Chap 4 Heat Transfer (PART 2)

BKF2422 HEAT TRANSFER

Chapter 2

Principles of steady-state heat

transfer in conduction

TOPIC OUTCOMES

It is expected that students will be able to:

• Solve problems using steady-state conduction

principles for one-dimensional solid conduction

heat transfer in parallel and series

• Calculate overall heat transfer coefficient to solve

problems related to combined conduction and

convection heat transfer mechanism.

• Solve the problem related to internal heat

generation and determine the critical thickness

and of insulation for a cylinder

• Apply shape factor to estimate the

multidimensional heat transfer

CONTENTS

• One Dimensional Conduction Heat

Transfer

– Conduction Through a Plane Wall

– Conduction Through Solids In Series

– Conduction Through Solids In Parallel

– Conduction Through a Hollow Cylinder

– Conduction Through a Multilayer Cylinders

– Conduction Through a Hollow Sphere

• Combined Conduction and Convection

CONDUCTION: FOURIER’S LAW

• Flux of conduction heat transfer can be calculated by Fourier’s Law

Fourier’s Law

q_x : heat-transfer rate in the x direction (SI: W or J/s; cgs: cal/s; Eng.: btu/h)

A : cross sectional are normal to the heat flow (m2₎

k : thermal conductivity ( SI: W/m. K; cgs: cal/s. cm. °C; Eng.: btu/h. °F. ft )

dT/dx : temperature different in the x direction

• The minus sign is required in Fourier’s equation because the heat transfer is positive in the direction from initial point 1 to the final point 2. Since the T1 > T2 (heat is transport from

high temperature to lower temperature region), minus sign is needed to make the value of heat rate positive.

dx dT k A q_x  

HEAT TRANSFER – CONDUCTION

• For steady state, the equation can be integrated,

This equation is basically a matter of putting in values to

solve.



_{1 2}



1 2 2 1 2 1

T

T

x

x

k

A

q

dT

k

dx

A

q

x T T x x x















HEAT TRANSFER – CONDUCTION

Conduction Through a Plane Wall.

The temperature various linearly with distance.

0 Δx Δx Distance,x (m) T₁ T₂ Temperature, (K) T₂ T₁ q

R

T

T

kA

x

T

T

q

1 2



1



2







EXERCISE 1

Calculate the heat loss per m

2

_{of surface area for an}

insulating wall composed of 25.4 mm thick fiber

insulating board, where the inside temperature is 352.7

K and the outside temperature is 297.1K.

From Table A.3 (pg 599), thermal conductivity for fiber

insulating board is 0.048 W/m.K.

1 2 2 1 2 ( ) 0.048 (352.7 297.1) 0.0254 105.1 / k T T q A x x q W m A      

Conduction Through Solids In Series.

T

₁

A B C

q T

₂

T

₃

Dx

_A

Dx

_B

Dx

_C

T

₄

• The rate of heat transfer,

where,

C

R

R

R

T

T

R

T

T

R

T

T

R

T

T

q

B A C B A





















1 2 2 3 3 4 1 4

A

k

x

R

A A A





A

k

x

R

B B B





A

k

x

R

C C C





EXERCISE 2

A cold storage room is constructed of an inner layer

of 12.7 mm of pine, a middle layer of 101.6 mm of

cork board and an outer layer of 76.2 concrete. The

wall surface temperature is 255.4 K inside the cold

room and 297.1 K at the outside surface of concrete.

The conductivites for pine, 0.151; cork board,

0.0433; and concrete, 0.762 W/m.K. Calculate the

heat loss in W for 1 m

2

_{and the temperature at the}

interface between wood and cork board.

Answer: (-16.48 W, 256.79 K)

SOLUTION

W

K

A

k

x

R

W

K

A

k

x

R

W

K

A

k

x

R

m

x

m

x

m

x

k

k

k

m

A

K

T

K

T

C C C B B B A A A C B A C B A

?

?

?

346

.

2

)

1

(

043

.

0

1016

.

0

0841

.

0

)

1

(

151

.

0

0127

.

0

0762

.

0

,

1016

.

0

,

0127

.

0

762

.

0

,

043

.

0

,

151

.

0

1

,

1

.

297

,

4

.

255

₄ 2 1

















































HEAT TRANSFER – CONDUCTION

• Conduction Through Solids In Parallel

A B C D E F G 1 2 2 3 2 3 2 3 ( ) ( ) ( ) ( ) ... T A B C D E F G C C A A B B D D A A B C D q q q q q q q q k A k A k A k A q T T T T T T T T x x x x                    

HEAT TRANSFER – CONDUCTION

Conduction Through A Hollow Cylinder.

T

₂

L

r

₂

q

r

₁

• The cross-sectional area normal to the heat flow

is, A =2

p

rL.

• The rate of heat transfer,

dr

dT

k

A

CONDUCTION THROUGH A HOLLOW CYLINDER



 



  2 1 2 1 2 r r T T dT k r dr L q dr dT k A q

p







 









₂2 ₁



1 1 2 1 2 1 2 1 2 ln ) ( 2 2 2 ln 2 2 ln r r r r L Lr Lr Lr Lr A A A A A_lm       p p p p p 1 2 2 1

r

r

T

T

kA

q

_lm







or





1 2



1 2

ln

2

T

T

r

r

L

k

q







p

Where:









kL r r kA r r R R T T kA r r T T q lm lm p 2 ln _{2 1} 1 2 2 1 1 2 2 1        

EXERCISE 3

A thick wall cylindrical tubing of hard rubber having

and

inside radius of 5 mm and an outside radius of 20

mm is

being used as a temporary cooling coil in a bath. Ice

water

is flowing rapidly inside, and the inside wall

temperature is

274.9 K. The outside temperature is at 297.1 K. A

total of

14.65 W must be removed from the bath by the

cooling

coil. How many m tubing are needed?(k=1.15

W/m.K)

1 2 2 1 2 ( ) ln( / ) 2 0.151 (274.9 297.1) ln(0.02 / 0.005) 15.2 / 14.65 0.964 15.2 / L q k T T r r L q x q W m L W length m W m p p       

HEAT TRANSFER – CONDUCTION

Conduction Through a Multilayer Cylinders.

Example, heat is being transferred through the walls of

an insulated pipe.

T₁ T₂ T₃ T₄ r₁ r2 _r3 r₄ q A B C

HEAT TRANSFER – CONDUCTION

• At steady-state, the heat-transfer rate q, be the same for

each layer.

• The rate of heat transfer,

where,

C B A C B A

R

R

R

T

T

R

T

T

R

T

T

R

T

T

q





















1 2 2 3 3 4 1 4





L

k

r

r

R

A A

p

2

ln

_{2 1}







L

k

r

r

R

B B

p

2

ln

_{3 2}







L

k

r

r

R

C C

p

2

ln

_{4 3}



EXERCISE 4

A thick walled tube of stainless steel (A) having a k =

21.63 W/m.K with dimensions of 0.0254 m ID and

0.0508 m OD is covered with a 0.0254 thick layer of

insulation (B), k = 0.2423 W/m.K. The inside wall

temperature of the pipe is 811 K and the outside is at

310.8 K. For a 0.305 m length pipe, calculate the heat

loss and also the temperature at the interface between

the metal and the insulation.

SOLUTION









A B A lm A A Alm R T T q W R R T T q W K A k r r R m A A A A A A m Lr A m r m r 2 1 3 1 1 2 2 1 2 1 2 2 2 1 1 2 1 7 . 331 / 01673 . 0 ) 0351 . 0 ( 63 . 21 0127 . 0 0351 . 0 0243 . 0 / 0487 . 0 ln 0243 . 0 0487 . 0 / ln , 0243 . 0 0127 . 0 305 . 0 2 2 ... 0254 . 0 2 0508 . 0 , 0127 . 0 2 0254 . 0                          p p

HEAT TRANSFER – CONDUCTION

Conduction Through a Hollow Sphere

T

₂

r

₂

q r

₁

T

₁

• The cross-sectional area normal to the heat flow

is, A = 4

p

r

2

_.

• The rate of heat transfer,

dr

dT

k

A





R

T

T

k

r

r

T

T

T

T

r

r

k

q

dT

k

r

dr

q

r r T T 2 1 2 1 2 1 2 1 2 1 2

4

1

1

1

1

4

4

2 1 2 1









































p

p

p

HEAT TRANSFER – CONDUCTION

Temperature Profile for Heat Transfer By

Convection From One Fluid To Another.

film Metal wall film

Warm liquid A Cold fluid B T₆ T₅ T₄ T₃ T₂ T1 q

HEAT TRANSFER – CONDUCTION

Region,

T

₁

– T

₂

: turbulent fluid flow. Mainly convective heat transfer.

