28
CHAPTER 3 STEADY-STATE ONE-DIMENSIONAL CONDUCTION

STEADY STATE

  • Upload
    boyhaha

  • View
    300

  • Download
    5

Embed Size (px)

DESCRIPTION

one dimension

Citation preview

Page 1: STEADY STATE

CHAPTER 3

STEADY-STATE ONE-DIMENSIONAL CONDUCTION

Page 2: STEADY STATE

• 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

Page 3: STEADY STATE

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

Page 4: STEADY STATE

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

Page 5: STEADY STATE

Temperature Distribution

• Apply to the general solution, the temperature profile, assuming

constant k:

• From Fourier’s 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

Page 6: STEADY STATE

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.

Page 7: STEADY STATE

Thermal Resistance

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

• Similarly for heat convection, Newton’s law of cooling applies:

And for radiation heat transfer:

Recall electric circuit theory - Ohm’s law for electrical resistance:

kAL

TTTT

L

kA

dx

dTkAq

ssssx

/

2,1,2,1,

hA

TTTThAq S

Sx/1

)()(

Ah

TTTTAhq

r

surssursrrad

/1

)()(

Resistance

Difference Potential=currentElectric

Page 8: STEADY STATE

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

Page 9: STEADY STATE

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

Page 10: STEADY STATE

Thermal Resistance for Composite Walls

Page 11: STEADY STATE

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

Page 12: STEADY STATE

Composite Walls

For resistances in series:

Rtot=R1+R2+…+Rn

For resistances in parallel:

1/Rtot=1/R1+1/R2+…+1/Rn

Page 13: STEADY STATE

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

Page 14: STEADY STATE

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/m2), 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∞ = 20°C, h = 50 W/m2·K, T1 = 30°C

Calculate the heat flux qo” that is required to

maintain the bonded surface at To = 60°C

Page 15: STEADY STATE

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/m⋅K

Hardwood, kB = 0.16 W/m⋅K

Gypsum, kC = 0.17 W/m⋅K

Insulation (glass fiber paper faced, 28 kg/m3), kD = 0.038W/m⋅K.

Page 16: STEADY STATE

Problem 3.20

A composite wall separates combustion gases at 2600°C

from a liq coolant at 100°C 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.

Page 17: STEADY STATE

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

adjoins a fluid at 40°c 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

Page 18: STEADY STATE

Radial Systems-Cylindrical Coordinates

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

dr

dT)rLπ2(k=

dr

dTkA=qr - -

Page 19: STEADY STATE

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

Page 20: STEADY STATE

Solve C1 and C2 and substitute into general solution: The conduction heat transfer rate can be calculated: - Fourier’s 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(

Page 21: STEADY STATE

Composite Walls

Page 22: STEADY STATE

The heat transfer rate may be expressed as

qr = T∞,1 - T ∞,4

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

2πr1Lh1 2πkAL 2πkBL 2πkcL 2πr4Lh4

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 = 2πr1L

44

1

3

41

2

31

1

21

1

1lnlnln

1

1

hr

r

r

r

k

r

r

r

k

r

r

r

k

r

h

U

CBA

Page 23: STEADY STATE

• Alternatively we can use:

A2 = 2πr2L , A3 = 2πr3L , etc

tRAUAUAUAU

144332211

Page 24: STEADY STATE

Radial Systems-Sphere Coordinates

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

• Integrate twice to obtain general soution

• To obtain C1 and C2 , boundary conditions:

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

• Apply to the general solution:

0=dr

dTkr

dr

d

r

1 22

( ) 21

C+ r

C-=rT

C+ r

C-=T

C+ r

C-=T

22

12,s

21

11,s

Solve C1 and C2 and substitute into general solution

Page 25: STEADY STATE

Sphere

• Fourier’s law:

dr

dTrk

dr

dTkAqr

)4( 2

Assuming constant k, qr = 4πk (Ts,1 – Ts,2) = (Ts,1 – Ts,2) (1/r1) – (1/r2) 1/ 4πk [1/r1 – 1/r2] Thermal resistance: Rt,cond = 1 [ 1/r1 - 1/r2 ] 4πk

Page 26: STEADY STATE

Problem 3.37

A thin electrical heater is wrapped around the outer surface of a long cylindrical tube whose inner surface is maintained at a temperature of 5°C. The tube wall has inner and outer radii of 25 and 75 mm respectively, and a thermal conductivity of 10 W/m.K. The thermal contact resistance between the heater and the outer surface of the tube (per unit length of the tube) is R’t,c=0.01 m.K/W. The outer surface of the heater is exposed to a fluid of temperature –10°C and a convection coefficient of h=100 W/m2 .K. Determine the heater power per unit length of tube required to maintain the heater at To=25°C.

Page 27: STEADY STATE

Problem 3.60

A spherical vessel used as a reactor for producing pharmaceuticals hs a 10mm thick stainless steel wall (k = 17w/m.k) and an inner diameter of 1m. The exterior surface of the vessel is exposed to ambient air (T∞ = 25°c) for which a convection coefficient of 6 w/m2k may be assumed.

a) During steady-state operation, an inner surface temp.of 50°c is maintained by energy generated within the reactor. What is the heat loss from the vessel?

b) If a 20mm thick layer of fiber glass insulation (k = 0.04 w/m.k) is applied to the exterior of the vessel and the rate of thermal energy generation is unchanged, what is the inner surface temp. of the vessel?

Page 28: STEADY STATE

Summary

• We obtained temperature distributions and thermal resistances for problems involving steady-state, one-dimensional conduction in orthogonal, cylindrical and spherical coordinates, without energy generation