1D STEADY STATE HEAT

1D STEADY STATE HEAT

CONDUCTION (1)

CONDUCTION (1)

Prabal Talukdar

Prabal Talukdar

Associate Professor

Department of Mechanical Engineering

Department of Mechanical Engineering

IIT Delhi

E-mail: [email protected]

p

Convection Boundary Condition

Convection 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 selected boundary 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

Radiative Boundary Condition

Radiative Boundary Condition

Heat conduction

at the surface in a selected direction

=

Radiation exchange at the surface in the same direction

Interface Boundary Conditions

Interface Boundary Conditions

The boundary conditions at an interface are based on the requirements that

(1) two bodies in contact must have the

h f

same 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

Generalized Boundary

Conditions

Heat transfer

to the surface

in all modes

Heat transfer

from the surface

in all modes

=

in all modes

in all modes

PTalukdar/Mech-IITD

Solution of steady heat

conduction equation

1D Cartesian

Differential Equation: Boundary Condition:

0

dx

T

d

2 2

=

T

( )

0

=

T

1 Integrate: C dT =

Applying the boundary condition to the general solution:

( )

x C₁x C₂ T = + 1 C dx = Integrate again: 0 0

( )

x

C

1

x

C

2

T

=

+

G l S l ti A bit C t t 1

T

Substituting: 2 1 1 C .0 C T = + 1 2 T C = PTalukdar/Mech-IITD

General Solution Arbitrary Constants 1 1 2 C2 T1 It cannot involve x or T(x) after the boundary condition is applied.

Cylindrical

- Spherical

Cylindrical

Spherical

Differential Equation: Differential Equation: 0 ) dr dT r ( dr d ₌ 0 ) dr dT r ( dr d 2 ₌ Integrate: 1 C dr dT r = Integrate: 1 2 _C dr dT r = dr Divide by r (r ≠ 0): C dT _{= 1} dr Divide by r2 (_r ≠ 0)_: 1 C dT r dr Integrate again:

( )

r C₁ ln r C₂ T = + 21 r dr = Integrate again: C PTalukdar/Mech-IITD

( )

r C₁ ln r C₂ T +

which is the general solution.

( )

1 _{C 2}

r C r

During steady one-dimensional heat conduction in a spherical (or heat 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

Heat Generation

Heat Generation

Under steady conditions, the energy

balance for this solid can be expressed as

Rate of heat 

Rate of energy 

=

transfer

from solid

hA

_s

(T

_s

‐T

_∞

)

generation within 

the solid

=

V

g&

s

(

s ∞

)

g

g V • PTalukdar/Mech-IITD s s hA g V T T = _∞ +

A large plane wall of thickness 2L (A = 2A

and V = 2LA

)

A large plane wall of thickness 2L (A

_s

= 2A

_wall

and V = 2LA

_wall

),

A long solid cylinder of radius r

_o

(A

_s

= 2πr

_o

L and V= πr

2

o

L),

A solid sphere of radius r

₀

(A

_s

= 4πr

2

o

L and V= 4/3πr

3o

)

• s s hA g V T T • ∞ + = PTalukdar/Mech-IITD

Under steady conditions, the

y

,

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 = r_o where T(r_o) = T_s yields

•

The maximum temperature

in a symmetrical solid with

uniform heat generation

occurs at its center

1-D plane wall

1 D plane wall

Energy balance

Energy 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 Q Q• _in − • _out = wall dt 0 dt dE_wall

= 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, Q_{cond, wall} constant.

dT

• Fourier’s law of heat conduction for the wall

t t

dx

dT

kA

Q

• _cond,wall

=

−

kAdT

d

Q

2 T L •

∫

∫

PTalukdar/Mech-IITD constant

kAdT

dx

Q

1 T T wall , cond 0 x=

∫

=

−

=

∫

Temp profile

Temp profile

T

T

kA

Q

• 1

−

2 (W)

L

kA

Q

_cond,wall

=

1 2 (W)

The rate of heat conduction through a

plane wall is proportional to the

average thermal conductivity the

average thermal conductivity, the

wall area, and the temperature

difference, but is inversely

i

l

h

ll hi k

proportional to the wall thickness

Temp profile

1 D steady state heat conduction equation

d (k dT ) 0

1 D steady state heat conduction equation

Integrate the above equation twice

Boundary conditions

0 ) dx k ( dx =

( )

x

C

₁

x

C

₂

T

=

+

T

)

0

(

T

and

_T

₍

_L

₎

_T

Boundary conditions

Apply the condition at x = 0 and L

1 , s

T

)

0

(

T

=

and

_T

₍

_L

₎

=

_T

_s,2 2 1 s

C

T

_s,1

=

C

₂ 1 , s 1 2 1 2 , s

C

L

C

C

L

T

T

=

+

=

+

1 2

T

T

−

1 1 , s 2 , s

C

L

T

T

=

1 1 , s 2 , s

T

_x

_T

T

)

x

(

T

=

−

+

PTalukdar/Mech-IITD 1 , s

T

x

L

)

x

(

T

=

+

Thermal Resistance Concept

Thermal Resistance Concept

Analogy between thermal and electrical resistance concepts

(W) wall 2 1 wall , cond

T

_R

T

Q

&

=

−

PTalukdar/Mech-IITD kA L R_wall = (o_C/W)

Convection Resistance

Convection Resistance

•

)

T

T

(

hA

Q

_convection

=

_{s s}

−

_∞ s i T T Q• = − ∞ (W) convection convection _R Q = convection

_hA

1

R

=

(W) (o_C/W) s convection

_hA

PTalukdar/Mech-IITD

Radiation Resistance

Radiation Resistance

(W) rad surr s surr s s rad 4 surr 4 s s rad A (T T ) h A (T T ) T _R T Q• = εσ − = − = − (K/W) s rad rad _h 1_A R =

Combined convection and radiation

(W/m2_K) ) T T )( T T ( ) T T ( A Q h _s2 _surr2 _{s surr} surr s s rad rad = ₋ = εσ + + • PTalukdar/Mech-IITD

Possible when T_∞ = T_surr (W/m2_K)

rad conv

combined h h

The thermal resistance network for heat transfer through a plane wall subjected to convection on both sides, and the electrical analogy

Network

subjected to convection on both sides

Network

subjected to convection on both sides

Rate of heat

convection into =

Rate of heat

convection from the Rate of heat

conduction

=

the wall through the wall wall

)

(

)

(

1 2 _{2 2 2} 1 1 1 ∞ ∞ •

−

=

−

=

−

=

h

A

T

T

L

T

T

kA

T

T

A

h

Q

L

A

h

T

T

kA

L

T

T

A

h

T

T

Q

2 2 2 2 1 1 1 1

1

1

∞ ∞ •

−

=

−

=

−

=

Adding the numerators and denominators yields

2 , 2 2 2 1 1 , 1 1 conv wall conv

R

T

T

R

T

T

R

T

T

_∞

−

₌

−

₌

−

_∞

=

g y total

R

T

T

Q

∞1 ∞2 •

−

=

(W) PTalukdar/Mech-IITD

kA

h

A

L

A

h

R

R

R

R

_{total conv wall conv}

2 1 2 , 1 ,

1

1

+

+

=

+

+

=

T

T

Q

∞1 ∞2 •

−

(W) total

R

Q

=

∞1 ∞2 (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 th

across any layer is proportional to the thermal resistance of the layer. The larger the resistance, the larger the temperature drop.p

R

Q

T

•

=

Δ

(o_C)

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

It is sometimes convenient to express heat transfer to express heat transfer through a medium in an analogous manner to

Newton’s law of cooling as

T Q& Δ

T

UA

Q

=

Δ

• (W) 1 UA = total R Q = total R The surface temperature of the wall can be

determined using the thermal resistance

T

T

T

T

Q

=

∞1

−

1

=

∞1

−

1

•

concept, but by taking the surface at which the temperature is to be determined as one of the terminal surfaces. Known

A

h

R

Q

conv 1 1 ,

1

=

=

PTalukdar/Mech-IITD Known