T

₂

– T

₃

: velocity gradient very steep. No turbulent flow,

(i.e. only laminar). Mainly conductive heat transfer.

T

₃

– T

₄

: conductive heat transfer.

T

₄

– T

₅

: no turbulent in film, mainly conductive heat transfer

T

₅

– T

₆

: turbulent flow, conductive heat transfer.

T

₁

– T

₂

and

T

₅

– T

₆

: different are small.

Convective coefficient for heat transfer through a fluid:

q = hA(T – T

_w

)

where,

h

= convective heat transfer coefficient.

T

= average temperature in fluid.

HEAT TRANSFER – CONDUCTION

Combined Convection and Conduction

and

Overall Coefficients.

• Heat flow with convective boundaries: plane

wall



_i

 

i_{A A}o

 

_o



R_i iR_A o R_o T T A h A k x A h T T q          1 1







 

 





i o



o A A i o i _{UA T T} h k x h T T A q        1 1 T_i T₁ T₂ T_o q h_i h_o x_A

HEAT TRANSFER – CONDUCTION

Heat flow with convective boundaries:

cylindrical wall with insulation.

T₁ T₂ T₃ T₄ r_O r_i r₁ h_o h_i A B



 





















_{o i}



_{o A A o} i i o o o o i A A i i o i i o o i i o o A A i o i i h A k A r r h A A U h A A A k A r r h U R T T T T A U T T A U q R T T A h A k r r A h T T q lm lm lm 1 1 1 1 1 1 4 1 4 1 4 1 4 1 4 1                        Similarly,

Overall heat transfer coefficient,

where,

HEAT TRANSFER

• Other way we can used,

where,



_{1 4}





_{1 4}



4 1

_U

_A

_T

_T

_U

_A

_T

_T

R

R

R

R

T

T

q

_{i i o o} o B A i























L

k

r

r

R

A i A

p

2

ln

₁







L

k

r

r

R

B o B

p

2

ln

₁





_{i i}



_{i i} i

A

h

h

Lr

R

1

2

1





p

o



_{o o}



_{o o}

A

h

h

Lr

R

1

2

1





p

A thick-walled tube of stainless steel (A)

having a k = 21.63

W/m.k with dimensions of 0.0254m ID and

0.0508m OD is

covered with a 0.0254m layer of asbestos (B)

insulation, k =

0.2423 W/m.k. The inside wall temperature of

the pipe is 811K

and the outside surface of the insulation is at

310.8K. For a

0.305m length of pipe, calculate the heat loss

and also the

temperature at the interface between the

metal and the

B A R R T T q    1 3

The resistances are

K/W 01673 . 0 ) 305 . 0 )( 63 . 21 ( 2 ) 0127 . 0 0254 . 0 ln( 2 ) ln( 2 ) ln( _{2/ 1 2/ 1}     p p p k L d d L k r r R A A A K/W 493 . 1 ) 305 . 0 . 0 )( 2423 . 0 ( 2 ) 0508 . 0 1016 . 0 ln( 2 ) ln( 2 ) ln( _{2/ 1 2/ 1}     p p p k L d d L k r r R B B B

The heat transfer rate is

B A R R T T q    1 3 W 7 . 331 493 . 1 01673 . 0 8 . 310 811     q

K

5

.

805

01673

.

0

811

7

.

331

2 2 2 1











T

T

R

T

T

q

A

CRITICAL THICKNESS OF INSULATION FOR

A CYLINDER

• If outer radius < r

_cr

: adding more insulation will

increase heat transfer rate

• If outer radius > r

_cr

: adding more insulation will

decrease heat transfer rate

h

k

r

)

_cr



(

₂

)

(

_{2 1} 0

A

T

T

h

q









0

2

1

2

0

1

1

/

ln

)

(

2

h

r

k

r

r

T

T

L

q







p

With insulation:

EXAMPLE

• An electric wire having a diameter of 1.5 mm covered

with a plastic insulation (thickness = 2.5mm) is exposed

to the air at 300K and ho = 20 W/m2.K. It is assumed

that the wire surface temperature is constant at 400K

and is not affected by the covering.

