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Solution manual heat and mass transfer a practical approach 3rd edition cengel CH09 3

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0 =+ Analysis a When the stack is exposed to 10 m/s winds, the heat transfer will be by forced convection... The rates of heat transfer from the vessel by natural convection, conduction,

Trang 1

Review Problems

9-99 A cold cylinder is placed horizontally in hot air The rates of heat transfer from the stack with and

without wind cases are to be determined

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

atmospheric pressure is 1 atm

Properties The properties of air at 1 atm and the film temperature of (Ts+T∞)/2 = (40+10)/2 = 25°C are (Table A-15)

2 5

K003356.0K)27325(

11

1

C W/m

02551

0

=+

Analysis (a) When the stack is exposed to 10 m/s winds, the heat transfer will be by forced convection We

have flow of air over a cylinder and the heat transfer rate is determined as follows:

020,64/sm10562.1

m)m/s)(0.110

(Re

7296.0()020,64(027.0PrRe027

0

C W/m87.45)8.179(m1.0

C W/m

02551.0

2 5

3 -1

2 2

3

10953.2)7296.0()

/sm10562.1(

)m1.0)(

K1040)(

K003356.0)(

m/s81.9(Pr)(

)10953.2(387.06.0Pr

/559.01

387.06

.0

2

27 / 8 16 / 9

6 / 1 6 2

27 / 8 16 / 9

6 / 1

Nu

m1.0

C W/m

02551

Q&

Trang 2

9-100 A spherical vessel is completely submerged in a large water-filled tank The rates of heat transfer

from the vessel by natural convection, conduction, and forced convection are to be determined

Assumptions 1 Steady operating conditions exist 2 The surface temperature is constant

Properties The properties of water at the film temperature of (Ts+T∞)/2 = (30+20)/2 = 25°C are (Table A-9)

1 - 3

2 7 3 3

K10247

skg/m10891

0

C W/m

607

0

kg/m997

Analysis (a) Heat transfer in this case will be by natural convection

The characteristic length in this case is L c = D = 0.3 m Then,

9 2

2 7

3 -1

3 2

2

3

10029.5)14.6()

/sm10937.8(

)m3.0)(

K2030)(

K10247.0)(

m/s81.9(Pr)(

)10029.5(589.02Pr

/469.01

589.0

16 / 9

4 / 1 9 9

/ 4 16 / 9

4 / 1

=+

×+

=+

2

2

m2827.0m)3.0(

C W/m0.293)8.144(m3.0

C W/m

607.0

(b) When buoyancy force is neglected, there will be no convection currents (since β = 0) and the heat

transfer will be by conduction Then Rayleigh number becomes zero (Ra = 0) The Nusselt number in this case is

Then

C W/m047.4)2(m3.0

C W/m

607

(c) In this case, the heat transfer from the vessel is by forced convection The properties of water at the free

stream temperature of 20°C are (Table A-9)

01.7Pr

kg/m.s10

798.0

/sm10004.1/

kg/m.s10

002.1

C W/m

0.598k

kg/m998

3 C

30

@

,

2 6 - 3 3

ρ

Trang 3

The Reynolds number is

760,59/sm10004.1

m)m/s)(0.3(0.2

10002.1)01.7()760,59(06.0)760,59(4.02

PrRe06.0Re4.02

4 / 1

3

3 4

0 3 / 2 5

0

4 / 1 4 0 3 / 2 5

0

hD

Nu

μμ

The heat transfer coefficient is

C W/m3.875)1.439(m3.0

C W/m

598

forced hA T T

Trang 4

9-101 A vertical cylindrical vessel looses heat to the surrounding air The rates of heat transfer from the

vessel with and without wind cases are to be determined

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

atmospheric pressure is 1 atm

Properties The properties of air at 1 atm and the film

temperature of (Ts+T∞)/2 = (60+0)/2 = 30°C are (Table A-15)

2 5

K003300.0K)27330(

11

1

C W/m

02588

0

=+

Analysis (a) When there is no wind heat transfer is by natural

convection The characteristic length in this case is the height

of the vessel, L c = L=3m.Then,

11 2

2 5

3 -1

2 2

3

10477.1)7282.0()

