Introduction of Heat Transfer (Week4 21 & 23 Sept)
(Week4, 21 & 23 Sept)
Ki-Bok Min, PhD
Assistant Professor
Department of Energy Resources Engineering S l N i l U i i
Seoul National University
Content of last lecture
• Heat Conduction
– Thermal conductivity Thermal conductivity Th l diff i it
x y z
q iq jq kq k T
– Thermal diffusivity
p
k
c
– Derivation of Heat Diffusion equation
T T T T
p
T T T T
k k k q c
x x y y z z t
– Solution methods
Today
• Steady state one-dimensional conduction
– A plane wallA plane wall
– Radial conduction in cylindrical coordinate
• Thermal conductivity measurement method
– Steady state methody – Transient method
Feed back on homework #1
• Distinction of formal and informal words.
– Like, a little bit, …
• Avoid using subjective, emotional words
– Veryy
• Proper tense (시제)
T t b ifi h l i
• Try to be more specific when you explain
– Use numbers if possible, take examples
B f bl f f il f l th
• Maintain a critical attitude
• 한국말을 영어로 직역하려고 하지말고, 이에 적합한 영어
Because of many problems of fossil fuel energy, the importance of renewable energy is increased.
,
표현이 무얼까라고 생각할 것.
• Run the spelling check – MS Word
Feed back on homework #1
• Why is the earth hot?
• Why is there a thermal gradient?
• Why is there a thermal gradient?
• Decay of radioactive element.
H t i t i f th th
• Hot interior of the earth
Feed back on homework #1
• What about the disadvantage of geothermal energy?
• Why don’t we use it now? Economic feasibility???
• Why don t we use it now? Economic feasibility???
• What about the safety of power plant in tectonic boundaries?
• Why doesn’t Korea build a geothermal power plant?
Homework #2
Thermal Resistance (열저항)
• Thermal resistance for conduction in a plane wall
– From Fourier’s lawFrom Fourier s law
Th h t
,1 ,2
s s
x x
T T
q q A kA dT kA
dx L
– Through rearrangement,
,1 ,2
s s
d
T T L
R
e Es,1 Es,2 L
R I A
Analogy with electrical resistance
– From Newton’s law of cooling (convection)
, t cond
x
R q kA I A
Thermal resistance for conduction 전도열저항
– From Newton s law of cooling (convection),
s
q q A hA T T
,
s 1
t conv
T T
R hA
, t conv
q hA
Thermal resistance for conduction 대류열저항
1D steady state solutions Pl ll
Plane wall
– 1D, steady state, no heat generation,
d dT 0 d k d
– From constant heat fluxdx dx
( )
T C C
– Applying BC,
1 2
( )
T x C x C
,1 ,2
(0) s ( ) s
T T and T L T
,2 ,1
s s
T T
C T and C
2 s,2 1
C T and C
L
Conduction
Stead state one dimensional cond ction Steady-state one dimensional conduction
– Temperature distribution,
,2 ,1 ,1
( ) ( s s ) x s
T x T T T
L
– Conduction heat transfer rate,
,2 ,1 ,1
s s s
L
( )
dT kA
qx kA (Ts,1 Ts,2)
q kA T T
dx L
1D steady state solutions
Composite all (la ered rock) Composite wall (layered rock)
– Equivalent thermal conductivity for a series composite wall (layered rock).
1 1 n Li
k L
k L
n Li1
eq i i
k L
k1 i
1 1 LA LB LC
A B C
eq A B C
k L k k k
Tn T1
T0
A1 …
L1 A2 L2
A3 L3
An Ln
Tn T1
T0
k11 k22 k33 knn
1D steady state solutions
di l d ti i li d i l ll radial conduction in cylindrical wall
– 1D, Steady state, no heat generation,
1 T 0
krkr 0 r r r
(2 )
r
T T
q kA k rL
– Heat transfer rate is a constant in the di l di ti
r r r
radial direction.
– Assuming constant k,
1 2
( ) ln
T r C r C
ln ln
Ts,1 C1 lnr1 C and T2 s,2 C1 ln r2 C2 T C r C and T C r C
1D steady state solutions
di l d ti i li d i l ll radial conduction in cylindrical wall
– Temperature distribution
associated with radial conduction through a cylindrical wall
through a cylindrical wall
,1 ,2
,2
1 2 2
( ) ln
ln( / )
s s
s
T T r
T r T
r r r
– Heat transfer rate is,1 2 2
,1 ,2
2
l ( / )
s s
r
Lk T T
q
– Thermal resistance
2 1
ln( / ) qr
r r
2 1
ln( / )r r R,
t cond 2
R Lk
1D steady state solutions
SSummary
Hydraulic conductivity measurement
Two groups of methods (Beardsmore & Cull 2001) Two groups of methods (Beardsmore & Cull, 2001)
• Steady-state method
– Divided-bar apparatusDivided bar apparatus
• Transient method
– Needle probe
Hydraulic conductivity measurement St d St t M th d
Steady State Method
– Use a ‘divided-bar apparatus’
– Measure the ‘k’ directly qx k T2 T1
L
y
– Takes long time to achieve thermal equilibrium More accurate than ‘transient method’
x L
– More accurate than ‘transient method’
– Rock sample in discs or cylindrical shape
– Top and bottom sections of the bar maintained at constant but different temperatures (warm
d t th t ! Wh ?) end at the top! Why?).
Beardsmore and Cull, 2001
Hydraulic conductivity measurement T i t M th d
Transient Method
– “Needle probe method” is the best known method.
– k can be deducted from the rate at which its T changed in
response to an applied heat response to an applied heat source.
Measure the thermal diffusivity – Measure the thermal diffusivity
(α)and thermal conductivity is calculated need to know ρρ and cp.
p
k
c
Single needle probe Dual-needle probe
– Less accurate than ‘steady state’
method
p
Beardsmore and Cull, 2001
Hydraulic conductivity measurement
T i t M th d (2) l
Transient Method (2) - an example
– With a line source of heat and a temperature sensor packed closely
packed closely,
l / 4
ln( ) /
k Q t T
– Ql : applied heat per unit length(W/m)
t : time (second) t : time (second) T: Temperature (K)
Fi d li it d bt i k – Find a linearity and obtain k
P= V x I = 5 x 0.25 = 1.25W Q=P/0 1(cm)=12 5W/m Q P/0.1(cm) 12.5W/m
Gradient between 60 & 200 sec is ~3.449 K = 12.5/4π x 3.449 = 3.43 W/mK
Data from Beardsmore and Cull, 2001
Heat Diffusion Equation
• Verbal description of heat diffusion equation
– The rate at which the temperature at a point is changing with time The rate at which the temperature at a point is changing with time is proportional to the rate at which the temperature gradient at that point is changing in the direction of heat flow. (Middleton and
il k 1999) wilcock, 1999)
p
T T T T
k k k q c
x x y y z z t
Last lecture
• Steady state one-dimensional conduction
– A plane wallA plane wall
– Radial conduction in cylindrical coordinate
• Thermal conductivity measurement method
– Steady state methody – Transient method
Ti d d ( i ) d i
• Time dependent (transient) conduction
• Convective heat transfer, thermal expansion & thermal stress Convective heat transfer, thermal expansion & thermal stress
(very briefly)
Conduction T i t t t Transient state
• If a solid body is suddenly subjected to a change in
environment, some time must elapse before an equilibrium temperature condition will prevail in the body. Transient
heating/cooling process takes place before equilibrium ( t d t t )
(steady state).
• Transient (or unsteady) problem: ( y) p
– Arise when boundary conditions are changed
E ) if f t t i lt d t t t h i t i – E.g.) if surface temperature is altered, temperature at each point in
the system will also begin to change The change will continue until a steady state temperature distribution is reached
until a steady state temperature distribution is reached.
Time dependent conduction i i fi it h lf bl
semi-infinite half space problem
• A single identifiable surface + a solid extends to infinity
• A sudden change of conditions is imposed at this surface A sudden change of conditions is imposed at this surface transient, 1D conduction occur within the solid.
S i i fi it lid (h lf ) id f l id li ti
• Semi-infinite solid (half-space) provides a useful idealization for many practical problems.
