Transcript Document

Heat Equations
of Change II
Outline
7. 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
Quiz – 2014.02.14
An electrically heated resistance wire has a diameter of 2
mm and a resistance of 0.10 ohm per foot of wire. The
thermal conductivity of the wire is 20 W/m·K. At a
current of 100 A, calculate the steady state temperature
difference between the center and surface of the wire.
TIME IS UP!!!
From the previous lecture…
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?
A Comparison Between Methods
Solution!
Energy Equation:
𝐷𝑇
𝜌𝐶𝑝
= − 𝛻 ∙ 𝑞 − 𝜏 ∙ 𝛻𝒗
𝐷𝑡
In spherical coordinates:
For solids with
generation
𝐷𝑇
𝜌𝑐𝑝
= 𝑘𝛻 2 𝑇 + 𝑔
𝐷𝑡
+𝑔
We are left with:
1 𝜕
𝜕𝑇
2
0=𝑘 2
𝑟
𝑟 𝜕𝑟
𝜕𝑟
Because T depends on r only:
𝑑
𝑑𝑇
𝑔𝑟 2
2
0=
𝑟
+
𝑑𝑟
𝑑𝑟
𝑘
+𝑔
From the
Energy Equation
A Comparison Between Methods
Solution! Another way!!
Differential Equation
of Heat Conduction:
𝛻 2𝑇
𝑔 1 𝜕𝑇
+ =
𝑘 𝛼 𝜕𝑡
In spherical coordinates:
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
We are left with:
1 𝜕
𝜕𝑇
𝑔
2
0= 2
𝑟
+
𝑟 𝜕𝑟
𝜕𝑟
𝑘
Because T depends on r only:
𝑑
𝑑𝑇
𝑔𝑟 2 From the Diff. Eqn
2
0=
𝑟
+
𝑑𝑟
𝑑𝑟
𝑘 of Heat Conduction
A Comparison Between Methods
Solution! Another way again!!
The Shell:
Overall Shell Heat Balance:
Rate of Heat IN:
4𝜋𝑟 2 𝑞𝑟
Rate of Heat OUT:
4𝜋𝑟 2 𝑞𝑟
Generation:
4𝜋𝑟 2 ∆𝑟 𝑔
𝑟
𝑟+∆𝑟
Adding the terms and dividing 4𝜋∆𝑟:
𝑟 2 𝑞𝑟
2
−
𝑟
𝑞𝑟
𝑟+∆𝑟
∆𝑟
𝑟
= 𝑔𝑟 2
𝑑 2
𝑟 𝑞𝑟 = 𝑔𝑟 2
𝑑𝑟
A Comparison Between Methods
Solution! Another way again!!
The Shell:
Overall Shell Heat Balance:
Rate of Heat IN:
4𝜋𝑟 2 𝑞𝑟
Rate of Heat OUT:
4𝜋𝑟 2 𝑞𝑟
Generation:
4𝜋𝑟 2 ∆𝑟 𝑔
𝑟
𝑟+∆𝑟
Inputting Fourier’s Law:
𝑑
𝑑𝑇
2
−
𝑟 𝑘
= 𝑔𝑟 2
𝑑𝑟
𝑑𝑟
𝑑 2
𝑟 𝑞𝑟 = 𝑔𝑟 2
𝑑𝑟
A Comparison Between Methods
Solution! Another way again!!
The Shell:
Overall Shell Heat Balance:
Rate of Heat IN:
4𝜋𝑟 2 𝑞𝑟
Rate of Heat OUT:
4𝜋𝑟 2 𝑞𝑟
Generation:
4𝜋𝑟 2 ∆𝑟 𝑔
𝑟
𝑟+∆𝑟
Inputting Fourier’s Law:
𝑑
𝑑𝑇
2
−
𝑟 𝑘
= 𝑔𝑟 2
𝑑𝑟
𝑑𝑟
2
𝑑
𝑑𝑇
𝑔𝑟
From the Overall
0=
𝑟2
+
𝑑𝑟
𝑑𝑟
𝑘 Shell Heat Balance
A Comparison Between Methods
Solution! Apparently…
Using any method below, we can
obtain the same ODE to solve!!
Increasing
range of
applicability of
the method
2
𝑑
𝑑𝑇
𝑔𝑟
0=
𝑟2
+
𝑑𝑟
𝑑𝑟
𝑘
From the
Energy Equation
2 From the Diff. Eqn
𝑑
𝑑𝑇
𝑔𝑟
0=
𝑟2
+
𝑑𝑟
𝑑𝑟
𝑘 of Heat Conduction
2
𝑑
𝑑𝑇
𝑔𝑟
From the Overall
0=
𝑟2
+
𝑑𝑟
𝑑𝑟
𝑘 Shell Heat Balance
A Comparison Between Methods
Solution!
The next step is to solve the ODE and
define boundary conditions.
Integrating twice:
𝑔𝑟 2 𝐶1
𝑇(𝑟) = −
− + 𝐶2
6𝑘
𝑟
Boundary conditions:
𝑎𝑡 𝑟 = 0,
𝑎𝑡 𝑟 = 𝑅, 𝑘
𝑞𝑟 𝑖𝑠 𝑓𝑖𝑛𝑖𝑡𝑒
𝑑𝑇
= ℎ 𝑇𝑎𝑖𝑟 − 𝑇 𝑅
𝑑𝑟
2
𝑑
𝑑𝑇
𝑔𝑟
0=
𝑟2
+
𝑑𝑟
𝑑𝑟
𝑘
From the
Energy Equation
2 From the Diff. Eqn
𝑑
𝑑𝑇
𝑔𝑟
0=
𝑟2
+
𝑑𝑟
𝑑𝑟
𝑘 of Heat Conduction
2
𝑑
𝑑𝑇
𝑔𝑟
From the Overall
0=
𝑟2
+
𝑑𝑟
𝑑𝑟
𝑘 Shell Heat Balance
A Comparison Between Methods
Solution!
Applying the
boundary
conditions:
𝑊
𝑘 = 10
𝑚𝐾
𝑊
ℎ = 10 2
𝑚 𝐾
𝑔𝑅2
𝑟
𝑇 𝑟 = 𝑇𝑎𝑖𝑟 +
1−
6𝑘
𝑅
𝑅 = 0.05 𝑚
𝑊
𝑔 = 800 3
𝑚
𝑇𝑎𝑖𝑟 = 20°𝐶
2
𝑔𝑅
+
3ℎ
Final Answer:
21.37°C
Outline
7. 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
1-D Transient Heat Conduction
Example!
Consider a flat slab of thickness L
with an initial temperature of T0.
It is later submerged in a large
fluid with temperature T1 (> T0).
Neglecting surface resistance
(infinite h), determine the
temperature at the center of the
slab at any time t.
1-D Transient Heat Conduction
Example!
What do we expect to
happen?
• The surface temp. is held at
T1. (because of infinite h)
• Temperature profile at any
time is symmetrical
• Unidirectional (x only) heat
flow only
• The temp. at the center, T0,
slowly approaches T1.
T0
T1
1-D Transient Heat Conduction
Example! Solution:
Differential Equation
of Heat Conduction:
With the assumptions:
𝛻2𝑇
𝑔 1 𝜕𝑇
+ =
𝑘 𝛼 𝜕𝑡
𝜕𝑇
𝜕2𝑇
=𝛼
𝜕𝑡
𝜕𝑥 2
Initial and Boundary conditions:
Initial:
Boundary:
T0
T1
1-D Transient Heat Conduction
Example! Solution:
To facilitate solving,
we will make this
dimensionless!
Implications:
2
𝜕𝑇
𝜕 𝑇
=𝛼
𝜕𝑡
𝜕𝑥 2
Let: Y = dimensionless temp.
𝑇1 − 𝑇
𝑌=
𝑇1 − 𝑇0
*The choice of definition
of Y is arbitrary.
𝜕𝑌
−1 𝜕𝑇
=
𝜕𝑡 𝑇1 − 𝑇0 𝜕𝑡
𝜕𝑌
−1 𝜕𝑇
=
𝜕𝑥 𝑇1 − 𝑇0 𝜕𝑥
𝜕2𝑌
−1 𝜕 2 𝑇
=
2
𝜕𝑥
𝑇1 − 𝑇0 𝜕𝑥 2
𝜕𝑌
𝜕2𝑌
− 𝑇1 − 𝑇0
= −𝛼 𝑇1 − 𝑇0
𝜕𝑡
𝜕𝑥 2
1-D Transient Heat Conduction
Example! Solution:
To facilitate solving,
we will make this
dimensionless!
Implications:
1
𝜕𝑋 = 𝜕𝑥
𝐿
2
𝜕𝑌
𝜕 𝑌
=𝛼
𝜕𝑡
𝜕𝑥 2
Let: X = dimensionless length
𝑥
𝑋=
𝐿
*The choice of definition
of X is arbitrary.
𝜕𝑋 2
1
=
𝜕𝑥
𝐿
2
1
= 2 𝜕𝑥 2
𝐿
𝜕𝑌 𝛼 𝜕 2 𝑌
= 2 2
𝜕𝑡 𝐿 𝜕𝑋
1-D Transient Heat Conduction
Example! Solution:
To facilitate solving,
we will make this
dimensionless!
