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CHAPTER 3
STEADY-STATE ONE-DIMENSIONAL CONDUCTION
• The term ‘one-dimensional’ refers to the fact that only one coordinate is needed to describe the spatial variation of the dependent variables.
• In one dimensional system, temperature gradients exists along only a single coordinate direction.
• 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
Steady-State, One-Dimensional Conduction
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
qx
1,T
1,sT
2,sT
2,T
x
x=0 x=L
11, ,hT
22, ,hT
Hot fluid
Cold fluid
Temperature Distribution • Heat diffusion equation in the x-direction for steady-state
conditions, with no energy generation:
• Integrate twice to obtain the general solution:
T(x) = C1x + C2
• To obtain C1 and C2, boundary conditions must be introduced
at x=0, T(0) = Ts,1
at x=L, T(L) = Ts,2
• Apply the condition at x=0 to the general solution
Ts,1 = C2
• Apply the condition at x=L to the general solution
Ts,2 = C1L + C2 = C1L + Ts,1
C1 = Ts,2 – Ts,1
L
0
dx
dTk
dx
d qx is constant
Temperature Distribution
• Apply to the general solution, the temperature profile, assuming
constant k:
• From Fourier’s Law, we can determine the conduction heat transfer rate:
qx = -kAdT = kA (Ts,1 – Ts,2)
dx L
And heat flux:
qx” = qx = k (Ts,1 – Ts,2)
A L
1,1,2, )()( sss TL
xTTxT
Problem 3.1
Consider the plane wall, separating hot and cold fluids at temp. T∞,1 and T∞,2 respectively. Using surface energy balances as boundary conditions at x=0 and x=L, obtain the temp. distribution within the wall and the heat flux in terms of T∞,1 , T∞,2 , h1 , h2 , k and L.
Thermal Resistance
• Based on the previous solution, the conduction hear transfer rate can be calculated:
• Similarly for heat convection, Newton’s law of cooling applies:
And for radiation heat transfer:
Recall electric circuit theory - Ohm’s law for electrical resistance:
kAL
TTTT
L
kA
dx
dTkAq
ssssx
/
2,1,2,1,
hA
TTTThAq S
Sx/1
)()(
Ah
TTTTAhq
r
surssursrrad
/1
)()(
Resistance
Difference Potential=currentElectric
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).
• The temperature difference is the “potential” or driving force for the heat flow
• The combinations of thermal conductivity, convection coefficient, thickness and area of material act as a resistance
R
Tq overall
Resistance
Force DrivingOverall
AhR
hAR
kA
LR
rradtconvtcondt
1,
1, ,,,
CONDUCTION CONVECTION RADIATION
Thermal Resistance for Plane Wall
In terms of overall temperature difference:
qx
1,T
1,sT
2,sT
2,T
x x=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
Thermal Resistance for Composite Walls
Thermal Resistance for Composite Walls
• Heat transfer rate for composite wall:
qx = T∞,1 - T ∞,4
∑Rt
= T∞,1 - T ∞,4
[(1/h1A) + (LA/kAA) + (LB/kBA) + (LC/kCA) + (1/h4A)]
-Alternatively:
qx= T∞,1 – Ts,1 = Ts,1 – T2 = T2 – T3 = … U = overall heat transfer coefficient
(1/h1A) (LA/kAA) (LB/kBA) ∆T = temperature difference (overall)
)]/1()/()/()/()/1[(
11
41 hkLkLkLhARU
CCBBAAtot
UAq
TRR
TUAq
ttot
x
1
Composite Walls
For resistances in series:
Rtot=R1+R2+…+Rn
For resistances in parallel:
1/Rtot=1/R1+1/R2+…+1/Rn
Contact Resistance
• In composite systems, the temperature drop across the interface between materials may be appreciable, due to surface roughness effects.
• This temperature change is attribute to thermal contact resistance:
"
",
x
BAct
q
TTR
See tables 3.1, 3.2 for typical values of Rt,c
Problem 3.4 In a manufacturing process, a transparent film is being bonded to a
substrate. To cure the bond at a temperature To, a radiant source is used to provide a heat flux qo” (W/m2), all of which is absorbed at the bonded surface. The back of the substrate is maintained at T1 while the free surface of the film is exposed to air at T∞ and a convection heat transfer coefficient, h.
a) Show a thermal circuit representing the steady
state heat transfer situation.
b) Assume the following conditions:
T∞ = 20°C, h = 50 W/m2·K, T1 = 30°C
Calculate the heat flux qo” that is required to
maintain the bonded surface at To = 60°C
Problem 3.15
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)
Given: PROPERTIES from Table A-3 (T ≈ 300K):
Hardwood siding, kA = 0.094 W/m⋅K
Hardwood, kB = 0.16 W/m⋅K
Gypsum, kC = 0.17 W/m⋅K
Insulation (glass fiber paper faced, 28 kg/m3), kD = 0.038W/m⋅K.
