steady state

CHAPTER 3

STEADY-STATE ONE-DIMENSIONAL CONDUCTION

The term one-dimensional refers to the fact that only one coordinate is needed to describe the spatial variation of the dependent variables.

In one dimensional system, temperature gradients exists along only a single coordinate direction.

In this chapter we will

- Learn how to obtain temperature profiles for common geometries with and without heat generation.

- Introduce the concept of thermal resistance and

thermal circuits

Steady-State, One-Dimensional Conduction

The Plane Wall

Consider a simple case of one-dimensional conduction in a plane

wall, separating two fluids of different temperature, without energy

generation

Temperature is a function of x

Heat is transferred in the x-direction

Must consider

Convection from hot fluid to wall

Conduction through wall

Convection from wall to cold fluid

qx

1,T

1,sT

2,sT

2,T

x

x=0 x=L

11, ,hT

22, ,hT

Hot fluid

Cold fluid

Temperature Distribution Heat diffusion equation in the x-direction for steady-state

conditions, with no energy generation:

Integrate twice to obtain the general solution:

T(x) = C1x + C2

To obtain C1 and C2, boundary conditions must be introduced

at x=0, T(0) = Ts,1

at x=L, T(L) = Ts,2

Apply the condition at x=0 to the general solution

Ts,1 = C2

Apply the condition at x=L to the general solution

Ts,2 = C1L + C2 = C1L + Ts,1

C1 = Ts,2 Ts,1

L

0

dx

dTk

dx

d qx is constant

Temperature Distribution

Apply to the general solution, the temperature profile, assuming

constant k:

From Fouriers Law, we can determine the conduction heat transfer rate:

qx = -kAdT = kA (Ts,1 Ts,2)

dx L

And heat flux:

qx = qx = k (Ts,1 Ts,2)

A L

1,1,2, )()( sss TL

xTTxT

Problem 3.1

Consider the plane wall, separating hot and cold fluids at temp. T,1 and T,2 respectively. Using surface energy balances as boundary conditions at x=0 and x=L, obtain the temp. distribution within the wall and the heat flux in terms of T,1 , T,2 , h1 , h2 , k and L.

Thermal Resistance

Based on the previous solution, the conduction hear transfer rate can be calculated:

Similarly for heat convection, Newtons law of cooling applies:

And for radiation heat transfer:

Recall electric circuit theory - Ohms law for electrical resistance:

kAL

TTTT

L

kA

dx

dTkAq

ssssx

/

2,1,2,1,

hA

TTTThAq SSx

/1

)()(

Ah

TTTTAhq

r

surssursrrad

/1

)()(

Resistance

Difference Potential=currentElectric

Thermal Resistance

We can use this electrical analogy to represent heat transfer problems using the concept of a thermal circuit (equivalent to an electrical circuit).

The temperature difference is the potential or driving force for the heat flow

The combinations of thermal conductivity, convection coefficient, thickness and area of material act as a resistance

R

Tq overall

Resistance

Force DrivingOverall

AhR

hAR

kA

LR

rradtconvtcondt

1,

1, ,,,

CONDUCTION CONVECTION RADIATION

Thermal Resistance for Plane Wall

In terms of overall temperature difference:

qx

1,T

1,sT

2,sT

2,T

x x=0

x=L

11, ,hT

22, ,hT

Hot fluid

Cold fluid

AhkA

L

AhR

R

TTq

tot

totx

21

2,1,

11

Ah

TT

kAL

TT

Ah

TTq

ssssx

2

2,2,2,1,

1

1,1,

/1//1

Thermal Resistance for Composite Walls

Thermal Resistance for Composite Walls

Heat transfer rate for composite wall:

qx = T,1 - T ,4

Rt

= T,1 - T ,4

[(1/h1A) + (LA/kAA) + (LB/kBA) + (LC/kCA) + (1/h4A)]

-Alternatively:

qx= T,1 Ts,1 = Ts,1 T2 = T2 T3 = U = overall heat transfer coefficient

(1/h1A) (LA/kAA) (LB/kBA) T = temperature difference (overall)

)]/1()/()/()/()/1[(

11

41 hkLkLkLhARU

CCBBAAtot

UAq

TRR

TUAq

ttot

x

1

Composite Walls

For resistances in series: Rtot=R1+R2++Rn

For resistances in parallel:

1/Rtot=1/R1+1/R2++1/Rn

Contact Resistance

In composite systems, the temperature drop across the interface between materials may be appreciable, due to surface roughness effects.

