Transcript 16b

Heat Equations
of Change I
Outline
So far…
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
Outline
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
Differential Equation for Heat Conduction
Consider a differential element balance:
Assumptions:
1. Solid conduction
thermal resistance only.
2. Constant density,
thermal conductivity and
specific heat.
Differential Equation for Heat Conduction
Consider a differential element balance:
Rate of HEAT in - out:
∆𝑦∆𝑧 𝑞𝑥
𝑥
− ∆𝑦∆𝑧 𝑞𝑥
𝑥+∆𝑥
Rate of HEAT generation:
𝑔 ∆𝑥∆𝑦∆𝑧
Rate of HEAT accumulation:
In Chem 16,
this is mcPdT
𝜕𝑇
𝜌𝑐𝑝
∆𝑥∆𝑦∆𝑧
𝜕𝑡
Differential Equation for Heat Conduction
Consider a differential element balance:
Heat Balance:
𝜕𝑇
∆𝑦∆𝑧 𝑞𝑥 − ∆𝑦∆𝑧 𝑞𝑥
+ 𝑔 ∆𝑥∆𝑦∆𝑧 = 𝜌𝑐𝑝
∆𝑥∆𝑦∆𝑧
𝜕𝑡
𝑥
𝑥+∆𝑥
Dividing by ∆𝑥∆𝑦∆𝑧 :
𝑞𝑥
𝑥
− 𝑞𝑥
∆𝑥
𝑥+∆𝑥
𝜕𝑇
+ 𝑔 = 𝜌𝑐𝑝
𝜕𝑡
Taking the limit ∆𝑥 → 0:
𝜕𝑞𝑥
𝜕𝑇
−
+ 𝑔 = 𝜌𝑐𝑝
𝜕𝑥
𝜕𝑡
Differential Equation for Heat Conduction
Consider a differential element balance:
𝜕𝑞𝑥
𝜕𝑇
−
+ 𝑔 = 𝜌𝑐𝑝
𝜕𝑥
𝜕𝑡
Substituting Fourier’s Law:
Noting that k is constant:
Extending to 3D space:
𝜕
𝜕𝑇
𝜕𝑇
−
−𝑘
+ 𝑔 = 𝜌𝑐𝑝
𝜕𝑥
𝜕𝑥
𝜕𝑡
𝜕2𝑇
𝜕𝑇
𝑘
+ 𝑔 = 𝜌𝑐𝑝
2
𝜕𝑥
𝜕𝑡
𝑘
𝛻 2𝑇
𝜕𝑇
+ 𝑔 = 𝜌𝑐𝑝
𝜕𝑡
Differential Equation for Heat Conduction
Consider a differential element balance:
𝑘
Recall the definition of
thermal diffusivity:
𝛻 2𝑇
𝜕𝑇
+ 𝑔 = 𝜌𝑐𝑝
𝜕𝑡
Measure of how
quickly a material
can carry heat away
from a source.
𝑘
𝛼=
𝜌𝑐𝑝
Dividing everything by k:
Differential Equation
for Heat Conduction
𝛻2𝑇
𝑔 1 𝜕𝑇
+ =
𝑘 𝛼 𝜕𝑡
Differential Equation for Heat Conduction
Differential Equation for Heat Conduction
𝛻 2𝑇
𝑔 1 𝜕𝑇
+ =
𝑘 𝛼 𝜕𝑡
Simplifications of the equation:
1) No heat generation:
2) Steady-state:
3) Steady-state & no
heat generation:
1 𝜕𝑇
𝛻 𝑇=
𝛼 𝜕𝑡
𝑔
2
𝛻 𝑇+ =0
𝑘
2
𝛻2𝑇 = 0
Fourier’s Second
Law of Conduction
Poisson’s Equation
Laplace’s Equation
Differential Equation for Heat Conduction
Differential Equation for Heat Conduction
𝛻 2𝑇
𝑔 1 𝜕𝑇
+ =
𝑘 𝛼 𝜕𝑡
The equation in different coordinate systems:
1) Rectangular:
2) Cylindrical:
3) Spherical:
𝜕 2 𝑇 𝜕 2 𝑇 𝜕 2 𝑇 𝑔 1 𝜕𝑇
+ 2+ 2+ =
2
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝑘 𝛼 𝜕𝑡
1 𝜕
𝜕𝑇
1 𝜕 2 𝑇 𝜕 2 𝑇 𝑔 1 𝜕𝑇
𝑟
+ 2 2+ 2+ =
𝑟 𝜕𝑟 𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
𝑘 𝛼 𝜕𝑡
1 𝜕
𝜕𝑇
1
𝜕
𝜕𝑇
1
𝜕 2 𝑇 𝑔 1 𝜕𝑇
2
𝑟
+ 2
sin 𝜃
+ 2 2
+ =
𝑟 2 𝜕𝑟
𝜕𝑟
𝑟 sin 𝜃 𝜕𝜃
𝜕𝜃
𝑟 sin 𝜃 𝜕𝜙 2 𝑘 𝛼 𝜕𝑡
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 heattransfer coefficient h.
Differential Equation for Heat Conduction
Example!
Assumptions (or given):
1. Steady-state
2. Unidirectional heat flow (x only)
3. Constant k, ρ, cP, and h.
Differential Equation
for Heat Conduction:
𝑔 1 𝜕𝑇
𝛻 𝑇+ =
𝑘 𝛼 𝜕𝑡
2
𝑑 2 𝑇 𝑔0
+
=0
2
𝑑𝑥
𝑘
Differential Equation for Heat Conduction
Example!
𝑑2 𝑇 𝑔0
+
=0
2
𝑑𝑥
𝑘
After 1st and 2nd integration:
𝑑𝑇
𝑔0
= − 𝑥 + 𝐶1
𝑑𝑥
𝑘
1 𝑔0 2
𝑇 𝑥 =−
𝑥 + 𝐶1 𝑥 + 𝐶2
2𝑘
Differential Equation for Heat Conduction
Example!
𝑑𝑇
𝑔0
= − 𝑥 + 𝐶1
𝑑𝑥
𝑘
1 𝑔0 2
𝑇 𝑥 =−
𝑥 + 𝐶1 𝑥 + 𝐶2
2𝑘
Boundary conditions:
𝑎𝑡 𝑥 = 0,
𝑎𝑡 𝑥 = 𝐿,
𝑇(0) = 𝑇0
𝑑𝑇
𝑘
= ℎ 𝑇∞ − 𝑇(𝐿)
𝑑𝑥
*The second B.C. denotes that the
heat leaving by conduction is equal to
the heat entering by convection.
Differential Equation for Heat Conduction
Example!
𝑑𝑇
𝑔0
= − 𝑥 + 𝐶1
𝑑𝑥
𝑘
1 𝑔0 2
𝑇 𝑥 =−
𝑥 + 𝐶1 𝑥 + 𝐶2
2𝑘
Applying B.C. 1: C2 = T0
Applying B.C. 2:
𝑘 −
𝑔0
𝑔0 2
𝐿 + 𝐶1 = ℎ 𝑇∞ +
𝐿 − 𝐶1 𝐿 − 𝑇0
𝑘
2𝑘
𝐶1 ℎ𝐿 + 𝑘 = ℎ 𝑇∞ − 𝑇0 −
𝑔0 𝐿
ℎ𝐿 + 2𝑘
2𝑘
Differential Equation for Heat Conduction
Example!
Applying B.C. 1: C2 = T0
Applying B.C. 2:
𝑔0
𝑔0 2
𝑘 − 𝐿 + 𝐶1 = ℎ 𝑇∞ +
𝐿 − 𝐶1 𝐿 − 𝑇0
𝑘
2𝑘
𝑔0 𝐿
𝐶1 ℎ𝐿 + 𝑘 = ℎ 𝑇∞ − 𝑇0 −
ℎ𝐿 + 2𝑘
2𝑘
ℎ 𝑇∞ − 𝑇0
𝑔0 𝐿 ℎ𝐿 + 2𝑘
𝐶1 =
−
ℎ𝐿 + 𝑘
2𝑘 ℎ𝐿 + 𝑘
𝐶2 = 𝑇0
1 𝑔0 2
𝑇 𝑥 =−
𝑥 + 𝐶1 𝑥 + 𝐶2
2𝑘
After substitution…
Differential Equation for Heat Conduction
Example!
