Heat Equations of Change I

Heat Equations of Change I

So far…

Outline

1. Heat Transfer Mechanisms

2. Conduction Heat Transfer

3. Convection Heat Transfer

4. Combined Heat Transfer

5. Overall Shell Heat Balances

6. Heat Equations of Change

6. Heat Equations of Change

7.1. Derivation of Basic Equations

7.1.1. Differential Equation for Heat Conduction

7.1.2. Energy Equation

7.1.3. Buckingham Pi Method

7.2. Unsteady-state Conduction

7.2.1. Gurney-Lurie Charts

7.2.2. Lumped Systems Analysis

Outline

Consider a differential element balance:

Differential Equation for Heat Conduction

Assumptions:1. Solid conduction

thermal resistance only.2. Constant density,

thermal conductivity and specific heat.


Rate of HEAT in - out:

Rate of HEAT generation:

(∆ 𝑦 ∆ 𝑧 )𝑞𝑥|𝑥− (∆ 𝑦 ∆ 𝑧 )𝑞𝑥|𝑥+∆ 𝑥

𝑔 (∆𝑥 ∆ 𝑦 ∆ 𝑧 )


Rate of HEAT accumulation:

𝜌𝑐𝑝𝜕𝑇𝜕𝑡 (∆𝑥 ∆ 𝑦 ∆𝑧 )

In Chem 16, this is mcPdT


Heat Balance:

Dividing by :

(∆ 𝑦 ∆ 𝑧 )𝑞𝑥|𝑥− (∆ 𝑦 ∆ 𝑧 )𝑞𝑥|𝑥+∆ 𝑥+𝑔 (∆ 𝑥 ∆ 𝑦 ∆ 𝑧 )=𝜌𝑐𝑝𝜕𝑇𝜕𝑡 (∆𝑥 ∆ 𝑦 ∆𝑧 )


𝑞𝑥 |𝑥−𝑞𝑥|𝑥+∆𝑥

∆ 𝑥 +𝑔=𝜌𝑐𝑝𝜕𝑇𝜕𝑡

Taking the limit :

−𝜕𝑞𝑥

𝜕 𝑥 +𝑔=𝜌𝑐𝑝𝜕𝑇𝜕𝑡


Substituting Fourier’s Law:


−𝜕𝑞𝑥

𝜕 𝑥 +𝑔=𝜌𝑐𝑝𝜕𝑇𝜕𝑡

− 𝜕𝜕 𝑥 (−𝑘 𝜕𝑇𝜕𝑥 )+𝑔=𝜌𝑐𝑝

𝜕𝑇𝜕𝑡

Noting that k is constant: 𝑘( 𝜕2𝑇𝜕 𝑥2 )+𝑔=𝜌𝑐𝑝𝜕𝑇𝜕𝑡

Extending to 3D space: 𝑘𝛻2𝑇 +𝑔=𝜌𝑐𝑝𝜕𝑇𝜕𝑡


Recall the definition of thermal diffusivity:


Dividing everything by k:

𝛼=𝑘𝜌𝑐𝑝


𝑘𝛻2𝑇 +𝑔=𝜌𝑐𝑝𝜕𝑇𝜕𝑡

Measure of how quickly a material

can carry heat away from a source.

𝛻2𝑇+𝑔𝑘=

1𝛼𝜕𝑇𝜕𝑡


Differential Equation for Heat Conduction 𝛻2𝑇+𝑔𝑘=


Simplifications of the equation:

1) No heat generation:

2) Steady-state:

3) Steady-state & no heat generation:

𝛻2𝑇=1𝛼𝜕𝑇𝜕𝑡

𝛻2𝑇+𝑔𝑘=0

𝛻2𝑇=0

Fourier’s Second Law of Conduction

Poisson’s Equation

Laplace’s Equation


Differential Equation for Heat Conduction 𝛻2𝑇+𝑔𝑘=


The equation in different coordinate systems:

1) Rectangular:

2) Cylindrical:

3) Spherical:

