Download - Chapter 3
Chapter 3 1
Chapter 3
One-Dimensional Steady-State Conduction
Chapter 3 2
One-Dimensional Steady-State Conduction
• Conduction problems may involve multiple directions and time-
dependent 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 3
The Plane Wall
Consider a simple case of one-dimensional conduction in a plane wall, separating two fluids of different temperature, without energy generation
• Temperature is a function of x
• Heat is transferred in the x-direction
Must consider– Convection from hot fluid to wall
– Conduction through wall
– Convection from wall to cold fluid
Begin by determining temperature distribution within the wall
qx
1,T
1,sT
2,sT
2,T
x
x=0 x=L
11, ,hT
22, ,hT
Hot fluid
Cold fluid
Chapter 3 4
Temperature Distribution
• Heat diffusion equation (eq. 2.4) in the x-direction for steady-state conditions, with no energy generation:
0
dx
dTk
dx
d
• Boundary Conditions: 2,1, )(,)0( ss TLTTT
• Temperature profile, assuming constant k:
1,1,2, )()( sss TL
xTTxT
Temperature varies linearly with x
qx is constant
(3.1)
Chapter 3 5
Thermal ResistanceBased on the previous solution, the conduction hear transfer rate can be calculated:
kAL
TTTT
L
kA
dx
dTkAq ss
ssx /2,1,
2,1,
Recall electric circuit theory - Ohm’s law for electrical resistance:
Similarly for heat convection, Newton’s law of cooling applies:
Resistance
e DifferencPotentialcurrent Electric
hA
TTTThAq S
Sx /1
)()(
And for radiation heat transfer:
Ah
TTTTAhq
r
surssursrrad /1
)()(
(3.2a)
(3.2b)
(3.2c)
Chapter 3 6
Thermal Resistance
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:
• We can use this electrical analogy to represent heat transfer problems using the concept of a thermal circuit (equivalent to an electrical circuit).
AhR
hAR
kA
LR
rradtconvtcondt
1,
1, ,,,
R
Tq overall
Resistance
Force DrivingOverall
Chapter 3 7
Thermal Resistance for Plane Wall
In terms of overall temperature difference:qx
1,T
1,sT
2,sT
2,T
xx=0 x=L
11, ,hT
22, ,hT
Hot fluid
Cold fluid
AhkA
L
AhR
R
TTq
tot
totx
21
2,1,
11
Ah
TT
kAL
TT
Ah
TTq ssssx
2
2,2,2,1,
1
1,1,
/1//1
Chapter 3 8
Composite Walls Express the following
geometry in terms of a an equivalent thermal circuit.
Chapter 3 9
Composite Walls What is the heat transfer rate for this system?
Alternatively
UAq
TRR
TUAq
ttot
x
1
where U is the overall heat transfer coefficient and T the overall temperature difference.
)]/1()/()/()/()/1[(
11
41 hkLkLkLhARU
CCBBAAtot
Chapter 3 10
Composite Walls
(a) Surfaces normal to the x-direction are isothermal
(b) Surfaces parallel to x-direction are adiabatic
For resistances in series: Rtot=R1+R2+…+Rn
For resistances in parallel:
Rtot=1/R1+1/R2+…+1/Rn
Chapter 3 11
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 12
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:
"",
x
BAct
q
TTR
See tables 3.1, 3.2 for typical values of Rt,c
Chapter 3 13
Alternative Conduction Analysis
• For steady-state conditions, no heat generation, one-dimensional heat transfer, qx is constant.
dx
dTxATkqx )()(
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:
T
T
x
xx
oo
dTTkxA
dxq )(
)(
Chapter 3 14
Example 3.3Consider a conical section fabricated from pyroceram. It is of circular cross section, with the diameter D=x, where =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
x1x2
x
Chapter 3 15
Radial Systems-Cylindrical Coordinates
Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures
Temperature distribution
Chapter 3 16
Temperature Distribution
• Heat diffusion equation (eq. 2.5) in the r-direction for steady-state conditions, with no energy generation:
01
dr
dTkr
dr
d
r
• Boundary Conditions: 2,21,1 )(,)( ss TrTTrT
• Temperature profile, assuming constant k:
2,221
2,1, ln)/ln(
)()( s
ss Tr
r
rr
TTrT
Logarithmic temperature distribution
(see previous slide)
• Fourier’s law: constdr
dTrLk
dr
dTkAqr )2(
Chapter 3 17
Thermal ResistanceBased on the previous solution, the conduction hear transfer rate can be calculated:
condt
ssssssx R
TT
Lkrr
TT
rr
TTLkq
,
2,1,
12
2,1,
12
2,1,
)2/()/ln()/ln(
2
In terms of equivalent thermal circuit:
)2(
1
2
)/ln(
)2(
1
22
12
11
2,1,
LrhkL
rr
LrhR
R
TTq
tot
totx
• Fourier’s law: constdr
dTrLk
dr
dTkAqr )2(
Chapter 3 18
Composite Walls
Express the following geometry in terms of a an equivalent thermal circuit.
Chapter 3 19
Composite Walls What is the heat transfer rate?
where U is the overall heat transfer coefficient. If A=A1=2r1L:
44
1
3
41
2
31
1
21
1
1lnlnln
11
hrr
rr
kr
rr
kr
rr
kr
h
U
CBA
alternatively we can use A2=2r2L, A3=2r3L etc. In all cases:
tRAUAUAUAU
144332211
Chapter 3 20
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 21
Spherical Coordinates
• 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?
• Fourier’s law:
dr
dTrk
dr
dTkAqr
)4( 2
Chapter 3 22
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 23
Equation 2.3-1 may be expressed in the integral form = -
For constant thermal conductivity, k
Q =
=
Generally, this equation can be written in terms of
Q = where
R =
2
1 24rr r
drQ
21 )(TT dTTk
21
21
/1/1
)(4
rr
TTk
2112
21( TT
rr
rr
condsphereR
TT
,
12
21
11
4
1
rrk
Chapter 3 24
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 25
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 26
Summary
• We obtained temperature distributions and thermal
resistances for problems involving steady-state, one-
dimensional conduction in orthogonal, cylindrical and
spherical coordinates, without energy generation
• Useful summary in Table 3.3