This temperature change is attribute to thermal contact resistance:

"

",

x

BAct

q

TTR

See tables 3.1, 3.2 for typical values of Rt,c

Problem 3.4 In a manufacturing process, a transparent film is being bonded to a

substrate. To cure the bond at a temperature To, a radiant source is used to provide a heat flux qo (W/m

2), all of which is absorbed at the bonded surface. The back of the substrate is maintained at T1 while the free surface of the film is exposed to air at T and a convection heat transfer coefficient, h.

a) Show a thermal circuit representing the steady

state heat transfer situation.

b) Assume the following conditions:

T = 20C, h = 50 W/m2K, T1 = 30C

Calculate the heat flux qo that is required to

maintain the bonded surface at To = 60C

Problem 3.15

Consider a composite wall that includes an 8-mm thick hardwood siding (A), 40-mm by 130-mm hardwood studs (B) on 0.65-m centers with glass fiber insulation (D) (paper faced, 28 kg/m3) and a 12-mm layer of gypsum (vermiculite) wall board (C).

What is the thermal resistance associated with a wall that is 2.5 m high by 6.5 m wide (having 10 studs, each 2.5 m high?)

(Note: Consider the direction of heat transfer to be downwards, along the x-direction)

Given: PROPERTIES from Table A-3 (T 300K): Hardwood siding, kA = 0.094 W/mK Hardwood, kB = 0.16 W/mK Gypsum, kC = 0.17 W/mK Insulation (glass fiber paper faced, 28 kg/m3), kD = 0.038W/mK.

Problem 3.20

A composite wall separates combustion gases at 2600C

from a liq coolant at 100C with gas and liq side convection

coefficients of 50 and 1000 W/m2K. The wall is composed of

a 10mm thick layer of beryllium oxide on the gas side. The

contact resistance between the oxide and the steel is 0.05

m2K/W.

a) What is the heat loss per unit surface area of the composite?

b) Sketch the temp. distribution from the gas to the liquid.

Assume temperature of beryllium oxide at 1500 K

and stainless steel at 1000 K.

Problem 3.22

Consider a plane composite wall that is composed of two

materials of thermal conductivities kA = 0.1 w/m.k and kB =

0.04 w/m.k and thicknesses LA = 10mm and LB = 20mm.

The contact resistance at the interface between the two

mterials is known to be 0.30 m2k/w. Material A adjoins a

fluid at 200c for which h = 10w/m2k and material B

adjoins a fluid at 40c for which h = 20w/m2k

a) What is the rate of heat transfer through a wall that is 2m high by 2.5m wide

b) Sketch the temp. distribution

Radial Systems-Cylindrical Coordinates

Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures

dr

dT)rL2(k=

dr

dTkA=qr - -

Radial Systems-Cylindrical Coordinates

Heat diffusion equation in the r-direction for steady-state conditions, with no energy generation (for cylinder):

Integrate twice to obtain general soution

T(r) = C1 ln r + C2

To obtain C1 and C2 , boundary conditions:

T(r1) = Ts,1 and T(r2) = Ts,2

Apply to the general solution:

Ts,1 = C1 ln r + C2

Ts,2 = C1 ln r + C2

01

dr

dTkr

dr

d

r

Solve C1 and C2 and substitute into general solution: The conduction heat transfer rate can be calculated: - Fouriers Law : In terms of equivalent thermal circuit:

2,221

2,1,ln

)/ln(

)()( s

ssT

r

r

rr

TTrT

condt

ssssssr

R

TT

Lkrr

TT

rr

TTLkq

,

2,1,

12

2,1,

12

2,1,

)2/()/ln()/ln(

2

)2(

1

2

)/ln(

)2(

1

22

12

11

2,1,

LrhkL

rr

LrhR

R

TTq

tot

totr

constdr

dTrLk

dr

dTkAqr )2(

Composite Walls

The heat transfer rate may be expressed as

qr = T,1 - T ,4

1 + ln (r2/r1) + ln (r3/r2) + ln (r4/r3) + 1

2r1Lh1 2kAL 2kBL 2kcL 2r4Lh4

The heat transfer rate may also be expressed in terms of an overall heat

transfer coefficient:

qr = T,1 - T ,4 = UA(T,1 - T ,4)

Rtot where U is the overall heat transfer coefficient. If A = A1 = 2r1L

44

1

3

41

2

31

1

2