Unit 3 Emotional Intelligence and Spiritual Intelligence.pdf
Conducción de calor en estado estacionario
1. 1D STEADY STATE HEAT
CONDUCTION (1)CONDUCTION (1)
Prabal TalukdarPrabal Talukdar
Associate Professor
Department of Mechanical EngineeringDepartment of Mechanical Engineering
IIT Delhi
E-mail: prabal@mech.iitd.ac.inp
PTalukdar/Mech-IITD
2. Convection Boundary ConditionConvection Boundary Condition
Heat conduction
at the surface in a
selected direction
=
Heat convection
at the surface in
the same direction
In writing the equations for convection
boundary conditions, we have selectedboundary conditions, we have selected
the direction of heat transfer to be the
positive x-direction at both surfaces. But
those expressions are equally applicable
h h t t f i i th it
PTalukdar/Mech-IITD
when heat transfer is in the opposite
direction
3. Radiative Boundary ConditionRadiative Boundary Condition
Heat conduction
at the surface in a
selected direction
=
Radiation exchange
at the surface in the
same direction
PTalukdar/Mech-IITD
4. Interface Boundary ConditionsInterface Boundary Conditions
The boundary conditions at an interface are
based on the requirements that
(1) two bodies in contact must have the
h fsame temperature at the area of contact
and
(2) an interface (which is a surface) cannot
store any energy, and thusy gy,
the heat flux on the two sides of an
interface must be the same
PTalukdar/Mech-IITD
6. Solution of steady heat
conduction equation1D Cartesian
Differential Equation: Boundary Condition:
0
dx
Td
2
2
=
( ) 10 TT =
Integrate:
C
dT
=
Applying the boundary condition to the general solution:
( ) 21 CxCxT +=
1C
dx
=
Integrate again:
00
( ) 21 CxCxT +=
G l S l ti A bit C t t
1T
Substituting:
211 C0.CT +=
12 TC =
PTalukdar/Mech-IITD
General Solution Arbitrary Constants 211 12 TC
It cannot involve x or T(x) after the
boundary condition is applied.
7. Cylindrical - SphericalCylindrical Spherical
Differential Equation:
Differential Equation:
0)
dr
dT
r(
dr
d
=
0)
dr
dT
r(
dr
d 2
=
Integrate:
1C
dr
dT
r =
Integrate:
1
2
C
dr
dT
r =
dr
Divide by r :)0( ≠r
CdT 1
=
dr
Divide by r2 :)0( ≠r
1CdT
rdr
Integrate again:
( ) 21 CrlnCrT +=
2
1
rdr
=
Integrate again:
C
PTalukdar/Mech-IITD
( ) 21 CrlnCrT +
which is the general solution.
( ) 2
1 C
r
C
rT +−=
8. During steady one-dimensional
heat conduction in a spherical (orheat conduction in a spherical (or
cylindrical) container, the total rate
of heat transfer remains constant,
but the heat flux decreases with
i i di
PTalukdar/Mech-IITD
increasing radius.
10. Heat GenerationHeat Generation
Under steady conditions, the energy
balance for this solid can be expressed as
Rate of heat Rate of energy
=
transfer
from solid
hAs(Ts‐T∞)
generation within
the solid
=
Vg&s( s ∞) g
gV
•
PTalukdar/Mech-IITD
s
s
hA
gV
TT ∞ +=
11. A large plane wall of thickness 2L (A = 2A and V = 2LA )A large plane wall of thickness 2L (As = 2Awall and V = 2LAwall),
A long solid cylinder of radius ro (As = 2πro L and V= πr2
o L),
A solid sphere of radius r0 (As = 4πr2
o L and V= 4/3πr3
o )
•
s
s
hA
gV
TT
•
∞ +=
PTalukdar/Mech-IITD
12. Under steady conditions, they ,
entire heat generated within the
medium is conducted through
the outer surface of the cylinder
The heat generated within this inner cylinder must
the outer surface of the cylinder.
g y
be equal to the heat conducted through the outer
surface of this inner cylinder
Integrating from r = 0 where T(0) = T0 to r = ro where T(ro) = Ts yields
PTalukdar/Mech-IITD
13. • The maximum temperature
in a symmetrical solid with
uniform heat generation
occurs at its center
PTalukdar/Mech-IITD
15. Energy balanceEnergy balance
Rate of heat
transfer into the =
Rate of change of
energy of the wall
Rate of heat
transfer out of the-
wall
gy
wall
dt
dE
QQ wall
outin =−
••
dt
0
dt
dEwall
= for steady operation
Therefore, the rate of heat transfer into the wall must be equal to the rate
of heat transfer out of it. In other words, the rate of heat transfer through
the wall must be constant, Qcond, wall constant.
dT•
Fourier’s law of heat conduction for the wall
t t
dx
dT
kAQ wall,cond −=
•
kAdTdQ
2TL •
∫∫
PTalukdar/Mech-IITD
constantkAdTdxQ
1TT
wall,cond
0x ==
∫−=∫
16. Temp profileTemp profile
TT
kAQ 21 −•
(W)
L
kAQ 21
wall,cond = (W)
The rate of heat conduction through a
plane wall is proportional to the
average thermal conductivity theaverage thermal conductivity, the
wall area, and the temperature
difference, but is inversely
i l h ll hi kproportional to the wall thickness
PTalukdar/Mech-IITD
17. Temp profile
1 D steady state heat conduction equation 0)
dT
k(
d1 D steady state heat conduction equation
Integrate the above equation twice
Boundary conditions
0)
dx
k(
dx
=
( ) 21 CxCxT +=
T)0(T and T)L(TBoundary conditions
Apply the condition at x = 0 and L
1,sT)0(T = and 2,sT)L(T =
21s CT = 21,s C
1,s1212,s TLCCLCT +=+=
12 TT −
1
1,s2,s
C
L
TT
=
1
1,s2,s
Tx
TT
)x(T +
−
=
PTalukdar/Mech-IITD
1,sTx
L
)x(T +=
18. Thermal Resistance ConceptThermal Resistance Concept
Analogy between thermal and
electrical resistance concepts
(W)
wall
21
wall,cond
R
TT
Q
−
=&
PTalukdar/Mech-IITD kA
L
R wall = (oC/W)
21. The thermal resistance network for heat transfer through a plane wall subjected
to convection on both sides, and the electrical analogy
PTalukdar/Mech-IITD
22. Network subjected to convection on both sidesNetwork subjected to convection on both sides
Rate of heat
convection into =
Rate of heat
convection from the
Rate of heat
conduction=
the wall wallthrough the wall
)()( 222
21
111 ∞∞
•
−=
−
=−= TTAh
L
TT
kATTAhQ
L
Ah
TT
kAL
TT
Ah
TT
Q
2
2221
1
11
11
∞∞
• −
=
−
=
−
=
Adding the numerators and denominators yields
2,
2221
1,
11
convwallconv R
TT
R
TT
R
TT ∞∞ −
=
−
=
−
=
g y
totalR
TT
Q 21 ∞∞
• −
= (W)
PTalukdar/Mech-IITD
AhkA
L
Ah
RRRR convwallconvtotal
21
2,1,
11
++=++=
23. TT
Q 21 ∞∞
• − (W)
totalR
Q 21 ∞∞
= (W)
The ratio of the temperature drop to the
thermal resistance across any layer is
constant, and thus the temperature drop
l i ti l t thacross any layer is proportional to the
thermal resistance of the layer. The larger
the resistance, the larger the temperature
drop.p
RQT
•
=Δ (oC)
This indicates that the temperature drop across
any layer is equal to the rate of heat transfer
times the thermal resistance across that layer
PTalukdar/Mech-IITD
times the thermal resistance across that layer
24. It is sometimes convenient
to express heat transferto express heat transfer
through a medium in an
analogous manner to
Newton’s law of cooling as
T
Q
Δ&
TUAQ Δ=
•
(W)
1
UA =
totalR
Q =
totalR
The surface temperature of the wall can be
determined using the thermal resistance TTTT
Q 1111 −
=
−
= ∞∞
•
concept, but by taking the surface at which the
temperature is to be determined as one of the
terminal surfaces.
Known
Ah
R
Q
conv
1
1,
1
==
PTalukdar/Mech-IITD
Known
