CHAPTER 3: One-Dimensional Steady-State Conduction • one primary direction in which heat transfers (generally the smallest dimension) • simple model • good representation for solving engineering problems
3.1
Plane Wall
3.1.1
hot fluid
cold fluid
T ∞ ,1 h1
T ∞ ,2 h2
x2=L
x1=0 x
For one-dimensional steady state conduction with no energy generation, the heat equation reduces to: & − E & + E & = E & E in out g st 1424 3
∂ ⎛ ∂T ⎞ ∂T & k q c + = ρ ⎟ ⎜ p ∂ x ⎝ ∂ x ⎠ ∂t Assume an isotropic medium ( k = = constant) and integrate to determine the x): temperature distribution T ( x k
d ⎛ dT ⎞
⎜
⎟=0
dx ⎝ dx ⎠ dT
= C 1 dx T ( x ) = C 1 x + C 2
3.1
Apply boundary conditions to solve for constants: T (0)=T s1 ; T (L)=T s2 T (0) = T s1 = C 2 T ( L) = T s 2 = C 1 L + T s1
∴ C 1 =
T s 2 − T s1 L
The resulting temperature distribution is: x
∴T ( x) = (T s 2 − T s1 ) + T s1 L
and varies linearly with x.
Applying Fourier’s law:
⎛ T − T s1 ⎞ = − kA⎜ s 2 ⎟ dx ⎝ L ⎠ dT ⎛ T − T s1 ⎞ q x′′ = −k = − k ⎜ s 2 ⎟ dx ⎝ L ⎠
• heat transfer rate: q x = q x′′ A = − kA • heat flux:
dT
Therefore, both the heat transfer rate and heat flux are independent of x.
Note: Alternative boundary conditions might involve convection if surface temperatures are unknown. T (0) = T s1 = C 2
e.g.,
dT
− k
dx
= h[T ( L) − T ∞ , 2 ] x = L
and T ( L) = C 1 L + C 2 ; solve for C 1
3.2
3.1.2 Thermal Resistance A resistance can be defined as the ratio of a driving potential to a corresponding transfer rate. e.g.,
Analogy: electrical resistance is to conduction of electricity as thermal resistance is to conduction of heat
⎛ T s 2 − T s1 ⎞ ⎟ L ⎝ ⎠
• thermal resistance for conduction:
qcond = − kA⎜
• thermal resistance for convection:
qconv = hA(T s − T ∞ )
• thermal resistance for radiation exchange:
4 4 qrad = εσ A T s − T sur
( = εσ A(T
) )(T
)
4 qrad = εσ A T s4 − T sur 2 s
2 − T sur
2 s
2 ) + T sur
2 = εσ A(T s − T sur )(T s + T sur )(T s2 + T sur ) = hr A(T s − T sur )
where
(
2 hr = εσ (T s + T sur ) T s2 + T sur
)
3.3
The equivalent thermal circuit for a plane wall
The total rate of heat transfer is:
3.1.3 The Composite Wall
• Equivalent thermal circuit analysis can also be used for complex systems such as composite walls • Often, it is convenient to define an overall heat transfer coefficient, U :
where ∆T is the overall temperature difference, such that:
3.4
3.1.4 Contact Resistance Realistically, the heat flux at the interface of two solid surfaces is continuous but the temperature is not. Temperature varies at the interface of two surfaces because of imperfect contact between the solids.
The temperature difference is attributed to a thermal contact resistance: T − T B Rt ′′,c = A q x′′ The equivalent thermal circuit for imperfect contact:
Thermal contact resistance can be reduced by (e.g.)
(Omit 3.2, An Alternative Conduction Analysis) 3.5
3.3
Radial Systems cold fluid
3.3.1
T ∞ ,2 , h ∞ ,2
L
hot fluid
T ∞ ,1 , h ∞ ,1
For one-dimensional, steady-state conditions with no energy generation, the heat equation reduces to: 1 d ⎛ dT ⎞ ⎜ kr ⎟ = 0 r dr ⎝ dr ⎠ Assume an isotropic medium and integrate to determine the temperature distribution, T (r ): k d ⎛ dT ⎞ ⎜ r ⎟ = 0 r dr ⎝ dr ⎠ dT
= C 1 dr T (r ) = C 1 ln r + C 2 r
Apply boundary conditions to solve for constants C 1 and C 2: T (r 1) = T s1 = C 1ln r 1 + C 2 T (r 2) = T s2 = C 1ln r 2 + C 2
T s , 2 − T s ,1
subtracting T s2 −T s1 :
C 1 =
substitute C 1 into T (r 1):
C 2 = T s ,1 −
ln (r 2 r 1 ) T s , 2 − T s ,1
ln (r 2 r 1 )
ln r 1
Note: The values of C 1 and C 2 can have other formulations, depending on the boundary conditions used to solve for T (r ).
