Solution manual heat and mass transfer a practical approach 3rd edition cengel CH07 3

7-86 The thickness of flat R-8 insulation in SI units is to be determined when the thermal conductivity of the material is known.. Analysis The thickness of flat R-8 insulation in m2.°C/

Special Topic: Thermal Insulation

7-78C Thermal insulation is a material that is used primarily to provide resistance to heat flow It differs

from other kinds of insulators in that the purpose of an electrical insulator is to halt the flow of electric current, and the purpose of a sound insulator is to slow down the propagation of sound waves

7-79C In cold surfaces such as chilled water lines, refrigerated trucks, and air conditioning ducts,

insulation saves energy since the source of “coldness” is refrigeration that requires energy input In this

case heat is transferred from the surroundings to the cold surfaces, and the refrigeration unit must now work harder and longer to make up for this heat gain and thus it must consume more electrical energy

7-80C The R-value of insulation is the thermal resistance of the insulating material per unit surface area

For flat insulation the R-value is obtained by simply dividing the thickness of the insulation by its thermal conductivity That is, R-value = L/k Doubling the thickness L doubles the R-value of flat insulation

7-81C The R-value of an insulation represents the thermal resistance of insulation per unit surface area (or

per unit length in the case of pipe insulation)

7-82C Superinsulations are obtained by using layers of highly reflective sheets separated by glass fibers in

an evacuated space Radiation between two surfaces is inversely proportional to the number of sheets used and thus heat loss by radiation will be very low by using this highly reflective sheets Evacuating the space between the layers forms a vacuum which minimize conduction or convection through the air space

7-83C Yes, hair or any other cover reduces heat loss from the head, and thus serves as insulation for the

head The insulating ability of hair or feathers is most visible in birds and hairy animals

7-84C The primary reasons for insulating are energy conservation, personnel protection and comfort,

maintaining process temperature, reducing temperature variation and fluctuations, condensation and corrosion prevention, fire protection, freezing protection, and reducing noise and vibration

7-85C The optimum thickness of insulation is the thickness that corresponds to a minimum combined cost

of insulation and heat lost The cost of insulation increases roughly linearly with thickness while the cost

of heat lost decreases exponentially The total cost, which is the sum of the two, decreases first, reaches a minimum, and then increases The thickness that corresponds to the minimum total cost is the optimum thickness of insulation, and this is the recommended thickness of insulation to be installed

7-86 The thickness of flat R-8 insulation in SI units is to be determined when the thermal conductivity of

the material is known

R-8

L

Assumptions Thermal properties are constant

Properties The thermal conductivity of the insulating material is given to

be k = 0.04 W/m⋅°C

Analysis The thickness of flat R-8 insulation (in m2.°C/W) is determined

from the definition of R-value to be

m 0.32

7-87E The thickness of flat R-20 insulation in English units is to be determined when the thermal

conductivity of the material is known

Assumptions Thermal properties are constant

R-20

L

Properties The thermal conductivity of the insulating material is given

to be k = 0.04 Btu/h⋅ft⋅°F

Analysis The thickness of flat R-20 insulation (in h⋅ft2⋅°F/Btu) is

determined from the definition of R-value to be

value= → L=Rvaluek=(20h.ft2.°F/Btu)(0.04Btu/h.ft.°F)=0.8 ft

k

L

R

7-88 A steam pipe is to be covered with enough insulation to reduce the exposed surface temperature to

30°C The thickness of insulation that needs to be installed 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 centerline and no variation in the axial

direction 3 Thermal properties are constant 4 The thermal contact resistance at the interface is negligible

Properties The thermal conductivities are given to be k = 52 W/m⋅°C for cast iron pipe and k = 0.038

W/m⋅°C for fiberglass insulation

Analysis The thermal resistance network for this problem involves 4 resistances in series The inner radius

of the pipe is r1 = 2.0 cm and the outer radius of the pipe and thus the inner radius of insulation is r2 = 2.3

cm Letting r3 represent the outer radius of insulation, the areas of the surfaces exposed to convection for a

L = 1 m long section of the pipe become

m)

in ( m2

=m)(12

2

m0.1257

=m)m)(102.0(2

2

3

2 3 3

3

3

2 1

1

r r

r L

W/m(80

11

2 2

R

i i

0.00043 C/W

)mC)(1 W/m

(522

)02.0/023.0ln(

2

)/ln(

1

1 2 pipe

R

)mC)(1 W/m

(0.0382

)023.0/ln(

2

)/ln(

3 3

2

2 3 insulation

r r R

R

ππ

2.138

1)mC)(2 W/m(22

11

3 2

3 2

A h R

R

o o

πNoting that all resistances are in series, the total resistance is

C/W )2.138/(

1)023.0/ln(

188.400043.009944

2 1

1)023.0/ln(

188.400043.009944.0[

C)22110(

3 3

R

T T

Q& i o

Noting that the outer surface temperature of insulation is specified to be 30°C, the rate of heat loss can also

be expressed as

3 3

C/W)1/(138.2

C)2230(

r r

R

T T

Setting the two relations above equal to each other and solving for r3 gives r3 = 0.0362 m Then the

minimum thickness of fiberglass insulation required is

7-89 EES Prob 7-88 is reconsidered The thickness of the insulation as a function of the maximum

temperature of the outer surface of insulation is to be plotted

Analysis The problem is solved using EES, and the solution is given below

7-90 A cylindrical oven is to be insulated to reduce heat losses The optimum thickness of insulation and

the amount of money saved per year are to be determined

Assumptions 1 Steady operating conditions exist 2 Heat transfer through the insulation is

one-dimensional 3 Thermal conductivities are constant 4 The thermal contact resistance at the interface is negligible 5 The surface temperature of the furnace and the heat transfer coefficient remain constant 6 The

surfaces of the cylindrical oven can be treated as plain surfaces since its diameter is greater than 1 m

Properties The thermal conductivity of insulation is given to be k = 0.038 W/m⋅°C

Analysis We treat the surfaces of this cylindrical furnace as plain surfaces since its diameter is greater than

1 m, and disregard the curvature effects The exposed surface area of the furnace is

2 2

2 side

=C)2790)(

mC)(70.69

W/m30()

=h A T T∞

Q& o o s

Noting that the plant operates 52×80 = 4160 h/yr, the

T∞

Rinsulation

Ts

Q=Q&Δt=(133.6kJ/s)(4160×3600s/yr)=2.001×109kJ/yr

The efficiency of the furnace is given to be 78 percent Therefore, to generate this much heat, the furnace must consume energy (in the form of natural gas) at a rate of

Qin =Q/ηoven =(2.001×109 kJ/yr)/0.78=2.565×109kJ/yr=24,314 therms/yr

since 1 therm = 105,500 kJ Then the annual fuel cost of this furnace before insulation becomes

AnnualCost=Qin×Unit cost=(24,314 therm/yr)($0.50/therm)=$12,157/yr

We expect the surface temperature of the furnace to increase, and the heat transfer coefficient to decrease somewhat when insulation is installed We assume these two effects to counteract each other Then the rate

of heat loss for 1-cm thick insulation becomes

W021,15C W/m30

1C

W/m

038.0

m01.0

C)2790)(

m69.70(1

)(

2 2

ins

ins conv ins total

−

=+

s

h k t

T T A R

R

T T R

T T

Q&

Also, the total amount of heat loss from the furnace per year and the amount and cost of energy

consumption of the furnace become

$1367/yr

=rm)($0.50/the therm/yr)

2734(costUnit Q

CostAnnual

therms2734

=kJ/yr102.884

=8kJ/yr)/0.710

249.2(/

kJ/yr102.249

=s/yr)3600kJ/s)(4160021

.15(

ins in,

8 8

oven ins ins in,

8 ins

=η

Q Q

t Q

Q &

Cost savings = Energy cost w/o insulation − Energy cost w/insulation = 12,157 − 1367 = $10,790/yr The unit cost of insulation is given to be $10/m2 per cm thickness, plus $30/m2 for labor Then the total cost of insulation becomes

InsulationCost=(Unit cost)(Surfacearea)=[($10/cm)(1cm)+$30/m2](70.69m2)=$2828

To determine the thickness of insulation whose cost is equal to annual energy savings, we repeat the calculations above for 2, 3, 15 cm thick insulations, and list the results in the table below

7-91 A cylindrical oven is to be insulated to reduce heat losses The optimum thickness of insulation and

the amount of money saved per year are to be determined

Assumptions 1 Steady operating conditions exist 2 Heat transfer through the insulation is

one-dimensional 3 Thermal conductivities are constant 4 The thermal contact resistance at the interface is negligible 5 The surface temperature of the furnace and the heat transfer coefficient remain constant 6 The

surfaces of the cylindrical oven can be treated as plain surfaces since its diameter is greater than 1 m

Properties The thermal conductivity of insulation is given to be k = 0.038 W/m⋅°C

Analysis We treat the surfaces of this cylindrical furnace as plain surfaces since its diameter is greater than

1 m, and disregard the curvature effects The exposed surface area of the furnace is

2 2

2 side

=C)2775)(

mC)(70.69

W/m30()

=h A T T∞

Q& o o s

Noting that the plant operates 52×80 = 4160 h/yr, the

annual heat lost from the furnace is

Q=Q&Δt=(101.794kJ/s)(4160×3600s/yr)=1.524×109kJ/yr

R o

T∞

Rinsulation

Ts

The efficiency of the furnace is given to be 78 percent

Therefore, to generate this much heat, the furnace must

consume energy (in the form of natural gas) at a rate of

Qin =Q/ηoven =(1.524×109kJ/yr)/0.78=1.954×109kJ/yr=18,526 therms/yr

since 1 therm = 105,500 kJ Then the annual fuel cost of this furnace before insulation becomes

AnnualCost=Qin×Unit cost=(18,526 therm/yr)($0.50/therm)=$9,263/yr

We expect the surface temperature of the furnace to increase, and the heat transfer coefficient to decrease somewhat when insulation is installed We assume these two effects to counteract each other Then the rate

of heat loss for 1-cm thick insulation becomes

C W/m30

1C

W/m

038.0

m01.0

C)2775)(

m69.70(1

)(

2 2

ins

ins conv ins total

−

=+

s

h k t

T T A R

R

T T R

T

T

Q&

Also, the total amount of heat loss from the furnace per year and the amount and cost of energy

consumption of the furnace become

$1041/yr

=rm)($0.50/the therm/yr)

2082(costUnit Cost

Annual

therms2082

=kJ/yr102.197

=8kJ/yr)/0.710

714.1(/

kJ/yr101.714

=s/yr)3600kJ/s)(4160445

.11(

ins in,

8 8

oven ins ins

in,

8 ins

InsulationCost=(Unit cost)(Surfacearea)=[($10/cm)(1cm)+$30/m2](70.69m2)=$2828

To determine the thickness of insulation whose cost is equal to annual energy savings, we repeat the calculations above for 2, 3, 15 cm thick insulations, and list the results in the table below

7-92E Steam is flowing through an insulated steel pipe, and it is proposed to add another 1-in thick layer

of fiberglass insulation on top of the existing one to reduce the heat losses further and to save energy and money It is to be determined if the new insulation will pay for itself within 2 years

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 centerline and no variation in the axial

direction 3 Thermal properties are constant 4 The heat transfer coefficients remain constant 5 The

thermal contact resistance at the interface is negligible

Properties The thermal conductivities are given to be k = 8.7 Btu/h ⋅ft⋅°F for steel pipe and k = 0.020

Btu/h⋅ft⋅°F for fiberglass insulation

Analysis The inner radius of the pipe is r1 = 1.75 in, the outer radius of the pipe is r2 = 2 in, and the outer

radii of the existing and proposed insulation layers are r3 = 3 in and 4 in, respectively Considering a unit

pipe length of L = 1 ft, the individual thermal resistances are determined to be

0364.0]ft)ft)(1(1.75/12F)[2

.Btu/h.ft(30

1)

2(

11

2 1

R

i i i

F/Btuh

00244.0)ftF)(1Btu/h.ft

(8.72

)75.1/2ln(

2

)/ln(

1

1 2 pipe

R

Current Case:

F/Btuh

227.3)ftF)(1Btu/h.ft

(0.0202

)2/3ln(

2

)/ln(

ins

2 3

F/Btuh

1273.0)]

ftft)(112/3(F)[2.Btu/h.ft(5

1)

2(

11

2 3

R

o o o

Then the steady rate of heat loss from the steam becomes

F/Btu.h )1273.0227.300244.00364.0(

F)85300(

ins pipe total

°+

++

°

−

=+++

−

=Δ

=

o i

o i

R R R R

T T R

T

Q&

Proposed Case:

F/Btuh

516.5)ftF)(1Btu/h.ft

(0.0202

)2/4ln(

2

)/ln(3 2

ins

F/Btuh

0955.0)]

ftft)(112/4(F)[2.Btu/h.ft(5

1)

2(

11

2 3

R

o o o

Then the steady rate of heat loss from the steam becomes

F/Btu.h )0955.0516.500244.00364.0(

F)85300(

ins pipe total

°+

++

°

−

=+++

−

=Δ

R R R R

T T R

T

Q&

7-93 The plumbing system of a plant involves some section of a plastic pipe exposed to the ambient air

The pipe is to be insulated with adequate fiber glass insulation to prevent freezing of water in the pipe The thickness of insulation that will protect the water from freezing under worst conditions is to be determined

Assumptions 1 Heat transfer is transient, but can be treated as steady at average conditions 2 Heat transfer

is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial

direction 3 Thermal properties are constant 4 The water in the pipe is stationary, and its initial temperature

is 15°C 5 The thermal contact resistance at the interface is negligible 6 The convection resistance inside

the pipe is negligible

Properties The thermal conductivities are given to be k = 0.16 W/m ⋅°C for plastic pipe and k = 0.035

W/m⋅°C for fiberglass insulation The density and specific heat of water are ρ = 1000 kg/m3 and cp = 4.18 kJ/kg.°C (Table A-15)

Analysis The inner radius of the pipe is r1 = 3.0 cm and the outer radius of the pipe and thus the inner

radius of insulation is r2 = 3.3 cm We let r3 represent the outer radius of insulation Considering a 1-m section of the pipe, the amount of heat that must be transferred from the water as it cools from 15 to 0°C is determined to be

kJ177.3

=C0)C)(15kJ/kg

kg)(4.18827

.2(

kg827.2m)]

(1m)(0.03)[

kg/m1000()(

total

2 3

2 1

°

−

°

=Δ

L r m

p

ππ

ρρV

Then the average rate of heat transfer

J300,177

(0.162

)03.0/033.0ln(

2

)/ln(

pipe

1 2 pipe

R

R

C/W )033.0/ln(

55.4)mC)(1 W/m

(0.0352

)033.0/ln(

2

)/ln(

3 3

2

2 3

r r

R

ππ

C/W5.188

1)mC)(2 W/m(30

11

3 2

3 2

A h R

R

o o

πThen the rate of average heat transfer from the water can be expressed as

m50.3 C/W]5.188/(

1)033.0/ln(

55.40948.0[

C)]

10(5.7[ W

0.821

3 3

total

°+

R

T T

Q& i ave o

Therefore, the minimum thickness of fiberglass needed to protect the pipe from freezing is

t = r3 - r2 = 3.50 − 0.033 = 3.467 m

which is too large Installing such a thick insulation is not practical, however, and thus other freeze

protection methods should be considered

7-94 The plumbing system of a plant involves some section of a plastic pipe exposed to the ambient air

The pipe is to be insulated with adequate fiber glass insulation to prevent freezing of water in the pipe The thickness of insulation that will protect the water from freezing more than 20% under worst conditions is to

be determined

Assumptions 1 Heat transfer is transient, but can be treated as steady at average conditions 2 Heat transfer

is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial

direction 3 Thermal properties are constant 4 The water in the pipe is stationary, and its initial temperature

is 15°C 5 The thermal contact resistance at the interface is negligible 6 The convection resistance inside

the pipe is negligible

Properties The thermal conductivities are given to be k = 0.16 W/m ⋅°C for plastic pipe and k = 0.035

W/m⋅°C for fiberglass insulation The density and specific heat of water are ρ = 1000 kg/m3 and Cp = 4.18 kJ/kg.°C (Table A-15) The latent heat of freezing of water is 333.7 kJ/kg

Analysis The inner radius of the pipe is r1 = 3.0 cm and the outer radius of the pipe and thus the inner

radius of insulation is r2 = 3.3 cm We let r3 represent the outer radius of insulation Considering a 1-m section of the pipe, the amount of heat that must be transferred from the water as it cools from 15 to 0°C is determined to be

kJ0.3667.1883.177

kJ7.188)kJ/kg7.333)(

kg827.2(2.02

0

kJ177.3

=C0)-C)(15kJ/kg

kg)(4.18827

.2(

kg827.2m)]

(1m)(0.03)[

kg/m1000()(

freezing cooling

total

freezing

total

2 3

2 1

=+

=+

Q

mh Q

T mc

Q

L r m

if p

ππ

ρρV

Then the average rate of heat

transfer during 60 h becomes

J000,366

(0.162

)03.0/033.0ln(

2

)/ln(

pipe

1 2 pipe

R

C/W )033.0/ln(

55.4)mC)(1 W/m

(0.0352

)033.0/ln(

2

)/ln(

3 3

2

2 3

r r R

ππ

C/W5

.188

1)mC)(2 W/m(30

11

3 2

3 2

A h R

R

o o

πThen the rate of average heat transfer from the water can be expressed as

m312.0 C/W]5.188/(

1)033.0/ln(

55.40948.0[

C)]

10(5.7[ W

1.694

3 3

total

°+

R

T T

Q& i avg o

Therefore, the minimum thickness of fiberglass needed to protect the pipe from freezing is

Properties Assuming a film temperature of Tf = 10°C for the

outdoors, the properties of air are evaluated to be (Table A-15)

1

C W/m

02439

0

2 5 -

Analysis Air flows along 8-m side The

Reynolds number in this case is

2

/sm10426.1

m)(8m/s )3600/100050

which is greater than the critical Reynolds number Thus we have

combined laminar and turbulent flow Using the proper relation for

Nusselt number, heat transfer coefficient is determined to be

C W/m

02439.0

096,10)7336.0(871)

10792.7(037.0Pr)871Re

037.0(

2

3 / 1 8

0 6 3

/ 1 8

0

k

h

k

L h

Nu

o

L o

The thermal resistances are

2

m32

=m)m)(84(

11

C/W1056.0m

32

C/W.m38.3)

38.3(

C/W0039.0)mC)(32 W/m8(

11

2 2

2 2

2 2

s

value insulation

s i i

A h

+

=++

=

∞

∞

C/W1105.0

C)622(

C/W1105.00010.01056.00039.0

2 1

total

o insulation i

total

R

T T Q

R R

R R

&

7-96 A car travels at a velocity of 60 km/h The rate of heat transfer from the bottom surface of the hot

automotive engine block is to be determined for two cases

Assumptions 1 Steady operating conditions exist 2 The critical Reynolds number is Recr = 5×105 3 Air is

an ideal gas with constant properties 4 The pressure of air is 1 atm 5 The flow is turbulent over the entire surface because of the constant agitation of the engine block 6 The bottom surface of the engine is a flat

surface

Properties The properties of air at 1 atm and the film

temperature of (T s + T∞)/2 = (75+5)/2 = 40°C are (Table A-15)

1

C W/m

02662

0

2 5 -

m)(0.7m/s )3600/100060

which is less than the critical Reynolds number But we

will assume turbulent flow because of the constant

agitation of the engine block

C W/m97.58)1551(m7.0

C W/m

02662.0

1551)

7255.0()10855.6(037.0PrRe037.0

2

3 / 1 8

0 5 3

/ 1 8 0

W/m97.58()

m6.0)(

92.0(

)(

4 4

4 2 8 -

4 4

=+

=+

= conv rad 1734 181

total Q Q

Q& & &

The gunk will introduce an additional resistance to heat dissipation from the engine The total heat transfer rate in this case can be calculated from

m)0.7m6.0)(

C W/m

3(

)m002.0(m)]

m)(0.7C)[(0.6 W/m97.58(

1

C5)-(751

°

=+

−

s s

s

kA

L hA

T T

Q&

The decrease in the heat transfer rate is

1734 − 1668 = 66 W (3.8%)

7-97E A minivan is traveling at 60 mph The rate of heat transfer to the van is to be determined

Assumptions 1 Steady operating conditions exist 2 The critical Reynolds number is Recr = 5×105 3 Radiation effects are negligible 4 Air flow is turbulent because of the intense vibrations involved 5 Air is

an ideal gas with constant properties 5 The pressure of air is 1 atm

Properties Assuming a film temperature of T f = 80°F, the

properties of air are evaluated to be (Table A-15E) Air V = 60 mph

T∞ = 90°F

L = 11 ft

Minivan 7290

0

Pr

/sft10697

1

FBtu/h.ft

01481

0

2 4 -

Analysis Air flows along 11 ft long side The

Reynolds number in this case is

/sft10697.1

ft)(11]ft/s )3600/528060[(

which is greater than the critical Reynolds number The air flow is assumed to be entirely turbulent because

of the intense vibrations involved Then the Nusselt number and the heat transfer coefficient are

determined to be

F.Btu/h.ft39.11)8461(ft

11

FBtu/h.ft

01481.0

8461)

7290.0()10704.5(037.0PrRe037.0

2

3 / 1 8

0 6 3

/ 1 8 0

k

h

k

L h

Nu

o

L o

0004.0)ftF)(240.8

Btu/h.ft39.11(

11

F/Btuh

0125.0)ft(240.8

F/Btu.h.ft3)

3(

F/Btuh

0035.0)ftF)(240.8

Btu/h.ft2.1(

11

2 2

2 2

2 2

s

value insulation

s i i

A h

R

Then the total thermal resistance and the heat transfer rate into the minivan are determined to be

Btu/h 1220

+

=++

=

∞

∞

F/Btuh

0164.0

F)7090(

F/Btuh

0164.00004.00125.00035.0

1 2

total

o insulation i

total

R

T T Q

R R

R R

&

7-98 Wind is blowing parallel to the walls of a house with windows The rate of heat loss through the

window is to be determined

Assumptions 1 Steady operating conditions exist 2 The critical Reynolds number is Recr = 5×105 3 Radiation effects are negligible 4 Air is an ideal gas with constant properties 5 The pressure of air is 1

atm

Properties Assuming a film temperature of 5°C, the

properties of air at 1 atm and this temperature are

evaluated to be (Table A-15)

1

C W/m

02401

0

2 5 -

Analysis Air flows along 1.8 m side The Reynolds

number in this case is

2

/sm10382.1

m)(1.8m/s )3600/100035(

C W/m

02401.0

1759)

7350.0(871)

10266.1(037.0Pr)871Re

037.0(

2

3 / 1 8

0 6 3

/ 1 8

0

=m)m)(1.58.1(

11

C/W0008.0)mC)(8.1 W/m

(0.78

m005.0

C/W0154.0)mC)(8.1 W/m8(

11

2 2

,

2

2 2

conv

s cond

s i i

conv

A h R

kA

L R

A h R

Then the total thermal resistance and the heat transfer rate through the 3 windows become

W 1116

+

=+

C)]

2(22[

C/W0215.00053.00008.00154.0

2 1

, ,

total

o conv cond i conv total

R

T T Q

R R R

R

&

7-99 A fan is blowing air over the entire body of a person The average temperature of the outer surface of

the person is to be determined

Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 The

pressure of air is 1 atm 4 The average human body can be treated as a 30-cm-diameter cylinder with an

exposed surface area of 1.7 m2

Properties We assume the film temperature to be 35°C The

properties of air at 1 atm and this temperature are (Table A-15)

0

Pr

/sm10655

1

C W/m

02625

0

2 5 -

/sm10655.1

m)m/s)(0.3(5

10063.917268

.0/4.01

)7268.0()10063.9(62.03.0

000,282

Re1

Pr/4.01

PrRe62.03.0

5 / 4 8 / 5 4 4

/ 1 3 / 2

3 / 1 5

0 4

5 / 4 8 / 5

4 / 1 3 / 2

3 / 1 5 0

+

×+

C W/m

02655

generated Q Q

Q& + & = &

Substituting values with proper units and then application of trial & error method or the use of an equation solver yields the average temperature of the outer surface of the person

C 36.2 K

×+

−

=

−+

−

∞

s

s s

s s s surr s

T

T T

T T hA T T A

)]

27332()[

7.1)(

02.18(])27340)[(

1067.5)(

7.1)(

9

0

(

90

)(

)(

W90

4 4 8

4 4

σε

7-100 The heat generated by four transistors mounted on a thin vertical plate is dissipated by air blown

over the plate on both surfaces The temperature of the aluminum plate is to be determined

Assumptions 1 Steady operating conditions exist 2 The critical Reynolds number is Recr = 5×105 3 Radiation effects are negligible 4 The entire plate is nearly isothermal 5 The exposed surface area of the transistor is taken to be equal to its base area 6 Air is an ideal gas with constant properties 7 The pressure

of air is 1 atm

Properties Assuming a film temperature

of 40°C, the properties of air are

evaluated to be (Table A-15)

1

C W/m

02662

0

2 5 -

m)(0.22m/s )60/250(

C W/m

02662.0

5.138)

7255.0()10386.5(664.0PrRe664.0

2

3 / 1 5

0 4 3

/ 1 5 0

=

°

×+

°

=+

C W/m75.16(

W)124(C

20)

s s

s s

hA

Q T T T

T hA

&

Discussion In reality, the heat transfer coefficient will be higher since the transistors will cause turbulence

in the air

7-101 A spherical tank used to store iced water is subjected to winds The rate of heat transfer to the iced

water and the amount of ice that melts during a 24-h period are to be determined

Assumptions 1 Steady operating conditions exist 2 Thermal resistance of the tank is negligible 3

Radiation effects are negligible 4 Air is an ideal gas with constant properties 5 The pressure of air is 1

atm

Properties The properties of air at 1 atm pressure and the free stream temperature of 30°C are (Table A-15)

7282.0Pr

kg/m.s10

729.1

kg/m.s10

872.1

/sm10608.1

C W/m

02588.0

5 C

m)(3.02m/s1000/3600)(25

10872.1)7282.0()10304.1(06.0)10304.1(4.02

PrRe06.0Re4.02

4 / 1

5

5 4

0 3

/ 2 6 5

0 6

4 / 1 4 0 3 / 2 5

0

×+

hD

Nu

μμ

m02.3

C W/m

02588

.7

=Δ

=Q t

Q &

Then the amount of ice that melts during this period becomes

kg 2014

kJ000,672

if if

h

Q m mh

Q

7-102 A spherical tank used to store iced water is subjected to winds The rate of heat transfer to the iced

water and the amount of ice that melts during a 24-h period are to be determined

Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 7 The

pressure of air is 1 atm

Properties The properties of air at 1 atm pressure and the free stream temperature of 30°C are (Table A-15)

kg/m.s10

872

1

/sm10608

1

C W/m

02588.0

5

2 5 -

Pr

kg/m.s10

729

m)(3.02m/s1000/3600)(25

10872.1)7282.0()10304.1(06.0)10304.1(4.02

PrRe06.0Re4.02

4 / 1

5

5 4

0 3

/ 2 6 5

0 6

4 / 1 4 0 3 / 2 5

0

×+

hD

Nu

μμ

m02.3

C W/m

02588

)( , 4 4,

, ,

rad + conv tank, from nk through ta

out s surr o out s surr o o sphere

in s out s

T T A T

T A h R

T T

Q

Q Q

Q

−+

15(4

m )50.151.1(4

5 2

R sphere

2 2

K)27325)[(

.K W/m1067.5)(

m65.28)(

75.0(

C))(30mC)(28.65

W/m05.9(C/W10

34

2

C0

4 ,

4 4

2 8 2

, 2

2 5

,

+

−+

×+

out s out

s

T

T T

Q&

7-103E A cylindrical transistor mounted on a circuit board is cooled by air flowing over it The maximum

power rating of the transistor is to be determined

Assumptions 1 Steady operating conditions exist 2 Radiation effects are negligible 3 Air is an ideal gas

with constant properties 4 The pressure of air is 1 atm

Properties The properties of air at 1 atm and the film

temperature of T f =(180+120)/2=150°F are (Table A-15)

2

FBtu/h.ft

01646

0

2 4 -

ft10099.2

ft)/12ft/s)(0.22(500/60

9.7271

7188.0/4.01

)7188.0()9.727(62.03

Re1

Pr/4.01

PrRe62.03

0

5 / 4 8 / 5

4 / 1 3 / 2

3 / 1 5

0

5 / 4 8 / 5 4

/ 1 3 / 2

3 / 1 5 0

Power transistor

D = 0.22 in

L = 0.25 in

and (13.72) 12.32Btu/h.ft F

ft)12/22.0(

FBtu/h.ft

01646

=

W (1

C)120180(ft)2ft)(0.25/1(0.22/12

F).Btu/h.ft32.12

(

)(

)(

2

W 0.26

= Btu/h 0.887

T T hA

Q& s s s

7-104 Wind is blowing over the roof of a house The rate of heat transfer through the roof and the cost of

this heat loss for 14-h period are to be determined

Assumptions 1 Steady operating conditions exist 2 The critical Reynolds number is Recr = 5×105 3 Air is

an ideal gas with constant properties 4 The pressure of air is 1 atm

Properties Assuming a film temperature of 10°C,

the properties of air are (Table A-15)

7336

0

Pr

/s m 10 426

1

C W/m

02439

0

2 5

-=

×

=

°

=

ν

k

Analysis The Reynolds number is

[ ] 7

2 5 2.338 10 /s m 10 426 1 m) (20 m/s ) 3600 / 1000 60 ( Re = × × × = = − ν VL L which is greater than the critical Reynolds number Thus we have combined laminar and turbulent flow Then the Nusselt number and the heat transfer coefficient are determined to be Q& Tsky = 100 K Air V = 60 km/h T∞= 10°C Tin = 20°C C W/m 0 31 ) 10 542 2 ( m 20 C W/m 02439 0 10 542 2 ) 7336 0 ]( 871 ) 10 338 2 ( 037 0 [ Pr ) 871 Re 037 0 ( 2 4 4 3 / 1 8 0 7 3 / 1 8 0 ° = × ° = = × = − × = − = = Nu L k h k hL Nu L In steady operation, heat transfer from the room to the roof (by convection and radiation) must be equal to the heat transfer from the roof to the surroundings (by convection and radiation), which must be equal to the heat transfer through the roof by conduction That is, rad + conv gs, surroundin to roof cond roof, rad + conv roof, to room Q Q Q Q&= & = & = & Taking the inner and outer surface temperatures of the roof to be T s,in and T s,out , respectively, the quantities above can be expressed as [ 4] , 4 4 2 8 2 , 2 2 4 , 4 , rad + conv roof, to room K) 273 ( K) 273 20 ( K W/m 10 67 5 )( m 300 )( 9 0 (

C ) )(20 m C)(300 W/m 5 ( ) (

) (

+

− +

× +

°

−

°

=

− +

−

=

−

in s

in s in

s room s in s room s i

T

T T

T A T

T A h

Q& ε σ

m 15 0 ) m 300 )(

C W/m

2

, , cond

roof,

out s in s out

s in s s

T T L

T T

kA

°

=

−

=

&

, 4 2 8 2

, 2 2

4 4 , ,

rad

+

conv

surr,

to

roof

K) 100 ( K) 273 (

.K W/m 10 67 5 )(

m 300 )(

9 0 (

C ) 10 )(

m C)(300 W/m 0 31 ( ) (

) (

− +

× +

°

−

°

=

− +

−

=

−

out s

out s surr

out s s surr out s s o

T

T T

T A T

T A h

Q& ε σ

Solving the equations above simultaneously gives

Q&=28,025 W=28.03 kW, T s,in =10.6°C,andT s,out =3.5°C

7-105 Steam is flowing in a stainless steel pipe while air is flowing across the pipe The rate of heat loss

from the steam per unit length of the pipe is to be determined

Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 The

pressure of air is 1 atm

Properties Assuming a film temperature of 10°C,

the properties of air are (Table A-15)

1

C W/m

02439

0

2 5 -

/sm10426.1

m)m/s)(0.116(4

The Nusselt number for flow across a cylinder is determined from

000,282

10254.317336

.0/4.01

)7336.0()10254.3(62.03

Re1

Pr/4.01

PrRe62.03

0

5 / 4 8 / 5 4 4

/ 1 3 / 2

3 / 1 5

0 4

5 / 4 8 / 5 4

/ 1 3 / 2

3 / 1 5 0

+

×+

Insulation

ε = 0.3

m116.0

C W/m0.02439 ⋅° = 2⋅°

Area of the outer surface of the pipe per m length of the pipe is

2

m3644.0)m1)(

m116.0

gs surroundin to surface insulation

and

Q

Q&= & = &

Using the thermal resistance network, heat transfer from the steam to the outer surface is expressed as

C/W874.3)m1)(

C W/m

038.0(2

)3.2/8.5ln(

2

)/ln(

C/W0015.0)m1)(

C W/m

15(2

)2/3.2ln(

2

)/ln(

C/W0995.0)m1(m)04.0()C W/m80(

11

2 3

1 2

2 ,

ππ

π

kL

r r R

kL

r r R

A h R

insulation

pipe

i i i conv

and

C/W )874.30015.00995.0(

C)250(

,

1 ins

+

−

insulation pipe

i conv

R R R

T T Q&

7-106 A spherical tank filled with liquid nitrogen is exposed to winds The rate of evaporation of the liquid

nitrogen due to heat transfer from the air is to be determined for three cases

Assumptions 1 Steady operating conditions exist 2 Radiation effects are negligible 3 Air is an ideal gas

with constant properties 4 The pressure of air is 1 atm

Properties The properties of air at 1 atm pressure and the free stream temperature of 20°C are (Table A-15)

7309.0

Pr

EES)(from kg/m.s10

023.5

kg/m.s10

825.1

/sm10516.1

C W/m

02514.0

6 C

Analysis (a) When there is no insulation,

D = Di = 4 m, and the Reynolds number is

2

/sm10516.1

m)(4m/s1000/3600)(40

10825.1)7309.0()10932.2(06.0)10932.2(4.02

PrRe06.0Re4.02

4 / 1

6

5 4

0 3

/ 2 6 5

0 6

4 / 1 4 0 3 / 2 5

0

×+

hD

Nu

μμ

Nitrogen tank -196°C

Do

D i

Wind 20°C

40 km/h

Insulation

m4

C W/m

02514

W/m66.14())(

()

kJ/s2.159

if if

h

Q m h

(b) Note that after insulation the outer surface temperature and diameter will change Therefore we need to

evaluate dynamic viscosity at a new surface temperature which we will assume to be -100°C At -100°C,

Noting that D = D

kg/m.s10

m)(4.1m/s1000/3600)(40

W7361)m81.52)(

C W/m71.11(

1m)

m)(2C)(2.05 W/m

(0.0354

m )205.2(

C)]

196(20[

14

m81.52)m1.4(

2 2

2 1

1 2

tan , tan

,

2 2

s

k s conv

insulation

k s s

hA r kr

r r

T T R

R

T T Q

kJ/s361.7

if if

h

Q m h

m)(4.04m/s1000/3600)(40

10825.1)7309.0()10961.2(06.0)10961.2(4.02

PrRe06.0Re4.02

4 / 1

5

5 4

0 3

/ 2 6 5

0 6

4 / 1 4 0 3 / 2 5

0

×+

hD

Nu

μμ

m04.4

C W/m

02514

C W/m73.10(

1m)

m)(2C)(2.02 W/m

(0.000054

m )202.2(

C)]

196(20[

14

m28.51)m04.4(

2 2

2 1

1 2

tan , tan

,

2 2

s

k s conv

insulation

k s s

hA r kr

r r

T T R

R

T T Q

kJ/s0274.0

if if

h

Q m h