Problem 3.20
A composite wall separates combustion gases at 2600°C
from a liq coolant at 100°C with gas and liq side convection
coefficients of 50 and 1000 W/m2K. The wall is composed of
a 10mm thick layer of beryllium oxide on the gas side. The
contact resistance between the oxide and the steel is 0.05
m2K/W.
a) What is the heat loss per unit surface area of the composite?
b) Sketch the temp. distribution from the gas to the liquid.
Assume temperature of beryllium oxide at 1500 K
and stainless steel at 1000 K.
Problem 3.22
Consider a plane composite wall that is composed of two
materials of thermal conductivities kA = 0.1 w/m.k and kB =
0.04 w/m.k and thicknesses LA = 10mm and LB = 20mm.
The contact resistance at the interface between the two
mterials is known to be 0.30 m2k/w. Material A adjoins a
fluid at 200°c for which h = 10w/m2k and material B
adjoins a fluid at 40°c for which h = 20w/m2k
a) What is the rate of heat transfer through a wall that is 2m high by 2.5m wide
b) Sketch the temp. distribution
Radial Systems-Cylindrical Coordinates
Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures
dr
dT)rLπ2(k=
dr
dTkA=qr - -
Radial Systems-Cylindrical Coordinates
• Heat diffusion equation in the r-direction for steady-state conditions, with no energy generation (for cylinder):
• Integrate twice to obtain general soution
T(r) = C1 ln r + C2
• To obtain C1 and C2 , boundary conditions:
T(r1) = Ts,1 and T(r2) = Ts,2
• Apply to the general solution:
Ts,1 = C1 ln r + C2
Ts,2 = C1 ln r + C2
01
dr
dTkr
dr
d
r
Solve C1 and C2 and substitute into general solution: The conduction heat transfer rate can be calculated: - Fourier’s Law : In terms of equivalent thermal circuit:
2,221
2,1,ln
)/ln(
)()( s
ssT
r
r
rr
TTrT
condt
ssssssr
R
TT
Lkrr
TT
rr
TTLkq
,
2,1,
12
2,1,
12
2,1,
)2/()/ln()/ln(
2
)2(
1
2
)/ln(
)2(
1
22
12
11
2,1,
LrhkL
rr
LrhR
R
TTq
tot
totr
constdr
dTrLk
dr
dTkAqr )2(
Composite Walls
The heat transfer rate may be expressed as
qr = T∞,1 - T ∞,4
1 + ln (r2/r1) + ln (r3/r2) + ln (r4/r3) + 1
2πr1Lh1 2πkAL 2πkBL 2πkcL 2πr4Lh4
The heat transfer rate may also be expressed in terms of an overall heat
transfer coefficient:
qr = T∞,1 - T ∞,4 = UA(T∞,1 - T ∞,4)
Rtot
where U is the overall heat transfer coefficient. If A = A1 = 2πr1L
44
1
3
41
2
31
1
21
1
1lnlnln
1
1
hr
r
r
r
k
r
r
r
k
r
r
r
k
r
h
U
CBA
• Alternatively we can use:
A2 = 2πr2L , A3 = 2πr3L , etc
tRAUAUAUAU
144332211
Radial Systems-Sphere Coordinates
• Heat diffusion equation in the r-direction for steady-state conditions, with no energy generation (for sphere):
• Integrate twice to obtain general soution
• To obtain C1 and C2 , boundary conditions:
T(r1) = Ts,1 and T(r2) = Ts,2
• Apply to the general solution:
0=dr
dTkr
dr
d
r
1 22
( ) 21
C+ r
C-=rT
C+ r
C-=T
C+ r
C-=T
22
12,s
21
11,s
Solve C1 and C2 and substitute into general solution
Sphere
• Fourier’s law:
dr
dTrk
dr
dTkAqr
)4( 2
Assuming constant k, qr = 4πk (Ts,1 – Ts,2) = (Ts,1 – Ts,2) (1/r1) – (1/r2) 1/ 4πk [1/r1 – 1/r2] Thermal resistance: Rt,cond = 1 [ 1/r1 - 1/r2 ] 4πk
Problem 3.37
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.
Problem 3.60
A spherical vessel used as a reactor for producing pharmaceuticals hs a 10mm thick stainless steel wall (k = 17w/m.k) and an inner diameter of 1m. The exterior surface of the vessel is exposed to ambient air (T∞ = 25°c) for which a convection coefficient of 6 w/m2k may be assumed.
a) During steady-state operation, an inner surface temp.of 50°c is maintained by energy generated within the reactor. What is the heat loss from the vessel?
b) If a 20mm thick layer of fiber glass insulation (k = 0.04 w/m.k) is applied to the exterior of the vessel and the rate of thermal energy generation is unchanged, what is the inner surface temp. of the vessel?
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