Steady Heat Transfer
February 14, 2007
Outline
Steady Heat Transfer with
Conduction and Convection
• Review last lecture
• Equivalent circuit analyses
Larry Caretto
Mechanical Engineering 375
– Review basic concept
– Application to series circuits with
conduction and convection
– Application to composite materials
– Application to other geometries
Heat Transfer
February 14, 2007
• Two-dimensional shape factors
2
Review Steady, 1-D, e& gen = 0
• Rectangular
k (TL − T0 )
Q&
q& = = −
A
L
Review Heat Generation
• Cylindrical shell
Q&
2πk (T2 − T1 )
=−
L
ln (r2 r1 )
Figure 2-21 from Çengel,
• Various
Heat and Mass Transfer
phenomena in
solids can
generate heat
• Define e& gen
as the heat
I 2 ρL
generated per
I 2ρ
I 2R
= A = 2
unit volume e&gen =
LA
V
A
per unit time
• Spherical shell k is an average thermal
4πk (T2 − T1 )
conductivity (or a constant
Q& =
value) if k is constant
1 / r1 − 1 / r2
T0, TL = temperatures at x = 0,L; T1, T2 =
temperatures at inner (r1) and outer(r2) radii
3
4
Plot of (T - T0)/(TL - T0) for Heat Generation in a Slab
Review Heat Generation II
2
1.8
• Temperature and heat flux equations
2k
+
e&genxL
2k
TemperatureDifference Ratio
T = T0 −
e&genx2
1.6
(T − T )x
− 0 L
L
e&gen(2x − L) k (T0 − TL )
+
q& =
2
L
1.4
H=0
1.2
H = .01
H = .1
H=1
1
H=2
H=5
0.8
H = 10
0.6
2
H=
0.4
0.2
Q& x =0 + E& gen = Q& x = L
L e&gen
k (TL − T0 )
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x/L
5
ME 375 – Heat Transfer
6
1
Steady Heat Transfer
February 14, 2007
Steady Heat Transfer Definition
Rectangular Steady Conduction
Figure 3-2 from
Çengel, Heat
and Mass
Transfer
• In steady heat transfer the temperature
and heat flux at any coordinate point do
not change with time
• Both temperature and heat transfer can
change with spatial locations, but not
with time
• Steady energy balance (first law of
thermodynamics) means that heat in
plus heat generated equals heat out
The heat
transfer is
constant
in this 1D
rectangle
for both
constant
& variable
k
dT
Q&
= q& = − k
A
dx
7
Thermal Resistance
Figure 2-63 from Çengel,
Heat and Mass Transfer
8
Thermal Resistance II
• Heat flow analogous to current
• Temperature difference analogous to
potential difference
• Both follow Ohm’s law with appropriate
resistance term
• Conduction
k A(T1 − T2 )
T −T
Q& =
⇒ Q& = 1 2
L
Rcond
• Convection
(
Q& = hA Ts − T f
)
⇒ Q& =
Ts − T f
Rconv
⇒ Rcond =
L
kA
⇒ Rconv =
1
hA
• Radiation
Rrad =
(
1
1
A1F12σ T13 + T23 + T22T1 + T12T2
)= Ah
1 rad
9
Where Does the Heat Go?
10
Where Does the Heat Go? II
Energy conservation
requires that conduction
heat through wall equals
the heat leaving the wall
by convection and
radiation
Figure 1-18
from Çengel,
Heat and
Mass
Transfer
Q&1 = Q& 2 + Q& 3
Figure 1-18 from
Çengel, Heat and
Mass Transfer
11
ME 375 – Heat Transfer
Figure 3-5 from Çengel,
Heat and Mass Transfer
12
2
Steady Heat Transfer
February 14, 2007
Parallel Resistances (T∞ = Tsurr)
1
1
1
=
+
Rtotal Rconv Rrad
T∞ =
•
1
Tsurr Rtotal
= As hconv + As hrad
Define total heat transfer
coefficient, htotal
Figure 3-5
from Çengel,
Heat and
Mass Transfer
htotal =
1
= hconv + hrad
As Rtotal
Combined Modes
q& = h(T∞1 − T1 )
Convection or
convection
plus radiation
L
Q& =
T∞1 − T∞ 2
1
1
L
+
+
Ah1 kA Ah2
⇒ q& =
Series
Resistance
Formula
Q&
T −T
= ∞1 ∞ 2
A 1 +L+ 1
h1 k h2
A is area normal
to heat flow
14
Q&
T −T
= ∞1 ∞ 2
1
L 1
A
+ +
h1 k h2
T1 = T∞1 −
Problem
A house has a 4 in thick brick wall with k = 0.6
Btu/hr·ft·oF. The interior temperature is 70oF and
the exterior temperature is 0oF. The inside and
outside convection plus radiation coefficients are
3 Btu/hr·ft2·oF and 4 Btu/hr·ft2·oF, respectively.
Find the heat flux through the wall.
Given: Wall with L = 4 in = 4/12 ft and k =0.6
Btu/hr·ft·oF has convection on two sides. T∞1 =
70oF, T∞2 = 0oF, h1 = 3 Btu/hr·ft2·oF and h2 = 4
Btu/hr·ft2·oF.
&
Q
Find: q& =
17
A
Figure 3-6 from
Çengel, Heat and
Mass Transfer
L
If you know h1, h2, L, k, T∞1,
and T∞2, but you do not know
T1 and T2, can you find the
heat flux?
Once you found the heat flux from the
information give, can you find T1 and T2?
q& =
15
ME 375 – Heat Transfer
q& = h(T2 − T∞ 2 )
Combined Modes III
Figure 3-6 from
Çengel, Heat and
Mass Transfer
T −T
T∞1 − T∞ 2
Q& = ∞1 ∞ 2 =
Rtotal
Rconv,1 + Rwall + Rconv,1
k (T1 − T2 )
L
All q& values are the same
Combined Modes II
L
q& =
Figure 3-6 from
Çengel, Heat and
Mass Transfer
13
A is area normal
to heat flow
Convection or
convection
plus radiation
Conduction
q&
h1
T2 = T∞ 2 +
q&
h2
16
Solution
A is area normal
to heat flow
q& =
L
Figure 3-6 from
Çengel, Heat and
Mass Transfer
Q&
T −T
70o F −0o F
= ∞1 ∞ 2 =
2
o
A 1 +L+ 1
hr ⋅ ft ⋅ F 4 hr ⋅ ft ⋅o F hr ⋅ ft 2 ⋅o F
+
+
ft
h1 k h2
3 Btu
12
0.6 Btu
4 Btu
q& =
Q& 61.5 Btu
=
A hr ⋅ ft 2
Find values of T1 and
T2. Can you check
these values?
18
3
Steady Heat Transfer
February 14, 2007
Solution II
A is area normal
to heat flow
T −T
q& = ∞1 1
1
h1
T −T
q& = 2 ∞ 2
1
h2
Solution III
Figure 3-6 from
Çengel, Heat and
Mass Transfer
L
q&
⇒ T1 = T∞1 − = 70o F −
h1
q&
⇒ T2 = T∞ 2 + =0o F −
h1
61.5 Btu
hr ⋅ ft 2
= 49.5o F
3 Btu
hr ⋅ ft 2 ⋅o F
61.5 Btu
hr ⋅ ft
= 15.4o F
4 Btu
19
hr ⋅ ft 2 ⋅o F
A is area normal
to heat flow
Figure 3-6 from
Çengel, Heat and
Mass Transfer
L
How can we check results below found from analysis of
overall problem and convection processes?
q& =
61.5 Btu
hr ⋅ ft 2
T1 = 49.5o F
T2 = 15.4o F
Analyze conduction step for consistency.
2
q& =
61.5 Btu
hr ⋅ ft 2
Composite Materials
(
0.6 Btu
49.5o F − 15.4o F
Q& k (T1 − T2 ) hr ⋅ ft ⋅o F
= =
=
4
A
L
ft
20
12
)
Composite Materials II
How
would
you
analyze
this
problem?
Figure 3-9 from Çengel, Heat and
Mass Transfer
21
Review Cylindrical Shell
For constant k
Q& r 2πk (T1 − T2 )
=
L
⎛r ⎞
ln⎜⎜ 2 ⎟⎟
⎝ r1 ⎠
Figure 250 from
Çengel,
Heat and
Mass
Transfer
R=
Q& r =
ME 375 – Heat Transfer
22
Cylindrical Shell with Convection
Q& =
A1 = 2πr1L
Rconv,1 =
⎛r ⎞
1
ln⎜ 2 ⎟
2πkL ⎜⎝ r1 ⎟⎠
T1 − T2
T −T
= 1 2
R
⎛r ⎞
1
ln⎜ 2 ⎟
23
2πkL ⎜⎝ r1 ⎟⎠
T∞1 − T∞ 2
Rconv,1 + Rcond + Rconv, 2
Rconv,2 =
A2 = 2πr2 L
Q& =
1
1
=
h1 A1 h1 2πr1L
1
1
=
h2 A2 h2 2πr2 L
T∞1 − T∞2
⎛r ⎞
1
1
1
ln⎜ 2 ⎟ +
+
h1 2πr1L 2πkL ⎜⎝ r1 ⎟⎠ h2 2πr2 L
Figure 3-25 from Çengel, Heat
and Mass Transfer
24
4
Steady Heat Transfer
February 14, 2007
Cylinder plus Convection Result
Q& =
Problem
T∞1 − T∞2
⎛r ⎞
1
1
1
+
ln⎜ 2 ⎟ +
h1 2πr1L 2πkL ⎜⎝ r1 ⎟⎠ h2 2πr2 L
• A hot-water pipe (k = 35 Btu/hr·ft·oF) in a
house, made of ¾ inch schedule 40 pipe
(OD = 1.050 in; ID = 0.824 in) is 40 ft
long and contains water at 120oF. The
air around the pipe is at 60oF. The heat
transfer coefficients inside and outside
the pipe are, respectively, 200 and 3
Btu/hr·ft2·oF. Determine the heat loss
Q&
2π(T∞1 − T∞2 )
from the pipe.
=
We can rearrange
this equation as
shown below
Q&
2π(T∞1 − T∞2 )
=
L
1 1 ⎛ r2 ⎞ 1
+ ln⎜ ⎟ +
h1r1 k ⎜⎝ r1 ⎟⎠ h2r2
Figure 3-25 from Çengel, Heat
and Mass Transfer
L
25
Solution
26
Solution II
60oF,
Given: T∞2 =
T∞1 =
120oF, r1 = ID/2 = 0.412 in,
r2 = OD/2 = 0.525 in, k =35
Btu/hr·ft·oF, L = 40 ft,
h1= 200
Btu/hr·ft2·oF,
h2= 3 Btu/hr·ft2·oF
Figure 3-25 from
Çengel, Heat and
Mass Transfer
1 1 ⎛ r2 ⎞ 1
+ ln⎜ ⎟ +
h1r1 k ⎜⎝ r1 ⎟⎠ h2r2
&
Find: Q
Q&
2π(T∞1 − T∞2 )
=
L
1 1 ⎛ r2 ⎞ 1
+ ln⎜ ⎟ +
h1r1 k ⎜⎝ r1 ⎟⎠ h227
r2
Given: T∞2 = 60oF, T∞1 =
1
1 12 in
hr ⋅ ft 2 ⋅o F
=
120oF, r1 = ID/2 = 0.412 in,
200 Btu 0.412 in ft
h1r1
r2 = OD/2 = 0.525 in, k =35
Btu/hr·ft·oF, L = 40 ft,
1 ⎛ r2 ⎞ hr ⋅ ft⋅o F ⎛ 0.525 in ⎞
⎟
ln⎜ ⎟ =
ln⎜
h1= 200 Btu/hr·ft2·oF,
k ⎜⎝ r1 ⎟⎠ 35 Btu ⎜⎝ 0.412 in ⎟⎠
2
o
h2= 3 Btu/hr·ft · F
hr ⋅ ft 2 ⋅o F
1
1 12 in
1,940 Btu
=
Find: Q& =
