(𝐹) )|𝑟+∆ 𝑟+(4 𝜋𝑟2∆𝑟 )𝑆𝑛=0


(4𝜋 𝑟2𝑞𝑟(𝐶 ))|𝑟− ( 4𝜋𝑟 2𝑞𝑟

(𝐶))|𝑟 +∆𝑟=0

No generation

here!

Taking :


(𝐹 ))=𝑆𝑛0[1+𝑏 ( 𝑟𝑅 (𝐹 ) )

2]𝑟 2Taking :


(𝐶))=0

Nuclear Heat Source



(𝐹) )|𝑟+∆ 𝑟+(4 𝜋𝑟2∆𝑟 )𝑆𝑛=0


(4𝜋 𝑟2𝑞𝑟(𝐶 ))|𝑟− ( 4𝜋𝑟 2𝑞𝑟

(𝐶))|𝑟 +∆𝑟=0

No generation

here!

Integrating: Integrating:

Nuclear Heat Source

Integrating: Integrating:

Boundary Conditions: Boundary Conditions:

𝑎𝑡 𝑟=0 ,𝑞𝑟 𝑖𝑠 𝑓𝑖𝑛𝑖𝑡𝑒 𝑎𝑡 𝑟=𝑅(𝐹 ) ,𝑞𝑟(𝐹 )=𝑞𝑟

(𝐶 )

𝐶1(𝐹 )=0 𝐶1

(𝐶)=𝑆𝑛0 ( 13+𝑏5 )𝑅(𝐹 )3

For the fissionable material For the Al cladding

Nuclear Heat Source

Inserting Fourier’s Law: Inserting Fourier’s Law:


Nuclear Heat Source


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:


Q by Convective Transport Q by Molecular Transport

W by Molecular Transport W by External Forces

Energy Generation

Steady-State!


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:

𝒆=( 12 𝜌𝑣2+𝜌 �̂�)𝒗+ [𝝅 ∙𝒗 ]+𝒒

We introduce something new to replace q:




𝝅=𝑝𝜹+𝝉Recall the molecular stress tensor:When dotted with v: [𝝅 ∙𝒗 ]=𝑝 𝒗+[𝝉 ∙𝒗 ]

Substituting into e:

𝒆=( 12 𝜌𝑣2+𝜌 �̂�)𝒗+𝑝𝒗+[𝝉 ∙𝒗 ]+𝒒




𝒆=( 12 𝜌𝑣2+𝜌 �̂�)𝒗+𝑝𝒗+[𝝉 ∙𝒗 ]+𝒒

Simplifying the boxed expression:

𝜌 �̂� 𝒗+𝑝 𝒗=𝜌 (�̂�+ 𝑝𝜌 )𝒗=𝜌 (𝑈+𝑝 �̂� ) 𝒗=𝜌 �̂� 𝒗

Finally: 𝒆=( 12 𝜌𝑣2+𝜌 �̂� )𝒗+[𝝉 ∙𝒗 ]+𝒒

Viscous Dissipation Source

Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:

𝑣 𝑧 (𝑥 )=𝑣𝑏( 𝑥𝑏 )



We now make a shell balance shown in red on the left.

Rate of Energy IN:

Rate of Energy OUT:

𝑊𝐿𝒆𝒙 |𝑥

𝑊𝐿𝒆𝒙 |𝑥+∆ 𝑥

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



We now make a shell balance shown in red on the left.

Rate of Energy IN:

Rate of Energy OUT:

𝑊𝐿𝒆𝒙 |𝑥

𝑊𝐿𝒆𝒙 |𝑥+∆ 𝑥


𝑊𝐿𝒆𝒙 |𝑥+∆ 𝑥−𝑊𝐿𝒆𝒙 |𝑥=0𝑑𝒆𝒙

𝑑𝑥 =0

𝒆𝒙=𝐶1




𝒆𝒙=𝐶1 𝒆=( 12 𝜌𝑣2+𝜌 �̂� )𝒗+[𝝉 ∙𝒗 ]+𝒒

Fourier’s Law:

Newton’s Law:

𝑞𝑥=−𝑘𝑑𝑇𝑑𝑥

𝜏𝑥𝑧=−𝜇𝑑𝑣 𝑧

𝑑𝑥




𝒆𝒙=𝐶1 −𝑘 𝑑𝑇𝑑𝑥 −𝜇𝑣𝑧

𝑑𝑣𝑧

𝑑𝑥 =𝐶1

Substituting the velocity profile:

Integrating:

−𝑘 𝑑𝑇𝑑𝑥 −𝜇𝑥 (𝑣𝑏

𝑏 )2

=𝐶1

𝑇=− 𝜇𝑘 ( 𝑣𝑏

𝑏 )2 𝑥22−𝐶1

𝑘 𝑥+𝐶2




Boundary Conditions:

After applying the B.C.:

𝑇 −𝑇 0

𝑇𝑏−𝑇0=12 [ 𝜇𝑣𝑏

2

𝑘 (𝑇 𝑏−𝑇 0 ) ]( 𝑥𝑏 )(1− 𝑥𝑏 )+𝑥𝑏


𝑏 )2 𝑥22−𝐶1

𝑘 𝑥+𝐶2




𝑇 −𝑇 0


2

𝑘 (𝑇 𝑏−𝑇 0 ) ]( 𝑥𝑏 )(1− 𝑥𝑏 )+𝑥𝑏

Q: So where is Sv?

𝑆𝑣=𝜇(𝑣𝑏

𝑏 )2


𝑏 )2 𝑥22−𝐶1

𝑘 𝑥+𝐶2

After applying the B.C.:



𝑇 −𝑇 0


2

𝑘 (𝑇 𝑏−𝑇 0 ) ]( 𝑥𝑏 )(1− 𝑥𝑏 )+𝑥𝑏Temperature

Profile:

New Dimensionless Number:Dim. Group Ratio Equation

Brinkman, Br viscous heat dissipation/ molecular heat transport


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 earth’s atmosphere.

Forms of Energy Generation : Degradation of electrical energy to heat

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