Post on 06-Feb-2018
+OUTLINE
u CONDUCTION: PLANE WALL
u CONDUCTION: MULTI LAYER PLANE WALL (SERIES)
u CONDUCTION: MULTI LAYER PLANE WALL (SERIES AND PARALLEL)
u MULTIPLE LAYERS WITH CONDUCTION AND CONVECTION
u CONDUCTION: CYLINDER
u CONDUCTION: SHPERE
u CRITICAL RADIUS OF INSULATION
+Conduction: Plane Wall
)( 12 TTxkAQ −Δ
−=
A plane wall
First consider the plane wall where a direct application of Fourier’s law may be made. Integration yields:
T1
T2
Q = heat rate in direction normal to surface
Δx = Wall thickness
T1, T2 = the wall face temperature
A = surface area
k = thermal conductivity
Δx
Q
+
composite wall
cc
BB
AA x
TTAkxTTAk
xTTAkQ
Δ
−−=
Δ
−−=
Δ
−−= 342312
The temperature gradients in the three materials are shown, and the heat flow may be written:
AkxAkxAkxTTQ
ccBBAA ///41
Δ+Δ+Δ
−=
Solving these three equations simultaneously, the heat flow is written
Note: the heat flow must be the same through all sections.
Heat flow through multilayer plane walls
+
composite wall
Note: the heat flow must be the same through all sections.
A relation quite like Ohm’s law in electric-circuit theory
Rth = the thermal resistances of the various materials
CBA RRRTTQ++
−= 41
AkxRn
ncond
Δ=
∑Δ
=th
overall
RTQ
Heat flow through multilayer plane walls
+ QUIZ 1
A composite wall is formed of a 2.5-cm copper plate, a 3.2-mm layer of asbestos, and a 5-cm layer of glass wool. The wall is subjected to an overall temperature difference of 560ºC. Calculate the heat flux through the composite structure.
kcopper = 385 W/m.ºC
kasbestos = 0.166 W/m.ºC
Kglass wool = 2.22 W/m.ºC
THERMAL RESISTANCE NETWORKS u THE GENERALIZED
FORM FOR THE THERMAL RESISTANCE NETWORK IS BASED ON THE ELECTRICAL ANALOGY
u FOR PARALLEL PATHS, THE DRIVING FORCES ARE THE SAME FOR THE SAME TERMINAL TEMPERATURES, AS PER FIGURE (3-19)
THERMAL RESISTANCE NETWORKS u TOTAL HEAT
TRANSFER
u RESISTANCE THROUGH EACH LAYER
u OVERALL EQUATION
u OVERALL RESISTANCE FOR PARALLEL FLOWS:
+
A
B
C
D
1.1 Heat flow through plane wall
DCB
cBA R
RRRRR
TTQ+⎥⎦
⎤⎢⎣
⎡
++
−= 41
AkxRn
ncond
Δ=
Akx
Akx
Akx
Akx
Akx
Akx
TTQ
D
D
C
C
B
B
C
C
B
B
A
A Δ+
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
Δ+
Δ
Δ⋅
Δ
+Δ
−=
2/2/
2/2/
41
+
Dr. Şaziye Balku
14 NEWTON’S LAW OF COOLING FOR CONVECTION HEAT TRANSFER RATE
)( ∞
•
−= TThAQ SSconv
conv
Sconv R
TTQ ∞• −
=
Sconv hAR 1
=
convR
h
Convection resistance of surface
(W)
(0C / W)
Convection heat transfer coefficient
MULTIPLE LAYERS WITH CONDUCTION AND CONVECTION u FOR A SERIES OF LAYERS WHERE
SYSTEM THE FLUX THROUGH EACH LAYER IS CONSTANT
MULTIPLE LAYERS
u THE FLUX THROUGH EACH LAYER IS THE SAME, SO:
u IN TERMS OF RESISTANCE THIS RELATIONSHIP BECOMES:
MULTIPLE LAYERS WITH CONDUCTION AND CONVECTION
MULTIPLE LAYERS
u IN OVERALL TERMS, CONSIDER THE DRIVING FORCE TO BE T∞1 - T∞2 AND THEN EXPRESS THE OVERALL RESISTANCE AS
u SO THE OVERALL HEAT TRANSFER CAN THEN BE EXPRESSED AS
MULTIPLE LAYERS WITH CONDUCTION AND CONVECTION
+
Dr. Şaziye Balku
18
HEAT CONDUCTION IN CYLINDERS
Steady-state heat conduction
Heat is lost from a hot-water pipe to the air outside in the radial direction.
Heat transfer from a long pipe is one dimensional
+
Dr. Şaziye Balku
19
A LONG CYLINDERICAL PIPE STEADY STATE OPERATION
drdTkAQ cylcond −=
•
,
Fourier’s law of conduction
=•
cylcondQ , constant
∫∫ ==
•
−=2
1
2
1
, T
TT
r
rr
cylcond kdTdrA
Q
rLA π2=
)/ln(2
12
21, rr
TTLkQ cylcond−
=•
π
cylcylcond R
TTQ 21,
−=
•
LkrrRcyl π2)/ln( 12=
+Heat flow through radial system
Multilayer cylinder
CBA krr
krr
krr
TTLQ )/ln()/ln()/ln()(2
342312
41
++
−=
π
Note: the heat flow, q must be the same through all layers!
+EXAMPLE: Combination Conduction and Convection
A thick–walled tube of stainless steel (A) having k= 21.63 W/m.K with dimension of 0.0254 m ID and 0.0508 m OD is covered with a 0.0254-m thick layer of insulation (B), k= 0.2423 W/m.K . The inside wall temperature of the pipe is 811 K and outside surface of the insulation is at 310.8 K. For a 0.305 m length of pipe, calculate the heat loss and also the temperature at the interface between metal and insulation.
+ 24
CONDUCTION IN SPHERES
24 rA π=
Rsph =r2 − r14πr1r2k
sphsphcond R
TTQ 21,
−=
•
FOR A SPHERICAL SYSTEM (HOLLOW BALL) THE SAME METHOD IS USED:
+
Dr. Şaziye Balku
25
CRITICAL RADIUS OF INSULATION
)2(1
2)/ln(
2
12
11
LrhLkrr
TTRRTTQconvins
ππ+
−=
+
−= ∞∞
•
0/ 2 =•
drQd
hkr cylindercr =,
Thermal conductivity
External convection heat transfer coefficient
show
CYLINDER
+
Dr. Şaziye Balku
26
CHOSING INSULATION THICKNESS
cr
cr
cr
rrrrrr
>
=
<
2
2
2
max
Before insulation check for critical radius
hkr spherecr2
, =
+EXAMPLE
An electric wire having diameter of 1.5 mm and covered with a plastic insulation (thickness = 2.5 mm) is exposed to air at 300 K and ho= 20 W/m2. K. The insulation has a k of 0.4 W/m. K. It is assumed that the wire surface temperature is constant at 400 K and it is not affected by the recovering.
a) Calculate the value of the critical radius
b) Calculate the heat loss per m of wire length with no insulation
c) Repeat (b) for insulation being present
+SUMMARY
u Specify appropriate form of the heat equation.
u Solve for the temperature distribution.
u Apply Fourier’s Law to determine the heat flux.
Simplest Case: One-Dimensional, Steady-State Conduction with
u Common Geometries:
Ø The Plane Wall: Described in rectangular (x) coordinate. Area
perpendicular to direction of heat transfer is constant (independent
of x).
Ø The Tube Wall: Radial conduction through tube wall.
Ø The Spherical Shell: Radial conduction through shell wall