• A) calculate the value of the critical radius

• B) calculate the heat loss per (m) of wire length with no

insulation

• C) repeat (b) for insulation being present

Convection: Heat transfer using movement of

fluids.

Heat transfer is considered as convection with the

presence of bulk fluid motion. Fluid motion

enhances heat transfer where the higher the fluid

velocity, the higher the rate of heat transfer.

 2 main classification of convective heat transfer;

1. Forced Convection :

fluid flow by pressure

differences, a pump, a fan and so on

2. Natural Convection:

motion of fluid results from

the density changes in heat transfer

The rate of heat transfer :

T_w = 80 o_C

T_o = 30 o_C

q



q



Ah(T

_w



T

_o

)

The convection coefficient is _{a measure of how effective}

a fluid is at carrying heat to and away from the surface. h = heat transfer coefficient

(W/m2_.K)

A= surface area (m2₎

CONVECTION HEAT TRANSFER

Metal wall Warm fluid A Cold fluid B q Turbulence absent T2 T3 Turbulence region T1

q = hA

(T-Tw)

FKKSA

FORCED CONVECTION INSIDE PIPES

 Forced convection – fluid forced to flow by pressure differences  Types of fluid, laminar or turbulent

– great effect on heat-transfer coefficient More turbulent– greater heat-transfer coefficient

 Reynolds number, N_Re  N_Re D_  where v = velocity of fluid (m/s)

 = viscosity of fluid (Pa.s)

= density of fluid (kg/m3₎

FKKSA

FORCED CONVECTION

where

= viscosity of fluid (Pa.s)

= density of fluid (kg/m3₎

k = thermal conductivity of fluid (W/m.K) c_P = heat capacity of fluid (J/kg.K)

h = heat transfer coefficient (W/m2_.K)

D = diameter of pipe (m)  Prandtl number, N_Pr Dimensionless numbers:  Nusselt number, N_Nu k μ c ρc k ρ μ N P P Pr   k hD N_Nu

FKKSA

LAMINAR FLOW INSIDE HORIZONTAL PIPE

where

D = inside diameter of pipe (m) L = length of pipe (m)

_b = viscosity of fluid at bulk temperature (Pa.s)

_w = viscosity of fluid at wall temperature (Pa.s) h_a = average heat transfer coefficient (W/m2_.K)

 N_Nu       a h_aD k 1.86 NReNPr D L         1 3  b _w         0.14 N_Re  2100 & N_ReN_{P r}  100 : L D

All physical properties at except ₂ bi  _w T bo T mean b T       _  q = h_aA∆T_a where 2 bo T w T bi T w T a ΔT      _        _  100 2100 Pr Re Re   L D N N N Limitations

FKKSA

TURBULENT FLOW INSIDE HORIZONTAL PIPE

where  N_NuhLD k 0.027 NRe 0.8 N_Pr13 b _w         0.14 N_Re  6000 , 0.7 ≤ N_{P r} ≤ 16000 & _DL  60: Rate of heat transfer is greater

c_P = heat capacity of fluid (J/kg.K)

D = inside diameter of pipe (m)

k = thermal conductivity of fluid (W/m.K)

 _b = viscosity of fluid at bulk average temperature (Pa.s)

h_L = heat transfer coefficient based on the log mean driving force ∆T_lm (W/m2_.K)

 _w = viscosity of fluid at wall temperature (Pa.s)

Many industrial heat transfer processes in the turbulent region

60 16000 7 . 0 6000 Pr Re     D L L D N N Limitations

EXAMPLE 4.5-1 Page 262: Heating of Air in

Turbulent Flow

Air at 206.8 kPa and an average of 477.6 K is being heated as it flows through a tube of 25.4mm inside diameter at velocity of 7.62 m/s. The heating medium of 488.7 K steam condensing on the outside of the tube. Since the heat-transfer coefficient of condensing steam is several thousand W/m2.K and the resistance of the metal wall is very small, it will be assumed that the surface wall temperature of the metal in contact with the air is 488.7 K. Calculate the heat-transfer coefficient for an L/D > 60 and also the heat-transfer flux q/A.

bo T K 7 . 488 Steam,Tw  steam o h h  L mm 5.4 2 air bi T kPa 8 . 206 m/s 62 . 7 K 6 . 477    P v T_ave 3 Pr 5 kg/m 74 . 0 W/m 03894 . 0 686 . 0 Pa.s 10 6 . 2 K 477.6 kPa, 101.32 at A.3, Appendix From            k N T T P b bm ave Pa.s 10 64 . 2 K 7 . 88 4 at A.3, Appendix From 5     w w T 





3 k Pa 8 . 2 0 6 2 1 1 2 1 2 1 2 kg/m 509 . 1 35 . 101 8 . 206 74 . 0 on dep end is , For                                                   T T T T P P P & T RT PM RT PM RT V m PM RT M m PV nRT PV

) 6000 ( 10 122 . 1 10 6 . 2 ) 509 . 1 )( 62 . 7 ( 10 4 . 25 4 5 3 Re         





D N









.K W/m 2 . 63 0264 . 0 0260 . 0 686 . 0 10 122 . 1 027 . 0 03894 . 0 ) 10 4 . 25 ( 027 . 0 2 14 . 0 3 1 8 . 0 4 3 14 . 0 3 1 Pr 8 . 0 Re                    L L w b L Nu h h N N k D h N  









2 W/m 1 . 701 6 . 477 7 . 488 2 . 63      h_L T_w T_bm A q

flow rent countercur flow parallel ho T hi T hi T T_ho co T co T ci T ci T 1 T  1 T  2 T  2 T  hi T hi T ho T ho T ci T ci T co T co T

EXAMPLE 4.5-4 Page 268: Heat Transfer Area and

Log Mean Temperature Difference

A heavy hydrocarbon oil which has a c_pm = 2.30kJ/kg is being cooled in a heat exchanger from 371.9 K to 349.7 K and flows inside the tube at a rate of 3630 kg/h. A flow of 1450kg water/h enters at 288.6K for cooling and lows outside the tube.

a) Calculate the water outlet temperature and heat-transfer area if the overall U_i = 340 W/m2.K and the streams are countercurrent

flow rent countercur (a) 1 T  2 T  hi T  K 1 . 397 K 7 . 349  ho T K 6 . 288  ci T co T kg/h 3630 oil,m  kg/h 1450 water,m 

 

    W 51490 3600 / ) 7 . 349 9 . 371 )( 3 . 2 ( 3630      _{ho hi} h p T T c m q      ? .K W/m 340 kJ/kg.K 187 . 4 kJ/kg.K 3 . 2 2     i i water p oil p A U c c

 

  K 1 . 319 ) 6 . 288 )( 187 . 4 ( 1450 51490      co co ci co c p T T T T c m q 

K 9 . 56 8 . 52 1 . 61 ln 8 . 52 1 . 61 ln 2 1 2 1                       T T T T T_lm 2 m 66 . 2 ) 9 . 56 ( 340 51490     i i lm i i A A T A U q K 1 . 61 6 . 288 7 . 349 K 8 . 52 1 . 319 9 . 371 2 1         T T

flow parallel (b) K 3 . 83 6 . 288 9 . 371 K 6 . 30 1 . 319 7 . 349 2 1         T T K 7 . 52 6 . 30 3 . 83 ln 6 . 30 3 . 83 ln 2 1 2 1                       T T T T T_lm 2 m 87 . 2 ) 7 . 52 ( 340 51490     i i lm i i A A T A U q forces. driving perature larger tem gives ws counterflo because occurs This w. counterflo for than area larger a is This

•Radiation heat transfer is the transfer of heat by

electromagnetic radiation

•Occur in solid, liquid and gas •Not require heat transfer

medium

•Fastest energy transfer

•Example: microwave, radar, cordless telephones

Absorptivity

• When thermal radiation (light waves) falls

upon a body, part is absorbed, part is

reflected into space and part is

transmitted through the body.

• BLACK BODY – one that absorb all radiant

energy and reflect none.

reflectivity/fraction reflected 0

1.0 absorbed n ty/fractio absorptivi    





• Kirchoff’s Law states at the same

temperature T

1

• For

• For a perfect black body with :

• Substances that have emissivity < than

1.0 are called

gray bodies

1 body, Black   body black of power emissive total surface a of power emissive total , Emissivity   B E E  1 body, Gray   1 1





 1   4 T A q 



4 T A q 

