etht grp 10 ,140080125005 006-007-008
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ETHT
Conduction Montu Faldu 140080125005Haresh Gajera 140080125006Vishal Gajjar 140080125007Sarthak Gokani 140080125008
Guided by : Dr. Manish Maheta
Many heat transfer problems require the understanding of the complete time history of the temperature variation. For example, in metallurgy, the heat treating process can be controlled to directly affect the characteristics of the processed materials. Annealing (slow cool) can soften metals and improve ductility. On the other hand, quenching (rapid cool) can harden the strain boundary and increase strength. In order to characterize this transient behavior, the full unsteady equation is needed:
2 21, or
kwhere = is the thermal diffusivityc
T Tc k T Tt t
Transient heat transfer with no internal resistance: Lumped Parameter Analysis
Solid
Valid for Bi<0.1
Total Resistance= Rexternal + Rinternal
GE: dTdt
hAmcp
T T BC: T t 0 Ti
Solution: let T T, therefore
ddt
hA
mcp
Lumped Parameter Analysis
ln
hA
mct
i
tmchA
i
pi
ii
p
p
eTTTT
e
tmchA
TT
Note: Temperature function only of time and not of space!
- To determine the temperature at a given time, or- To determine the time required for the temperature to reach a specified value.
)exp(T0
tcVhA
TTTT
tL
BitLLc
kk
hLtcVhA
ccc
c2
11
ck
Thermal diffusivity: (m² s-1)
Lumped Parameter Analysis
Lumped Parameter Analysis
tL
Foc
2
khLBi C
T = exp(-Bi*Fo)
Define Fo as the Fourier number (dimensionless time)
and Biot number
The temperature variation can be expressed as
thickness2La with wallaplane is solid the when) thickness(half cL
sphere is solid the whenradius) third-one(3cL
cylinder.a is solid the whenradius)- (half2or
cL, examplefor
problem thein invloved solid theof size the torealte:scale length sticcharacteria is c Lwhere
L
or
Graphical Representation of the One-Term Approximation:The Heisler ChartsMidplane Temperature:
Change in Thermal Energy Storage
Temperature Distribution
Assumptions in using Heisler charts:•Constant Ti and thermal properties over the body•Constant boundary fluid T by step change•Simple geometry: slab, cylinder or sphere
),(''. trgqt
H
),(.. trgTktTC p
Incorporation of the constitutive equation into the energy equation above yields:
Dividing both sides by Cp and introducing the thermal diffusivity of the material given by
smm
sm
Ck
p
2
Thermal DiffusivityThermal diffusivity includes the effects of properties like mass
density, thermal conductivity and specific heat capacity. Thermal diffusivity, which is involved in all unsteady heat-
conduction problems, is a property of the solid object. The time rate of change of temperature depends on its numerical
value. The physical significance of thermal diffusivity is associated with
the diffusion of heat into the medium during changes of temperature with time.
The higher thermal diffusivity coefficient signifies the faster penetration of the heat into the medium and the less time required to remove the heat from the solid.
pp CtrgT
Ck
tT
),(..
This is often called the heat equation.
pCtrgT
tT
),(..
For a homogeneous material:
pCtxgT
tT
),(2
This is a general form of heat conduction equation.
Valid for all geometries.
Selection of geometry depends on nature of application.
xqxxq
yyq
yqzzq
zq
),(. txgTktTC p
For an isotropic and homogeneous material:
),(2 txgTktTC p
):,,(2
2
2
2
2
2
tzyxgzT
yT
xTk
tTC p
):,,(12
2
2
2
2 tzrgzTT
rrTr
rk
tTC p
):,,(sin1sin
sin11
2
2
2222
2 trgTr
Trr
Trrr
ktTC p
X
Y
zyxkk ,,
),(. txgTktTC p
),,,( tzyxgz
zTk
yyTk
xxTk
tTC p
),,,(2
2
2
2
2
2
tzyxgzTk
zT
zk
yTk
yT
yk
xTk
xT
xk
tTC p
More service to humankind than heat transfer rate calculations
Steady-State One-Dimensional Conduction
Assume a homogeneous medium with invariant thermal conductivity ( k = constant) :
• For conduction through a large wall the heat equation reduces to:
),,,(2
2
tzyxgxTk
xT
xk
tTC p
),,,(2
2
tzyxgxTk
tTC p
One dimensional Transient conduction with heat generation.
02
2
dx
TdA
0),,,(2
2
tzyxg
xTk
No heat generation
211 CxCTCdxdT
Apply boundary conditions to solve for constants: T(0)=Ts1 ; T(L)=Ts2
211 CxCTCdxdT
The resulting temperature distribution is:
and varies linearly with x.
Applying Fourier’s law:
heat transfer rate:
heat flux:
Therefore, both the heat transfer rate and heat flux are independent of x.
Wall Surfaces with Convection
2112
2
0 CxCTCdxdT
dxTdA
Boundary conditions:
110
)0(
TThdxdTk
x
22 )(
TLThdxdTk
Lx
Wall with isothermal Surface and Convection Wall
2112
2
0 CxCTCdxdT
dxTdA
Boundary conditions:
1)0( TxT
22 )(
TLThdxdTk
Lx
Electrical Circuit Theory of Heat TransferThermal ResistanceA resistance can be defined as the ratio of a
driving potential to a corresponding transfer rate.
iVR
Analogy:Electrical resistance is to conduction of electricity as thermal resistance is to conduction of heat.The analog of Q is current, and the analog of the temperature difference, T1 - T2, is voltage difference. From this perspective the slab is a pure resistance to heat transfer and we can define
qTR
RTq thth
WKmW
KmmkAL
LTTkA
TTq
TRss
ss
condth /1.
212
21
WKmW
KmhATThA
TTq
TRs
s
convth /1.1
2
2
WKmW
KmAhTTAh
TTq
TRrsurrsr
surrs
radth /1.1
2
2
The composite Wall The concept of a thermal
resistance circuit allows ready analysis of problems such as a composite slab (composite planar heat transfer surface).
In the composite slab, the heat flux is constant with x.
The resistances are in series and sum to Rth = Rth1 + Rth2.
If TL is the temperature at the left, and TR is the temperature at the right, the heat transfer rate is given by
21 thth
RL
th RRTT
RTq
Wall Surfaces with Convection
2112
2
0 CxCTCdxdT
dxTdA
Boundary conditions:
110
)0(
TThdxdTk
x
22 )(
TLThdxdTk
LxRconv,1 Rcond Rconv,2
T1 T2
Heat transfer for a wall with dissimilar materials
For this situation, the total heat flux Q is made up of the heat flux in the two parallel paths:
Q = Q1+ Q2 with the total resistance given by:
Composite Walls
The overall thermal resistance is given by
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