/sm10608.1(

)m3)(

K060)(

K003300.0)(

m/s81.9(Pr)(

/ 1 11

4

/

1

35 thusand 1.0

<

156.0)7282.0/10477.1(

)3(3535

Gr

L D Gr

.528)

10477.1(1.01

C W/m

02588

(b) When the vessel is exposed to 20 km/h winds, the heat transfer will be by forced convection We have

flow of air over a cylinder and the heat transfer rate is determined as follows:

5 2

/sm10608.1

m)m/s)(1.03600

/10020(

7282 0 ( ) 10 455 3 ( 027 0 Pr Re 027

0

Nu = 0 805 1 / 3 = × 5 0 805 1 / 3 = (from Table 7-1)

C W/m07.18)1.698(m0.1

C W/m

02588.0

Trang 5

9-102 A solid sphere is completely submerged in a large pool of oil The rates of heat transfer from the

sphere by natural convection, conduction, and forced convection are to be determined

Assumptions 1 Steady operating conditions exist 2 The surface temperature is constant

Properties The properties of oil are given in problem statement

Analysis (a) For conduction heat transfer, β = 0

and Ra = 0 Then the Nusselt number is

Nu = 2

Then

m5.0

C W/m

22

W/m22.0

C)J/kgs)(1880kg/m

010.0(

kg/m010.0

)kg/mm)(888m/s)(0.5(1.5

010.0)45.85()600,66(06.0)600,66(4.02

PrRe06.0Re4.02

4 / 1 4

0 3

/ 2 5

0

4 / 1 4 0 3 / 2 5

0

+

=

s k

hD

Nu

μμ

The heat transfer coefficient is

C W/m4.662)1506(m5.0

C W/m

22

(c) Heat transfer in this case will be by natural convection Note that the fluid properties in this case are to

be evaluated at the film temperature of 40ºC The characteristic length in this case is L c = D = 0.5 m Then,

10 2

2

3 -1

2 2

3

10522.3)5.65()

/sm876/007.0(

)m5.0)(

K2060)(

K00070.0)(

m/s81.9(Pr)

C)J/kgs)(1965kg/m

007.0(

)10522.3(589.02Pr

/469.01

589.0

16 / 9

4 / 1 10 9

/ 4 16 / 9

4 / 1

=+

×+

=+

+

Nu

C W/m2.105)4.250(m5.0

C W/m

21

Discussion The heat transfer from the sphere by forced convection is much greater than that by natural convection and the heat transfer by natural convection is much greater than that by conduction

Trang 6

9-103E A small cylindrical resistor mounted on the lower part of a vertical circuit board The approximate

surface temperature of the resistor is to be determined

Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 The local atmospheric pressure is 1 atm 4 Radiation effects are negligible 5 Heat transfer through the connecting

wires is negligible

Properties The properties of air at 1 atm and the

anticipated film temperature of (Ts+T∞)/2 =

(220+120)/2 = 170°F are (Table A-15E)

Q&

Resistor 0.1 W

D = 0.2 in

Air

T∞ = 120°F

1 -

2 3

R001587.0R)460170(

11

0

FBtu/h.ft

01692

0

=+

properties and h We will check the accuracy of this guess later and repeat the calculations if necessary

The characteristic length in this case is the diameter of resistor, L c = D=0.2in Then,

8.343)7161.0()

/sft10222.0(

)ft12/2.0)(

R120220)(

R001587.0)(

ft/s2.32(Pr)(

2 2 3

3 -1

2 2

)8.343(387.06

.0Pr

/559.01

387.06

.0

2

27 / 8 16 / 9

6 / 1 2

27 / 8 16 / 9

6 / 1

Nu

2 2

2

2

ft00175.04/ft)12/2.0(2)ft12/3.0)(

ft12/2.0(4/2

F.Btu/h.ft138.2)105.2(ft12/2.0

FBtu/h.ft

01692.0

=+

=+

π

A

Nu D

°

=+

F.Btu/h.ft138.2(

Btu/h )412.31.0(F

120)

s s

s

Q T T T

T hA

&

which is sufficiently close to the assumed temperature for the evaluation of properties Therefore, there is

no need to repeat calculations

Trang 7

9-104 An ice chest filled with ice at 0°C is exposed to ambient air The time it will take for the ice in the

chest to melt completely is to be determined for natural and forced convection cases

Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 Heat transfer from the base of the ice chest is disregarded 4 Radiation effects are negligible 5 Heat transfer coefficient is the same for all surfaces considered 6 The local atmospheric pressure is 1 atm

Properties The properties of air at 1 atm and the

anticipated film temperature of (Ts+T∞)/2 =

(15+20)/2 = 17.5°C are (Table A-15)

Ice chest, 0°C

3 cm

Q&

Air, T∞= 20°C

30 cm1

-2 5

K003442.0K)2735.17(

11

1

C W/m

02495

0

=+

properties and h We will check the accuracy of this guess later and repeat the calculations if necessary The characteristic length for the side surfaces is the height of the chest, L c = L = 0.3 m Then,

7 2

2 5

3 -1

2 2

3

10495.1)7316.0()

/sm10493.1(

)m3.0)(

K1520)(

K003442.0)(

m/s81.9(Pr)(

.0

492.01

)10495.1(387.0825.0Pr

492.01

Ra387.0825

.0

2

27 / 8 16 / 9

6 / 1 7 2

27 / 8 16 / 9

6 / 1

×+

+

=

Nu

C W/m923.2)15.35(m3.0

C W/m

02495

2m64.0)m4.0)(

m4.0()m4.0)(

m3.0(

C W/m923.2(

1)

m64.0)(

C W/m

033.0(

m03.0

C)020(1

2 2

2

, ,

=+

s s

i s o

conv

wall

i s

hA kA L

T T R

(2.923

W23.10C

20

)

s s

s

Q T T T T hA

&

which is almost identical to the assumed value of 15°C used in the evaluation of properties and h

Therefore, there is no need to repeat the calculations

Trang 8

The rate at which the ice will melt is

kg/s10066.3kJ/kg7.333

kJ/s1023

⎯→

⎯Δ

kg/s10066.3

(T

C

s+T∞)/2 = (19+20)/2 = 19.5°C are (Table A-15)

1 -

2 5

K00342.0K)2735.19(

11

1

C W/m

0251

0

=+

)m4.0(m/s)3600/100050(

7.362)

7310.0()700,367(664.0PrRe664

C W/m

0251

C W/m76.22(

1)

m64.0)(

C W/m

033.0(

m03.0

C)020(1

2 2

2

, ,

=+

s s

i s o

conv

wall

i s

hA kA L

T T R

(22.76

W43.13C

20

)

s s

s

Q T T T T hA

&

which is almost identical to the assumed value of 19°C used in the evaluation of properties and h

Therefore, there is no need to repeat the calculations Then the rate at which the ice will melt becomes

kg/s10025.4kJ/kg7.333

kJ/s1043

⎯→

⎯Δ

10025.4

5

m

m t t

m

m

&

&

Trang 9

9-105 An electronic box is cooled internally by a fan blowing air into the enclosure The fraction of the

heat lost from the outer surfaces of the electronic box is to be determined

Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 Heat transfer from the base surface is disregarded 4 The pressure of air inside the enclosure is 1 atm

Properties The properties of air at 1 atm and the film temperature of

(Ts+T∞)/2 = (32+15)/2 = 28.5°C are (Table A-15)

1 -

2 5

K003317.0K)2735.28(

11

1

C W/m

02577

0

=+

Analysis Heat loss from the horizontal top surface:

The characteristic length in this case is 0.125m

)]

m5.0()m5.0[(

2

)m5.0

2 5

3 -1

2 2

3

10275.1)7286.0()

/sm10594.1(

)m125.0)(

K2532)(

K003317.0)(

m/s81.9(Pr)(

2

m25.0)m5.0(

C.W/m741.3)15.18(m125.0

C W/m

02577.0

Nu L

k h

and Q&top =hA top(T sT∞)=(3.741 W/m2.°C)(0.25m2)(32−25)°C=6.55 W

Heat loss from vertical side surfaces:

The characteristic length in this case is the height of the box L c = L =0.15 m Then,

6 2

2 5

3 -1

2 2

3

10204.2)7286.0()

/sm10594.1(

)m15.0)(

K2532)(

K003317.0)(

m/s81.9(Pr)

.0

492.01

)10204.2(387.0825.0Pr

492.01

Ra387.0825

.0

2

27 / 8 16 / 9

6 / 1 7 2

27 / 8 16 / 9

6 / 1

×+

m15.0(4

C.W/m530.3)55.20(m15.0

C W/m

02577.0

k h

and Q&side =hA side(T sT∞)=(3.530 W/m2.°C)(0.3m2)(32−25)°C=7.41 W

The radiation heat loss is

W95.17])K27325()K27332)[(

.K W/m1067.5)(

m3.025.0)(

75.0(

)(

4 4

4 2 8 2

4 4

=+

−+

×+

+

W200

W)95.1741.755

6

(

f

Trang 10

9-106 A spherical tank made of stainless steel is used to store iced water 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 3 Thermal resistance of the tank is negligible 4 The local atmospheric pressure is 1 atm

Properties The properties of air at 1 atm and the film temperature of

(Ts+T∞)/2 = (0+20)/2 = 10°C are (Table A-15)

1.5 cmIced water

Di = 6 m 0°C

Q&

T s = 0°C

T∞ = 20°C

1 -

2 5

K003534.0K)27310(

11

1

C W/m

02439

0

=+

2 5

3 -1

2 2

3

10485.5)7336.0()

/sm10426.1(

)m03.6)(

K020)(

K003534.0)(

m/s81.9(Pr)

)10485.5(589.02Pr

/469.01

589.0

16 / 9

4 / 1 11 9

/ 4 16 / 9

4 / 1

=+

×+

=+

+

Nu

2 2

2

2

m2.114m)03.6(

C W/m596.1)5.394(m03.6

C W/m

02439.0

A

Nu D

k

h

and

W3646C)020)(

m2.114)(

C W/m596.1()

=+

=

=+

−+

W759,11])K2730()K27320)[(

.K W/m1067.5)(

m2.114)(

1(

)(

4 4

4 2 8 2

4 4

total

surr s s rad

Q

T T A Q

&

& ε σ

(b) The total amount of heat transfer during a 24-hour period is

kJ/day10

331.1)s/h3600h/day24)(

kJ/s4.15

=Q t

Q &

Then the amount of ice that melts during this period becomes

kg 3988

kJ10331

if

Q m mh

Trang 11

9-107 A double-pane window consisting of two layers of glass separated by an air space is considered The

rate of heat transfer through the window and the temperature of its inner surface are to be determined

Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 Radiation effects are negligible 4 The pressure of air inside the enclosure is 1 atm

Properties We expect the average temperature of the air gap to be roughly the average of the indoor and

outdoor temperatures, and evaluate The properties of air at 1 atm and the average temperature of

(T∞1+T∞2)/2 = (20 +0)/2 = 10°C are (Table A-15)

1 -

2 5

K003534.0K)27310(

11

1

C W/m

02439

0

=+

Analysis We “guess” the temperature difference across the air gap to be

15°C = 15 K for use in the Ra relation The characteristic length in this

case is the air gap thickness, L c = L = 0.03 m Then,

4 2

2 5

3 -1

2 2

3 2

)/sm10426.1(

)m03.0)(

K15)(

K003534.0)(

m/s81.9(Pr)

03.0

m2.1)7336.0()10065.5(42.0Pr

42

0

3 0 012

0 4

/ 1 4 3

0 012 0 4 /

Nu

C W/m688.1)076.2(m03.0

C W/m

02439

=+

++

=

++

+

=+

++

)4.2)(

25(

1)4.2)(

688.1(

1)

4.2)(

78.0(

)003.0(2)4.2)(

10

(

1

020

11

2

, ,

, ,

,

, ,

s o s air s glass s i

i s o

conv air conv glasses cond i conv

o i

A h A h A k

t A h

T T R

R R

R

T T Q&

Check: The temperature drop across the air gap is determined from

C0.16)mC)(2.4 W/m688.1(

W65

→Δ

=

s

Q T T hA

&

which is very close to the assumed value of 15°C used in the evaluation of the Ra number

Trang 12

9-108 An electric resistance space heater filled with oil is placed against a wall The power rating of the

heater and the time it will take for the heater to reach steady operation when it is first turned on are to be determined

Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 Heat transfer from the back, bottom, and top surfaces are disregarded 4 The local atmospheric pressure is 1 atm

Properties The properties of air at 1 atm and the film temperature of

-2 5

K003096.0K)27350(

11

1

C W/m

02735

0

=+

Analysis Heat transfer from the top and bottom surfaces are

said to be negligible, and thus the heat transfer area in this case

consists of the three exposed side surfaces The characteristic

length is the height of the box, L c = L = 0.5 m Then,

8 2

2 5

3 -1

2 2

3

10244.4)7228.0()

/sm10798.1(

)m5.0)(

K2575)(

K 003096.0)(

m/s81.9(Pr)(

.0

492.01

)10244.4(387.0825.0Pr

492.01

Ra387.0825

.0

2

27 / 8 16 / 9

6 / 1 8 2

27 / 8 16 / 9

6 / 1

×+

m15.0(2)m8.0)(

m5.0

(

C.W/m179.5)68.94(m5.0

C W/m

02735.0

=+

.K W/m1067.5)(

m55.0)(

8.0(

)(

4 4

4 2 8 2

4 4

=+

−+

Q& ε σ

Then the total rate of heat transfer, thus the power rating of the heater becomes

W 311.5

=+

J10514.4C)2575)(

CJ/kg

2006)(

kg45()

kJ10514

Q

Q t t

Trang 13

9-109 A horizontal skylight made of a single layer of glass on the roof of a house is considered The rate of

heat loss through the skylight is to be determined for two cases

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

atmospheric pressure is 1 atm

Properties The properties of air at 1 atm and the anticipated film

temperature of (Ts+T∞)/2 = (-4-10)/2 = -7°C are (Table A-15)

ε = 0.9

t = 0.5 cm

1 -

2 5

K003759.0K)2737(

11

1

C W/m

02311

0

=+

Analysis We assume radiation heat transfer inside the house to be negligible We start the calculations by

“guessing” the glass temperature to be -4°C for the evaluation of the properties and h We will check the accuracy of this guess later and repeat the calculations if necessary The characteristic length in this case is

m)2.5+m2(1

m)m)(2.51

3 -1

2 2

3 2

)/sm10278.1(

)m357.0)(

K )10(4)[

K003759.0)(

m/s81.9(Pr)(

m1

(

C.W/m467.3)56.53(m357.0

C W/m

02311.0

A

Nu L

3

K])27330()2734)][(

27330()2734)[(

.K W/m1067.5(9

0

))(

(

2

3 2 2

4 2 8

2 2

=

+

−++

−+

−++

×

=

++

=

sky s sky s

Then the combined convection and radiation heat transfer coefficient outside becomes

2 ,combined = o + rad =3.467+3.433=6.90W/m

h

Again we take the glass temperature to be -4°C for the evaluation of the properties and h for the inner

surface of the skylight The properties of air at 1 atm and the film temperature of T f = (-4+20)/2 = 8°C are (Table A-15)

1 -

2 5

K003559.0K)2738(

11

1

C W/m

02424

0

=+

2 2

3 2

)/sm10408.1(

)m357.0)(

K )4(20)[

K003559.0)(

m/s81.9(Pr)(

Nu

Trang 14

C W/m

02424

1C

W/m

78.0

m005.0C W/m998.11

C)]

10(20)[

m5.2(1

1

)(

2 2

2

glass

, ,

,

, ,

=

++

h k

t h

T T A

R R

R

T T Q

glass i

out room s

o combined glas

cond i conv

o i s skylight

&

Using the same heat transfer coefficients for simplicity, the rate of heat loss through the roof in the case of R-5.34 construction is determined to be

W5.12C W/m90.6

1C/W

.m34.5C W/m998.11

C)]

10(20)[

m5.2(1

1

)(

2

2 2

2 ,

,

, ,

=

°+

°+

=

++

h

R h

T T A

R R R

T T Q

glass i

out room s

o combined cond

i conv

o i s roof

&

Therefore, a house loses 115/12.5 ≅ 9 times more heat through the skylights than it does through an insulated wall of the same size

