Download - Heat Equations of Change I
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Heat Equations of Change I
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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
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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
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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.
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Differential Equation for Heat Conduction
Rate of HEAT in - out:
Rate of HEAT generation:
(∆ 𝑦 ∆ 𝑧 )𝑞𝑥|𝑥− (∆ 𝑦 ∆ 𝑧 )𝑞𝑥|𝑥+∆ 𝑥
𝑔 (∆𝑥 ∆ 𝑦 ∆ 𝑧 )
Consider a differential element balance:
Rate of HEAT accumulation:
𝜌𝑐𝑝𝜕𝑇𝜕𝑡 (∆𝑥 ∆ 𝑦 ∆𝑧 )
In Chem 16, this is mcPdT
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Differential Equation for Heat Conduction
Heat Balance:
Dividing by :
(∆ 𝑦 ∆ 𝑧 )𝑞𝑥|𝑥− (∆ 𝑦 ∆ 𝑧 )𝑞𝑥|𝑥+∆ 𝑥+𝑔 (∆ 𝑥 ∆ 𝑦 ∆ 𝑧 )=𝜌𝑐𝑝𝜕𝑇𝜕𝑡 (∆𝑥 ∆ 𝑦 ∆𝑧 )
Consider a differential element balance:
𝑞𝑥 |𝑥−𝑞𝑥|𝑥+∆𝑥
∆ 𝑥 +𝑔=𝜌𝑐𝑝𝜕𝑇𝜕𝑡
Taking the limit :
−𝜕𝑞𝑥
𝜕 𝑥 +𝑔=𝜌𝑐𝑝𝜕𝑇𝜕𝑡
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Differential Equation for Heat Conduction
Substituting Fourier’s Law:
Consider a differential element balance:
−𝜕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
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Differential Equation for Heat Conduction
Recall the definition of thermal diffusivity:
Consider a differential element balance:
Dividing everything by k:
𝛼=𝑘𝜌𝑐𝑝
Differential Equation for Heat Conduction
𝑘𝛻2𝑇 +𝑔=𝜌𝑐𝑝𝜕𝑇𝜕𝑡
Measure of how quickly a material
can carry heat away from a source.
𝛻2𝑇+𝑔𝑘=
1𝛼𝜕𝑇𝜕𝑡
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Differential Equation for Heat Conduction
Differential Equation for Heat Conduction 𝛻2𝑇+𝑔𝑘=
1𝛼𝜕𝑇𝜕𝑡
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
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Differential Equation for Heat Conduction
Differential Equation for Heat Conduction 𝛻2𝑇+𝑔𝑘=
1𝛼𝜕𝑇𝜕𝑡
The equation in different coordinate systems:
1) Rectangular:
2) Cylindrical:
3) Spherical:
𝜕2𝑇𝜕𝑥2
+𝜕2𝑇𝜕𝑦 2
+𝜕2𝑇𝜕 𝑧 2
+𝑔𝑘=
1𝛼𝜕𝑇𝜕𝑡
1𝑟𝜕𝜕𝑟 (𝑟 𝜕𝑇𝜕𝑟 )+ 1𝑟2
𝜕2𝑇𝜕𝜃2
+𝜕2𝑇𝜕 𝑧2
+𝑔𝑘=
1𝛼𝜕𝑇𝜕𝑡
1𝑟2
𝜕𝜕𝑟 (𝑟2 𝜕𝑇𝜕𝑟 )+ 1
𝑟2 sin𝜃𝜕𝜕𝜃 (sin𝜃 𝜕𝑇𝜕𝜃 )+ 1
𝑟 2sin 2𝜃𝜕2𝑇𝜕𝜙2
+𝑔𝑘=
1𝛼𝜕𝑇𝜕𝑡
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Differential Equation for Heat Conduction
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.
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Differential Equation for Heat Conduction
Example!
Assumptions (or given):1. Steady-state2. Unidirectional heat flow (x only)3. Constant k, ρ, cP, and h.
