Introduction of Heat Transfer (Week4 21 & 23 Sept)

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 E^s,1 E^s,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),





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 flux^{dx 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

q_x  kA  (T_s,1 T_s,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 L_i

k  L





1

eq i i

k L



1 i

1 1  L_A L_B L_C 

 ^{A B C} 

eq A B C

k  L k  k  k 

 

T_n T₁

T₀

A₁ …

L₁ A₂ L₂

A₃ L₃

A_n L_n

T_n T₁

T₀

k₁1 k²₂ k₃³ kⁿ_n

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

T_s,1  C₁ lnr₁ C and T_{2 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 ^q_x ^{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 c_p.

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,







k  Q   t T

– Q_l : 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(   ) T_i

( , 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}^



0



( 0) _s T   T

2





2 s

C  T

2 1

0

exp( ) _s

T  C



Semi-infinite half space problem D i ti (4)

Derivation (4)

( ) _i

T     T ^{1 2}

0

exp( )

i s

T C u du T







( ) ( ) ( )

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

 



;

 

( , ) 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

t_T=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. t_T  d² /

Convective heat transfer

• What will be the factors that make these two cases different?

Saturated Porous media - Fluid NOT flowing

T₀=100°C T_L=15°C

Saturated Porous media Fluid Flowing

T₀=100°C T_L=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

References used for this lecture

• Beardsmore and Cull, 2001, Crustal Heat Flow – A guide to measurement and modelling, Cambridge Univ Press

• Somerton WH, 1992, Thermal properties and temperature- related behavior of rock/fluid systems, Elsevier

related behavior of rock/fluid systems, Elsevier

• Middleton GV and Wilcock PR, 1999, Mechanics in the Earth

d E i t l S i C b id U i P

and Environmental Sciences, Cambridge Univ Press