Transport Phenomena II: Heat and Mass Transport
CENG 3220, HKUST, Fall 2023
3. Differential Equations of
Heat Transfer
Yoonseob Kim
yoonseobkim@ust.hk
http://yoonseobkim.com/
Y. Kim | 1
Problem – composite
The outside walls of a house are constructed of a 4-in. layer of
brick, 1/2 in. of celotex, an air space 3 and 5/8 in. thick, and 1/4
in. of wood paneling. If the outside surface of the brick is at 30 ºF
and the inner surface of the paneling at 75 ºF, what is the heat
flux if
a. the air space is assumed to transfer heat by conduction only?
b. the equivalent conductance of the air space is 1.8 Btu/h ft ºF?
c. the air space is filled with glass wool?
𝑘𝑏𝑟𝑖𝑐𝑘 = 0.38 Btu/hft℉
𝑘𝑐𝑒𝑙𝑜𝑡𝑒𝑥 = 0.028 Btu/hft℉
𝑘𝑎𝑖𝑟 = 0.015 Btu/hft℉
𝑘𝑤𝑜𝑜𝑑 = 0.12 Btu/hft℉
𝑘𝑔𝑙𝑎𝑠𝑠 𝑤𝑜𝑜𝑙 = 0.025 Btu/hft℉
Y. Kim | 2
Problem solving skills
1. Identify the “Knowns”
2. Identify “To find”
3. Draw “Schematic”, if not given
4. Set appropriate “Assumptions”
5. “Analysis”
Set your governing equation
Find the conditions (either initial or boundary)
Be careful about unit conversion
Y. Kim | 3
Solution – composite
𝑉1 = 𝑇𝑖 ; 𝑉2 = 𝑇𝑜
𝑅1 = 𝑅𝑖 ; 𝑅2 = 𝑅𝑔𝑙𝑎𝑠𝑠 ; 𝑅3 = 𝑅𝑎𝑖𝑟 ; 𝑅4 = 𝑅𝑔𝑙𝑎𝑠𝑠 ; 𝑅5 = 𝑅𝑜 ;
∆𝑇
𝑞=
σ𝑅
𝑞𝑥 =
𝑇ℎ − 𝑇𝑐
𝐿
1
𝐿1
𝐿
1
(
+
+ 2 + 3 +
)
ℎℎ 𝐴
𝑘1 𝐴 𝑘2 𝐴 𝑘3 𝐴 ℎ𝑐 𝐴
Y. Kim | 4
Solution – composite
a. the air space is assumed to transfer heat by conduction only?
𝑞
75 − 30
45
=
=
𝐴 ( 1/3 + 1/24 + 29/96 + 1/48 )
(0.875 + 1.49 + 20.14 + 0.174)
0.38
0.028 0.015 0.12
Btu
= 1.98
h ft2
**beware the unit conversion, from inch to feet
𝑘𝑏𝑟𝑖𝑐𝑘 = 0.38 Btu/hft℉
𝑘𝑐𝑒𝑙𝑜𝑡𝑒𝑥 = 0.028 Btu/hft℉
𝑘𝑎𝑖𝑟 = 0.015 Btu/hft℉
𝑘𝑤𝑜𝑜𝑑 = 0.12 Btu/hft℉
𝑘𝑔𝑙𝑎𝑠𝑠 𝑤𝑜𝑜𝑙 = 0.025 Btu/hft℉
Y. Kim | 5
Solution – composite
b. the equivalent conductance of the air space is 1.8 Btu/h ft ºF?
𝑅𝑎𝑖𝑟
∆𝑥 29/96
=
=
= 0.167
𝑘
1.8
෍ 𝑅 = 0.875 + 1.49 + 0.167 + 0.174 = 2.706
𝑞
45
=
= 16.63 𝐵𝑡𝑢/ℎ𝑓𝑡2
𝐴 2.706
Y. Kim | 6
Solution – composite
c. the air space is filled with glass wool?
𝑅𝑔𝑙𝑎𝑠𝑠𝑤𝑜𝑜𝑙 =
29/96
= 12.1
0.025
෍ 𝑅 = 14.64
𝑞
45
=
= 3.07𝐵𝑡𝑢/ℎ𝑓𝑡2
𝐴 14.64
Y. Kim | 7
So we want to know heat transfer.
The amount of heat transfer.
It is very easy when the temperature information is given.
𝑞
= −𝑘∇𝑇
𝐴
Then we only need the heat equations.
What if, the numbers – temperature, are not given???
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Goal of this chapter
We now generate the fundamental heat equations,
so, energy equations, then temperature profiles,
for a differential control volume from a first-law-of
thermodynamics approach.
Y. Kim | 9
Our approach to the problems
1. Conservation – heat → Temperature profile
2. Appropriate assumptions – 1D, S-S, etc…
3. Balancing equations
4. Heat (Fourier’s) equations – governing equations
1. Find solutions – Some math. 1st ODE, 2nd ODE
2. Apply B.C.s
3. Complete the solutions
5. Plotting, if necessary.
**The same for mass: Concentration profile then mass (Fick’s) equation
Y. Kim | 10
Balancing energy/mass equation
General balance equation:
In – Out + Generation – Consumption = Accumulation
For a system that does not do the “work” (consume):
In – Out + Generation = Accumulation
For a system without a chemical reaction:
In – Out = Accumulation
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Steady-State and Unsteady-State
Steady-state and unsteady-state processes describe the time interval
that a process occurs over. Steady-state refers to the time where the
variable of interest doesn't change. Unsteady-state is when the variable
of interest changes over time.
Note: Transience is used interchangeably with unsteady-state.
Properties
Steady-state
Unsteady-state
Time
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Law of Energy Conservation
“The change of energy in a system as a function of time is
equal to the amount of energy which enters the system minus
the amount of energy which leaves the system.”
ΔU = Q – W
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Conservation of Energy
Q – W = ΔU
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑜𝑓 ℎ𝑒𝑎𝑡 𝑡𝑜 𝑐𝑜𝑛𝑡𝑟𝑜𝑙
− 𝑏𝑦 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
Figure. A differential control volume
𝑣𝑜𝑙𝑢𝑚𝑒 𝑓𝑟𝑜𝑚 𝑖𝑡𝑠
𝑜𝑛 𝑖𝑡𝑠 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠
𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛𝑡𝑜
𝑟𝑎𝑡𝑒 𝑜𝑓
𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
= 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑢𝑒 −
+ 𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓
𝑡𝑜 𝑓𝑙𝑢𝑖𝑑 𝑓𝑙𝑜𝑤
𝑑𝑢𝑒 𝑡𝑜 𝑓𝑙𝑢𝑖𝑑 𝑓𝑙𝑜𝑤
𝑒𝑛𝑒𝑟𝑔𝑦
𝛿𝑄 𝛿𝑊𝑠 𝛿𝑊𝜇
𝑃
𝜕
−
−
=ඵ 𝑒+
𝜌(v. n)𝑑𝐴 +
ම 𝑒𝜌𝑑𝑉
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝜌
𝜕𝑡
𝑐.𝑠.
𝑐.𝑣.
Heat generation
Energy transfer
Energy accumulation
Work rate
Viscous work rate
Y. Kim | 14
General differential equations for energy transfer
Conservation of energy
𝛿𝑄 𝛿𝑊𝑠 𝛿𝑊𝜇
𝑃
𝜕
−
−
=ඵ 𝑒+
𝜌(v. n)𝑑𝐴 +
ම 𝑒𝜌𝑑𝑉
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝜌
𝜕𝑡
𝑐.𝑠.
𝑐.𝑣.
Heat generation
Energy transfer
Work rate
Energy accumulation
Viscous work rate
Figure. A differential
control volume
Heat generation
(1)
**We have the
Fourier’s law here.
𝑞:ሶ volumetric rate of thermal energy generation having units W/m3
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Work rate
The shaft work rate or power term will be taken as zero for our present
purposes. This term is specifically related to work done by some effect
within the control volume that, for the differential case, is not present.
The power term is thus evaluated as:
(2)
𝛿𝑊𝑠
=0
𝑑𝑡
Viscous work rate
The viscous work rate, occurring at the control surface, is formally
evaluated by integrating the dot product of the viscous stress and the
velocity over the control surface. As this operation is tedious, we shall
express the viscous work rate as Λ ∆𝑥 ∆𝑦 ∆𝑧, where Λ is the viscous
work rate per unit volume. Therefore,
(3)
𝛿𝑊𝜇
= Λ ∆𝑥 ∆𝑦 ∆𝑧,
𝑑𝑡
Y. Kim | 16
Energy transfer
The surface integral includes all energy transfer
across the control surface due to fluid flow. All terms
associated with the surface integral have been
defined previously. The surface integral is
𝑃
𝑒+
𝜌(𝐯. 𝐧)𝑑𝐴
𝜌
ඵ
𝑐.𝑠.
= 𝜌𝑣𝑥
+ 𝜌𝑣𝑦
+ 𝜌𝑣𝑧
𝑣2
𝑃
+ 𝑔𝑦 + 𝑢 + อ
2
𝜌
𝑣2
𝑃
+ 𝑔𝑦 + 𝑢 + อ
2
𝜌
− 𝜌𝑣𝑥
𝑥+∆𝑥
− 𝜌𝑣𝑦
𝑦+∆𝑦
𝑣2
𝑃
+ 𝑔𝑦 + 𝑢 + อ
2
𝜌
− 𝜌𝑣𝑧
𝑧+∆𝑧
𝑣2
𝑃
+ 𝑔𝑦 + 𝑢 + อ
2
𝜌
∆𝑦∆𝑧
𝑥
𝑣2
𝑃
+ 𝑔𝑦 + 𝑢 + อ
2
𝜌
𝑣2
𝑃
+ 𝑔𝑦 + 𝑢 + อ
2
𝜌
(4)
∆𝑥∆𝑧
𝑦
∆𝑥∆𝑦
𝑧
*Specific total energy, e, can include the kinetic, potential, and internal energy contributions,
𝑣2
𝑒 = + 𝑔𝑦 + 𝑢
2
See Ch 6. Conservation of Energy: Control-Volume Approach, for more information.
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Energy accumulation
The energy accumulation term, relating the variation in total energy within
the control volume as a function of time, is
𝜕
𝜕 𝑣2
ම 𝑒𝜌𝑑𝑉 =
+ 𝑔𝑦 + 𝑢 𝜌∆𝑥 ∆𝑦 ∆𝑧
𝜕𝑡 𝑐.𝑣.
𝜕𝑡 2
(5)
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Conservation of energy
𝛿𝑄 𝛿𝑊𝑠 𝛿𝑊𝜇
𝑃
𝜕
−
−
=ඵ 𝑒+
𝜌(v. n)𝑑𝐴 +
ම 𝑒𝜌𝑑𝑉
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝜌
𝜕𝑡
𝑐.𝑠.
𝑐.𝑣.
Combine equations (1) through (5). Performing this combination and dividing
throughout by the volume of the element, we have
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Evaluated in the limit as ∆𝑥, ∆𝑦 and ∆𝑧 approach zero, this equation becomes
General
Equation
After expansion, rearrangement, and introducing Substantial derivative;
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With Continuity equation: 𝜵 ⋅ 𝜌v +
𝜕𝜌
𝜕𝑡
=0
Equation*
𝐷v
And an equation 𝜌 𝐷𝑇 = 𝜌𝑔 − 𝜵𝑃 + 𝜇𝜵𝟐 v, valid for incompressible flow of a fluid with
constant 𝜇, the second term on the right-hand side of the above equation* becomes
Also, for incompressible flow, the first term on the right-hand side of equation* becomes
Substitution of the two equation into equation * results in,
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The previous equation further reduces to
The function
may be expressed in terms of the viscous portion of
the normal- and shear-stress terms in equations from Fluid transport.
For the case of incompressible flow, it is written as
Finally, the energy equation becomes
is a function of fluid viscosity and shear-strain rates, and is positivedefinite. The effect of viscous dissipation is always to increase internal
energy at the expense of potential energy or stagnation pressure.
Y. Kim | 22
General conservation law. The 1st law of thermodynamics:
Q – W = ΔU
Conservation law for a differential control volume:
𝛿𝑄 𝛿𝑊𝑠 𝛿𝑊𝜇
𝑃
𝜕
−
−
=ඵ 𝑒+
𝜌(v. n)𝑑𝐴 +
ම 𝑒𝜌𝑑𝑉
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝜌
𝜕𝑡 𝑐.𝑣.
𝑐.𝑠.
Simplified energy equation for us to use:
Y. Kim | 23
Special forms of the differential energy equations
The applicable forms of the energy equation for some commonly
encountered situations follow.
In every case, the dissipation term is considered negligibly small.
𝑞:ሶ volumetric rate of thermal energy generation (W/m3)
**In this course, we study cases under certain conditions that we can get
analytical solutions.
Y. Kim | 24
Special forms of the differential energy equations
I. For an incompressible fluid without energy sources and with
constant 𝑘
𝐷𝑇
𝜌𝑐𝑣
= 𝑘𝜵2 𝑇
(6)
𝐷𝑡
II. For isobaric flow without energy sources and with constant 𝑘,
𝐷𝑇
𝜌𝑐𝑣
= 𝑘𝜵2 𝑇
(7)
𝐷𝑡
Y. Kim | 25
Special forms of the differential energy equations
III. In a situation where there is no fluid motion, all heat transfer is by
conduction. If this situation exists, as it most certainly does in
solids
where 𝑐𝑣 ≃ 𝑐𝑝 , the energy equation becomes:
(8)
𝐷𝑇
𝜌𝑐𝑝
= ∇ ∙ 𝑘∇𝑇 + 𝑞ሶ
𝐷𝑡
This equation applies in general to heat conduction.
Y. Kim | 26
Special forms of the differential energy equations
If the thermal conductivity is constant, the energy equation is
𝜕𝑇
𝑞ሶ
2
= 𝛼𝜵 𝑇 +
𝜕𝑡
𝜌𝑐𝑝
(9)
cp and cv: Specific heat at constant pressure and constant volume, respectively.
𝑘
has been symbolised as 𝛼 and is designated as thermal diffusivity.
𝜌𝑐𝑝
𝛼 has dimension of 𝐿2 /𝑡 (𝑚2 /𝑠 or 𝑓𝑡 2 /ℎ)
Y. Kim | 27
General heat conduction
equation with constant k:
𝜕𝑇
𝑞ሶ
= 𝛼𝜵2 𝑇 +
𝜕𝑡
𝜌𝑐𝑝
(9)
➢ If the conducting medium contains no heat sources, this equation reduces
to Fourier field equation;
𝜕𝑇
(10)
= 𝛼𝜵2 𝑇
𝜕𝑡
➢ For a system in which heat sources present but there is no time variation
(meaning the steady-state), the equation reduces to the Poisson equation
𝑞ሶ
2
𝜵 𝑇+ =0
𝑘
(11)
➢ The final form of the heat-conduction equation to be presented applies to a
steady-state situation without heat sources. For this case, the temperature
distribution must satisfy the Laplace equation
𝜵2 𝑇 = 0
**Remember,
𝛼=
𝑘
𝜌𝑐𝑝
(12)
Y. Kim | 28
Each of equations (9) through (12) has been written in general form, thus each
applies to any orthogonal coordinate system. Writing the Laplacian operator,
𝛁2 , in the appropriate form will accomplish the transformation to the desired
coordinate system. The Eq (10) 𝜕𝑇
can be written as
= 𝛼𝜵2 𝑇
𝜕𝑡
In cartesian coordinates;
𝜕𝑇
𝜕2𝑇 𝜕2𝑇 𝜕2𝑇
=𝛼
+
+
𝜕𝑡
𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2
In cylindrical coordinates;
𝜕𝑇
𝜕 2 𝑇 1 𝜕𝑇 1 𝜕 2 𝑇 𝜕 2 𝑇
=𝛼
+
+ 2 2+ 2
2
𝜕𝑡
𝜕𝑟
𝑟 𝜕𝑟 𝑟 𝜕𝜃
𝜕𝑧
In spherical coordinates;
2𝑇
𝜕𝑇
1 𝜕
𝜕𝑇
1
𝜕
𝜕𝑇
1
𝜕
=𝛼 2
𝑟2
+ 2
sin 𝜃
+ 2 2
𝜕𝑡
𝑟 𝜕𝑟
𝜕𝑟
𝑟 sin 𝜃 𝜕𝜃
𝜕𝜃
𝑟 sin 𝜃 𝜕𝜙 2
**For Laplacian operator, see Appendix B
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Summary so far,
𝑄 − 𝑊 = 𝛥𝑈
First law of thermodynamics
Conservation of energy, for a controlled volume
𝛿𝑄 𝛿𝑊𝑠 𝛿𝑊𝜇
𝑃
𝜕
−
−
=ඵ 𝑒+
𝜌(v. n)𝑑𝐴 +
ම 𝑒𝜌𝑑𝑉
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝜌
𝜕𝑡
𝑐.𝑠.
𝑐.𝑣.
Energy equation, for a controlled volume
• No heat source, no dissipation, and constant k
𝜕𝑇
𝜕𝑡
= 𝛼𝜵2 𝑇
(Fourier field equation)
• Heat sources, no time variation (steady-state), no dissipation, and constant k
𝜵2 𝑇
𝑞ሶ
𝑘
+ = 0 (Poisson equation)
• Steady-state, no heat sources, no dissipation, and constant k
𝜵2 𝑇 = 0
(Laplace equation)
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Energy equation
• Temperature is our interest, because T is heat/energy.
• T (dimension, time) → for 1D → T (x, t)
• With steady-state assumption, T (x)
• Temperature profile over x-direction can be obtained
• For unsteady-state, T (x, t)
• We keep both ∇2T and DT/Dt.
Y. Kim | 31
Energy equation
For a practical point, the above equation gives us an
equation for a temperature profile.
OK. But then why do we care?
Because, the above equation, the temperature profile, can be
applied into the Fourier equations to get the heat rate or flux.
𝑞
= −𝑘∇𝑇
𝐴
**What about convection and radiation?
Y. Kim | 32
1. Conservation – heat → Temperature profile
2. Appropriate assumptions – 1D, S-S, etc…
3. Balancing equations
For example,
rate of energy
rate of energy
rate of energy out
conduction into + generation within − conduction out
the element
of the element
the element
rate of
accumulation of
=
energy within
the element
3.1. Conduction equations – governing equations
1.
2.
3.
Find solutions – Some math. 1st ODE, 2nd ODE
Apply B.C.s
Complete the solutions
3.2. Other mechanisms can be combined
4. Plotting, if necessary.
**The same for mass: Concentration profile then mass (Fick’s) equation
Y. Kim | 33
Commonly encountered boundary conditions
**Important for problem solving
In solving one of the differential equations developed thus far,
the existing physical situation will dictate the appropriate initial
or boundary conditions, or both, which the final solutions must
satisfy.
Initial conditions refer specifically to the values of T and v at
the start of the time interval of interest.
Initial conditions may be as simply specified as stating that
𝑻ȁ𝒕=𝒐 = 𝑻𝒐 (a constant), or more complex if the temperature
distribution at the start of time measurement is some function
of the space variables.
Y. Kim | 34
Boundary conditions refer to the values of T and v existing at
specific positions on the boundaries of a system, that is, for
given values of the significant space variables.
Frequently encountered boundary conditions for temperature
are the case of
isothermal boundaries, along which the temperature is
constant, and
insulated boundaries, across which no heat conduction
occurs, where according to the Fourier rate equation, the
temperature derivative normal to the boundary is zero.
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Boundary conditions at the interface
One situation often existing at a solid boundary is the equality between
heat transfer (flux) to the surface by conduction and that leaving the
surface by convection. This condition is illustrated in right Figure.
**What assumptions are used here?
At the left-hand surface, the boundary
condition is:
𝜕𝑇
ℎℎ 𝑇ℎ − 𝑇ቚ
= −𝑘 ቤ
𝜕𝑥 𝑥=𝑜
𝑥=𝑜
At the right-hand surface:
ℎ𝑐 𝑇 ቚ
𝑥=𝐿
− 𝑇𝑐
𝜕𝑇
= −𝑘 ቤ
𝜕𝑥 𝑥=𝐿
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Problem – Cylindrical coordinate
1. The Fourier field equation (conduction mechanism without
heat source) in cylindrical coordinates is
𝜕𝑇
𝜕𝑡
=𝛼
𝜕2 𝑇
𝜕𝑟 2
1 𝜕𝑇
+
𝑟 𝜕𝑟
1 𝜕2 𝑇
+ 2 2
𝑟 𝜕𝜃
−
𝜕2 𝑇
𝜕𝑧 2
(r, θ, z)
**Remember,
• No heat source, no dissipation, and constant k
𝜕𝑇
𝜕𝑡
= 𝛼𝜵2 𝑇
(Fourier field equation)
a. What form does this equation reduce to for the case of
steady-state, radial heat transfer?
b. Given the boundary conditions, obtain T profile.
𝑇 = 𝑇𝑖 at 𝑟 = 𝑟𝑖
𝑇 = 𝑇𝑜 at 𝑟 = 𝑟𝑜
c. Generate an expression for the heat flow rate, 𝑞𝑟 , using
the result from part b.
Y. Kim | 37
Solution – Cylindrical coordinate
a. In cylindrical coordinates:
𝜕𝑇
𝜕 2 𝑇 1 𝜕𝑇 1 𝜕 2 𝑇 𝜕 2 𝑇
=𝛼
+
+
−
𝜕𝑡
𝜕𝑟 2 𝑟 𝜕𝑟 𝑟 2 𝜕𝜃 2 𝜕𝑧 2
𝑑 2 𝑇 1 𝑑𝑇
1𝑑
𝑑𝑇
+
= 0 or
𝑟
=0
2
𝑑𝑟
𝑟 𝑑𝑟
𝑟 𝑑𝑟 𝑑𝑟
b. From above equation we get:
𝑑𝑇
𝑟
= 𝐶1
𝑑𝑟
𝐶1
𝑑𝑇 = 𝑑𝑟
𝑟
On integration
𝑇 = 𝐶1 𝑙𝑛𝑟 + 𝐶2
Using Boundary Conditions:
𝐵𝐶1. 𝑇 = 𝑇𝑖 at 𝑟 = 𝑟𝑖
𝐵𝐶2. 𝑇 = 𝑇𝑜 at 𝑟 = 𝑟𝑜
𝑇𝑖 = 𝐶1 𝑙𝑛𝑟𝑖 + 𝐶2
𝑇𝑜 = 𝐶1 𝑙𝑛𝑟𝑜 + 𝐶2
𝑇𝑖 − 𝑇𝑜
𝐶1 = −
𝑙𝑛 𝑟𝑜 /𝑟𝑖
𝐶2 = 𝑇𝑖 − 𝐶1 𝑙𝑛𝑟𝑖
𝑟
𝑟𝑖
𝑇 = 𝑇𝑖 − 𝑇𝑖 − 𝑇𝑜
𝑟
𝑙𝑛 𝑜
𝑟𝑖
𝑙𝑛
Y. Kim | 38
Solution – Cylindrical coordinate
𝑟
𝑙𝑛
𝑟𝑖
𝑇 = 𝑇𝑖 − 𝑇𝑖 − 𝑇𝑜
𝑟𝑜
𝑙𝑛
𝑟𝑖
c. We know that
𝑑𝑇
𝑑𝑇
𝑞𝑟 = −𝑘𝐴
= −𝑘 2𝜋𝑟𝐿
𝑑𝑟
𝑑𝑟
= −𝑘 2𝜋𝐿 𝐶1
2𝜋𝑘𝐿
=
𝑟𝑜 𝑇𝑖 − 𝑇𝑜
ln
𝑟𝑖
(r, θ, z)
We had T profile as below:
𝑇 = 𝐶1 𝑙𝑛𝑟 + 𝐶2
We also obtained C1 as below
𝑇𝑖 − 𝑇𝑜
𝐶1 = −
𝑙𝑛 𝑟𝑜 /𝑟𝑖
Y. Kim | 39
Problem – Cylinder with heat generation
Heat is generated in a cylindrical fuel rod in a nuclear reactor
according to the relationship
where 𝑞ሶ is the volumetric heat generation rate, kW/m3, and ro is
the outside cylinder radius.
Develop the equation that expresses the temperature difference
between the rod center line and its surface.
Y. Kim | 40
Solution – Cylinder with heat generation
**Remember,
• Heat sources, no time variation (steadystate), no dissipation, and constant k
𝑞ሶ
𝑘
𝜵2 𝑇 + = 0 (Poisson equation)
ro
How does the three dimensions in a cylinder work?
Poisson equation reduces to
2
**Note the change from partial derivative to the total derivative
Y. Kim | 41
Solution – Cylinder with heat generation
**Useful tips for problem solving.
1. Assume the nuclear reactor
constantly generate heat.
2. Symmetry condition:
ro
0
Y. Kim | 42
Solution – Cylinder with heat generation
0
𝑇0
𝑟0
𝑞𝑚𝑎𝑥
ሶ
𝑟
𝑟3
න 𝑑𝑇 +
න
− 2 𝑑𝑟 = 0
𝑘
2
4𝑟0
𝑇𝑐
0
**To will be a function of r, we can plot it.
Y. Kim | 43
Problem – Sphere with heat generation
Heat is generated in a spherical fuel element according to the
relationship
3
𝑟
𝑞ሶ = 𝑞ሶ 𝑚𝑎𝑥 1 −
𝑟𝑜
where 𝑞ሶ is the volumetric heat generation rate, kW/m3 and
𝑟𝑜 is the radius of the sphere.
Develop the equation that expresses the temperature
difference between the centre of the sphere and its surface.
Y. Kim | 44
Solution – Sphere with heat generation
**Remember,
• Heat sources, no time variation (steady-state), no dissipation, and constant k
𝑞ሶ
𝑘
How does the three dimensions in a sphere work?
𝜵2 𝑇 + = 0 (Poisson equation)
Heat generation in a sphere :
Fourier field equation reduces to:
1 𝜕
𝜕𝑇
𝑞ሶ 𝑚𝑎𝑥
𝑟
2
𝑟
+
1
−
𝑟2 𝜕𝑟
𝜕𝑟
𝑘
𝑟𝑜
3
=0
First integration yields:
𝑟2
𝜕𝑇
𝑞ሶ 𝑚𝑎𝑥 𝑟 3
𝑟6
+
−
= 𝐶1
𝜕𝑟
𝑘
3 6𝑟𝑜3
𝐶1 = 0 due to symmetry. dT/dr=0 @ r=0
𝜕𝑇
𝑞ሶ 𝑚𝑎𝑥 𝑟 3
𝑟6
2
𝑟
+
−
=0
𝜕𝑟
𝑘
3 6𝑟𝑜3
**The symmetry condition
is useful for cylindrical
and spherical systems.
Y. Kim | 45
Solution – Sphere with heat generation
**Remember,
• Heat sources, no time variation (steady-state), no dissipation, and constant k
𝑞ሶ
𝑘
How does the three dimensions in a sphere work?
𝜵2 𝑇 + = 0 (Poisson equation)
Second integration yields:
𝑇𝑜
𝑞ሶ 𝑚𝑎𝑥 𝑟𝑜 𝑟
𝑟4
න 𝑑𝑇 +
න
− 3 𝑑𝑟 = 0
𝑘 0 3 6𝑟𝑜
𝑇𝑐
𝑞ሶ 𝑚𝑎𝑥 2 2
𝑇𝑐 − 𝑇𝑜 =
𝑟
𝑘 15 𝑜
Y. Kim | 46
Problem – convection and radiation
A 1-in.-thick steel plate measuring 10 in. in diameter is heated from
below by a hot plate, its upper surface exposed to air at 80 °F. The
heat-transfer coefficient on the upper surface is 5 Btu/h ft2 °F and k for
steel is 25 Btu/h ft °F:
a. How much heat must be supplied to the lower surface of the steel if
its upper surface remains at 160 °F? (Include radiation.)
b. What are the relative amounts of energy dissipated from the upper
surface of the steel by convection and radiation?
Tair = 80 °F; h = 5 Btu/h ft2 °F
Tupper = 160 °F; k = 25 Btu/h ft °F
Y. Kim | 47
Solution – convection and radiation
𝑞
= ℎ∆𝑇
𝐴
q: rate of convective heat transfer (Watts or Btu/h),
A: area normal to the heat flow (m2 or ft2)
h: convective heat transfer coefficient (W/m2·K or Btu/h·ft2·°F)
∆T: temperature difference between the surface and the fluid (K or °F)
Convection:
𝑞
= 𝜎𝑇 4
𝐴
T is the absolute temperature, in K or °R; and
𝜎 is the Stefan–Boltzmann constant,
which is equal to 5.676 x10–8 W/m2·K4 or 0.1714 X 10-8 Btu/h·ft2·°R4
** conversion T (°R) = T (°F) + 459.67
Radiation :
A,
80
B,
61.2 and 38.3 percent by convection and radiation, respectively.
Y. Kim | 48