𝜕2𝑇𝜕𝑥2

+𝜕2𝑇𝜕𝑦 2

+𝜕2𝑇𝜕 𝑧 2

+𝑔𝑘=


1𝑟𝜕𝜕𝑟 (𝑟 𝜕𝑇𝜕𝑟 )+ 1𝑟2

𝜕2𝑇𝜕𝜃2

+𝜕2𝑇𝜕 𝑧2

+𝑔𝑘=


1𝑟2

𝜕𝜕𝑟 (𝑟2 𝜕𝑇𝜕𝑟 )+ 1

𝑟2 sin𝜃𝜕𝜕𝜃 (sin𝜃 𝜕𝑇𝜕𝜃 )+ 1

𝑟 2sin 2𝜃𝜕2𝑇𝜕𝜙2

+𝑔𝑘=



Example!

Determine the steady-state temperature distribution and the heat flux in a slab in the region 0 ≤ x ≤ L for thermal conductivity k and a uniform heat generation in the medium at a rate of g0 when the boundary surface at x = 0 is kept at a uniform temperature T0 and the boundary surface at x = L dissipates heat by convection into an environment at a constant temperature T∞ with a heat-transfer coefficient h.


Example!

Assumptions (or given):1. Steady-state2. Unidirectional heat flow (x only)3. Constant k, ρ, cP, and h.

𝛻2𝑇+𝑔𝑘=


Differential Equation for Heat Conduction:

𝑑2𝑇𝑑𝑥2

+𝑔0𝑘 =0


Example!

𝑑2𝑇𝑑𝑥2

+𝑔0𝑘 =0

After 1st and 2nd integration:

𝑑𝑇𝑑𝑥 =−

𝑔0𝑘 𝑥+𝐶1

𝑇 (𝑥 )=− 12𝑔0𝑘 𝑥2+𝐶1𝑥+𝐶2


Example!

Boundary conditions:

𝑇 (𝑥 )=− 12𝑔0𝑘 𝑥2+𝐶1𝑥+𝐶2

𝑎𝑡 𝑥=0 , 𝑇 (0)=𝑇0

𝑎𝑡 𝑥=𝐿 , 𝑘 𝑑𝑇𝑑𝑥 =h (𝑇 ∞−𝑇 (𝐿))

*The second B.C. denotes that the heat leaving by conduction is equal to the heat entering by convection.




Example!

Applying B.C. 1: C2 = T0

Applying B.C. 2:

𝑘(− 𝑔0𝑘 𝐿+𝐶1)=h(𝑇 ∞+𝑔02𝑘 𝐿

2−𝐶1𝐿−𝑇 0)

𝑇 (𝑥 )=− 12𝑔0𝑘 𝑥2+𝐶1𝑥+𝐶2



𝐶1 (h𝐿+𝑘 )=h (𝑇∞−𝑇0 )−𝑔0𝐿2𝑘

(h𝐿+2𝑘 )


Example!

Applying B.C. 1: C2 = T0

Applying B.C. 2:

𝑘(− 𝑔0𝑘 𝐿+𝐶1)=h(𝑇 ∞+𝑔02𝑘 𝐿

2−𝐶1𝐿−𝑇 0)𝐶1 (h𝐿+𝑘 )=h (𝑇∞−𝑇0 )−

𝑔0𝐿2𝑘

(h𝐿+2𝑘 )

𝐶1=h (𝑇 ∞−𝑇0 )

(h𝐿+𝑘)−𝑔0𝐿2𝑘 ( h𝐿+2𝑘

h𝐿+𝑘 )𝐶2=𝑇0

𝑇 (𝑥 )=− 12𝑔0𝑘 𝑥2+𝐶1𝑥+𝐶2

After substitution…


Example!

𝑇 (𝑥 )=− 12𝑔0𝑘 𝑥2+𝐶1𝑥+𝐶2

After substitution…

𝑇 (𝑥 )=− 12𝑔0𝑘 𝑥2+

h (𝑇 ∞−𝑇 0 )(h𝐿+𝑘 )

𝑥− 𝑔0𝑥 𝐿2𝑘 ( h𝐿+2𝑘

h𝐿+𝑘 )+𝑇 0

Further manipulation into a desired form:

𝑇 (𝑥 )−𝑇0=−12𝑔0𝑘 𝑥2+(𝑇 ∞−𝑇 0

1+ 𝑘h𝐿 ) 𝑥𝐿 − 𝑔0𝑥𝐿2𝑘 ( 1+

2𝑘h𝐿

1+ 𝑘h𝐿

)


Example!

Manipulating into a desired form even further:

𝑇 (𝑥 )−𝑇0=−12𝑔0𝑘 𝑥2+(𝑇 ∞−𝑇 0

1+ 𝑘h𝐿 ) 𝑥𝐿 − 𝑔0𝑥𝐿2𝑘 ( 1+

2𝑘h𝐿

1+ 𝑘h𝐿

)

𝑇 (𝑥 )−𝑇0=(𝑇 ∞−𝑇0

1+ 𝑘h𝐿 ) 𝑥𝐿 − 𝑔0 𝐿

2

2𝑘 [( 1+ 2𝑘h𝐿1+ 𝑘h𝐿 )( 𝑥𝐿 )−( 𝑥𝐿 )

2]Now, we introduce a new dimensionless number…


Manipulating into a desired form even further:

𝑇 (𝑥 )−𝑇0=(𝑇 ∞−𝑇0

1+ 𝑘h𝐿 ) 𝑥𝐿 − 𝑔0 𝐿

2

2𝑘 [( 1+ 2𝑘h𝐿1+ 𝑘h𝐿 )( 𝑥𝐿 )−( 𝑥𝐿 )

2]

Dim. Group Ratio EquationBiot, Bi convection at body’s surface/

conduction within the body

Finally: 𝑇 (𝑥 )−𝑇0=( 𝑇∞−𝑇0

1+1 /𝐵𝑖 ) 𝑥𝐿 − 𝑔0𝐿2

2𝑘 [( 1+2/𝐵𝑖1+1/𝐵𝑖 )( 𝑥𝐿 )−( 𝑥𝐿)2]


Example!

Special cases of the problem:I. The Biot Number approaches infinity.

𝑇 (𝑥 )−𝑇0=( 𝑇∞−𝑇0

1+1 /𝐵𝑖 ) 𝑥𝐿 − 𝑔0𝐿2

2𝑘 [( 1+2/𝐵𝑖1+1/𝐵𝑖 )( 𝑥𝐿 )−( 𝑥𝐿)2]

In this case, the boundary conditions should have been:

𝑎𝑡 𝑥=0 , 𝑇 (0)=𝑇0

𝑎𝑡 𝑥=𝐿 , 𝑇 (𝐿 )=𝑇 ∞

*When Bi approaches infinity, then the heat transfer coefficient, h, approaches infinity also.


Example!

Special cases of the problem:I. The Biot Number approaches infinity.

𝑇 (𝑥 )−𝑇0=( 𝑇∞−𝑇0

1+1 /𝐵𝑖 ) 𝑥𝐿 − 𝑔0𝐿2

2𝑘 [( 1+2/𝐵𝑖1+1/𝐵𝑖 )( 𝑥𝐿 )−( 𝑥𝐿)2]

The resulting equation when is:

𝑇 (𝑥 )−𝑇0=(𝑇 ∞−𝑇 0 )

𝐿 𝑥+𝑔0 𝐿2

2𝑘 [ 𝑥𝐿 −( 𝑥𝐿 )2]

Recall the result when g0 is zero!


Example!

Special cases of the problem:II. The Biot Number approaches zero.

𝑇 (𝑥 )−𝑇0=( 𝑇∞−𝑇0

1+1 /𝐵𝑖 ) 𝑥𝐿 − 𝑔0𝐿2

2𝑘 [( 1+2/𝐵𝑖1+1/𝐵𝑖 )( 𝑥𝐿 )−( 𝑥𝐿)2]

In this case, the boundary conditions should have been:

𝑎𝑡 𝑥=0 , 𝑇 (0)=𝑇0

𝑎𝑡 𝑥=𝐿 , 𝑑𝑇𝑑𝑥=0

*When Bi approaches zero, then the heat transfer coefficient, h, approaches zero also.


Example!

Special cases of the problem:II. The Biot Number approaches zero.

𝑇 (𝑥 )−𝑇0=( 𝑇∞−𝑇0

1+1 /𝐵𝑖 ) 𝑥𝐿 − 𝑔0𝐿2

2𝑘 [( 1+2/𝐵𝑖1+1/𝐵𝑖 )( 𝑥𝐿 )−( 𝑥𝐿)2]

The resulting equation when is:

𝑇 (𝑥 )−𝑇0=𝑔0 𝐿2

2𝑘 [2 𝑥𝐿−( 𝑥𝐿 )2]

Q: What does dT/dx = 0 imply?


A 10-cm diameter nickel-steel sphere has a thermal conductivity, k = 10 W/m-K. Within the sphere, 800 W/m3 of heat is being generated. The surrounding air is at 20°C and the heat transfer coefficient from the surroundings to the surface of the sphere is 10 W/m2-K. What is the temperature at the center of the sphere?

Exercise!

Consider a differential volume element:

Energy Equation

Recall: Combined Energy Flux

𝒆=( 12 𝜌𝑣2+𝜌 �̂�)𝒗+ [𝝅 ∙𝒗 ]+𝒒

Recall: First Law of Thermodynamics

Energy Equation

Rate of Increase in KE and Internal Energy: (Accumulation)

Rate of Energy IN – OUT:

Rate of Work Done by External Forces, g:


Energy Equation

Combining them:


Expanding the combined energy flux term…

Energy Equation

THE ENERGY EQUATION


Energy Equation

The complete form of the Energy Equation


Energy Equation

If we subtract the mechanical energy balance from the energy equation:


THE EQUATION OF CHANGE FOR INTERNAL ENERGY

Energy Equation

If we subtract the mechanical energy balance from the energy equation:


THE EQUATION OF CHANGE FOR INTERNAL ENERGY

𝜕𝜕𝑡 ( 𝜌𝑈 )+𝛻 ∙ 𝜌 �̂� 𝒗=𝜌 𝐷𝑈

𝐷𝑡

Energy Equation

Putting the internal energy in substantial derivative form:


By absorbing the pressure force term, U becomes H.Since then at constant pressure:

𝜌𝐶𝑝𝐷𝑇𝐷𝑡 =− (𝛻 ∙𝑞)− (𝜏 ∙𝛻 𝒗 )

Convenient Form!

Energy Equation

Special Cases of the Energy Equation: 𝜌𝐶𝑝

𝐷𝑇𝐷𝑡 =− (𝛻 ∙𝑞)− (𝜏 ∙𝛻 𝒗 )

1. Fluid at constant pressure and small velocity gradients. 𝜌𝑐𝑝

𝐷𝑇𝐷𝑡 =𝑘𝛻2𝑇

R:

C:

S:

Energy Equation

Special Cases of the Energy Equation: 𝜌𝐶𝑝

𝐷𝑇𝐷𝑡 =− (𝛻 ∙𝑞)− (𝜏 ∙𝛻 𝒗 )

2. For solids 𝜌𝑐𝑝𝜕𝑇𝜕𝑡 =𝑘𝛻2𝑇

3. With Heat Generation (simply added)

Fourier’s Second Law of Conduction

𝜌𝑐𝑝𝐷𝑇𝐷𝑡 =𝑘𝛻2𝑇 +𝑔

Energy Equation

Example! 𝜌𝐶𝑝𝐷𝑇𝐷𝑡 =− (𝛻 ∙𝑞)− (𝜏 ∙𝛻 𝒗 )

A solid cylinder in which heat generation is occurring uniformly as g W/m3 is insulated on the ends. The temperature of the surface of the cylinder is held constant at Tw K. The radius of the cylinder is r = R m. Heat flows only in the radial direction. Using the Energy Equation only, derive the temperature profile at steady-state if the solid has a constant k.

Energy Equation


Using the solids special case with cylindrical coordinates:

This can be rewritten as:

Energy Equation


From here on, the solution is just the same as with the electrical wire:

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Heat Equations of Change I