LECTURE 6
CHAPTER 4
Transient Conduction
1
Objectives
• Assess when the spatial variation of temperature is negligible, and temperature
varies nearly uniformly with time, making the simplified lumped system analysis
applicable.
• Obtain analytical solutions for transient one-dimensional conduction problems in
rectangular, cylindrical, and spherical geometries using the method of separation
of variables, and understand why a one-term solution is usually a reasonable
approximation.
• Solve the transient conduction problem in large mediums using the similarity
variable, and predict the variation of temperature with time and distance from the
exposed surface.
• Construct solutions for multi-dimensional transient conduction problems using the
product solution approach.
© McGraw-Hill Education
2
Introduction
The temperature of a body, (𝑇) in general, varies with time, (𝑡) as well as
position. In rectangular coordinates, this variation is expressed as 𝑇 (𝑥, 𝑦, 𝑧, 𝑡),
where (𝑥, 𝑦, 𝑧) indicate variation in the 𝑥 −, 𝑦 −, and 𝑧 − directions, and t
indicates variation with time. In this chapter, we consider the variation of
temperature with time as well as position in one- and multidimensional
systems. We start this chapter with the analysis of lumped systems in which the
temperature of a body varies with time but remains uniform throughout at any
time. Then we consider the variation of temperature with time as well as
position for one-dimensional heat conduction problems such as those
associated with a large plane wall, a long cylinder, and sphere.
3
Lumped System Analysis
1.
In heat transfer analysis, some bodies are
observed to behave like a “lump” whose interior
temperature remains uniform at any times during
a heat transfer process.
2.
The temperature of such bodies can be taken to
be a function of time only, 𝑇(𝑡).
3.
Heat
transfer
analysis
that
utilizes
this
idealization is known as lumped system analysis.
𝐿𝑢𝑚𝑝𝑒𝑑 𝑠𝑦𝑠𝑡𝑒𝑚 𝑎𝑛𝑎𝑙𝑦𝑠𝑖𝑠: 𝑇 (𝑡)
A small copper ball can be modeled
as a lumped system, but a roast
beef cannot.
4
Introduction
▪ Required data:
1- Time 𝑡 required for cooling.
2- Amount of heat transfer 𝑄 during cooling time 𝑡.
5
1- The Lumped Capacitance Equation
1.
Consider a body of arbitrary shape of mass m, volume
V, surface area 𝐴𝑠 , density 𝜌, and specific heat cp
initially at a uniform temperature 𝑇𝑖 .
2.
At time 𝑡 = 0, the body is placed into a medium at
temperature 𝑇∞ , and heat transfer takes place between
the body and its environment, with a heat transfer
coefficient ℎ. For the sake of discussion, we assume
that 𝑇𝑖 > 𝑇∞ .
3.
We assume lumped system analysis: The temperature
remains uniform within the body at all times and
changes with time only, T(t).
4.
An energy balance of the solid for the time interval 𝑑𝑡
can be expressed as.
6
1- The Lumped Capacitance Method
𝐸ሶ 𝑖𝑛 + 𝐸ሶ𝑔𝑛 = 𝐸ሶ 𝑜𝑢𝑡 + 𝐸ሶ 𝑠𝑡𝑟
𝐸ሶ 𝑖𝑛 = 0
&
𝐸ሶ𝑔𝑛 = 0
𝐸ሶ 𝑜𝑢𝑡 + 𝐸ሶ 𝑠𝑡𝑟 = 0
ℎ. 𝐴𝑠 𝑇 − 𝑇∞
Assume:
𝑑𝑇
+ 𝑚. 𝐶
=0
𝑑𝑡
𝑑𝜃
𝑑𝑇
𝜃 = 𝑇 − 𝑇∞ →→
=
𝑑𝑡
𝑑𝑡
𝜃𝑖 = 𝑇𝑖 − 𝑇∞
𝑑𝜃
ℎ. 𝐴𝑠 . 𝜃 + 𝜌. 𝑉. 𝐶
=0
𝑑𝑡
𝑑𝜃
𝜌. 𝑉. 𝐶.
= −ℎ. 𝐴𝑠 . 𝜃
𝑑𝑡
7
1- The Lumped Capacitance Method
𝑑𝜃
ℎ. 𝐴𝑠
=−
𝑑𝑡 →→
𝜃
𝜌. 𝑉. 𝐶
𝜃
𝑑𝜃
ℎ. 𝐴𝑠 𝑡
න
=−
න 𝑑𝑡
𝜌. 𝑉. 𝐶 0
𝜃𝑖 𝜃
𝑡𝑖𝑚𝑒1 > 𝑡𝑖𝑚𝑒2 > 𝑡𝑖𝑚𝑒3
𝜃
ℎ. 𝐴𝑠
ℎ. 𝐴𝑠
𝑙𝑛 = −
𝑡−0 =−
𝑡
𝜃𝑖
𝜌. 𝑉. 𝐶
𝜌. 𝑉. 𝐶
1
1
1
>
>
𝑏1 𝑏2 𝑏3
𝑏3 > 𝑏2 > 𝑏1
𝜃
𝑇 − 𝑇∞
ℎ. 𝐴𝑠
=
= 𝑒𝑥𝑝 −
𝑡 = 𝑒 −𝑏𝑡
𝜃𝑖 𝑇𝑖 − 𝑇∞
𝜌. 𝑉. 𝐶
The lumped equation enables us to determine:
1-The temperature 𝑇(𝑡) of a body at time 𝑡.
2- The time 𝑡 required for the temperature to reach a specified value 𝑇(𝑡).
ℎ. 𝐴𝑠
𝜏=𝑏=
𝜌. 𝑉. 𝐶
(1Τ𝑠)
where: time constant: 1Τ𝜏 = 1Τ𝑏 =
𝜌.𝑉.𝐶
ℎ.𝐴𝑠
(𝑠)
8
2- To determine the total energy transfer Q:
Once the temperature 𝑇 𝑡 at time 𝑡 is available, the rate of convection
heat transfer between the body and its environment at that time can be
determined from Newton’s law of cooling as:
𝑞 = ℎ. 𝐴𝑠 𝑇 𝑡 − 𝑇∞
𝑊
The total amount of heat transfer between the body and the surrounding
medium over the time interval 𝑡 = 0 𝑡𝑜 𝑡 is simply the change in the
energy content of the body:
𝑞 = 𝑚. 𝑐𝑃 𝑇 𝑡 − 𝑇𝑖
𝑘𝐽
The maximum heat transfer between the body and its surroundings is:
𝑞𝑚𝑎𝑥 = 𝑚. 𝑐𝑃 𝑇∞ − 𝑇𝑖
𝑘𝐽
9
Criteria for Lumped System Analysis
▪
▪
The Biot Number: The first of many dimensionless
parameters to be considered.
ℎ. 𝐿𝑐
Definition: 𝐵𝑖 =
𝑘
h : Convection coefficient
k : thermal conductivity of the solid
𝐿𝑐 : characteristic length of solid ( V/As )
▪
ℎ. 𝐿𝑐
𝐿/𝑘. 𝐴 𝑅𝑡,𝑐𝑜𝑛𝑑
Physical Interpretation: 𝐵𝑖 =
=
=
𝑘
1/ℎ. 𝐴 𝑅𝑡,𝑐𝑜𝑛𝑣𝑒
▪ Lumped Capacitance Method: 𝐵𝑖 ≤ 0.1
10
Criteria for Lumped System Analysis
A small Biot number represents small resistance to heat conduction, and thus small
temperature gradients within the body.
1- Lumped system analysis is exact when Bi = 0
2- And approximate when Bi > 0.
3- Of course, the smaller the Bi number, the more
accurate the lumped system analysis.
4-It is generally accepted that lumped system
analysis is applicable if Bi ≤ 0.1
ℎ. 𝐿𝑐
𝐵𝑖 =
𝑘
𝐼𝐹 𝐵𝑖 ≤ 0.1
𝜃
𝑇 − 𝑇∞
ℎ. 𝐴𝑠
=
= 𝑒𝑥𝑝 −
𝑡
𝜃𝑖 𝑇𝑖 − 𝑇∞
𝜌. 𝑉. 𝐶
Effect of Biot number on steady-state
temperature distribution in a plane
wall with surface convection
11
Criteria for Lumped System Analysis
The Lumped system analysis is to define a characteristic length as:
𝐵𝑖 =
where:
𝐿𝑐 =
𝑉
𝐴𝑠
=
ℎ.𝐿𝑐
𝑘
=
𝐿/𝑘.𝐴
1/ℎ.𝐴
∆𝒙
𝟐
for plan wall
𝒓𝒐
𝟐
for cylinder
𝒓𝒐
𝟑
for sphere
=
𝑅𝑡,𝑐𝑜𝑛𝑑
𝑅𝑡,𝑐𝑜𝑛𝑣𝑒
12
Criteria for Lumped System Analysis
▪ The characteristic length for plan wall is:
𝑉
∆𝑥. 𝐴𝑠 ∆𝑥
𝐿𝑐 =
=
=
𝐴𝑠
2𝐴𝑠
2
▪ The characteristic length for plan wall
insulated at one surface is:
𝑉
∆𝑥. 𝐴𝑠
𝐿𝑐 =
=
= ∆𝑥
𝐴𝑠
𝐴𝑠
13
Lumped Capacitance Method
1-Temperature Measurement by Thermocouples The temperature of a gas stream is to
be measured by a thermocouple whose junction can be approximated as a 1-mmdiameter sphere. The properties of the junction are 𝑘 = 35 𝑊/𝑚. 𝐾, 𝜌 = 8500 𝑘𝑔/𝑚3 ,
and 𝐶𝑃 = 320 𝐽/𝑘𝑔. 𝐾 , and the convection heat transfer coefficient between the
junction and the gas is ℎ = 210 𝑊/𝑚2 . 𝐾. Determine:
How long it will take for the thermocouple to read 99 % of the initial temperature
difference (𝑇𝑖 − 𝑇∞ ).
𝜌 = 8500 𝑘𝑔/𝑚3 𝐶𝑃 = 320 𝐽/𝑘𝑔. 𝐾.
Solution:
The junction: 𝑘 = 35 𝑊/𝑚. 𝐾, 𝜌 = 8500 𝑘𝑔/𝑚3 , and 𝐶𝑃 = 320 𝐽/𝑘𝑔. 𝐾
𝐺𝑎𝑠: ℎ = 210 𝑊/𝑚2 . 𝐾
4 3
𝜋𝑅
𝑉
𝑅 0.001
3
𝐿𝑐 =
=
= =
= 1.67𝑥10−4 𝑚
2
𝐴𝑠 4𝜋𝑅
3
6
ℎ. 𝐿𝑐 210 𝑥 1.67𝑥10−4
𝐵𝑖 =
=
= 0.001 < 0.1
𝑘
35
14
Lumped Capacitance Method
Therefore, lumped system analysis is applicable:
In order to read 99 % of the initial temperature difference 𝑇𝑖 − 𝑇∞ between the junction and the
gas, we must have:
𝑇(𝑡) − 𝑇∞
= 0.01
𝑇𝑖 − 𝑇∞
For example, when 𝑇𝑖 = 0𝑜 𝐶 and 𝑇∞ = 100𝑜 𝐶, a thermocouple is considered to have read 99 %
of this applied temperature difference when its reading indicates:
𝑇 𝑡 = 0.99(100 − 0) = 99𝑜 𝐶
The value of the exponent b is:
𝑏=
ℎ. 𝐴𝑠
ℎ
210
−1
=
=
=
0.462
𝑠
𝜌. 𝐶𝑃 . 𝑉 𝜌. 𝐶𝑃 . 𝐿𝑐 8500 𝑥 320 𝑥 1.67𝑥10−4
We now substitute these values into Eq. 4–4 and obtain:
𝑇(𝑡) − 𝑇∞
= 𝑒 −𝑏𝑡 → 0.01 = 𝑒 −0.462 𝑡
𝑇𝑖 − 𝑇∞
𝑡 = 10 𝑠
15
Lumped Capacitance Method
EXAMPLE 2: Polyvinyl chloride automotive body panels (𝑘 = 0.092 𝑊/𝑚. 𝐾, 𝑐𝑝 =
1.05 𝑘𝐽/𝑘𝑔. 𝐾, 𝜌 = 1714 𝑘𝑔/𝑚3 ), 1 − 𝑚𝑚 thick, emerge from an injection molder at
120𝑜 𝐶. They need to be cooled to 40𝑜 𝐶 by exposing both sides of the panels to 20𝑜 𝐶
air before they can be handled. If the convective heat transfer coefficient is ℎ =
15 𝑊/𝑚2 . 𝐾 and radiation is not considered, the time that the panels must be exposed to
air before they can be handled is:
(a) 1.6 min
(b) 2.4 min
(c) 2.8 min
(d) 3.5 min
(e) 4.2 min
Answer (a) 1.6 min
Solution:
Body panels: ∆𝑥 = 0.001𝑚
𝑘 = 0.092 𝑊/𝑚. 𝐾,
𝑐𝑝 = 1.05 𝑘𝐽/𝑘𝑔. 𝐾,
𝜌 = 1714 𝑘𝑔/𝑚3
𝑇𝑖 = 120𝑜 𝐶 → 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑡𝑜 → 𝑇 = 40𝑜 𝐶
Air: 𝑇∞ = 20𝑜 𝐶
ℎ = 15 𝑊/𝑚2 . 𝐾
16
Lumped Capacitance Method
Analysis The characteristic length of the junction and the Biot number are:
∆𝑥 0.001
∆𝑥 = 0.001 𝑚 → 𝐿𝑐 =
=
= 0.0005
2
2
𝑉
𝐿𝑐 =
= 0.0005 𝑚
𝐴𝑠
ℎ. 𝐿𝑐 15 𝑥 0.0005
𝐵𝑖 =
=
= 0.0815 < 0.1
𝑘
0.092
Since,𝐵𝑖 < 0.1 the lumped system analysis is applicable.
𝑇 − 𝑇∞
−ℎ. 𝐴𝑠
= 𝑒𝑥𝑝
𝑡
𝑇𝑖 − 𝑇∞
𝜌. 𝑉. 𝑐𝑝
40 − 20
−15 𝑥 𝐴𝑠
= 𝑒𝑥𝑝
𝑡
→
120 − 20
1714 𝑥 𝑉 𝑥 1050
40 − 20
−15
= 𝑒𝑥𝑝
𝑡
120 − 20
1714 𝑥 0.0005 𝑥 1050
The time that the panels must be exposed to air before they can be handled is:
𝑡 = 96.55 sec = 1.6 𝑚𝑖𝑛
17
Lumped Capacitance Method
EXAMPLE 3: A 10 𝑐𝑚-inner diameter, 30 𝑐𝑚-long can filled with water ( 𝜌 =
1000 𝑘𝑔/𝑚3 , 𝑐𝑝 = 4180 𝐽/𝑘𝑔. 𝐾, 𝑘 = 5.6 𝑊/𝑚. 𝐾) initially at 25𝑜 𝐶 is put into a
household refrigerator at 3𝑜 𝐶. The heat transfer coefficient on the surface of the can is
ℎ = 14 𝑊/𝑚2 . 𝐾. Assuming that the temperature of the water remains uniform during
the cooling process, the time it takes for the water temperature to drop to 5𝑜 𝐶 is:
(a) 0.55 ℎ
(b) 1.17 ℎ
(c) 2.09 ℎ
(d) 3.6 ℎ
(e) 4.97 ℎ
Solution:
For cylinder:
long can: 𝐿 = 0.3 𝑚, 𝐷 = 0.10 𝑚 → 𝑟𝑜 = 0.05 𝑚
𝜌 = 1000 𝑘𝑔/𝑚3 ,
𝑐𝑝 = 4118 𝐽/𝑘𝑔. 𝐾,
𝑘 = 5.6 𝑊/𝑚. 𝐾
𝑇𝑖 = 25𝑜 𝐶 →→→ 𝑐𝑜𝑜𝑙𝑒𝑑 𝑡𝑜 𝑇 = 5𝑜 𝐶
𝑡 =? ? ? ?
Refrigerator: 𝑇∞ = 3𝑜 𝐶, ℎ = 14 𝑊/𝑚2 . 𝐾
18
Lumped Capacitance Method
𝐴𝑠 = 𝜋. 𝐷. 𝐿 = 𝜋 𝑥 0.10 𝑥 0.3 = 0.094248 𝑚2
𝐷2
(0.10)2
𝑉=𝜋
𝐿=𝜋
𝑥 0.3 = 0.002356194 𝑚3
4
4
𝑟
0.05
For cylinder: 𝐿𝑐 = 𝑜 =
= 0.025 𝑚
2
2
ℎ. 𝐿𝑐 14 𝑥 0.025
𝐵𝑖 =
=
= 0.0625 < 0.1
𝑘
5.6
𝑏=
ℎ. 𝐴𝑠
14 𝑥 0.094248
=
= 0.00013397
𝜌. 𝑉. 𝑐𝑝 1000 𝑥 0.002356194 𝑥 4180
𝑇 − 𝑇∞
−ℎ. 𝐴𝑠
𝑇 − 𝑇∞
= 𝑒𝑥𝑝
𝑡 →→
= 𝑒𝑥𝑝 −𝑏. 𝑡 →→
𝑇𝑖 − 𝑇∞
𝜌. 𝑉. 𝑐𝑝
𝑇𝑖 − 𝑇∞
5−3
= 𝑒𝑥𝑝 −0.00013397 𝑥 𝑡
25 − 3
−0.00013397𝑥 𝑡 = −2.3978 →→
𝑡 = 17898 𝑠 = 4.97 ℎ
19
20
21
22
23
24
25
26
27
Consider a steel pipeline (AISI 1010) that is 1 m in diameter and has a wall thickness of 40
mm. The pipe is heavily insulated on the outside, and, before the initiation of flow, the walls
of the pipe are at a uniform temperature of 20C. With the initiation of flow, hot oil at 60C is
pumped through the pipe, creating a convective condition corresponding to h 500 W/m2 K
at the inner surface of the pipe.
1. What are the appropriate Biot and Fourier numbers 8 min after the initiation of flow?
2. At t 8 min, what is the temperature of the exterior pipe surface covered by the insulation?
3. What is the heat flux q(W/m2) to the pipe from the oil at t 8min?
4. How much energy per meter of pipe length has been transferred from the oil to the pipe at
t 8min?
Assumption
1. Pipe wall can be approximated as plane wall,
since thickness is much less than diameter.
2. Constant properties.
3. Outer surface of pipe is adiabatic.
28
1. What are the appropriate Biot and Fourier numbers 8 min after the initiation of flow?
2. At t 8 min, what is the temperature of the exterior pipe surface covered by the insulation?
3. What is the heat flux q(W/m2) to the pipe from the oil at t 8min?
4. How much energy per meter of pipe length has been transferred from the oil to the pipe at
t 8min?
29
1. What are the appropriate Biot and Fourier numbers 8 min after the initiation of flow?
2. At t 8 min, what is the temperature of the exterior pipe surface covered by the insulation?
3. What is the heat flux q(W/m2) to the pipe from the oil at t 8min?
4. How much energy per meter of pipe length has been transferred from the oil to the pipe at
t 8min?
30
1. What are the appropriate Biot and Fourier numbers 8 min after the initiation of flow?
2. At t 8 min, what is the temperature of the exterior pipe surface covered by the insulation?
3. What is the heat flux q(W/m2) to the pipe from the oil at t 8min?
4. How much energy per meter of pipe length has been transferred from the oil to the pipe at
t 8min?
31
1. What are the appropriate Biot and Fourier numbers 8 min after the initiation of flow?
2. At t 8 min, what is the temperature of the exterior pipe surface covered by the insulation?
3. What is the heat flux q(W/m2) to the pipe from the oil at t 8min?
4. How much energy per meter of pipe length has been transferred from the oil to the pipe at
t 8min?
32
Example
33
Example
Assumptions:
1. One-dimensional conduction in r.
2. Constant properties
34
Example
35
Example
36