Using Newton’s law of cooling, the glass temperature corresponding to a heat transfer rate of 115

W is calculated to be –3.3°C, which is sufficiently close to the assumed value of -4°C Therefore, there is

no need to repeat the calculations

Trang 15

9-110 A solar collector consists of a horizontal copper tube enclosed in a concentric thin glass tube Water

is heated in the tube, and the annular space between the copper and glass tube is filled with air The rate of heat loss from the collector by natural convection is to be determined

Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 Radiation effects are negligible 3 The pressure of air in the enclosure is 1 atm

Properties The properties of air at 1 atm and the average temperature

of (Ti+To)/2 = (60+32)/2 = 46°C are (Table A-15)

Do =9 cm

Di =5 cm, Ti = 60°C Air space

Glass cover

To = 32°C

1 -

2 5

K003135.0K)27346(

11

1

C W/m

02706

0

=+

Analysis The characteristic length in this case is the

distance between the two cylinders

cm22

cm )59(

c

D D

L

and,

100,16)7238.0()

/sm10760.1(

)m02.0)(

K3260)(

K003135.0)(

m/s81.9(Pr)(

2 2 5

3 -1

2 2

m05.0

m09.0ln)

(

ln

5 7/5 - 7/5

3

-4

5 5 / 3 5 / 3 3

c

i o

D D

L

D

D F

[(0.1303)(16,100)] 0.05811 W/m.C7238

.0861.0

7238.0)C W/m

02706.0(386.0

)Ra(Pr861.0

Pr386

.0

4 / 1 4

/ 1

4 / 1 cyl

4 / 1 eff

k

Then the heat loss from the collector per meter length of the tube becomes

W 17.4

m09.0ln

)C W/m

05811.0(2)(ln

o i

i o

T T D D k Q&

Trang 16

9-111 A solar collector consists of a horizontal tube enclosed in a concentric thin glass tube is considered

The pump circulating the water fails The temperature of the aluminum tube when equilibrium is

established is to be determined

Assumptions 1 Steady operating conditions exist 2 Air is an ideal

gas with constant properties 3 The local atmospheric pressure is 1

atm

Properties The properties of air at 1 atm and the

anticipated film temperature of (Ts+T∞)/2 = (33+30)/2 =

31.5°C are (Table A-15)

2 5

K003284.0K)2735.31(

11

1

C W/m

02599

0

=+

Analysis This problem involves heat transfer from the aluminum tube to the glass cover, and from the outer

surface of the glass cover to the surrounding ambient air When steady operation is reached, these two heat transfers will be equal to the rate of heat gain That is,

length)meter (per

W 20

Q& & &

Now we assume the surface temperature of the glass cover to be 33°C We will check this assumption later

on, and repeat calculations with a better assumption, if necessary

The characteristic length for the outer diameter of the glass cover L c = D o =0.07 m Then,

700,91)7278.0()

/sm10622.1(

)m07.0)(

K3033)(

K003284.0)(

m/s81.9(Pr)(

2 2 5

3 -1

2 2

)700,91(387.06

.0Pr

/559.01

387.06

.0

2

27 / 8 16 / 9

6 / 1 2

27 / 8 16 / 9

6 / 1

Nu

2m2199.0m)1)(

m07.0

C W/m

02599

m2199.0)(

C W/m832.2()

s s

2 2

)K27320()K273(

.K W/m1067.5)(

m2199.0)(

1(

C)30)(

m2199.0)(

C W/m832.2(

×+

rad conv total

T T

Q Q

Q& & &

Its solution is T glass =33.34°C, which is sufficiently close to the assumed value of 33°C

Trang 17

Now we will calculate heat transfer through the air layer between aluminum tube and glass cover We will assume the aluminum tube temperature to be 45°C and evaluate properties at the average temperature of

(Ti+To)/2 = (45+33.34)/2 = 39.17°C are (Table A-15)

1 -

2 5

K003203.0K)27317.39(

11

1

C W/m

02656

0

=+

/sm10694.1(

)m01.0)(

K34.3345)(

K003203.0)(

m/s81.9(Pr)(

2 2 5

3 -1

2 2

3 2

m05.0

m07.0ln)

(

ln

5 3/5 - 3/5

3

-4

5 5 / 3 5 / 3 3

c

i o

D D

L

D

D F

[(0.08085)(926.5)] 0.02480 W/m.C7257

.0861.0

7257.0)C W/m

02656.0(386.0

)Ra(Pr861.0

Pr386

.0

4 / 1 4

/ 1

4 / 1 cyl

4 / 1 eff

k

The heat transfer expression is

C)34.33(

m05.0

m07.0ln

)C W/m

02480.0(2)(ln

2

2 1

T T

T D D

4 4

2 2

)K27334.33()K273(

.K W/m1067.5)(

m1571.0)(

1(

)(

m1571.0m)1(m)05.0(

+

−+

i s

T

T T A Q

L D A

σε

ππ

&

The expression for the total rate of heat transfer is

4 2 8

2)(5.67 10 W/m K ( 273K) (33.34 273K)m

1571.0)(

1(

C)34.33(

m05.0

m07.0ln

)C W/m

02480.0(2 W

20

+

−+

×+

T

T

Q Q Q

π

&

&

&

Its solution is T tube =46.3°C,

which is sufficiently close to the assumed value of 45°C Therefore, there is no need to repeat the

calculations

Trang 18

9-112E The components of an electronic device located in a horizontal duct of rectangular cross section is

cooled by forced air The heat transfer from the outer surfaces of the duct by natural convection and the average temperature of the duct are to be determined

Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant properties 3 The local atmospheric pressure is 1 atm 4 Radiation effects are negligible 5 The thermal resistance of the duct is

22 cfm

100°F

Air 80°F

Properties The properties of air at 1 atm and the

anticipated film temperature of (Ts+T∞)/2 =

(120+80)/2 = 100°F are (Table A-15E)

1 -

2 3

R001786.0R)460100/(

0

FBtu/h.ft

01529

0

=+

Analysis (a) Using air properties at the average temperature of (85+100))/2 = 92.5°F and 1 atm for the

forced air, the mass flow rate of air and the heat transfer rate by forced convection are determined to be

m& =ρV&=(0.07186lbm/ft3)(22ft3/min)=1.581lbm/min

Q&forced =m&c p(T outT in)=(1.581×60lbm/h)(0.2405Btu/lbm.°F)(100−85)°F=342.1Btu/h

Noting that radiation heat transfer is negligible, the rest of the 180 W heat generated must be dissipated by natural convection,

Btu/h 272

Q& & &

(b) We start the calculations by “guessing” the surface temperature to be 120°F for the evaluation of the properties and h We will check the accuracy of this guess later and repeat the calculations if necessary

Horizontal top surface: The characteristic length is 0.2222ft

ft)6/12+ft2(4

ft)ft)(6/124

2 2

3

10599.5)726.0()

/sft101809.0(

)ft2222.0(R)80120)(

R001786.0)(

ft/s2.32(Pr)(

c top

A A

Nu L

k h

ft4(

F.Btu/h.ft016.1)77.14(ft2222.0

FBtu/h.ft

01529.0

Horizontal bottom surface: The Nusselt number for this geometry and orientation can be determined

from

386.7)10599.5(27.027

FBtu/h.ft

01529

c bottom

Vertical side surfaces: The characteristic length in this case is the height of the duct, L c = L = 6 in Then,

6 2

2 3

3 -1

2 2

3

10379.6)726.0()

/sft101809.0(

)ft5.0(R)80120)(

R001786.0)(

ft/s2.32(Pr)(

Trang 19

.0

492.01

)10379.6(387.0825.0Pr

492.01

Ra387.0825

.0

2

27 / 8 16 / 9

6 / 1 6 2

27 / 8 16 / 9

6 / 1

×+

ft4(2

F.Btu/h.ft843.0)57.27(ft

5.0

FBtu/h.ft

01529.0

k h

Then the total heat loss from the duct can be expressed as

)](

)()

()

=++

Q&total &top &bottom &side top bottom side s

Substituting and solving for the surface temperature,

F)80](

FBtu/h

)4843.025082.02016.1[(

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