3.6
The resulting temperature distribution is:
∴ T ( r ) =
T s ,1 − T s, 2
ln (r 1 r 2 )
⎛ r ⎞ ⎟⎟ + T s, 2 ⎝ r 1 ⎠
ln ⎜⎜
and varies logarithmically with r . Applying Fourier’s law: • heat transfer rate: ⎡ T s , 2 − T s ,1 ⎤ 1 1 dT q r = −kA = − k (2π rL )⎢ ⎥ dr ⎣ ln(r 2 r 1 ) ⎦ (r r 1 ) r 1
⎡ T − T s ,1 ⎤ ∴ q r = −2π Lk ⎢ s , 2 ⎥ ⎣ ln(r 2 r 1 ) ⎦ • heat flux: q ′r ′ =
qr A
=
− 2π Lk (T s , 2 − T s ,1 )
(2π rL )ln(r 2
r 1 )
k ⎡ T s , 2 − T s ,1 ⎤
=− ⎢ ⎥ r ⎣ ln (r 2 r 1 ) ⎦
Thus, the heat transfer rate qr is constant and independent of r . The heat flux q′r ′ is a function of the radius. The equivalent thermal circuit for a cylinder
• thermal resistance for conduction: • thermal resistance for convection:
⎡ T s ,1 − T s , 2 ⎤ ⎥ ⎣ ln (r 1 r 2 ) ⎦
qcond = 2π Lk ⎢
qconv = 2π rLh(T s − T ∞ )
3.7
3.3.2 Sphere (assume hollow center)
r 1
r 2
For one-dimensional, steady-state conditions with no energy generation, the heat equation reduces to: 1 d ⎛ 2 dT ⎞ ⎜ kr ⎟=0 2 dr ⎠ r dr ⎝ Assume an isotropic medium and integrate to determine the temperature difference T (r ): k d ⎛ 2 dT ⎞ ⎜ r ⎟=0 r 2 dr ⎝ dr ⎠ r 2
dT dr
= C 1
T ( r ) = C 2 −
C 1 r
Apply boundary conditions: T (r 1)=T s1 and T (r 2)=T s2 The resulting temperature distribution is:
⎡ 1 − (r 1 r ) ⎤ ∴T (r ) = T s ,1 − (T s,1 − T s , 2 )⎢ ⎥ ⎣1 − (r 1 r 2 )⎦ The corresponding heat transfer rate is:
∴ q r =
4π k T s ,1 − T s , 2
⎛ 1 ⎞ ⎛ 1 ⎞ ⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ ⎝ r 1 ⎠ ⎝ r 2 ⎠
• thermal resistance for conduction: 1 ⎛ 1 1 ⎞ ⎜ − ⎟ Rt ,cond = 4π k ⎜⎝ r 1 r 2 ⎠⎟ 3.8
3.5
Conduction with Thermal Energy Generation Consider the effect of a process occurring within a medium such as thermal & , e.g., conversion of electrical to thermal energy energy generation, E g
3.5.1 Plane Wall
T s,1
T s,2
T ∞ ,1 , h1
T ∞ ,2 , h2 q& - L
x
L
For one-dimensional, steady-state conduction in an isotropic medium: d ⎛ dT ⎞ ⎜ k ⎟ + q& = 0 dx ⎝ dx ⎠
∴T ( x) = −
q&
2 x + C 1 x + C 2 2k
1
Note: The temperature distribution is parabolic in x. Applying Fourier’s law:
⎛ q& ⎞ = −kA⎜ − x + C 1 ⎟ dx ⎝ k ⎠ dT ⎛ q& ⎞ = −k ⎜ − x + C 1 ⎟ q x′′ = − k dx ⎝ k ⎠
• heat transfer rate: q x = q x′′ A = − kA • heat flux:
dT
Thus, dT/dx is a function of x, and therefore both the heat transfer rate and heat flux are dependent on x for a medium with energy generation .
3.9
Case #1: T s,1 > T s,2 T ∞,1 > T ∞,2 h1 ≠ h2
T ∞ ,1 , h1
boundary conditions: T (-L) = T s,1 T ( L) = T s,2
T ∞ ,2 , h2 q& - L
x
L
From 1 and the above boundary conditions: 2 2 q& L ⎡ x ⎤ T s , 2 − T s ,1 ⎛ x ⎞ T s ,1 + T s , 2 ∴ T ( x) = ⎜ ⎟+ ⎢1 − 2 ⎥ + 2k ⎣ L ⎦ 2 2 ⎝ L ⎠
To determine where the temperature is a maximum:
Case #2: T s,1 = T s,2 = T s T ∞,1 = T ∞,2 = T ∞ h1 = h2 = h
boundary conditions: T (-L) = T s = T ( L)
T ∞ ,1 , h1
T ∞ ,2 , h2 q& - L
x
L
From 1 and the above boundary conditions: 2 2 q& L ⎡ x ⎤ ∴T ( x) = ⎢1 − 2 ⎥ + T s 2k ⎣ L ⎦
3.10
Case #3: at x = 0, adiabatic surface at x = L, T = T s
boundary conditions: dT dx x=0
=0
(i.e., q ′′ = 0)
T ∞ , , h q& x L
T ( L) = T s
The insulated wall is analogous to a symmetric wall, and the same temperature distribution is found as Case #2: 2 2 q& L ⎡ x ⎤ ∴ T ( x) = ⎢1 − ⎥ + T s 2k ⎣ L2 ⎦
3.11
3.5.2 Radial Systems T ∞ , h
T s
q&
r o
L
For one-dimensional, steady-state conduction in an isotropic medium: k d ⎛ dT ⎞ ⎜ r ⎟ + q& = 0 r dr ⎝ dr ⎠
∴ T (r ) = −
q&
4k
r 2 + C 1 ln r + C 2
Apply boundary conditions to solve for the constants C 1 and C 2: dT dr r o = 0
=0
T ( r o ) = T s
∴ C 1 = 0 ∴ C 2 =
q&
r o2 + T s
4k The resulting temperature distribution is:
q&r o2 ⎛ r 2 ⎞ ⎜1 − ⎟ + T s ∴T (r ) = − 4k ⎜⎝ r o2 ⎟
and varies parabolically in r .
Caution…. Note that it is incorrect to use thermal resistance concepts within a medium when the heat transfer rate is NOT constant (energy generation). 3.12
3.6
Heat Transfer from Extended Surfaces
• involves conduction through a solid medium as well as convection and/or radiation energy transfer • goal is to enhance heat transfer between a solid and a fluid qconv = hA(T s − T ∞ )
Possibilities: - increase heat transfer coefficient - increase surface temperature - decrease fluid temperature - increase surface area The most common way to enhance heat transfer is by increasing the surface area for convection via an extension from a solid medium: fins Consider a long, slender “pin fin” attached to a base:
T ∞ , h D T b
x
L
3.13
3.6.1 General Conduction Analysis
Assume one-dimensional conditions in x−direction, i.e., the fin is thin, thus
Assume steady-state conditions, isotropic medium, neglect radiation, and there is no energy generation & = E & + E & − E & E st in g out
0 = q x − [q x+ dx + dqconv ] q x = − kAc ( x ) q x + dx = q x +
dT
dx dq x dx
dx = − kAc ( x )
dT dx
− k
d ⎛ dT ⎞ ⎜ Ac ( x ) ⎟dx dx ⎝ dx ⎠
dqconv = hdAs [T ( x ) − T ∞ ]
then, d ⎛ dT ⎞ h dAs (T − T ∞ ) = 0 ⎜ Ac ⎟− dx ⎝ dx ⎠ k dx
finally,
where, Ac, As, and T can be functions of x. 3.14
3.6.2 Fins of Uniform Cross-Sectional Area Consider a straight fin: at the base ( x = 0), T (0)=T b Ac is constant As = Px, P is the perimeter canceling and rearranging appropriate terms from the general equation: d 2T dx 2
−
hP Ac k
(T − T ∞ ) = 0
define
then,
Boundary conditions: at x = 0; T (0)=T b at x = L (refer to each case, A.-D.; same as Table 3.4): A. Convection at tip;
− kAc
dT dx x = L
qcond = qconv
= hAc [T ( L) − T ∞ ]
final solution is:
∴
θ θ b
=
T − T ∞ cosh[m( L − x )] + (h mk ) sinh[m( L − x )] T b − T ∞
cosh (mL ) + (h mk ) sinh (mL )
3.15
B. Adiabatic fin tip; dT dx x = L
=0
final solution is:
∴
θ θ b
=
cosh[m( L − x )] cosh (mL)
C. Fixed temperature at tip; T ( x=L) = T L
final solution is: ∴
θ θ b
=
(θ L
θ b ) sinh (mx ) + sinh[m( L − x )] sinh (mL )
D. Infinitely long fin;
T ( x→∞ ) = T ∞
final solution is:
∴
θ θ b
= e − mx
3.16
Analyze Case B (adiabatic tip) to determine other information about the fin:
• total heat transfer from the fin: q f = − kAc
dT dx x = 0
= − kAc
d θ dx x = 0
Note: the same solution can be obtained by: q f =
∫
A f
h[T ( x) − T ∞ ]dA f
where A f is the surface area of the fin.
• fin effectiveness: ε f =
q f qb
=
• fin efficiency, η f : ratio of the actual fin heat transfer rate to the rate predicted for an isothermal fin η f =
q f qb
=
iso
3.17
Regardless of the tip boundary condition, the solution will provide the temperature distribution through the fin, θ / θ b. Due to the decreasing temperature through the fin, the conduction heat transfer (locally) decreases as a result of continuous convection losses
The total heat transfer by the fin:
T ∞ , h tip
T b base
3.18