𝛻2𝑇+𝑔𝑘=
1𝛼𝜕𝑇𝜕𝑡
Differential Equation for Heat Conduction:
𝑑2𝑇𝑑𝑥2
+𝑔0𝑘 =0
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Differential Equation for Heat Conduction
Example!
𝑑2𝑇𝑑𝑥2
+𝑔0𝑘 =0
After 1st and 2nd integration:
𝑑𝑇𝑑𝑥 =−
𝑔0𝑘 𝑥+𝐶1
𝑇 (𝑥 )=− 12𝑔0𝑘 𝑥2+𝐶1𝑥+𝐶2
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Differential Equation for Heat Conduction
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.
𝑑𝑇𝑑𝑥 =−
𝑔0𝑘 𝑥+𝐶1
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Differential Equation for Heat Conduction
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
𝑑𝑇𝑑𝑥 =−
𝑔0𝑘 𝑥+𝐶1
𝐶1 (h𝐿+𝑘 )=h (𝑇∞−𝑇0 )−𝑔0𝐿2𝑘
(h𝐿+2𝑘 )
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Differential Equation for Heat Conduction
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…
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Differential Equation for Heat Conduction
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
)
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Differential Equation for Heat Conduction
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…
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Differential Equation for Heat Conduction
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]
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Differential Equation for Heat Conduction
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.
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Differential Equation for Heat Conduction
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!
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Differential Equation for Heat Conduction
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.
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Differential Equation for Heat Conduction
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?
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Differential Equation for Heat Conduction
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!
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Consider a differential volume element:
Energy Equation
Recall: Combined Energy Flux
𝒆=( 12 𝜌𝑣2+𝜌 �̂�)𝒗+ [𝝅 ∙𝒗 ]+𝒒
Recall: First Law of Thermodynamics
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Energy Equation
Rate of Increase in KE and Internal Energy: (Accumulation)
Rate of Energy IN – OUT:
Rate of Work Done by External Forces, g:
Consider a differential volume element:
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Energy Equation
Combining them:
Consider a differential volume element:
Expanding the combined energy flux term…
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Energy Equation
THE ENERGY EQUATION
Consider a differential volume element:
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Energy Equation
The complete form of the Energy Equation
Consider a differential volume element:
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Energy Equation
If we subtract the mechanical energy balance from the energy equation:
Consider a differential volume element:
THE EQUATION OF CHANGE FOR INTERNAL ENERGY
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Energy Equation
If we subtract the mechanical energy balance from the energy equation:
Consider a differential volume element:
THE EQUATION OF CHANGE FOR INTERNAL ENERGY
𝜕𝜕𝑡 ( 𝜌𝑈 )+𝛻 ∙ 𝜌 �̂� 𝒗=𝜌 𝐷𝑈
𝐷𝑡
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Energy Equation
Putting the internal energy in substantial derivative form:
Consider a differential volume element:
By absorbing the pressure force term, U becomes H.Since then at constant pressure:
𝜌𝐶𝑝𝐷𝑇𝐷𝑡 =− (𝛻 ∙𝑞)− (𝜏 ∙𝛻 𝒗 )
Convenient Form!
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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:
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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
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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.
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Energy Equation
Example! 𝜌𝐶𝑝𝐷𝑇𝐷𝑡 =− (𝛻 ∙𝑞)− (𝜏 ∙𝛻 𝒗 )
Using the solids special case with cylindrical coordinates:
This can be rewritten as:
![Page 37: Heat Equations of Change I](https://reader036.vdocuments.us/reader036/viewer/2022062323/56815ed4550346895dcd696d/html5/thumbnails/37.jpg)
Energy Equation
Example! 𝜌𝐶𝑝𝐷𝑇𝐷𝑡 =− (𝛻 ∙𝑞)− (𝜏 ∙𝛻 𝒗 )
From here on, the solution is just the same as with the electrical wire: