ejercicios transferencia de calor.doc

7/28/2019 ejercicios transferencia de calor.doc

1/15

Heat andMass

Transfer(ME 209)

"Solved Problems"

Part 1

References

Heat Transfer "A Practical Approach" by Yunus Cengel

Fundamentals of Heat and Mass Transfer by Incropera

Heat Transfer by J.P. Holman

1(Consider a 1.2-m-high and 2-m-wide double-pane window consisting of two 3-mm-thick layers of glass (k =0.78 W/mC) separated by a 12-mm-wide stagnant air


2/15

space (k = 0.026 W/m C). Determine the steady rate of heat transfer through this

double-pane window and the temperature of its inner surface for a day during

which the room is maintained at 24C while the temperature of the outdoors is-5C. Take the convection heat transfer coefficients on the inner and outer surfaces

of the window to be 10 W/m2 C and 25 W/m2

C respectively.2(A 2-m1.5-m section of wall of an industrial furnace burning natural gas is not

insulated, and the temperature at the outer surface of this section is measured to be

80C. The temperature of the furnace room is 30C, and the combined convection

and radiation heat transfer coefficient at the surface of the outer furnace is 10

W/m2C. It is proposed to insulate this section of the furnace wall with glass wool

insulation (k = 0.038 W/mC) in order to reduce the heat loss by 90 percent.

Assuming the outer surface temperature of the metal section still remains at about

80C; determine the thickness of the insulation that needs to be used.

3(Water is boiling in a 25-cm-diameter aluminum pan (k = 237 W/m C) at 95C.Heat is transferred steadily to the boiling water in the pan through its 0.5-cm-thick

flat bottom at a rate of 800 W. If the inner surface temperature of the bottom of the

pan is 108C, determine (a) the boiling heat transfer coefficient on the inner surface

of the pan, and (b) the outer surface temperature of the bottom of the pan.

4(Two 5-cm-diameter, 15cm-long aluminum bars (k = 176 W/mC) with groundsurfaces are pressed against each other with a pressure of 20 atm (h = 11,400

W/m2C). The bars are enclosed in an insulation sleeve and, thus, heat transfer

from the lateral surfaces is negligible. If the top

and bottom surfaces of the two-bar system aremaintained at temperatures of 150C and 20C,

respectively, determine (a) the rate of heat transfer

along the cylinders under steady conditions and

(b) the temperature drop at the interface.

5(An electric current is passed through a wire 1 mm in diameter and 10 cm long. The

wire is submerged in liquid water at atmospheric pressure, and the current is

increased until the water boils. For this situation h = 5000 W/m2C, and the water

temperature will be 100 C. How much electric power must be supplied to the wire

to maintain the wire surface at 114 C?

6(Steam at 320C flows in a cast iron pipe (k = 80 W/m C)

whose inner and outer diameters are 5 cm and 5.5 cm,

respectively. The pipe is covered with 3-cm-thick glass

wool insulation with k = 0.05 W/m C. Heat is lost to the

surroundings at 5C by natural convection and radiation,

with a combined heat transfer coefficient of h2= 18 W/m

C. Taking the heat transfer coefficient inside the pipe to

be h1= 60 W/m2 C, determine the rate of heat loss from

the steam per unit length of the pipe. Also determine the temperature drops across

the pipe shell and the insulation.7(


3/15

8(Consider a large 3-cm-thick stainless steel plate (k = 15.1 W/m C) in which heat is

generated uniformly at a rate of 5 105 W/m3. Both sides of the plate are exposed

to an environment at 30C with a heat transfer coefficient of 60 W/m C. Explain

where in the plate the highest and the lowest temperatures will occur, and

determine their values.9(In a nuclear reactor, 1-cm-diameter cylindrical uranium rods cooled by water from

outside serve as the fuel. Heat is generated uniformly in the rods (k= 29.5 W/m

C) at a rate of 7107 W/m3. If the outer surface temperature of rods is 175C,

determine the temperature at their center.

10) Consider a long resistance wire of radius r1 = 0.2 cm and thermal conductivity

kwire = 15 W/m C in which heat is generated uniformly as a result of resistance

heating at a constant rate of qv = 50 W/m3. The wire is embedded in a 0.5-cm-thick

layer of ceramic whose thermal conductivity is kceramic = 1.2 W/m C. The outer

surface temperature of the ceramic layer is measured to be45C, .and is surrounded by air at 30 C with heat transfer

coefficient is of 10 W/m2C. Determine the temperatures at

the center of the resistance wire and the interface of the wire

and the ceramic layer under steady conditions.

11) Steam in a heating system flows through tubes whose outer

diameter is 3 cm and whose walls are maintained at a temperature of 120C. Circular

aluminum fins (k=180 W/m C) of outer diameter 6 cm and constant thickness of 2

mm are attached to the tube. Thespace between the fins is 3 mm, and thus there are

200 fins per meter length of the tube. Heat is transferred to the surrounding air at25C, with a combinedheat transfer coefficient of h = 60 W/m2 C. Determine the

increase in heat transfer from the tube per meter of its length as a result of adding fins.

(Fin efficiency = 95%).

12)A hot surface at 100C is to be cooled by attaching 3-cm-long, 0.25-cm-diameter

aluminum pin fins (k =237 W/m C) to it, with a center-to-center distance of 0.6 cm.

The temperature of the surrounding medium is 30C, and the heat transfer coefficient

on the surfaces is 35 W/m2C. Determine the rate of heat transfer from the surface for

a 1-m 1-m section of the plate. Also determine the overall effectiveness of the fins. .

(Fin efficiency = 95.9%).13) Consider steady two-dimensional heat transfer in a

long solid body whose cross section is given in the

figure. The temperatures at the selected nodes and the

thermal conditions at the boundaries are as shown.

The thermal conductivity of the body is k= 45 W/m

C, and heat is generated in the body uniformly at a

rate ofqv= 6 106 W/m3. Using the finite difference

method with a mesh size of x = y= 5.0 cm,

determine the temperatures at nodes:


4/15

14) Consider steady two-dimensional heat transfer in a long solid body whose cross

section is given in the figure. The measured temperatures at

selected points of the outer surfaces are as shown. The

thermal conductivity of the body is k = 45 W/m C, and

there is no heat generation. Using the finite differencemethod with a mesh size of x = y= 2.0 cm, determine

the temperatures at the indicated points in the medium.

15) Consider steady two-dimensional heat transfer in a

long solid bar whose cross section is given in the Figure 1

(a) and (b). The measured temperatures at selected points of the outer surfaces are as

shown. The thermal conductivity of the body is k = 20 W/m C, and there is no heat

generation. Using the finite difference method with a mesh size of x = y= 1.0 cm.

Determine the temperatures at the indicated points in the medium

FIGURE (1)

16) Consider steady two-dimensional heat transfer in a

long solid body whose cross section is given in the

figure. The temperatures at the selected nodes and the

thermal conditions on the boundaries are as shown.

The thermal conductivity of the body is k = 180

W/mC, and heat is generated in the body uniformly

at a rate ofqv= 107 W/m3. Using the finite difference

method with a mesh size of x = y= 10 cm,

determine the temperatures at nodes 1, 2, 3, and 4.

17) Consider steady two-dimensional heat transfer in an L-shaped solid body whose

cross section is given in the figure. The thermal conductivity of the body is k = 45

W/m C, and heat is generated in the body at a rate of qv= 5 106 W/m3. The right

surface of the body is insulated, and the bottom surface is maintained at a uniform

temperature of 120C. The entire top surface is

subjected to convection with ambient air at T = 30C

with a heat transfer coefficient of h = 55 W/m2 C,

and the left surface is subjected to heat flux at a

uniform rate of 8000 W/m2. The nodal network of the


5/15

problem consists of 13 equally spaced nodes with x = y= 1.5 cm. Five of the nodes

are at the bottom surface and thus their temperatures are known. Obtain the finite

difference equations at the remaining eight nodes.

Solutions

1) A double-pane window consists of two 3-mm thick layers of glass separated by a 12-mm wide

stagnant air space. For specified indoors and outdoors temperatures, the rate of heat loss

through the window and the inner surface temperature of the window are to be determined.

Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor

temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since

any significant temperature gradients will exist in the direction from the indoors to the outdoors.

3Thermal conductivities of the glass and air are constant. 4 Heat transfer by radiation is

negligible.

Properties The thermal conductivity of the glass and air are given to be kglass = 0.78 W/mC

and kair = 0.026 W/mC.

Analysis The area of the window and the individual resistances are


6/15

A = =( . ( .12 2 2 4m) m) m2

C/W2539.0

0167.01923.0)0016.0(20417.02

C/W0167.0)m4.2(C).W/m25(

11

C/W1923.0)m4.2(C)W/m.026.0(

m012.0

C/W0016.0

)m4.2(C)W/m.78.0(

m003.0

C/W0417.0)m4.2(C).W/m10(

11

2,211,

o

2o22

2,o

22

22

21

1glass31

221

1,i

=

+++=+++=

====

=

===

=

====

=

===

convconvtotal

conv

air

conv

RRRRR

AhRR

Ak

LRR

Ak

LRRR

AhRR

The steady rate of heat transfer through window glass then becomes

W114=

=

=

C/W2539.0

C)]5(24[21

totalR

TTQ

The inner surface temperature of the window glass can be determined from

C19.2==

= =C/W)W)(0.0417114(C24o1,11

1,

11conv

conv

RQTTR

TTQ

2) An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the

heat loss through that section of the wall by 90 percent. The thickness of the insulation that

needs to be used is to be determined. Also, the length of time it will take for the insulation to

pay for itself from the energy it saves will be determined.

Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2Thermal

conductivities are constant. 3The furnace operates continuously. 4The given heat transfercoefficient accounts for the radiation effects.

Properties The thermal conductivity of the glass wool insulation is given to be k= 0.038

W/mC.

AnalysisThe rate of heat transfer without insulation is

A = =(2 3m)(1.5 m) m2

( ) ( . ( )(80 )Q hA T T s= = = 10 3 30 1500W / m C) m C W2 2

In order to reduce heat loss by 90%, the new heat transfer rate and thermal

resistance must be

.

( ).

Q

QT

RR

T

Qtotaltotal

= =

= = =

=

010 1500 150

80 30

1500333

W W

C

WC / W

and in order to have this thermal resistance, the thickness of insulation must be

cm3.4==

=

+

=

+=+=

m034.0

C/W333.0)mC)(3W/m.038.0()mC)(3.W/m10(

1

1

222

conv

L

L

kA

L

hARRR insulationtotal

Air

R1

R2

R3

Ro

Ri

T1

T2

Insulation

Ro

T

Rinsulation

Ts L


7/15

3) Heat is transferred steadily to the boiling water in an aluminum pan. The inner surface

temperature of the bottom of the pan is given. The boiling heat transfer coefficient and the

outer surface temperature of the bottom of the pan are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since the

thickness of the bottom of the pan is small relative to its diameter. 3 The thermal conductivity

of the pan is constant.Properties The thermal conductivity of the aluminum pan is given to be k= 237 W/mC.

Analysis (a) The boiling heat transfer coefficient is

222

m0491.04

m)25.0(

4===

DAs

C.W/m1254

2 =

=

=

=

C)95108)(m0491.0(

W800

)(

)(

2TTA

Qh

TThAQ

ss

ss

(b) The outer surface temperature of the bottom of the pan is

C108.3=

=+=

=

)mC)(0.0491W/m.237(

m)005.0W)(800(+C108

21,,

,,

kA

LQTT

L

TTkAQ

innersouters

innersouters

4) Two cylindrical aluminum bars with ground surfaces are pressed against each other

in an insulation sleeve. For specified top and bottom surface temperatures, the rate of

heat transfer along the cylinders and the temperature drop at the interface are to be

determined.

Assumptions 1 Steady operating conditions exist. 2

Heat transfer is one-dimensional in the axial directionsince the lateral surfaces of both cylinders are well-

insulated. 3Thermal conductivities are constant.

Properties The thermal conductivity of aluminum bars

is given to be k= 176 W/mC. The contact

conductance at the interface of aluminum-aluminum

plates for the case of ground surfaces and of 20 atm 2

MPa pressure is hc= 11,400 W/m2

C (Table 3-2).

Analysis (a) The thermal resistance network in this

case consists of two conduction resistance and the

contact resistance, and are determined to be

C/W0447.0/4]m)(0.05C)[.W/m400,11(

1122

ccontact =

==

cAhR

RL

kAplate 2

m

(176 W/ m. C)[ (0.05 m) / 4]C / W= =

=

01504341

..

Then the rate of heat transfer is determined to be

W142.4=+

=

+

=

=

C/W)4341.020447.0(

C)20150(

2 barcontacttotal RR

T

R

TQ

Therefore, the rate of heat transfer through the bars is 142.4 W.

95C

108C600

W0.5

cm

Ri

Rglass

Ro

T1

T2

Bar Bar

Interface


8/15

(b) The temperature drop at the interface is determined to be

C6.4=== C/W)W)(0.04474.142(contactinterface RQT

6)

7(An electric hot water tank is made of two concentric cylindrical metal sheets with

foam insulation in between. The fraction of the hot water cost that is due to the heat

loss from the tank and the payback period of the do-it-yourself insulation kit are to be

determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with

time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the

center line and no variation in the axial direction. 3Thermal conductivities are

constant. 4The thermal resistances of the water tank and the outer thin sheet metal

shell are negligible. 5 Heat loss from the top and bottom surfaces is negligible.

Properties The thermal conductivities are given to be k= 0.03 W/mC for foam

insulation and k= 0.035 W/mC for fiber glass insulation


9/15

Analysis We consider only the side surfaces of the water heater for simplicity, and

disregard the top and bottom surfaces (it will make difference of about 10 percent).

The individual thermal resistances are

2m89.2m)2(m)46.0( === LDA oo

C/W029.0)m89.2(C).W/m12(

1122

=

==oo

oAh

R

C/W40.037.0029.0

C/W37.0)m2(C).W/m03.0(2

)20/23ln(

2

)/ln(2

12

=+=+=

=

==

foamototal

foam

RRR

kL

rrR

The rate of heat loss from the hot water tank is

(55 )

Q

T T

R

w

total=

=

=

2 27

70

C

0.40 C / W W

The amount and cost of heat loss per year are

Q Q t= = = ( . . 0 07 6132kW)(365 24 h / yr) kWh / yr

17.5%===

==

1752.0280$

056.49$

056.49$kWh)/08.0($kWh)2.613(=cost)itenergy)(UnofAmount(EnergyofCost

f

If 3 cm thick fiber glass insulation is used to wrap the entire tank, the individual

resistances becomes

2m267.3m)2(m)52.0( === LDA oo

C/W026.0)m267.3(C).W/m12(

11 o2o2===

oo

oAh

R

C/W676.0279.0371.0026.0

C/W279.0)m2(C).W/m035.0(2

)23/26ln(

2

)/ln(

C/W371.0)m2(C).W/m03.0(2

)20/23ln(

2

)/ln(

22

23

21

12

=++=++=

===

=

=

=

fiberglassfoamototal

fiberglass

foam

RRRR

Lk

rr

R

Lk

rrR

The rate of heat loss from the hot water heater in this case is

W42.41C/W0.676

C)2755(2=

=

=

total

w

R

TTQ

Tw

Ro

T2

Rfoam

Tw

Rfibergla

ss

Ro

T2

Rfoam


10/15

8) Both sides of a large stainless steel plate in which heat is generated uniformly are exposed to

convection with the environment. The location and values of the highest and the lowest

temperatures in the plate are to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2

Heat transfer is one-dimensional since the plate is large relative to its thickness, and there is

thermal symmetry about the center plane 3Thermal conductivity is constant. 4 Heat generation

is uniform.

PropertiesThe thermal conductivity is given to be k=15.1 W/mC.

Analysis The lowest

temperature will occur at

surfaces of plate while the

highest temperature will occur

at the midplane. Their values

are determined directly from

C155=

+=+=

C.W/m60

m)015.0)(W/m105(C30

2

35

h

LgTTs

C158.7=

+=+=

C)W/m.1.15(2

m)015.0)(W/m105(C155

2

2352

k

LgTT so

9) A nuclear fuel rod with a specified surface temperature is used as the fuel in a nuclear

reactor. The center temperature of the rod is to be determined.

Assumptions 1 Heat transfer is steady since there is no

indication of any change with time. 2 Heat transfer is one-

dimensional since there is thermal symmetry about the center

line and no change in the axial direction. 3Thermal

conductivity is constant. 4 Heat generation in the rod is

uniform.

PropertiesThe thermal conductivity is given to be k= 29.5

W/mC.

Analysis The center temperature of the rod is determined

from

C545.8=

+=+=

C)W/m.5.29(4

m)025.0)(W/m107(C175

4

2372

k

rgTT oso

10)

T

=30C

h=60

W/m2.C

2L=3 cm

k

gT

=30C

h=60

W/m2.C

g175

C

Uranium

rod


11/15

11)


12/15

12) The number of fins, finned and unfinned surface areas, and heat transfer rates from those

areas are

/n = =1

0 006 0 00627777

m

m) m)

2

( . ( .

W2107C)30100)(m86.0)(CW/m35()(

W700,15

C)30100)(m68.6)(C.W/m35(959.0

)(

m86.04

)0025.0(277771

4277771

m68.64

)0025.0()03.0)(0025.0(277774

27777

2o2unfinnedunfinned

22

finfinmaxfin,finfinned

222

unfinned

222fin

===

==

==

=

=

=

=

+=

+=

TThAQ

TThAQQ

DA

DDLA

b

b

Then the total heat transfer from the finned plate becomes

W17.8==+=+= W1078.12107700,15 4unfinnedfinnedfintotal, QQQ

The rate of heat transfer if there were no fin attached to the plate would be

A

Q hA T T b

no fin2

no fin no fin2 2

m m m

W / m C m C W

= =

= = =

( )( )

( ) ( . )( )( )

1 1 1

35 1 100 30 2450

Then the fin effectiveness becomes

7.27===2450

17800

finno

finfin

Q

Q

13) T T T T T g l

kleft top right bottom nodenode

+ + + + =4 0

2

where

C5.93CW/m214

)m05.0)(W/m108( 2362

02

node=

==

k

lg

k

lg

The finite difference equations for boundary nodes are obtained by applying an energy

balance on the volume elements and taking the direction of all heat transfers to be

towards the node under consideration:

04-200240290260:(interior)3Node

04-290325290350:(interior)2Node

02

)(325

2

290240

2:)convection(1Node

20

3

20

2

20

1111

=++++

=++++

=++

+

+

k

lgT

k

lgT

k

lgTThl

l

Tlk

l

Tkl

l

Tlk

where C20,W/m108C,.W/m50C,W/m.45 362 ==== Tghk

Substituting, T1 = 280.9C, T2 = 397.1C, T3 = 330.8C,

(b) The rate of heat loss from the bottom surface through a 1-m long section is


13/15

W1808=+++=

+++=

==

C20)/2]-(32520)-(280.920)-(24020)/2-m)[(2001mC)(0.05W/m50(

)325)(2/()()240()200)(2/(

)(

2

1

surface,element,

TlhTThlThlTlh

TThAQQ

m

mm

m

m

14) 4/)(04 bottomrighttopleftnode

2node

nodebottomrighttopleft TTTTTk

lgTTTTT +++==++++

There is symmetry about the horizontal, vertical, and

diagonal lines passing through the midpoint, and thus

we need to consider only 1/8th of the region. Then,

8642

9731

TTTT

TTTT

===

===

Therefore, there are there are only 3 unknown nodal

temperatures, 531 and,, TTT , and thus we need only

3 equations to determine them uniquely. Also, we can

replace the symmetry lines by insulation and utilize the

mirror-image concept when writing the finite difference

equations for the interior nodes.

225

152

21

4/4:(interior)3Node

4/)2200(:(interior)2Node


TTT

TTT

TT

==

++=

++=

Solving the equations above simultaneously gives

C190

C185

=====

====

86542

9731

TTTTTTTTT

15) 4/)(04 bottomrighttopleftnode

2node

nodebottomrighttopleft TTTTTk

lgTTTTT +++==++++

(a) There is symmetry about the insulated surfaces as well as about the diagonal line. Therefore

23 TT = , and 421 and,, TTT are the only 3 unknown nodal temperatures. Thus we need only 3

equations to determine them uniquely. Also, we can replace the symmetry lines by insulation

and utilize the mirror-image concept when writing the finite difference equations for the interior

nodes.


4/)2200(:(interior)2Node4/)180180(:(interior)1Node

324

142

321

TTT

TTTTTT

+=

++=

+++=

Also, 23 TT =


C185

C190

=

===

1

432

T

TTT

(b) There is symmetry about the insulated surface as well as the diagonal line. Replacing the

symmetry lines by insulation, and utilizing the mirror-image concept, the finite difference

equations for the interior nodes can be written as

18

0

20

0

180

150180200180

150

150180200180

150

18

0

20

0

18

0

12

3

456

78

9

Insulat

ed

18

0

20

0

150180

200

3

1 2

4

Insulat

ed


14/15





324

243

142

321

TTT

TTTT

TTT

TTT

++=

=++=

+++=

+++=


C128.6

C122.9

====

43

21

TT

TT

16) 042

nodenodebottomrighttopleft =++++

k

lgTTTTT

There is symmetry about a vertical line passing through the middle of the region, and thus weneed to consider only half of the region. Then,

4321 and TTTT ==

Therefore, there are there are only 2 unknown nodal temperatures, T1 and T3, and thus we need

only 2 equations to determine them uniquely. Also, we can replace the symmetry lines by

insulation and utilize the mirror-image concept when writing the finite difference equations for

the interior nodes.

04200150:(interior)3Node

04120100:(interior)1Node

2

341

2

132

=++++

=++++

k

lg

TTT

k

lgTTT

Noting that 4321 and TTTT == and substituting,

0CW/m180

m))(0.1W/m10(3350

0CW/m180

m))(0.1W/m10(3220

237

31

237

13

=

++

=

++

TT

TT

The solution of the above system is

C439.0

C411.5

==

==

43

21

TT

TT

100

120

140

120120

3

1 2

4

Insulat

ed

100

120

140

100100100

100

12

0

15

0

3

1 2

0.1

m

g

200200200

200

12

0

15

0

4


15/15

17) 042

0nodebottomrighttopleft =++++

k

lgTTTTT

We observe that all nodes are boundary nodes except node 5 that is an interior node. Therefore,

we will have to rely on energy balances to obtain the finite difference equations. Using energy

balances, the finite difference equations for each of the 8 nodes are obtained as follows:

Node 1: 0422

)(22

2

01412

1 =+

+

++ l

gl

TTlk

l

TTlkTT

lh

lqL

Node 2: 0222

)(2

0252321

2 =+

+

+

+l

gl

TTkl

l

TTlk

l

TTlkTThl

Node 3: 0422

)(2

03632

3 =+

+

+l

gl

TTlk

l

TTlkTThl

Node 4: 02

120

22

2

045441

=+

+

+

+l

gl

TTkl

l

Tlk

l

TTlklqL

Node 5: 04120

2

05624 =++++k

lgTTTT

Node 6: 04

3

2

120

2)(

2

06766563

6 =+

+

+

+

+l

gl

TTlk

l

Tkl

l

TTkl

l

TTlkTThl

Node 7: 02

120

22)(

2

077876

7 =+

+

+

+l

gl

Tkl

l

TTlk

l

TTlkTThl

Node 8: 04

120

22)(

2

2

0887

8 =+

+

+l

gl

Tlk

l

TTlkTT

lh

where ,W/m8000,W/m105 2360 == Lqg l = 0.015 m, k= 45 W/mC, h = 55 W/m2

C,

and T

=30

C. This system of 8 equations with 8 unknowns is the finite difference formulation ofthe problem.

(b) The 8 nodal temperatures under steady conditions are determined by solving the 8

equations above simultaneously with an equation solver to be

T1 =163.6C, T2 =160.5C, T3 =156.4C, T4 =154.0C, T5 =151.0C, T6

=144.4C,

T7 =134.5C, T8 =132.6C

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Documents

ejercicios transferencia de calor.doc