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 and Overall Heat Transfer Coefficient

CONDUCTION: FOURIER’S LAW• Flux of conduction heat transfer can be calculated by

Fourier’s LawFourier’s Law

qx : 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

dTk

A

qx

HEAT TRANSFER – CONDUCTION

• For steady state, the equation can be integrated,

This equation is basically a matter of putting in values to solve.

21

12

2

1

2

1

TTxx

k

A

q

dTkdxA

q

x

T

T

x

x

x

HEAT TRANSFER – CONDUCTIONConduction Through a Plane Wall.

The temperature various linearly with distance.

0 ΔxΔx Distance,x (m)

T1

T2

Temperature, (K)

T2

T1

q

R

TT

kAx

TTq 2121

EXERCISE 1

Calculate the heat loss per m2 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 Tq

A x x

qW m

A

Conduction Through Solids In Series.

T1 A B C

q T2 T3

DxA DxB DxC T4

• The rate of heat transfer,

where,

CRRR

TT

R

TT

R

TT

R

TTq

BACBA

41433221

Ak

xR

A

AA

Ak

xR

B

BB

Ak

xR

C

CC


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 m2 and the temperature at the interface between wood and cork board.

Answer: (-16.48 W, 256.79 K)

SOLUTION

W

K

Ak

xR

W

K

Ak

xR

W

K

Ak

xR

mxmxmx

kkk

mAKTKT

C

CC

B

BB

A

AA

CBA

CBA

??

?

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 2

41


• 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 CA A B B D DA

A B C D

q q q q q q q q

k Ak A k A k Aq T T T T T T T T

x x x x


Conduction Through A Hollow Cylinder.T2 L

r2

q

r1

• The cross-sectional area normal to the heat flow is, A =2prL.


dr

dTk

A

q

CONDUCTION THROUGH A HOLLOW CYLINDER

2

1

2

12

r

r

T

TdTk

r

dr

L

q

dr

dTk

A

q

p

12

12

12

12

12

12

ln

)(2

22ln

22

ln

rr

rrL

LrLr

LrLr

AA

AAAlm

p

pp

pp

12

21

rr

TTkAq lm

or

21

12ln

2TT

rr

Lkq

p

Where:

kL

rr

kA

rrR

R

TT

kArr

TTq

lm

lm

p2

ln 1212

21

12

21

EXERCISE 3A thick wall cylindrical tubing of hard rubber having

andinside 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( / )

20.151 (274.9 297.1)

ln(0.02 / 0.005)

15.2 /

14.650.964

15.2 /

Lq k T T

r r

Lq x

qW m

L

Wlength m

W m

p

p


Conduction Through a Multilayer Cylinders.

Example, heat is being transferred through the walls of an insulated pipe.

T1

T2

T3

T4

r1

r2r3

r4q

A

B

C


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


where,

CBACBA RRR

TT

R

TT

R

TT

R

TTq

41433221

Lk

rrR

A

Ap2

ln 12

Lk

rrR

B

Bp2

ln 23

Lk

rrR

C

Cp2

ln 34

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.

Answer: (331.7 W, 805.5 K)

SOLUTION

A

BA

lmA

A

Alm

R

TTq

WRR

TTq

WKAk

rrR

mAA

AAA

AmLrA

mrmr

21

31

12

2

12

12

2

2

11

21

7.331

/01673.0)0351.0(63.21

0127.0

0351.00243.0/0487.0ln

0243.00487.0

/ln

,0243.00127.0305.022

......0254.02

0508.0,0127.0

2

0254.0

pp


Conduction Through a Hollow Sphere

T2 r2

q r1 T1

• The cross-sectional area normal to the heat flow is, A = 4pr2.

• The rate of heat transfer,dr

dTk

A

q

R

TT

krr

TTTT

rr

kq

dTkr

drq r

r

T

T

21

21

2121

21

2

41111

4

4

2

1

2

1

p

p

p



Temperature Profile for Heat Transfer By Convection From One Fluid To Another.

film filmMetal wall

Warm liquid A

Cold fluid B

T6

T5

T4

T3

T2 T1

q

HEAT TRANSFER – CONDUCTIONRegion,T1 – T2: turbulent fluid flow. Mainly convective heat transfer.T2 – T3: velocity gradient very steep. No turbulent flow,

(i.e. only laminar). Mainly conductive heat transfer.T3 – T4: conductive heat transfer.T4 – T5: no turbulent in film, mainly conductive heat transferT5 – T6: turbulent flow, conductive heat transfer.T1 – T2 and T5 – T6 : different are small.

Convective coefficient for heat transfer through a fluid:q = hA(T – Tw)

where,h = convective heat transfer coefficient.T = average temperature in fluid.Tw = temperature of wall in contact.


Combined Convection and Conduction and Overall Coefficients.

• Heat flow with convective boundaries: plane wall

oAi

oi

oAAi

oi

RRR

TT

AhAkxAh

TTq

11

oi

oAAi

oi TTUAhkxh

TTAq

11

TiT1

T2

To

q

hi

ho

xA


Heat flow with convective boundaries: cylindrical wall with insulation.

T1

T2

T3

T4

rO

ri

r1ho

hi

A

B

oAAoioiio

o

ooiAAiioi

i

ooii

ooAAioii

hAkArrhAAU

hAAAkArrhU

R

TTTTAUTTAUq

R

TT

AhAkrrAh

TTq

lm

lm

lm

1

1

1

1

11

414141

4141

Similarly,

Overall heat transfer coefficient,

where,

and

HEAT TRANSFER

• Other way we can used,

where,

414141 TTAUTTAU

RRRR

TTq ooii

oBAi

Lk

rrR

A

iA

p2

ln 1

Lk

rrR

B

oB

p2

ln 1

iiii

iAhhLr

R1

2

1

p oooo

oAhhLr

R1

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 and0.0508m OD is

covered with a 0.0254m layer of asbestos (B)insulation, k =

0.2423 W/m.k. The inside wall temperature ofthe pipe is 811K

and the outside surface of the insulation is at310.8K. For a

0.305m length of pipe, calculate the heat lossand also the

temperature at the interface between themetal and the

BA RR

TTq

31

The resistances are

K/W 01673.0

)305.0)(63.21(2

)0127.0

0254.0ln(

2

)ln(

2

)ln( 1/21/2

p

pp Lk

dd

Lk

rrR

AA

A

K/W 493.1

)305.0.0)(2423.0(2

)0508.0

1016.0ln(

2

)ln(

2

)ln( 1/21/2

p

pp Lk

dd

Lk

rrR

BB

B

The heat transfer rate is

BA RR

TTq

31

W7.331

493.101673.0

8.310811

q

K 5.805

01673.0

8117.331

2

2

21

T

T

R

TTq

A

CRITICAL THICKNESS OF INSULATION FOR A CYLINDER

• If outer radius < rcr: adding more insulation will increase heat transfer rate

• If outer radius > rcr: adding more insulation will decrease heat transfer rate

h

kr cr )( 2

)( 120 TTAhq

02

12

01

1/ln

)(2

hrk

rr

TTLq

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

a) 20 mm b) 9.42W c) 32.98 W

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

CONVECTION HEAT TRANSFER

The rate of heat transfer :

Tw = 80 oC

To = 30 oCq

q Ah(Tw To)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)

Fluid flow

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, NRe

NReD

wherev = velocity of fluid (m/s)

= viscosity of fluid (Pa.s)

= density of fluid (kg/m3)

D = diameter of pipe (m)

FKKSA

FORCED CONVECTION

where

= viscosity of fluid (Pa.s)

= density of fluid (kg/m3)

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

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

h = heat transfer coefficient (W/m2.K)

D = diameter of pipe (m)

Prandtl number, NPr

Dimensionless numbers:

Nusselt number, NNu

k

μc

ρck

ρμ

N P

P

Pr

khDN

Nu

FKKSA

LAMINAR FLOW INSIDE HORIZONTAL PIPE

whereD = 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)

ha = average heat transfer coefficient (W/m2.K)

NNu

a

haD

k1.86 N

ReN

Pr

DL

13

b

w

0.14

NRe 2100 & NReNP r 100 :L

D

All physical properties at except w2

biT

boT

mean bT

q = haA∆Ta where 2

boT

wT

biT

wT

aΔT

100

2100

PrRe

Re

L

DNN

N

Limitations

FKKSA

TURBULENT FLOW INSIDE HORIZONTAL PIPE

where

NNu

hLD

k0.027 N

Re

0.8N

Pr

13 b

w

0.14

NRe 6000 , 0.7 ≤ NP r ≤ 16000 & 60:DL

Rate of heat transfer is greater

cP = 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)

hL = heat transfer coefficient based on the log mean driving force

∆Tlm (W/m2.K)

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

Many industrial heat transfer processes in the turbulent region

60

160007.0

6000

Pr

Re

D

L

L

DN

N

Limitations

EXAMPLE 4.5-1 Page 262: Heating of Air inTurbulent 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.

boT

K 7.488 Steam, wT

steamo hh

L

mm 5.42air

biTkPa 8.206

m/s 62.7

K 6.477

P

v

Tave

3

Pr

5

kg/m 74.0

W/m03894.0

686.0

Pa.s 106.2

K 477.6 kPa, 101.32at A.3,Appendix From

k

N

TTP

b

bmave

Pa.s 1064.2

K 7.884at A.3,Appendix From

5

w

wT

3

kPa 8.206

21

1

2

1

212

kg/m 509.135.101

8.20674.0

on depend is

,For

TT

T

T

P

P

P & TRT

PM

RTPMRTV

mPM

RTM

mPVnRTPV

)6000(10122.1

106.2

)509.1)(62.7(104.25

4

5

3

Re

DN

.K W/m2.63

0264.0

0260.0686.010122.1027.0

03894.0

)104.25(

027.0

2

14.0

3

18.043

14.0

3

1

Pr

8.0

Re

L

L

w

bLNu

h

h

NNk

DhN

2 W/m1.701

6.4777.4882.63

bmwL TThA

q

flowrent countercur

flow parallel

hoThiT

hiT hoT

coT

coTciT

ciT

1T

1T

2T

2T

hiT

hiT

hoT

hoT

ciT

ciT

coT

coT

EXAMPLE 4.5-4 Page 268: Heat Transfer Area andLog Mean Temperature Difference

A heavy hydrocarbon oil which has a cpm = 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 Ui = 340 W/m2.K and the streams are

countercurrent

b) Repeat for parallel flow

flowrent countercur (a)

1T

2T

hiTK 1.397

K7.349hoT

K 6.288ciT

coT

kg/h 3630 oil, m

kg/h 1450 water, m

W51490

3600/)7.3499.371)(3.2(3630

hihohp TTcmq

?

.K W/m340

kJ/kg.K 187.4

kJ/kg.K 3.2

2

i

i

waterp

oilp

A

U

c

c

K 1.319

)6.288)(187.4(145051490

co

co

cicocp

T

T

TTcmq

K 9.56

8.52

1.61ln

8.521.61

ln2

1

21

T

T

TTTlm

2m 66.2

)9.56(34051490

i

i

lmii

A

A

TAUq

K 1.616.2887.349

K 8.521.3199.371

2

1

T

T

flow parallel (b)

K 3.836.2889.371

K 6.301.3197.349

2

1

T

T

K 7.52

6.30

3.83ln

6.303.83

ln2

1

21

T

T

TTTlm

2m 87.2

)7.52(34051490

i

i

lmii

A

A

TAUq

forces. driving peraturelarger tem

gives wscounterflo because occurs This w.counterflofor than arealarger 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.

0reflectedn ty/fractioreflectivi

1.0absorbedn ty/fractioabsorptivi

• Kirchoff’s Law states at the same temperature T1

• For

• For a perfect black body with :

• Substances that have emissivity < than 1.0 are called gray bodies

1 body,Black

bodyblack ofpower emissive total

surface a ofpower emissive total ,Emissivity

BE

E

1 body,Gray

11

1 4TAq

4TAq

Documents

Chap 4 Heat Transfer (PART 2)