𝜕𝑌 𝛼 𝜕 2 𝑌
= 2
𝜕𝑡 𝐿 𝜕𝑋 2
Let: τ = dimensionless time
𝛼𝑡
𝜏= 2
𝐿
*The choice of definition
of τ is arbitrary.
Implications:
𝛼
𝜕𝜏 = 2 𝜕𝑡
𝐿
𝜕𝑌 𝜕 2 𝑌
=
𝜕𝜏 𝜕𝑋 2
1-D Transient Heat Conduction
Example! Solution:
DIMENSIONLESS
FORM:
𝜕𝑌
𝜕2𝑌
=
𝜕𝜏
𝜕𝑋 2
Previous IC & BC:
Easier to solve and no
worries with units!
New IC & BC:
𝑌 = 1,
𝑌 = 0,
𝑌 = 0,
𝑇1 − 𝑇
𝑌=
𝑇1 − 𝑇0
𝑥
𝑋=
𝐿
𝜏 = 0,
𝜏 = 𝜏,
𝜏 = 𝜏,
𝛼𝑡
𝜏= 2
𝐿
𝑋=𝑋
𝑋=0
𝑋=1
1-D Transient Heat Conduction
A new dimensionless number…
Dim. Group
Fourier, Fo
Ratio
Rate of heat conduction/
Rate of heat storage
Rate of heat
conduction:
𝑘
𝐿3
Rate of heat
storage:
𝜌𝑐𝑝 𝐿3 /𝑡
𝑇1 − 𝑇
𝑌=
𝑇1 − 𝑇0
𝑥
𝑋=
𝐿
Equation
𝛼𝑡
𝐿2
𝛼𝑡
𝜏= 2
𝐿
1-D Transient Heat Conduction
Going back…
DIMENSIONLESS
FORM:
𝜕𝑌
𝜕2𝑌
=
𝜕𝜏
𝜕𝑋 2
T0
New IC & BC:
𝑌 = 1,
𝑌 = 0,
𝑌 = 0,
𝜏 = 0,
𝜏 = 𝜏,
𝜏 = 𝜏,
𝑇1 − 𝑇
𝑌=
𝑇1 − 𝑇0
T1
𝑋=𝑋
𝑋=0
𝑋=1
𝑥
𝑋=
𝐿
𝛼𝑡
𝜏= 2
𝐿
1-D Transient Heat Conduction
Going back…
DIMENSIONLESS
FORM:
New IC & BC:
𝑌 = 1,
𝑌 = 0,
𝑌 = 0,
𝜕2𝑌
𝜕𝑌
=
𝜕𝜏
𝜕𝑋 2
𝜏 = 0,
𝜏 = 𝜏,
𝜏 = 𝜏,
Solution: (A Fourier series)
4
𝑌(𝜏, 𝑋) =
𝜋
∞
2 𝜋2 𝜏
−
2𝑛+1
𝑒
𝑛=0
𝑇1 − 𝑇
𝑌=
𝑇1 − 𝑇0
2𝑛 + 1
sin 2𝑛 + 1 𝜋𝑋
𝑥
𝑋=
𝐿
𝛼𝑡
𝜏= 2
𝐿
𝑋=𝑋
𝑋=0
𝑋=1
Gurney-Lurie Charts
Gurney-Lurie Charts
- plots of the dimensionless
temperature Y against Fo with
varying Bi and X for different
geometries.
- Each point in the curves are
solutions to the PDE involving
heat conduction + convection.
𝑇1 − 𝑇
𝑌=
𝑇1 − 𝑇0
𝑥
𝑋=
𝐿
𝛼𝑡
𝜏= 2
𝐿
Gurney-Lurie Charts
𝑇1 − 𝑇
𝑌=
𝑇1 − 𝑇0
For flat plates with convection
Geankoplis, Figure 5.3-6
𝛼𝑡
𝜏= 2
𝐿
Bi
Gurney-Lurie Charts
𝑇1 − 𝑇
𝑌=
𝑇1 − 𝑇0
For long cylinders with convection
Geankoplis, Figure 5.3-8
𝛼𝑡
𝜏= 2
𝐿
Bi
Gurney-Lurie Charts
𝑇1 − 𝑇
𝑌=
𝑇1 − 𝑇0
For solid spheres with convection
Geankoplis, Figure 5.3-10
𝛼𝑡
𝜏= 2
𝐿
Bi
Gurney-Lurie Charts
How do we compute the Biot Numbers for
different geometries?
Recall: Biot Number
ℎ𝑥1 Characteristic length:
𝐵𝑖 =
Volume/Surface Area
𝑘
From a while ago…
Recall this 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/m2K. What is the temperature at the center of the
sphere?