hr
h2r2
3 Btu 0.525 in ft
Q& =
(
Composite
Cylindrical
Shell II
Composite
Cylindrical Shell
)
2π(T∞1 − T∞2 )L
2π 120oF − 60oF (40 ft )
=
o
1 1 ⎛ r2 ⎞ 1
+ ln⎜⎜ ⎟⎟ +
(0.146 + 0.007 + 7.619) hr ⋅ ft⋅ F
h1r1 k ⎝ r1 ⎠ h2r2
Btu 28
Figure 3-26 from
Çengel, Heat and
Mass Transfer
1
1
=
h2 A4 h2 2πr4 L
Figure 3-26 from Çengel, Heat
and Mass Transfer
ME 375 – Heat Transfer
29
1 ⎛ r2 ⎞ 1 ⎛ r3 ⎞
ln⎜ ⎟
ln⎜ ⎟
1
1
=
k1L ⎜⎝ r1 ⎟⎠ k2 L ⎜⎝ r2 ⎟⎠
h1 A1 h1 2πr1L
1 ⎛ r4 ⎞
ln⎜ ⎟
k3 L ⎜⎝ r3 ⎟⎠
30
5
Steady Heat Transfer
February 14, 2007
Composite
Cylindrical
Shell III
Figure 3-26
from Çengel,
Heat and Mass
Transfer
Another Problem
• Insulation with k = 0.2 Btu/hr·ft·oF is to be
added to the pipe in the previous
example problem. Determine the heat
transfer if the insulation is one inch thick.
Q& =
Q& =
T∞1 − T∞2
⎛r ⎞
⎛ r2 ⎞
⎛r ⎞
1
1
1
1
1
+
ln⎜ 4 ⎟ +
ln⎜ 3 ⎟ +
ln⎜ ⎟ +
h1 2πr1L 2πk1L ⎜⎝ r1 ⎟⎠ 2πk2 L ⎜⎝ r2 ⎟⎠ 2πk3 L ⎜⎝ r3 ⎟⎠ h2 2πr4 L
Q& =
T∞1 − T∞ 2
⎛ r2 ⎞
⎛r ⎞
1
1
1
1
+
ln⎜ ⎟ +
ln⎜ 3 ⎟ +
h1 2πr1L 2πk1L ⎜⎝ r1 ⎟⎠ 2πk2 L ⎜⎝ r2 ⎟⎠ h2 2πr3 L
2πL(T∞1 − T∞ 2 )
1
1 ⎛ r2 ⎞ 1 ⎛ r3 ⎞ 1
+ ln⎜ ⎟ + ln⎜ ⎟ +
h1r1 k1 ⎜⎝ r1 ⎟⎠ k2 ⎜⎝ r2 ⎟⎠ h2r3
31
Another Problem II
Another Problem III
Q& =
Unchanged resistances from previous
example 1 1 ⎛ r2 ⎞ 0.153 hr ⋅ ft⋅o F
h1r1
+
ln⎜ ⎟ =
k1 ⎜⎝ r1 ⎟⎠
Know all terms
from previous
example except
32
these two
Btu
=
New and modified resistances
1 ⎛ r3 ⎞ hr ⋅ ft⋅o F ⎛ 1.525 in ⎞ 5.332 hr ⋅ ft⋅o F
⎟=
ln⎜ ⎟ =
ln⎜
k2 ⎜⎝ r2 ⎟⎠ 0.2 Btu ⎜⎝ 0.525 in ⎟⎠
Btu
2πL(T∞1 − T∞2 )
1
1 ⎛ r2 ⎞ 1 ⎛ r3 ⎞ 1
+ ln⎜ ⎟ + ln⎜ ⎟ +
h1r1 k1 ⎜⎝ r1 ⎟⎠ k2 ⎜⎝ r2 ⎟⎠ h2r3
(
)
2π 120oF − 60oF (40 ft )
(0.146 + 0.007 + 5.332 + 2.623) hr ⋅ ft⋅
o
F
=
1860 Btu
hr
Btu
• Insulation and outer convection resistances
are largest
hr ⋅ ft 2 ⋅o F
1
1 12 in 2.623 hr ⋅ ft⋅o F
=
=
h2r3
Btu
3 Btu 1.525 in ft
– Inner convection and pipe conduction negligible
– Outer convection resistance less with insulation
33
34
Effect of Insulation Thickness
Insulation Increases Q& ?
2500
• Why does initial amount of insulation
increase heat transfer?
2000
1500
Heat loss
(Btu/hr)
1000
500
0
0
1
2
3
4
5
6
7
8
9
10
– Tradeoff of two resistances
– Added insulation adds conduction
resistance
– Added insulation also increases outer
radius which decreases the outer
convection resistance 1/(houterAouter) =
1/(houter2πrouterL)
Thickness (in)
35
ME 375 – Heat Transfer
36
6
Steady Heat Transfer
February 14, 2007
Resistances for Pipe Insulation
Radius for Maximum Q&
12
10
Q& =
8
Resistance
6
(hr·ft·F/Btu)
2
0
0
0.5
1
1.5
2
ro = outer radius
ko = thermal conductivity
of outer layer
⎛1 1
1 ⎞⎟
2πL(T∞1 − T∞2 )⎜
−
⎜ k r h r2 ⎟
dQ&
2 o ⎠
⎝ o o
=−
=0
&:
For maximum Q
2
dro
⎛
⎛ ro ⎞ 1 ⎞
1
⎟
⎜ Rother + ln⎜ ⎟ +
⎜
ko ⎜⎝ ri ⎟⎠ h2ro ⎟⎠
⎝
Insultation
Resistance
Convection
Resistance
Total
Resistance
4
2πL(T∞1 − T∞ 2 )
1 ⎛r ⎞ 1
Rother + ln⎜⎜ o ⎟⎟ +
ko ⎝ ri ⎠ h2 ro
2.5
3
&
ro = ko/h2 for maximum Q
Insulation Thickness (in)
37
Radius for Maximum Q&
Spherical Shell with Convection
&
• ro = ko/h2 for maximum Q
• In the example problem h2 = 3
Btu/hr·ft2·oF, and ko = 0.2 Btu/hr·ft·oF so
&
ro = 0.0667 ft = 0.8 in for maximum Q
• Pipe radius was 0.525 in; ro = 0.8 in
gives an insulation thickness of 0.275 in
• Note that ro = ko/h2 does not depend on
ri and is usually larger than ri
&
• There is no radius for minimum Q
39
Spherical Shell Result
Rsph
Q& =
A2 = 4πr22 L
4π(T∞1 − T∞ 2 )
1
r −r
1
+ 2 1+
2
kr1r2 h2 r22
h1r1
Figure 3-25 from Çengel, Heat
and Mass Transfer
ME 375 – Heat Transfer
Q& =
A1 = 4πr12 L
Rsph
T∞1 − T∞ 2
Rconv,1 + Rsph + Rconv, 2
Rconv,1 =
1
1
=
h1 A1 h1 4πr12
Rconv,2 =
1
1
=
h2 A2 h2 4πr22
A2 = 4πr22 L
Figure 3-25 from Çengel, Heat
and Mass Transfer
Rsph
1 1
−
r1 r2
=
4πk
40
Conduction Shape Factors
T∞1 − T∞ 2
Q& =
1 1
−
1
1
r1 r2
+
+
2
4
k
π
h1 4πr1
h2 4πr22
A1 = 4πr12 L
38
41
• Simplified analysis
– for multidimensional geometries with each
surface at a uniform temperature
– Use shape factor, S, whose equation is
found from tables like Çengel Table 3-7
& = kS(T – T )
– Basic equation: Q
1
2
– S must have dimensions of length
• Equations for S depend on parameters in the
different geometries
42
7
Steady Heat Transfer
February 14, 2007
Buried Pipe Shape Factor
Example Shape Factor
3.5
3.0
S/L
2.5
2.0
1.5
1.0
0.5
0.0
0
From Table 7-1 in Çengel, Heat
and Mass Transfer
ME 375 – Heat Transfer
10
20
30
40
50
60
70
80
90
100
z/D
43
44
8