1 𝑔0 2
𝑇 𝑥 =−
𝑥 + 𝐶1 𝑥 + 𝐶2
2𝑘
After substitution…
1 𝑔0 2 ℎ 𝑇∞ − 𝑇0
𝑔0 𝑥𝐿 ℎ𝐿 + 2𝑘
𝑇 𝑥 =−
𝑥 +
𝑥−
+ 𝑇0
2𝑘
ℎ𝐿 + 𝑘
2𝑘
ℎ𝐿 + 𝑘
Further manipulation into a desired form:
1 𝑔0 2
𝑇∞ − 𝑇0
𝑇 𝑥 − 𝑇0 = −
𝑥 +
𝑘
2𝑘
1+
ℎ𝐿
2𝑘
𝑥 𝑔0 𝑥𝐿 1 + ℎ𝐿
−
𝐿
2𝑘 1 + 𝑘
ℎ𝐿
Differential Equation for Heat Conduction
Example!
1 𝑔0 2
𝑇∞ − 𝑇0
𝑇 𝑥 − 𝑇0 = −
𝑥 +
𝑘
2𝑘
1+
ℎ𝐿
2𝑘
𝑥 𝑔0 𝑥𝐿 1 + ℎ𝐿
−
𝐿
2𝑘 1 + 𝑘
ℎ𝐿
Manipulating into a desired form even further:
𝐿2
𝑇∞ − 𝑇0 𝑥 𝑔0
𝑇 𝑥 − 𝑇0 =
−
𝑘 𝐿
2𝑘
1+
ℎ𝐿
2𝑘
1+
ℎ𝐿
𝑘
1+
ℎ𝐿
Now, we introduce a new dimensionless number…
𝑥
𝑥
−
𝐿
𝐿
2
Differential Equation for Heat Conduction
Dim. Group
Biot, Bi
Ratio
convection at body’s surface/
conduction within the body
Equation
ℎ𝐿
𝑘
Manipulating into a desired form even further:
𝐿2
𝑇∞ − 𝑇0 𝑥 𝑔0
𝑇 𝑥 − 𝑇0 =
−
𝑘 𝐿
2𝑘
1+
ℎ𝐿
Finally:
2𝑘
1+
ℎ𝐿
𝑘
1+
ℎ𝐿
𝑇∞ − 𝑇0 𝑥 𝑔0 𝐿2
𝑇 𝑥 − 𝑇0 =
−
1 + 1 𝐵𝑖 𝐿
2𝑘
𝑥
𝑥
−
𝐿
𝐿
1 + 2 𝐵𝑖
1 + 1 𝐵𝑖
2
𝑥
𝑥
−
𝐿
𝐿
2
Differential Equation for Heat Conduction
Example!
𝑇∞ − 𝑇0 𝑥 𝑔0 𝐿2
𝑇 𝑥 − 𝑇0 =
−
1 + 1 𝐵𝑖 𝐿
2𝑘
1 + 2 𝐵𝑖
1 + 1 𝐵𝑖
Special cases of the problem:
I. The Biot Number approaches infinity.
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.
𝑥
𝑥
−
𝐿
𝐿
2
Differential Equation for Heat Conduction
Example!
𝑇∞ − 𝑇0 𝑥 𝑔0 𝐿2
𝑇 𝑥 − 𝑇0 =
−
1 + 1 𝐵𝑖 𝐿
2𝑘
1 + 2 𝐵𝑖
1 + 1 𝐵𝑖
Special cases of the problem:
I. The Biot Number approaches infinity.
The resulting equation when 𝐵𝑖 → ∞ is:
𝑇∞ − 𝑇0
𝑔0 𝐿2 𝑥
𝑥
𝑇 𝑥 − 𝑇0 =
𝑥+
−
𝐿
2𝑘 𝐿
𝐿
Recall the result when g0 is zero!
2
𝑥
𝑥
−
𝐿
𝐿
2
Differential Equation for Heat Conduction
Example!
𝑇∞ − 𝑇0 𝑥 𝑔0 𝐿2
𝑇 𝑥 − 𝑇0 =
−
1 + 1 𝐵𝑖 𝐿
2𝑘
1 + 2 𝐵𝑖
1 + 1 𝐵𝑖
Special cases of the problem:
II. The Biot Number approaches zero.
In this case, the
boundary conditions
should have been:
𝑎𝑡 𝑥 = 0, 𝑇(0) = 𝑇0
𝑑𝑇
𝑎𝑡 𝑥 = 𝐿,
=0
𝑑𝑥
*When Bi approaches zero, then the heat
transfer coefficient, h, approaches zero also.
𝑥
𝑥
−
𝐿
𝐿
2
Differential Equation for Heat Conduction
Example!
𝑇∞ − 𝑇0 𝑥 𝑔0 𝐿2
𝑇 𝑥 − 𝑇0 =
−
1 + 1 𝐵𝑖 𝐿
2𝑘
1 + 2 𝐵𝑖
1 + 1 𝐵𝑖
Special cases of the problem:
II. The Biot Number approaches zero.
The resulting equation when 𝐵𝑖 → 0 is:
𝑔0 𝐿2 𝑥
𝑥
𝑇 𝑥 − 𝑇0 =
2 −
2𝑘
𝐿
𝐿
Q: What does dT/dx = 0 imply?
2
𝑥
𝑥
−
𝐿
𝐿
2
Differential Equation for Heat Conduction
Exercise!
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?
Energy Equation
Consider a differential
volume element:
Recall: Combined Energy Flux
1 2
𝒆=
𝜌𝑣 + 𝜌𝑈 𝒗 + 𝝅 ∙ 𝒗 + 𝒒
2
Recall: First Law of Thermodynamics
Energy Equation
Consider a differential
volume element:
Rate of Increase in KE
and Internal Energy:
(Accumulation)
Rate of Energy IN – OUT:
Rate of Work Done
by External Forces, g:
Energy Equation
Consider a differential
volume element:
Combining them:
Expanding the combined energy flux term…
Energy Equation
Consider a differential
volume element:
THE ENERGY EQUATION
Energy Equation
Consider a differential
volume element:
The complete form of the Energy Equation
Energy Equation
Consider a differential
volume element:
If we subtract the mechanical energy balance
from the energy equation:
THE EQUATION OF CHANGE FOR INTERNAL ENERGY
Energy Equation
Consider a differential
volume element:
If we subtract the mechanical energy balance
from the energy equation:
THE EQUATION OF CHANGE FOR INTERNAL ENERGY
𝜕
𝐷𝑈
𝜌𝑈 + 𝛻 ∙ 𝜌𝑈𝒗 = 𝜌
𝜕𝑡
𝐷𝑡
Energy Equation
Consider a differential
volume element:
Putting the internal energy in substantial
derivative form:
By absorbing the pressure force term, U becomes H.
Since
D𝐻
𝐷𝑡
=
𝐷𝑇
𝜌𝐶𝑝
𝐷𝑡
then at constant pressure:
𝐷𝑇
𝜌𝐶𝑝
= − 𝛻 ∙ 𝑞 − 𝜏 ∙ 𝛻𝒗
𝐷𝑡
Convenient
Form!
Energy Equation
Special Cases of the
Energy Equation:
𝐷𝑇
𝜌𝐶𝑝
= − 𝛻 ∙ 𝑞 − 𝜏 ∙ 𝛻𝒗
𝐷𝑡
1. Fluid at constant pressure and
small velocity gradients.
R:
C:
S:
𝐷𝑇
𝜌𝑐𝑝
= 𝑘𝛻 2 𝑇
𝐷𝑡
Energy Equation
Special Cases of the
Energy Equation:
𝐷𝑇
𝜌𝐶𝑝
= − 𝛻 ∙ 𝑞 − 𝜏 ∙ 𝛻𝒗
𝐷𝑡
2. For solids 𝜌𝑐 𝜕𝑇 = 𝑘𝛻 2 𝑇
𝑝
𝜕𝑡
Fourier’s Second
Law of Conduction
3. With Heat Generation (simply added)
𝐷𝑇
𝜌𝑐𝑝
= 𝑘𝛻 2 𝑇 + 𝑔
𝐷𝑡
Energy Equation
Example!
𝐷𝑇
𝜌𝐶𝑝
= − 𝛻 ∙ 𝑞 − 𝜏 ∙ 𝛻𝒗
𝐷𝑡
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 steadystate if the solid has a constant k.
Energy Equation
Example!
𝐷𝑇
𝜌𝐶𝑝
= − 𝛻 ∙ 𝑞 − 𝜏 ∙ 𝛻𝒗
𝐷𝑡
Using the solids special case with
cylindrical coordinates:
This can be
rewritten as:
Energy Equation
Example!
𝐷𝑇
𝜌𝐶𝑝
= − 𝛻 ∙ 𝑞 − 𝜏 ∙ 𝛻𝒗
𝐷𝑡
From here on, the solution is just
the same as with the electrical wire: