Heat Transfer: Exercises - eBooks and textbooks from bookboon.com

Example 1.2 A 20 mm diameter copper pipe is used to carry heated water, the external surface of the pipe is subjected to a convective heat transfer coefficient of h  6 W / m 2 K , find [r]

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Download free books at

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Chris Long & Naser Sayma

Heat Transfer: Exercises

Download free eBooks at bookboon.com

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Heat Transfer: Exercises

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Preface

Worked examples are a necessary element to any textbook in the sciences, because they reinforce the theory (i.e the principles, concepts and methods) Once the theory has been understood, well chosen examples can be used, with modification, as a template to solve more complex, or similar problems

This work book contains examples and full solutions to go with the text of our e-book (Heat Transfer,

by Long and Sayma) The subject matter corresponds to the five chapters of our book: Introduction to Heat Transfer, Conduction, Convection, Heat Exchangers and Radiation They have been carefully chosen with the above statement in mind Whilst compiling these examples we were very much aware

of the need to make them relevant to mechanical engineering students Consequently many of the

problems are taken from questions that have or may arise in a typical design process The level of

difficulty ranges from the very simple to challenging Where appropriate, comments have been added which will hopefully allow the reader to occasionally learn something extra We hope you benefit

from following the solutions and would welcome your comments

Christopher Long

Naser Sayma

Brighton, UK, February 2010

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1 Introduction

Example 1.1

The surface temperature on the inside of the wall is 16oC and that on the outside is 6oC Find the heat flux through the wall and the total heat loss through it

The minus sign indicates heat flux from inside to outside

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7

Example 1.2

A 20 mm diameter copper pipe is used to carry heated water, the external surface of the pipe is

q A q

Qconv  conv  conv 2   360  2    0 01  22 6 /

For radiation, assuming black body behaviour:

 4 4

f s

A

q

Q rad  rad    

loss by (black-body) radiation is seen to be comparable to that by convection

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8

Example 1.3

A plate 0.3 m long and 0.1 m wide, with a thickness of 12 mm is made from stainless steel (

K m

W

the plate is heated by an electrical heater (also 0.3 m by 0.1 m) positioned on the underside of the plate and the temperature of the plate adjacent to the heater is maintained at 100oC A voltmeter and ammeter are

connected to the heater and these read 200 V and 0.25 A, respectively Assuming that the plate is

perfectly insulated on all sides except the top surface, what is the convective heat transfer coefficient?

Solution

Heat flux equals power supplied to electric heater divided by the exposed surface area:

2

/7.16663

.01.0

25.0

L W

I

V A

293 75 371

293 75

371 10 67 5 7

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9

Example 1.4

An electronic component dissipates 0.38 Watts through a heat sink by convection and radiation (black

transfer coefficient is 6 W/m2 K, and the heat sink has an effective area of 0.001 m2 ?

610

T dT

df

668.22

9.25556

1067

5

3

4 8 1

s s

dT df

f T

9 2555 300

6 300 10

67 5

9 2555 46

324 6 46 324 10

67 5 46

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10

The difference between the last two iterations is small, so:

C K

T s0 323 50

The value of 300 K as a temperature to begin the iteration has no particular significance other than being above the ambient temperature

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T r

Q Q

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Q r

T r r k z

r r

T z r k

t

T cr z z

T r k z

r r

Dividing by r, noting that r can be taken outside the brackets in the second term because it is not a

function of z Also dividing by k since the thermal conductivity is constant:

t

T k

c z

T r

r r

T r

2

which gives the required outcome:

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13

t

T z

T r

2

Example 2.2

layer of plastic (k = 1 W/m K, and thickness of 3 mm), and an outer layer of stainless steel (k = 16

W/m K, and thickness of 1 mm) Sandwiched between these two layers is a layer of insulation material with k = 0.07 W/m K Find the width of the insulation that is required to reduce the convective heat loss to 15 W/m2

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) 20 ( 25

s i

i p

p

L k

L

h

U

333.0

11

s i

i p

p

L k

001

0 1

003

0 12

1 333 0

1 07 0 1

1 333

.

0

1

o s

s p

p i i

L k

L h k

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15

when the water temperature is 15oC, the outside air temperature is -10oC, the water side heat transfer coefficient is 30 kW/m2 K and the outside heat transfer coefficient is 20 W/m2 K

outer diameter of 300 mm, and thermal conductivity of k = 0.05 W/m K

T T

Lk

Q   



)/ln(

T

2

2   2Adding the three equations on the right column which eliminates the wall temperatures gives:

2

1 2 1

1 /

ln

1

2

r h k

r

r r

h

T T L Q

o i

1 50

05 0 / 052 0

ln 05 0 30000

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2 1

1 ) / ln(

/ ln

r h k

r

r k

r

r r

h

T

T L

Q

o ins

i

o i

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1 05

0

) 052 0 / 15 0

ln(

50

05 0 / 052 0

ln 05 0 30000

K Calculate the heat flow through the pipe Also calculate the heat flow through the pipe when a layer

of insulation, k = 0.1 W/m K and 50 mm radial thickness is wrapped around the pipe

1 /

ln

1

2

r h k

r

r r

h

T T L Q

o i

1 16

47 / 50

ln 2000 047

2 1

1 ) / ln(

/ ln

r h k

r

r k

r

r r

h

T T L Q

o ins

i

o i

1 1

0

) 50 / 100

ln(

16

47 / 50

ln 2000 047

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18

ins

o i



1 0

) 50 / 100 ln( 100 80 20

A diagram of a heat sink to be used in an electronic application is shown below There are a total of 9

each 60 mm long, 40 mm wide and 1 mm thick The spacing between adjacent fins, s, is 3 mm The

air temperature T f is 20oC Under these conditions, the external heat transfer coefficient h is 12 W/m2

K The fin may be assumed to be sufficiently thin so that the heat transfer from the tip can be

neglected The surface temperature T, at a distance, x, from the base of the fin is given by:

mL

x L m T

Determine the total convective heat transfer from the heat sink, the fin effectiveness and the

fin efficiency

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dT kA







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20

Then

T b T f

m mL

x L

m dx

cosh

)(sinh

Thus

 b f

c x

c

mL

mL kA

dx

dT kA

w

P2(  )2(0.040.001)0.082

2 6

10400001.004

40 175

082 0

in theoccur d that woulrate

fer Heat trans

ratefer heat transFin



1040

12

03

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21

Fin efficiency:

re temperatu base

at the were area fin the all if ed transferr be

would Heat that

fin

h the fer throug heat trans

Lt Lt wL

wL

2 3

1092.4)001.004.0(06

12

03

For the fin of example 4.5, a fan was used to improve the thermal performance, and as a result, the

to predict the rate of change of temperature with time Using the lumped mass approximation given below, calculate the time taken,  , for the heat sink to cool from 60oC to 30oC

T

f i f

0005.0402

f

T T hA

mC ln



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2030ln40

2

001.09002700ln

f

T T

T T h

Ct





Example 2.7

The figure below shows part of a set of radial aluminium fins (k = 180 W/m K) that are to be fitted to

a small air compressor The device dissipates 1 kW by convecting to the surrounding air which is at

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Solution

Consider a single fin:

m t

w

P2(  )2( 0050.03)0.07

2 6

1015003.0005

150 180

07 0

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24

Example 2.8

An air temperature probe may be analysed as a fin Calculate the temperature recorded by a probe of length

an actual air temperature of 50oC and the surface temperature at the base of the probe is 60oC

Solution

(from the full fin equation) is given by:

mL mk

h mL

x L m mk

h x L m T

) ( sinh )

( cosh

h mL T

T

tip b

tip

sinh cosh

 Ttip  T (no error)

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D P D

A   2/ 4 ,  

1 2

/ 1 2

2 /

1

235.59003

.019

5044

k

D

h kA

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A design of an apartment block at a ski resort requires a balcony projecting from each of the 350

separate apartments The walls of the building are 0.3 m wide and made from a material with k = 1

W/m K Use the fin approximation to examine the implications on the heat transfer for two separate suggestions for this design In each case, the balcony projects 2 m from the building and has a length

-5oC; the overall (convective + radiative) heat transfer coefficient on the inside of the building is 8

a) A balcony constructed from solid concrete and built into the wall, 0.2 m thick, k = 2 W/m K

b) A balcony suspended from 3 steel beams, k = 40 W/m K, built into the wall and projecting out by 2

m each of effective cross sectional area A e 0 m.01 2, perimetre P0.6m(The actual floor of the balcony in this case may be considered to be insulated from the wall

c) No balcony

Solution

a) For the concrete balcony

Treat the solid balcony as a fin

b

o

t h

B  / 2

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27

12

1

Not that Bi is not << 1, thus 2D analysis would be more accurate However, treating it as a fin will

give an acceptable result for the purpose of a quick calculation

m t

H

P2(  )2(40.2)8.4

2

8.02.0

.02

4.8

20 1/2

2 / 1

c b o

c

A

Pk h T

T A Pk h

2 / 1

1

Also assuming 1-D conduction through the wall:

) ( T T1

h

) ( T1 T2

1

) (

c b

i

i o b

Pk h

A k

L

h

T T

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28

From (4)

2 2

/

24.820

8

02

1 1

3

0 8

1 20 ( 5 ) 1

h k

L

h

T T

q

o w i

i o

The difference for one balcony is Ac( 77 2  52 6 )  0 8  24 6  19 7 W

For 350 apartments, the difference is 6891 W

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29

As before, however, in this case Bi << 1 because k s kb

1121

.040

6

20 1/2

2 / 1

1

) (

c s

i

i o b

Pk h

A k

L

h

T T

2 /

40620

01

040

3

081

)5(

In free convection, the heat transfer coefficient varies with the surface to fluid temperature difference

temperature ratio with time will be given by

SpecificMass

Area,

and hinit= the heat transfer coefficient at t = 0 Use this expression to determine the time taken for an

from the equation ( eht) which assumes a constant value of heat transfer coefficient

Solution

Heat transfer by convection = rate of change of internal energy

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30

dt

T T d mC T

)

We know that h  G (  Ts Tf)n

Where G is a constant

(Note that this relation arises from the usual Nusselt/Grashof relationship in free convection; for

example: Nu  0 Gr.1 Pr1 / 3in turbulent flow or Nu 0.54GrPr1 / 4for laminar flow)

Equation 1 then becomes:

dt

T T d A

mC T

T T

T

f s

n

f

s

) (

T T

T T d dt

mC

GA

) (

t f s

n f

s

T T mC

GnAt T

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m 27500.040.0020.22

J K m mC

87022.0

04

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4

/

1

1 2

0

4

4 / 1

For the equation  eht

which assumes that the heat transfer coefficient is independent of surface-to-fluid temperature

difference

s h

101.216

2.0ln

will occur?

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K kg kJ T T

K) and k = 0.03186 W/m K

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a)

95 6 603

0

4183 10

002 1

0

4208 10

22 3 965

k

C k

0

1005 10

563 1 19

373 10

46

1 110

10 46

T T

Cp

/ 7 1007

373 10

98 3 373 10

58 2 917 0 10

98 3 10

58 2 917

0

7 1007 10

18

2

e)

0261 0 10

0081 0

139 10

0

4183 10

002 1

0

4208 10

22 3 965

k

C k

0

1005 10

563 1 19

373 10

46

1 110

10 46

T T

Cp

/ 7 1007

373 10

98 3 373 10

58 2 917 0 10

98 3 10

58 2 917

0

7 1007 10

18

2

e)

0261 0 10

0081 0

139 10

0

4183 10

002 1

0

4208 10

22 3 965

k

C k

0

1005 10

563 1 19

373 10

46

1 110

10 46

T T

Cp

/ 7 1007

373 10

98 3 373 10

58 2 917 0 10

98 3 10

58 2 917

0

7 1007 10

0081 0

139 10

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37

0067 0 86

1369 10

0

2035 10

36 8

d) The roof of a coach 6 m long, travelling at 100 km/hr in air ( = 1.2 kg/m3 and  = 1.8 x 105

kg/m s)

a valve guide of diameter 10 mm in a 1.6 litre, four cylinder four stroke engine running at

port diameter of 25 mm)

f)

0067 0 86

1369 10

0

2035 10

36 8

d) The roof of a coach 6 m long, travelling at 100 km/hr in air ( = 1.2 kg/m3 and  = 1.8 x 105

kg/m s)

a valve guide of diameter 10 mm in a 1.6 litre, four cylinder four stroke engine running at

port diameter of 25 mm)

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39

Characteristic length is r not D

7

5 1 12 10 10

26 3

3 3 471 59 2

D m

Re   2 

400 10

97.102.0

05.04

8 1

6 8 27 2 1

second

s kg

2

1 60

3600 4

10 6 1 2

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40

6869025

.01056.3

Re  )

Example 3.3

Calculate the appropriate Grashof numbers and state if the flow is laminar or turbulent for the following:

10 84 1 10

8 1 291

6 0 57 81 9 2

84.1

2 0 4

0 Perimeter

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