Transcript Chapter 3
Chapter 3
One-Dimensional Steady-State Conduction
Chapter 3
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One-Dimensional Steady-State Conduction
• Conduction problems may involve multiple directions and timedependent conditions
• Inherently complex – Difficult to determine temperature distributions
• One-dimensional steady-state models can represent accurately
numerous engineering systems
• In this chapter we will
Learn how to obtain temperature profiles for common geometries
with and without heat generation.
Introduce the concept of thermal resistance and thermal circuits
Chapter 3
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The Plane Wall
Consider a simple case of onedimensional conduction in a plane
wall, separating two fluids of different
temperature, without energy
generation
Cold fluid
T ,1
T,2 , h2
Ts ,1
• Temperature is a function of x
Ts , 2
• Heat is transferred in the x-direction
Must consider
– Convection from hot fluid to wall
T,1,h1
qx
T , 2
– Conduction through wall
– Convection from wall to cold fluid
Begin by determining temperature
distribution within the wall
Chapter 3
Hot fluid
x=0
x=L
x
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Temperature Distribution
• Heat diffusion equation (eq. 2.4) in the x-direction for steady-state
conditions, with no energy generation:
d dT
k
0
dx dx
• Boundary Conditions:
qx is constant
T (0) Ts,1, T ( L) Ts,2
• Temperature profile, assuming constant k:
x
T ( x) (Ts,2 Ts,1 ) Ts,1 (3.1)
L
Temperature varies linearly with x
Chapter 3
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Thermal Resistance
Based on the previous solution, the conduction hear transfer rate can
be calculated:
Ts,1 Ts,2
dT kA
Ts,1 Ts,2
qx kA
dx L
L / kA
(3.2a)
Similarly for heat convection, Newton’s law of cooling applies:
(TS T )
qx hA(TS T )
1 / hA
(3.2b)
And for radiation heat transfer:
qrad
(Ts Tsur )
hr A(Ts Tsur )
1 / hr A
(3.2c)
Recall electric circuit theory - Ohm’s law for electrical resistance:
Electriccurrent
Chapter 3
PotentialDifference
Resistance
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Thermal Resistance
• We can use this electrical analogy to represent heat transfer problems
using the concept of a thermal circuit (equivalent to an electrical circuit).
OverallDrivingForce Toverall
q
Resistance
R
Compare with equations 3.2a-3.2c
The temperature difference is the “potential” or driving force for the
heat flow and the combinations of thermal conductivity, convection
coefficient, thickness and area of material act as a resistance to this
flow:
Rt ,cond
Chapter 3
L
1
1
, Rt ,conv
, Rt ,rad
kA
hA
hr A
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Thermal Resistance for Plane Wall
Cold fluid
T ,1
T,2 , h2
Ts ,1
qx
Ts , 2
T,1,h1
Hot fluid
qx
x=0
x
T , 2
x=L
T,1 Ts ,1
1 / h1 A
L / kA
Ts ,2 T,2
1 / h2 A
In terms of overall
temperature difference:
qx
Rtot
Chapter 3
Ts ,1 Ts ,2
T,1 T,2
Rtot
1
L
1
h1 A kA h2 A
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Composite Walls
Express the following
geometry in terms of a
an equivalent thermal
circuit.
Chapter 3
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Composite Walls
What is the heat transfer rate for this system?
Alternatively
qx UAT
Rtot
T
1
Rt
q
UA
where U is the overall heat transfer coefficient and T the overall
temperature difference.
1
1
U
Rtot A [(1 / h1 ) ( LA / k A ) ( LB / kB ) ( LC / kC ) (1 / h4 )]
Chapter 3
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Composite Walls
(a) Surfaces normal to the xdirection are isothermal
For resistances in series:
Rtot=R1+R2+…+Rn
For resistances in parallel:
Rtot=1/R1+1/R2+…+1/Rn
Chapter 3
(b) Surfaces parallel to xdirection are adiabatic
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Example (Problem 3.15 textbook)
Consider a composite wall that includes an 8-mm thick hardwood
siding (A), 40-mm by 130-mm hardwood studs (B) on 0.65-m centers
with glass fiber insulation (D) (paper faced, 28 kg/m3) and a 12-mm
layer of gypsum (vermiculite) wall board (C).
What is the thermal resistance associated with a wall that is 2.5 m
high by 6.5 m wide (having 10 studs, each 2.5 m high?)
(Note: Consider the direction of heat transfer to be downwards, along
the x-direction)
Chapter 3
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Contact Resistance
The temperature drop
across the interface
between materials may be
appreciable, due to surface
roughness effects, leading
to air pockets. We can
define thermal contact
resistance:
Rt",c
T A TB
q "x
See tables 3.1, 3.2 for
typical values of Rt,c
Chapter 3
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Alternative Conduction Analysis
When area varies in the x direction and k is a function of temperature,
Fourier’s law can be written in its most general form:
dT
qx k (T ) A( x)
dx
• For steady-state conditions, no heat generation, one-dimensional heat
transfer, qx is constant.
qx
Chapter 3
x
xo
dx
A( x)
T
k (T )dT
To
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Example 3.3
Consider a conical section fabricated from pyroceram. It is of circular
cross section, with the diameter D=ax, where a=0.25. The small end
is at x1=50 mm and the large end at x2=250 mm. The end
temperatures are T1=400 K and T2=600 K, while the lateral surface is
well insulated.
1. Derive an expression for the temperature distribution T(x) in symbolic
form, assuming one-dimensional conditions. Sketch the temperature
distribution
2. Calculate the heat rate, qx, through the cone.
T2
T1
x2
Chapter 3
x1
x
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Radial Systems-Cylindrical Coordinates
Consider a hollow cylinder, whose inner and outer surfaces are
exposed to fluids at different temperatures
Temperature distribution
Chapter 3
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Temperature Distribution
• Heat diffusion equation (eq. 2.5) in the r-direction for steady-state
conditions, with no energy generation:
1 d dT
kr
0
r dr dr
• Fourier’s law:
qr kA
• Boundary Conditions:
dT
dT
k (2rL)
const
dr
dr
T ( r1 ) Ts,1, T ( r2 ) Ts,2
• Temperature profile, assuming constant k:
(Ts ,1 Ts ,2 )
r
T (r)
ln Ts ,2 Logarithmic temperature distribution
(see previous slide)
ln(r1 / r2 ) r 2
Chapter 3
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Thermal Resistance
Based on the previous solution, the conduction hear transfer rate can
be calculated:
• Fourier’s law:
qx
dT
dT
qr kA
k (2rL)
const
dr
dr
2Lk Ts ,1 Ts ,2
ln(r2 / r1 )
Ts,1 Ts,2
ln(r2 / r1 ) /(2Lk )
Ts ,1 Ts ,2
Rt ,cond
In terms of equivalent thermal circuit:
qx
Rtot
Chapter 3
T,1 T,2
Rtot
1
ln(r2 / r1 )
1
h1 (2r1L)
2kL
h2 (2r2 L)
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Composite Walls
Express the following
geometry in terms of a
an equivalent thermal
circuit.
Chapter 3
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Composite Walls
What is the heat transfer rate?
where U is the overall heat transfer coefficient. If A=A1=2r1L:
1
U
1 r1
r
r
r
r
r
r 1
ln 2 1 ln 3 1 ln 4 1
h1 k A r1 k B r2 kC r3 r4 h4
alternatively we can use A2=2r2L, A3=2r3L etc. In all cases:
1
U1 A1 U 2 A2 U 3 A3 U 4 A4
Rt
Chapter 3
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Example (Problem 3.37 textbook)
A thin electrical heater is wrapped around the outer surface of a long
cylindrical tube whose inner surface is maintained at a temperature of
5°C. The tube wall has inner and outer radii of 25 and 75 mm
respectively, and a thermal conductivity of 10 W/m.K. The thermal
contact resistance between the heater and the outer surface of the
tube (per unit length of the tube) is R’t,c=0.01 m.K/W. The outer
surface of the heater is exposed to a fluid of temperature –10°C and a
convection coefficient of h=100 W/m2 .K.
Determine the heater power per unit length of tube required to
maintain the heater at To=25°C.
Chapter 3
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Spherical Coordinates
• Fourier’s law:
dT
qr kA
dr
2 dT
k (4r )
dr
• Starting from Fourier’s law, acknowledging that qr is constant,
independent of r, and assuming that k is constant, derive the equation
describing the conduction heat transfer rate. What is the thermal
resistance?
Chapter 3
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For steady-state, one dimensional conditions with no heat generation;
The appropriate form of Fourier’s equation is
Q = -k A dT/dr
= -k(4πr2) dT/dr
Note that the cross sectional area normal to the heat flow is
A= 4πr2
(instead of dx) where r is the radius of the sphere
Chapter 3
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Equation 2.3-1 may be expressed in the integral form
T2
Q r 2 dr
= - T1 k (T )dT
r1
4
r2
For constant thermal conductivity, k
4k (T 1 T 2)
Q=
1 / r1 1 / r 2
=
r1r 2
(T 1 T 2
r 2 r1
Generally, this equation can be written in terms of
Q = T 2 T1
Rsphere, cond
where
R =
Chapter 3
1 1 1
4k r1 r 2
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Example:
Consider a hollow steel sphere of inside radius r1 = 10 cm and outside
radius, r2 = 20 cm. The thermal conductivity of the steel is k = 10 W/moC.
The inside surface is maintained at a uniform temperature of T1 = 230 oC
and the outside surface dissipates heat by convection with a heat transfer
coefficient h = 20 W/m2oC into an ambient at T = 30oC. Determine the
thickness of asbestos insulation (k=0.5 W/mK) required to reduce the heat
loss by 50%.
Chapter 3
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Example (Problem 3.69 textbook)
One modality for destroying malignant tissue involves imbedding a
small spherical heat source of radius ro within the tissue and
maintaining local temperatures above a critical value Tc for an
extended period. Tissue that is well removed from the source may be
assumed to remain at normal body temperature (Tb=37°C).
Obtain a general expression for the radial temperature distribution in
the tissue under steady-state conditions as a function of the heat rate
q.
If ro=0.5 mm, what heat rate must be supplied to maintain a tissue
temperature of T>Tc=42°C in the domain 0.5<r<5 mm? The tissue
thermal conductivity is approximately 0.5 W/m.K.
Chapter 3
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Summary
• We obtained temperature distributions and thermal
resistances for problems involving steady-state, onedimensional conduction in orthogonal, cylindrical and
spherical coordinates, without energy generation
• Useful summary in Table 3.3
Chapter 3
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