week_4_heat_transfer_lecture.pdf
Transcript of week_4_heat_transfer_lecture.pdf
Lecture Notes for CO2 (Part 2) 1-D STEADY STATE HEAT
CONDUCTION
Wan Azmi bin Wan Hamzah Universiti Malaysia Pahang
Week – 3/4
Course Outcome 2 (CO2)
Students should be able to understand and
evaluate one-dimensional heat flow and in
different geometries
2
Lesson Outcomes from CO2 (Part 2)
To derive the equation for temperature
distribution in various geometries
Thermal Resistance concept – to derive
expression for various geometries
To evaluate the heat transfer using thermal
resistance in various geometries
To evaluate the critical radius of insulation
To evaluate heat transfer from the rectangular fins
3
STEADY HEAT CONDUCTION IN PLANE WALLS
for steady operation
In steady operation, the rate of heat transfer
through the wall is constant.
Fourier’s law of
heat conduction
Heat transfer through the wall can be modeled as
steady and one-dimensional.
4
Under steady conditions, the
temperature distribution in a plane
wall is a straight line: dT/dx = const.
Once the rate of heat conduction is
available, the temperature T(x) at
any location x can be determined by
replacing T2 by T, and L by x.
5
rate of heat transfer electric current
thermal resistance electrical resistance
temperature difference voltage difference
Concept of Thermal Resistance
Conduction resistance of the wall:
Thermal resistance of the wall against
heat conduction.
Ohm’s Law
Heat conduction through wall
Increasing of R value will decrease
the Q value and vice versa
6
Schematic for convection resistance
at a surface.
Newton’s law of cooling for convection
Convection resistance of the
surface: Thermal resistance of the
surface against heat convection.
When the convection heat transfer coefficient is very large (h → ),
the convection resistance becomes zero and Ts T.
That is, the surface offers no resistance to convection.
• A surface exposed to the surrounding air might involves convection and radiation simultaneously.
• Total heat transfer at the surface is determined by adding ( subtracting if opposite direction) the radiation and convection components
• The convection and radiation resistances are parallel to each other.
• When Tsurr≈T∞, the radiation effect can properly be accounted for by replacing h in the convection resistance relation by hcombined = hconv+hrad
7
Schematic for convection and
radiation resistances at a surface.
8
Radiation resistance of the surface:
Thermal resistance of a surface against
radiation.
Radiation heat transfer coefficient
surrssurrssurrss
surrss
surrssrad TTTT
TTA
TTA
TTAh
22
44radQ
9
Consider steady one-dimensional heat transfer through
a plane wall that is exposed to convection on both sides.
10 The thermal resistance network in electrical analogy.
11
U is the overall heat
transfer coefficient
Temperature drop The temperature drop is proportional to thermal resistance of the layer
The temperature drop across a layer is
proportional to its thermal resistance.
from
rearrange to
12
Multilayer Plane Walls
• Often walls are made of several layers of different materials. The thermal resistance
concept can still be used for these composite walls.
• This is done by developing a total thermal resistance for the wall.
• The rate of steady heat transfer through this two-layer composite wall can be
expressed by:
13
Example of heat transfer to
wall 1
14
15
16
The roof of a house consists of a 15-cm-thick concrete slab
(k = 2 W/m·0C) that is 15 m wide and 20 m long. The
convection heat transfer coefficients on the inner and outer
surfaces of the roof are 5 and 12 W/m2 0C, respectively.
On a clear winter night, the ambient air is reported to be at
10 0C, while the night sky temperature is 100 K. The house
and the interior surfaces of the wall are maintained at a
constant temperature of 20 0C. The emissivity of both
surfaces of the concrete roof is 0.9. Considering both
radiation and convection heat transfers, determine the rate
of heat transfer through the roof, and the inner surface
temperature of the roof.
If the house is heated by a furnace burning natural gas with
an efficiency of 80 percent, and the price of natural gas is
$1.20/therm (1therm=105,500 kJ of energy content),
determine the money lost through the roof that night during
a 14 hours period.
Problem
Assumptions: 1. Steady operating conditions exist, 2
The emissivity and thermal conductivity of the roof are
constant.
Properties: The thermal conductivity of the concrete is k
= 2 W/m⋅°C. The emissivity of both surfaces of the roof is
0.9.
In steady operation, heat transfer from the room to the
roof (by convection and radiation) must be equal to the
heat transfer from the roof to the surroundings (by
convection and radiation), that must be equal to the heat
transfer through the roof by conduction.
17
radconv g,surroundin toroofcond roof,radconv roof, toroom QQQQ
• Taking the inner and outer surface temperatures of the roof to be Ts,in and
Ts,out , respectively
18
4
,
48
,
4
ins,
4
room,radconv roof, toroom
27327320300105679.0203005
)(
insins
insroomi
TT
TTATTAhQ
0.153002
L
out s,in s,out s,in s,
cond roof,
TTTTkAQ
44
,
8
,
44
outs,,radconv surr., toroof
100273300105679.01030012
)(
outsouts
surroutso
TT
TTATTAhQ
• The total amount of natural gas consumption during a 14-hour period is
• The money lost through the roof
19
therms36.22
kJ 105500
therms1
0.80
s606014kJ440.37
80.080.0
tQQQ total
gas
$26.8ms$1.20/ther therms22.36 lost Money
Solving the equations above simultaneously gives
C 1.2T C, 3.7T W,37440 0
outs,
0
ins, Q
Consider a house that has a 10-m × 20-m base and a 4-m-high wall.
All four walls of the house have an R of 2.31 m2 0C/W. The two 10-m × 4-m walls have no
windows. The third wall has five windows made of 0.5-cm thick glass (k = 0.78 W/m · C),
1.2 m×1.8 m in size. The fourth wall has the same size and number of windows, but they
are double-paned with a 1.5-cm-thick stagnant air space (k=0.026 W/m·0C) enclosed
between two 0.5 cm-thick glass layers.
The thermostat in the house is set at 24 0C and the average temperature outside at that
location is 8 0C during the seven-month long heating season. Disregarding any direct
radiation gain or loss through the windows and taking the heat transfer coefficients at the
inner and outer surfaces of the house to be 7 and 18 W/m2 0C, respectively, determine the
average rate of heat transfer through each wall.
If the house is electrically heated and the price of electricity is $0.08/kWh, determine the
amount of money this household will save per heating season by converting the single-pane
windows to double-pane windows.
20
Problem
• Assumptions: 1 Steady heat transfer , 2 Heat transfer is one-dimensional,
3 Thermal conductivities of the glass and air are constant. 4 Heat transfer
by radiation is disregarded.
• Properties: k = 0.026 W/m⋅°C for air, and 0.78 W/m⋅°C for glass.
• Analysis:
21
C/W 001389.0
41018
11
C/W 05775.0410
31.2/R
C/W 003571.04107
11
0
0
wall
0
AhR
A
kL
kA
L
AhR
o
o
wallwall
i
i
W1.25506271.0
824TQ
C/W 06271.0001389.005775.0003571.0R
21
0
total
total
owalli
R
T
RRR
Solution
:
22
W5224003063.0
824
R
T-TQ
C/W 003063.0000694.0000583.0001786.0R
C/W 000694.042018
11
C/W 000583.0R002968.0
15
033382.0
115
1
R
1
C/W 002968.08.12.178.0
005.0
C/W 033382.08.1125420
31.2/R
C/W 001786.04207
11
total
21
0
total
0
0
eq
eq
0
0
wall
0
oeqi
o
o
glasswall
glass
glass
wallwall
i
i
RRR
AhR
RR
kA
LR
A
kL
kA
L
AhR
23
C/W 020717.0R27303.0
15
033382.0
115
1
R
1
C/W 27303.0267094.0002968.022R
C/W 267094.08.12.1026.0
015.0R
C/W 002968.08.12.178.0
005.0
C/W 033382.08.1125420
31.2/R
C/W 001786.04207
11
0
eq
eq
0
window
0
air
0
0
wall
0
glasswall
airglass
air
air
glass
glass
wallwall
i
i
RR
RR
Ak
L
kA
LR
A
kL
kA
L
AhR
24
W690023197.0
824
R
T-TQ
C/W 023197.0000694.0020717.0001786.0R
C/W 000694.042018
11
total
21
0
total
0
oeqi
o
o
RRR
AhR
W45346905224pane doublepane single QQQsave
kWhr 22851hr 24307kW 1000
4534
tQQ savesave
1818$0.08 22851
energy ofcost unit savedEnergy savingsMoney
25
Problem
• Assumptions: 1. steady state, 2. one-dimensional, 3. Thermal conductivities are
constant. 4 Thermal contact resistances at the interfaces are disregarded.
• Properties: kA = kF = 2, kB = 8, kC = 20, kD = 15, kE = 35 W/m⋅°C.
• Analysis :
26 C/W 16.0
04.08
05.0
C/W 06.004.020
05.0
C/W 04.012.02
01.0
0
3
0
42
0
1
B
B
C
C
A
A
kA
LRR
kA
LRRR
kA
LRR
Solution
27
C/W 25.012.02
06.0
C/W 05.006.035
1.0
C/W 11.006.015
1.0
0
7
0
6
0
5
F
F
E
E
D
D
kA
LRR
kA
LRR
kA
LRR
C/W349.0025034.0025.004.0
C/W034.005.0
1
11.0
1111
C/W025.00.06
1
0.16
1
0.06
11111
0
7211
0
2
652
0
1
4321
RRRRR
R R
RR
RRR
RR
equ, midequ, midtotal
equ, mid
equ, mid
equ, mid
equ, mid
28
W 572349.0
100300
R
TTQ
total
21
W 1091.112.0
85572Q 5
total
C 263065.0572300RT T R
T-TQ
meet F and DB,point at the re temperatuThen the
C/W 0.0650.0250.04RR
0
total,11
total,1
1
0
mid1equ,1total,1
Q
R
29
C 14325.0572RQTR
TQ 0
7
7
30
HEAT CONDUCTION IN CYLINDERS AND SPHERES
Heat is lost from a hot-water pipe to
the air outside in the radial direction,
and thus heat transfer from a long
pipe is one-dimensional.
Heat transfer through the pipe
can be modeled as steady
and one-dimensional.
The temperature of the pipe
depends on one direction only
(the radial r-direction) and can
be expressed as T = T(r).
31
A long cylindrical pipe (or spherical
shell) with specified inner and outer
surface temperatures T1 and T2.
is the conduction resistance of the cylinder layer.
32 is the conduction resistance of the spherical layer.
A spherical shell
with specified
inner and outer
surface
temperatures T1
and T2.
SPHERES
33
The thermal resistance network for
a cylindrical (or spherical) in case
of convection from in and out
sides.
for a cylindrical layer, and
for a spherical layer
where
34
Multilayered Cylinders and Spheres
The thermal resistance
network for heat transfer
through a three-layered
composite cylinder
subjected to convection
on both sides.
35
Once heat transfer rate Q has been
calculated, the interface temperature
T2 can be determined from any of the
following two relations: