A- surface area

C

HAPTER

2

–

C

ONDUCTION

Consider a wall where the surface temperatures are such that T1>T2

qx = flow rate of heat in the x

direction (vector!) [units of Watts]

qx is dependent on THREE factors:

1. A - surface area perpendicular

to the flow of heat (i.e. y x z),

2. k – the capacity of material to transfer heat across its body

(thermal conductivity)

3. ∆T _L – temperature gradient across the thickness of the wall, L

T1

T2

qx

L

These three factors are related

together through FOURIER’S LAW:

dx

dT

kA

q

=

−

Q: why the negative sign?

Or, in terms of heat flux q” (q” = q/A)

_dx

dT

k

q

"

=

−

Although we are only considering 1D heat flow in this case, Fourier’s Law may be expressed in 3 dimensions:













+

+

−

=

∇

−

=

dz

dT

k

dy

dT

j

dx

dT

i

k

T

k

q

"

where ! is the del operator

NOTE: Fourier’s law is typically

written individually for each dimension over which heat flow occurs.

heat transfer always going to go from hot to cold. heat transfer is always a vector quantity

The driving force of conduction is the temperature gradient

dx

dT

[o_{C/m or K/m]}

Average gradient:

L

T

T

run

rise

dx

dT

−

=

=

here

_dx

<

0

dT

, thus qx>0

Instantaneous

gradient: may be some function of x

If T(x)=Px2_+Qx+R

qx = -kA(dT/dx)

= -kA(2Px+Q)

T

L

T

T

dT

dx

How can you determine if the temperature gradient is linear?

• Is there an internal source of heat? • Is the material through which heat

is flowing homogeneous?

o Is k constant?

• Does the surface area

perpendicular to the heat flow change with dx?

o Increasing cross-sectional area

= lower flux, more uniform

temperature as a function of x

r

q

wall T

x

T

q

H

EAT

T

RANSFER

P

ROPERTIES

Thermal Conductivity (k): (W/m•K)

High k values are good conductors Low k values are better insulators

e.g.

Material Temp

(K) (W/mK) k Material Temp (K) (W/mK) k

Brick 300 0.72 Air 300 0.026

Cork 300 0.039 Copper 300 401

Glass 300 1.4 Aluminum 300 237

More values available in Appendix A

k is often given as a constant, but it does vary with T–

get a value valid for

the T range you are

looking at.

A more accurate relationship is:

(

aT

)

k

k

=

1

+

where ko and a are constants.

! a can be negative or positive

depending on the material.

For blends of materials A and B, use a weighted average to determine k:

(

)

eff

ak

a

k

k

=

+

1

−

(a = fraction of material A in the blend)

For porous media consisting of a mixture of a stationary solid and a fluid, k depends on the geometry of the porous media (size distribution of pores, pore shape, etc.)

Minimum k:heat conducts sequentially through a fluid region of length ɛL and a solid region of length (1- ɛ)L: (fluid forms long, random fingers in solid)

q Solid

ks

Pore (fluid) kf

εL

(1−ε)L

(

)

f s

eff

k k

k

ε ε ₊

− =

1

1

min ,

Maximum k: heat can conduct through either a continuous fluid region of cross-section ɛw or a continuous solid region of

cross-section (1-ɛ)w: (interconnected solid)

(

)

f

eff k k

k _,max = ε + 1− ε

Choose best approximation based on the material you are using.

For solids with uniformly distributed, non-contacting spherical pores

(ɛ<0.25), use Maxwell equation:

q Solid - ks

Pore (fluid) kf εL

(1−ε)L

(

)

(

)

f s

s f

f s

s f

eff k

k k

k k

k k

k k

k

    

  

− +

+

− −

+ =

ε ε

2

2 2

min ,

Thermal Diffusivity (α): (m2/s)

p

c k

ρ α =

k = thermal conductivity (W/m•K)

cp = heat capacity of material (J/kg•K)

ρ = density of material (kg/m3₎

Heat Capacity = the ability of a material to store heat (per kg)

Density = volume per kg

ρcp is the volumetric heat capacity (per m3 _{instead of per kg)}

Q: What would ρcp and α be for a

good heat conductor? A bad heat conductor?

H

EAT

D

IFFUSION

E

QUATION

Consider a control volume (CV) as shown below (no bulk motion):

Heat balance on the control volume:

   

  =    

  +    

  −    

 

on accumulati

heat of

rate generation

heat of

rate out

flow

heat of

rate in

flow

heat of

rate

or, E&in − E&out + E&g = E&st

where

(

)

_dx

dT

dz

dy

k

q

=

−

⋅

qx q_x+dx

qz

qz+dz

qy

qy+dy

dx

dz dy

First, consider transfer (Ein and Eout) To determine qx+dx, we must use the

Taylor Expansion:

( )

Taking only the first term of the

expansion (since ∆x = dx is small):

( )

q

q_{x dx x x}

∂ ∂ + =

+

Thus, in the x direction, the difference

      −       = − out flow heat of in flow heat of rate rate E

E&_in &_out is given as:

( )

x T k x dx q x q

q_{x x dx x} ∂ ∂ ∂

     ∂ ∂ ∂ ∂ + = ∂ ∂ − = − ₊

In parallel, _y x y z

T k y q

q_{y y dy }∂ ∂ ∂

     ∂ ∂ ∂ ∂ = − ₊

qz qz dz _z k T_z ∂x∂y∂z

Summing up over all dimensions:

(

) (

)

=

− _{out x y z x+dx y+dy z+dz}

in E q q q q q q

E& &

dxdydz z T k z y T k y x T k

x _

          ∂ ∂ ∂ ∂ +       ∂ ∂ ∂ ∂ +       ∂ ∂ ∂ ∂

The energy source term Eg can be written as:

Eg heat_timecontent = q• dxdydz

∂ ∂ = & & ) ( ) (

where q is the rate of heat generation per unit volume (units W/m3₎

Heat generation may occur via several different mechanisms, including:

" Electrical resistance, P = I2R [W]

" Nuclear reaction

" Chemical reaction (∆Hrxn )

" Direct thermal heating (furnaces)

Assuming no phase changes (i.e. only

sensible heat effects are present), the energy storage (or accumulation)

term is given as (again in parallel to the other derivations):

(

)

E_{st p}

∂ ∂ = ∂ ∂ = ρ ) ( ) ( &

Substituting each of the derived terms in the energy balance E&in − E&out + E&g = E&st

and dividing out the δxδyδz term:

= +             ∂ ∂ ∂ ∂ +       ∂ ∂ ∂ ∂ +       ∂ ∂ ∂ ∂ q z T k z y T k y x T k

x & _t

T c_p

∂ ∂

ρ

General Heat Diffusion Equation

Solutions for other coordinates are:

Cylindrical Coordinates (r,θ , z)

Spherical Coordinates (r, θ ,ϕ )

Z r θ r θ ϕ







+













∂

∂

∂

∂

+













∂

∂

∂

∂

θ

θ

ϕ

T

k

r

r

T

kr

r

r

2 2

sin

1

1

r p ∂

∂ = +          ∂ ∂ ∂ ∂

ρ

ϕ

ϕ

ϕ

ϕ

sin 1 2 = +             ∂ ∂ ∂ ∂ +       ∂ ∂ ∂ ∂ +       ∂ ∂ ∂ ∂ q z T k z T k r r T kr r

r 2 θ θ &

S

IMPLIFYING

A

SSUMPTIONS

1. Semi-Infinite Assumption

Heat flow is always a vector quantity – it has a magnitude and direction

dx

dT

kA

q

=

−

dy

dT

kA

q

=

−

_dz

dT

kA

q

=

−

(x-axis) (y-axis) (z-axis)

But, do all the dimensions always matter to a given problem? # NO!

In cases where heat transfer in one or more dimension is negligible relative to heat transfer in another dimension, the semi-infinite assumption may be used to simplify the problem

# neglect edges of large slabs # neglect small contact areas

# neglect smaller cross-section areas

Example: Consider our wall thought experiment but add the floor and the ceiling. Find the wall temperature.

Ceiling = 16°C

air = 20°C outside = -10°C

Floor = 18°C

Over 98% of the wall’s cross-sectional area, the wall is in contact with only the inside and outside air, ∆∆∆∆T = 30°C

At the ceiling contact (1% of the wall’s cross-sectional area), ∆∆∆∆T = 4°C

At the floor contact (1% of the wall’s cross-sectional area), ∆∆∆∆T = 2°C

Does heat flow between the ceiling,

the floor, the wall, and the air? # YES

Is it significant in the context of the problem # NOT REALLY

wall

2. Steady State Assumption

This assumption is useful in situations in which there is a steady heat flow through a plane wall or slab

INSIDE OUTSIDE

air = 20°C air = -10°C

While the inside wall feels cool to our hand, the temperature is constant. Outside the wall will be >-10°C since the wall is heated from the inside.

The process is at steady-state since none of the temperatures vary as a function of time.

In this case, δT/δt = 0 such that

0

= +

   

 

   

 

∂ ∂ ∂

∂

+

   

 

∂ ∂ ∂

∂

+

   

 

∂ ∂ ∂

∂ ⋅

p

c q z

T z

y T y

x T

x

ρ

α

wall

There are two conditions under which you can apply this assumption:

1) An external driving force is acting to either add or remove heat from the surfaces of interest.

# In the case of the wall, the

furnace adds heat as it flows out of the room to keep the room

temperature constant while the air flow continuously replaces the air adjacent to the wall to keep the outdoor temperature constant

! TRUE STEADY STATE

2) When T1 and T2 do change over time, if the rate of change is

sufficiently slow or the time period we are considering is sufficiently short, we can assume T1 and T2 to

be constant over this period

! “PSEUDO” STEADY STATE

K

EY

P

HRASES

T

O

C

HECK

F

OR

FOR STEADY STATE ASSUMPTION

TRUE STEADY STATE:

“with a uniform temperature gradient” “with a constant heat flux”

“with constant surface temperatures”

PSEUDO-STEADY STATE:

“find the instantaneous temperature” “find the instantaneous heat flow”

“find the temperature after x minutes”

FOR SEMI-INFINITE ASSUMPTION “thin”

“slab” “wall”

“narrow” “plane”

“negligible thickness” “neglect edge effects”

P

ROBLEM

A

SSESSMENT

x & _t

T c_p

∂ ∂

ρ

the result will be T(x,y,z,t) (T profile)

ASK YOURSELF:

1. Steady State? ∂_∂T_t = 0

2. No Heat Generation?

q

&

=

0

3. 2-D Heat Flow? ∂_∂T_z = 0

4. 1-D Heat Flow? , _∂ = 0

∂ ∂ ∂ z T y T

5. k constant? 2

2 x T k x T k x ∂ ∂ =             ∂ ∂ ∂ ∂ Then, solve the simplified equation for

T and solve

_dx

dT

kA

q

=

−

MUST USE Heat Diffusion Equation

" To determine the temperature

profile of a system (T(x,y,z,t)) –

integrate over significant variable(s)

" To determine heat flows if k or A

vary along a dimensional axis

Use Heat Diffusion Equation OR Energy Balance Ein – Eout + Eg = Est

" To determine heat flows in cases

with volumetric heat generation

" To account for non steady-state

heat flow within the object (Est ≠ 0)

You only need Fourier’s Law if:

" Heat flow is steady state AND

" Cross-sectional area is constant AND

" k is constant over full volume AND

" Temperature profile is known (i.e.

T(x)=ax2 _{+ bx + cz) or assumed to} be a certain shape (line, parabola...)

! T(x,y,z,t) solved for you!

B

OUNDARY

C

ONDITIONS

From calculus, remember that every time you integrate a function, one integration constant is generated.

If 2 0 2

=

∂ ∂

x T

# C1

x T

= ∂

∂

# T = C1x + C2

To define the temperature profile (i.e. to find the integration constants C1

and C2), we need to know T(x) at two points at the boundary of our object (i.e. at x=0 and x=L in a wall or x=0 and x=ro in a cylinder)

! 2 equations in 2 unknowns

e.g. If T=T1 at x=0 and T=T2 at x=L

x=0 # T(x=0)=T1= C2

x=L # T(x=L)=T2=C1L+C2 #

Therefore, the temperature profile is:

L

T

T

C

=

−

1 1

2 2

1

)

(

x

T

L

T

T

C

x

C

x

T

=

+

=

−

+

The heat equation has three 2nd

derivatives as a function of position (d2_T/dx2_{, d}2_T/dy2_{, d}2_T/dz2_{) and one 1}st

derivative as a function of time(dT/dt)

= +

    

  

   

 

∂ ∂

+

   

 

∂ ∂

+

   

 

∂ ∂

q z

T y

T x

T

k &

2 2

2 2

2 2

t T c_p

∂ ∂

ρ

Thus, to solve for T(t), we need 1

initial condition (generally, T or q at

t=0, the start of transient heat flow) to find 1 integration constant

" Initial Conditions # time

oDefined in problem statement

To solve for T(x,y,z), we need 2

boundary conditions per dimension of heat flow (T or q defined at x=0 and

x=L) to find 2 integration constants

" Boundary Conditions #### position

oDefined in problem but typically

one of three common conditions

Typical Boundary Conditions

1) Constant surface temperature (Dirichlet condition)

T(0,t) = Ts

2) Constant surface heat flux (Neumann condition)

(a) Finite heat flux at surface - heater

(b) Adiabiatic/insulated surface

3) Convection surface condition

Ts

T(x,t) x

T(x,t) x

qs”

x "

0

s x

q dx

dT

k =

−

=

0

0

=

=

x

dx dT

x

T(0,t)

T∞, h

( (0, )) 0

t T T

h dx

dT k

x

−

=

− _∞

=

T(x,t)

T(x,t)

EXAMPLE

A plane wall has constant

properties, no internal heat generation, and is initially at a uniform

temperature of 20°C. The back of the wall is attached to a thin electric

heater, backed by insulation, which is initially off. Suddenly, the outer surface is heated by a fluid at 50°C with a convection coefficient of 1000W/m2_K. 1) Write the simplified heat diffusion equation to find the temperature distribution in the wall a few seconds after the fluid flow begins.

2) Specify the boundary and initial conditions.

3) Sketch the temperature profile throughout the system (a) before heat or fluid flow; (b) with fluid flow but no heater; (c) with heater and fluid flow (assume the q” of heater is higher than q” of fluid). Draw curves just after the heater/flow is started, in the middle of the heating, and at steady state.

4) Sketch the heat flux at x=L as a function of time under the same conditions.