2 2
1
T T
x t
Time dependent conduction i i fi it h lf bl
semi-infinite half space problem
• Derivation for case (1)
2 0
( , ) 2
exp( )
4
s
i s
T x t T x
u du erf erf
T T t
( s i) k T T q s q t
Semi-infinite half space problem D i ti (1)
Derivation (1)
– Existence of similarity variable, η
– Partial differential equation with x, t ordinary differential q y equation with η
4 x
t
– We first transform the pertinent differential operator,
1 4
T dT dT
x d x t d
2 2
2 2
1 4
T d T d T
x d x x t d
2 4
T dT x dT
t d t t t d
Semi-infinite half space problem D i ti (2)
Derivation (2)
– Substituting into 1D diffusion equation,
2T 1 T
2
T 2 T
Boundary conditions become;
x2 t
T2 2 T
– Boundary conditions become;
( 0) s T T (0, ) s
T t T
( , )
( 0)
T x t Ti
T x T
T( ) Ti
( , 0) i
T x T Both the IC and the interior BC correspond to the single requirement
Semi-infinite half space problem D i ti (3)
Derivation (3)
– Now T is uniquely defined by η. Let’s solve T now. T(η) may be obtained by separating variables, such that
( / )
d T
– By integration,
( / )
( / ) 2 d T
T d
y g ,
I t ti d ti
2 2
1 1
ln( / ) dT exp( )
dT d C or C
d
– Integrating a second time,
– T C1
exp(u du2) C2 where u is dummy.0
( 0) s T T
2
2 s
C T
2 1
0
exp( ) s
T C
u du TSemi-infinite half space problem D i ti (4)
Derivation (4)
( ) i
T T 1 2
0
exp( )
i s
T C u du T
C1 2(Ti Ts) 2 2( ) ( ) ( )
T t T T d T
– Note that error function is defined as;
x
2 0
( , ) ( i s) exp( ) s
T x t T T u du T
– For example,
– The final solution can be also described as;
2 0
( ) 2 exp( )
x
erf x u du
erf (0) 0, erf ( ) 1;
( , ) 2 2
exp( )
4
T x t Ts x
u du erf erf
T T t
0 4
i s
T T t
Semi-infinite half space problem D i ti (5)
Derivation (5)
– Surface heat flux may be obtained by applying Fourier’s law,
( )
( )
T d erf d
q k k T T
0 0
( )
s i s
x
q k k T T
x d dx
2 2 1
( ) exp( )
s s i 4
q k T T
4 0
s s i
t
( s i)
s
k T T q t
t
Time dependent conduction i i fi it h lf bl
semi-infinite half space problem
( , )
4
s
i s
T x t T x
T T erf t
( s i)
s
k T T
q t
1/ 2 2
0 0
2 ( / )
( ) q t exp x q x x
T x t( , ) T exp erfc
4 4
T x t Ti erfc
k t k t
( ) 2
T x t T x hx h t x h t
2
( , )
4 exp 2
s
i s
T x t T x hx h t x h t
erf erfc
T T t k k t k
Time dependent conduction i i fi it h lf bl
semi-infinite half space problem
Time dependent conduction i i fi it h lf bl
semi-infinite half space problem
– For granite from Forsmark, Sweden
Sweden,
– k = 3.58 W/mK, ρ: 1000 kg/m3, cp: 796 (J/kg·K) α = 4.5x10-6 m2/sec
m2/sec
– Homework#2 Q3. Reproduce this graph.
Time dependent conduction
ith fi d T t b th d a cases with fixed T at both ends
tT=0.1
– Thermal relaxation time: a characteristic time for heat to diffuse th h th l (Middl t d Wil k 1999) d thi k f
Middleton and Wilcock, 1999
through the layer(Middleton and Wilcock,1999). d: thickness of layer, α: diffusivity. tT d2 /
Convective heat transfer
• What will be the factors that make these two cases different?
Saturated Porous media - Fluid NOT flowing
T0=100°C TL=15°C
Saturated Porous media Fluid Flowing
T0=100°C TL=15°C
- Fluid Flowing
• These will be covered after we deal with fluid flow in rock
Summary
• Transient conduction problem
– Temperature changes with timeTemperature changes with time
• Coupled process associated with thermal transfer
– Convective heat transfer
– Thermal expansion and thermal stressp
Fl id fl i k k
• Fluid flow in rock - next week
– Porous rock – Fractured rock