CHAPTER 3_Part 1
One-Dimensional, Steady-State
Conduction without
Thermal Energy Generation
Instructor
Dr. Salman Abbasi
Methodology
Methodology of a Conduction Analysis
• Specify appropriate form of the heat equation.
• Solve for the temperature distribution.
• Apply Fourier’s law to determine the heat flux.
Simplest Case: One-Dimensional, Steady-State Conduction with No Thermal Energy
Generation.
• Common Geometries:
– The Plane Wall: Described in rectangular (x) coordinate. Area
perpendicular to direction of heat transfer is constant (independent of x).
– The Tube Wall: Radial conduction through tube wall.
– The Spherical Shell: Radial conduction through shell wall.
Plane Wall
The Plane Wall
• Consider a plane wall between two fluids of different temperature:
• Heat Equation:
d dT
k =0 (3.1)
dx dx
• Implications:
Heat flux ( qx ) is independent of x.
Heat rate ( qx ) is independent of x.
• Boundary Conditions: T ( 0 ) = Ts ,1, T ( L ) = Ts,2
• Temperature Distribution for Constant k :
T ( x ) = Ts ,1 + (Ts ,2 − Ts ,1 )
x
(3.3)
L
Plane Wall (cont.)
• Heat Flux and Heat Rate:
= (Ts ,1 − Ts ,2 )
dT k
qx = − k (3.5)
dx L
= (Ts ,1 − Ts ,2 )
dT kA
q x = − kA (3.4)
dx L
T
• Thermal Resistances Rt = and Thermal Circuits:
q
L
Conduction in a plane wall: Rt ,cond = (3.6)
kA
1
Convection: Rt ,conv = (3.9)
hA
Thermal circuit for plane wall with adjoining fluids:
1 L 1
Rtot = + + (3.12)
h1 A kA h 2 A
T,1 − T,2
qx = (3.11)
Rtot
Plane Wall (cont.)
• Thermal Resistance for Unit Surface Area:
L 1
Rt,cond = Rt,conv =
k h
Units: Rt K/W Rt m2 K/W
• Radiation Resistance:
1 1
Rt ,rad = Rt,rad =
hr A hr
(
hr = (Ts + Tsur ) Ts2 + Tsur
2
) (1.9)
• Contact Resistance:
TA − TB Rt,c
, =
Rtc Rt ,c =
qx Ac
Values depend on: Materials A and B, surface finishes, interstitial conditions, and
contact pressure (Tables 3.1 and 3.2)
Plane Wall (cont.) • Composite Wall with Negligible
Contact Resistance:
T,1 − T,4
qx = (3.14)
Rt
For the temperature distribution
shown, kA > kB < kC.
1 1 LA LB LC 1 Rtot
Rt = Rtot = + + + + =
A h1 k A k B kC h4 A
• Overall Heat Transfer Coefficient (U) :
A modified form of Newton’s law of cooling to encompass multiple resistances
to heat transfer.
qx = UAToverall (3.17)
1
Rtot = (3.19)
UA
Plane Wall (cont.)
• Series – Parallel Composite Wall:
Assuming isothermal
surfaces perpendicular
to x-direction.
Assuming adiabatic
surfaces parallel
to x-direction.
• Note departure from one-dimensional conditions for k F kG .
• Circuits based on assumption of isothermal surfaces normal to x direction or
adiabatic surfaces parallel to x direction provide approximations for qx .
Tube Wall (cont.)
• Heat Flux and Heat Rate:
qr = − k
dT
=
k
dr r ln ( r2 / r1 )
(Ts,1 − Ts,2 ) [W/m2]
2 k
qr = 2 rqr =
ln ( r2 / r1 )
( Ts ,1 − Ts ,2 ) [W/m]
2 Lk
qr = 2 rLqr =
ln ( r2 / r1 )
(Ts,1 − Ts,2 ) [W] (3.32)
• Conduction Resistance:
ln ( r2 / r1 )
Rt ,cond = [K/W] (3.33)
2 Lk
ln ( r2 / r1 )
Rt,cond = [m K/W]
2 k
Tube Wall (cont.)
• Composite Wall with
Negligible Contact
Resistance
T,1 − T,4
qr =
Rtot
(
= UA T,1 − T,4 ) (3.35)
Note that
For the temperature distribution
−1
UA = Rtot shown, kA > kB > kC.
is a constant independent of radius,
but U itself is tied to specification of an interface.
−1
U i = ( Ai Rtot ) (3.37)
Spherical Shell
Spherical Shell
• Heat Equation for Constant k:
1 d 2 dT
2 dr
r =0
r dr
What does the form of the heat equation tell us about the variation of
qr with r ? Is this result consistent with conservation of energy?
• Temperature Distribution:
T ( r ) = Ts ,1 − (Ts ,1 − Ts ,2 )
( )
1 − r1/ r
(
1 − r1 / r 2 )
Spherical Shell (cont.)
• Heat Flux, Heat Rate and Thermal Resistance:
dT
qr = −k = 2
k
dr r (1 / r1 ) − (1 / r2 )
( Ts ,1 − Ts ,2 )
4 k
qr = 4 r 2qr =
(1 / r1 ) − (1 / r2 )
( Ts ,1 − Ts ,2 ) (3.40)
Rt ,cond =
(1 / r1 ) − (1 / r2 ) (3.41)
4 k
• Composite Shell:
Toverall
qr = = UAToverall
Rtot
UA = Rtot −1 Constant
−1
U i = ( Ai Rtot ) Depends on Ai