From a while ago…
Recall this exercise:
A 10-cm diameter nickel-steel sphere has a thermal
conductivity, k = 10 W/m-K. Within the sphere, 800
W/m3of heat is being generated.The sphere, initially at
T0 = 30°C, is suddenly submerged into a fluid…
The surrounding air is at 𝑻∞ = 𝟐𝟎°𝑪and 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 across time?
From a while ago…
To solve the new problem:
Differential Equation
of Heat Conduction:
𝑔 1 𝜕𝑇
𝛻 𝑇+ =
𝑘 𝛼 𝜕𝑡
2
In spherical coordinates:
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
Steps:
1. Turn the remaining PDE into
dimensionless form.
2. Solve analytically for T(r, t).
But there is an
easier way!
Lumped Systems Analysis
Let’s assume that the sphere is too
small for conduction to matter. The
temperature distribution inside the
sphere can, therefore, be assumed
uniform!
Heat Balance:
𝑅𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑓𝑙𝑜𝑤 𝑖𝑛𝑡𝑜
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑓
𝑠𝑜𝑙𝑖𝑑 𝑜𝑓 𝑣𝑜𝑙𝑢𝑚𝑒 𝑉 𝑡ℎ𝑟𝑜𝑢𝑔ℎ = 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓
𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑠𝑢𝑟𝑓𝑎𝑐𝑒𝑠 𝐴
𝑠𝑜𝑙𝑖𝑑 𝑜𝑓 𝑣𝑜𝑙𝑢𝑚𝑒 𝑉
Initial Condition:
𝑇 𝑡 = 𝑇0 ,
𝑡=0
𝐴ℎ 𝑇∞ − 𝑇 𝑡
𝑑𝑇(𝑡)
= 𝜌𝑐𝑝 𝑉
𝑑𝑡
𝑑𝑇(𝑡)
𝐴ℎ
=
𝑇∞ − 𝑇 𝑡
𝑑𝑡
𝜌𝑐𝑝 𝑉
Lumped Systems Analysis
Initial Condition:
𝑇 𝑡 = 𝑇0 ,
𝑡=0
𝑑𝑇(𝑡)
𝐴ℎ
=
𝑇∞ − 𝑇 𝑡
𝑑𝑡
𝜌𝑐𝑝 𝑉
𝑑𝑇 𝑡
𝐴ℎ
=
𝑑𝑡
𝑇∞ − 𝑇(𝑡) 𝜌𝑐𝑝 𝑉
Integrating and
plugging the IC:
𝑇∞ − 𝑇(𝑡)
𝐴ℎ𝑡
− ln
=
𝑇∞ − 𝑇0
𝜌𝑐𝑝 𝑉
Rearranging:
𝑇(𝑡) − 𝑇∞
−𝐴ℎ𝑡
= exp
𝑇0 − 𝑇∞
𝜌𝑐𝑝 𝑉
Lumped Systems Analysis
Temperature Profile at the center
of the sphere across time:
𝑇(𝑡) − 𝑇∞
= 𝑒 −𝐵𝑖𝐹𝑜
𝑇0 − 𝑇∞
Rearranging:
−𝐵𝑖 𝐹𝑜
−ℎ 𝑉 𝐴
𝑘
𝑘
𝑡
𝜌𝑐𝑝 𝑉 𝐴
𝑇(𝑡) − 𝑇∞
−𝐴ℎ𝑡
= exp
𝑇0 − 𝑇∞
𝜌𝑐𝑝 𝑉
2
Lumped Systems Analysis
Temperature Profile at the center
of the sphere across time:
𝑇(𝑡) − 𝑇∞
= 𝑒 −𝐵𝑖𝐹𝑜
𝑇0 − 𝑇∞
This Lumped System Analysis is
sufficiently accurate only when
Bi < 0.1.
Checking:
Given:
𝑊
𝑘 = 10
𝑚𝐾
𝑊
ℎ = 10 2
𝑚 𝐾
𝑊
2 𝑘 0.05 𝑚
𝑚
𝐵𝑖 =
= 0.0167 < 0.1
𝑊
3
10
𝑚𝐾
10
𝑅 = 0.05 𝑚
𝑇0 = 30°𝐶
𝑇∞ = 20°𝐶
Valid!!
Lumped Systems Analysis
For review:
A person is found dead at 5 PM in a room where T = 20°C. The
temperature of the body was measured at 25°C when found. The heat
transfer coefficient is estimated to be 0.8 W/m2K. Estimate the time of
death assuming:
(1) The body can be modeled as a 30-cm diameter, 1.7-m long cylinder.
(2) The thermal properties of the body and the heat transfer are
constant.
(3) The body temperature was 37°C at the time of death.
(4) Since the human body has almost the same properties of water at
the average temperature of 31°C: k = 0.617 W/mK, density = 996
kg/m3, and cp = 4178 J/kgK.
(5) Radiation effects are negligible.