Solution manual heat and mass transfer a practical approach 3rd edition cengel CH09 2

Natural Convection from Finned Surfaces and PCBs 9-53C Finned surfaces are frequently used in practice to enhance heat transfer by providing a larger heat transfer surface area.. 9-54C

Natural Convection from Finned Surfaces and PCBs

9-53C Finned surfaces are frequently used in practice to enhance heat transfer by providing a larger heat

transfer surface area Finned surfaces are referred to as heat sinks in the electronics industry since they provide a medium to which the waste heat generated in the electronic components can be transferred effectively

9-54C A heat sink with closely packed fins will have greater surface area for heat transfer, but smaller heat

transfer coefficient because of the extra resistance the additional fins introduce to fluid flow through the interfin passages

9-55C Removing some of the fins on the heat sink will decrease heat transfer surface area, but will increase

heat transfer coefficient The decrease on heat transfer surface area more than offsets the increase in heat transfer coefficient, and thus heat transfer rate will decrease In the second case, the decrease on heat transfer coefficient more than offsets the increase in heat transfer surface area, and thus heat transfer rate will again decrease

9-56 An aluminum heat sink of rectangular profile oriented vertically is used to cool a power transistor

The average natural convection heat transfer coefficient is to be determined

Assumptions 1 Steady operating

conditions exist 2 Air is an ideal gas with

constant properties 3 Radiation heat

transfer from the sink is negligible 4 The

entire sink is at the base temperature

b =1.52 cm

9.68 cm

Power transistoHeat sink

Analysis The total surface area of

the heat sink is

2

2

2

m021465.0006835.001463.0

m006835.0)m0762.0)(

m0317.0()m0762.0)(

m0145.0)(

4(

m01463.0)m0762.0)(

m0048.0)(

2()m0152.0)(

m0762.0)(

6)(

2(2

=+

=+

=

=+

=

=+

=

=

unfinned fins

total

unfinned

fins

A A A

A

nLb A

Then the average natural convection heat transfer coefficient becomes

C W/m

m021465.0(

W15)

()

(

2

T T A

Q h

T T hA

Q

s total s

total

&

&

9-57 Aluminum heat sinks of rectangular profile oriented vertically are used to cool a power transistor A shroud is placed very close to the tips of fins The average natural convection heat transfer coefficient is to

be determined

b =1.52 cm

9.68 cm

Power transistorHeat sink

Shroud

Assumptions 1 Steady operating

conditions exist 2 Air is an ideal

gas with constant properties 3

Radiation heat transfer from the

sink is negligible 4 The entire sink

is at the base temperature

Analysis The total surface area of

the shrouded heat sink is

2 2

2 2

m035486.0014752.0006835.0013898.0

m014752.0)m0762.0)(

m0968.0)(

2(

m006835.0)m0762.0)(

m0317.0()m0762.0)(

m0145.0)(

4(

m013898.0)m0152.0)(

m0762.0)(

6)(

2(2

=+

+

=+

fins total

shroud

unfinned

fins

A A

A A

A

A

nLb A

Then the average natural convection heat transfer coefficient becomes

C W/m

m035486.0(

W15)

()

(

2

T T A

Q h

T T hA

Q

s total s

total

&

&

9-58E A heat sink with equally spaced rectangular fins is to be used to cool a hot surface The optimum fin spacing and the rate of heat transfer from the heat sink are to be determined

Assumptions 1 Steady operating

conditions exist 2 Air is an ideal gas with

constant properties 3 The atmospheric

pressure at that location is 1 atm 4 The

thickness t of the fins is very small

relative to the fin spacing S so that Eqs

9-32 and 9-33 for optimum fin spacing are

Properties The properties of air at 1 atm

and 1 atm and the film temperature of

(Ts+T∞)/2 = (180+78)/2=129°F are (Table

A-15E)

1 -

2 3

R001698.0R)460129(

11

0

FBtu/h.ft

01597

0

=+

2 3

3 -1

2 2

3 2

)/sft101975.0(

)ft12/8)(

R78180)(

R001698.0)(

ft/s2.32(Pr)(

ft12/8714.2714

2

4 / 1 7 4

/ 1

Ra

L S

The heat transfer coefficient for this optimum spacing case is

F.Btu/h.ft8578.0ft

02433.0

FBtu/h.ft

01597.0307.1307

The number of fins and the total heat transfer surface area is

fins1608.02916.0

6

=+

=+

=

t S

w

n

2ft2.226

=ft)12/2.1(ft)(0.08/1216

2ft)12/8(ft)(0.08/12

16

ft)ft)(1.2/1212

/8(16222

×

×+

= nLH ntL ntH

A s

Then the rate of natural convection heat transfer becomes

Btu/h 196

6

≅

=

≅+

=

s

w t s

w

n

2ft2.8

=ft)ft)(1.2/1212

/8(212

= nLH

A s

Btu/h 245

9-59E EES Prob 9-58E is reconsidered The effect of the length of the fins in the vertical direction on the optimum fin spacing and the rate of heat transfer by natural convection is to be investigated

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

100 125 150 175 200 225 250 275 300

9-60 A heat sink with equally spaced rectangular fins is to be used to cool a hot surface The optimum fin height and the rate of heat transfer from the heat sink are to be determined

Assumptions 1 Steady operating

conditions exist 2 Air is an ideal gas

with constant properties 3 The

atmospheric pressure at that location

is 1 atm

Properties The properties of air at 1

atm and 1 atm and the film

temperature of (Ts+T∞)/2 = (85+25)/2

= 55°C are (Table A-15)

1 -

2 5

K003049.0K)27355(

11

1

C W/m

02772

0

=+

2 5

3 -1

2 2

3

10214.2)7215.0()

/sm10847.1(

)m18.0)(

K2585)(

K003049.0)(

m/s81.9(Pr)(

m18.0714.2714

2

4 / 1 7 4

The heat transfer coefficient for this optimum fin spacing case is

C W/m087.5m007122.0

C W/m

02772.0307.1307

The criteria for optimum fin height H in the literature is given by H= hA c/pk (not in the text) where

A c /p ≅ t/2 for rectangular fins Therefore,

m 0.00379

177(2

m)C)(0.001

W/m087.5(2

15.0

≅

=

≅+

=

s

w t S

w

n

2m0.02865

=m)m)(0.0037918

.0(212

Natural Convection inside Enclosures

9-61C We would recommend putting the hot fluid into the upper compartment of the container In this case

no convection currents will develop in the enclosure since the lighter (hot) fluid will always be on top of the heavier (cold) fluid

9-62C We would disagree with this recommendation since the air space introduces some thermal resistance

to heat transfer The thermal resistance of air space will be zero only when the convection coefficient approaches infinity, which is never the case However, when the air space is eliminated, so is its thermal resistance

9-63C Yes, dividing the air space into two compartments will retard air motion in the air space, and thus slow down heat transfer by natural convection The vinyl sheet will also act as a radiation shield and reduce heat transfer by radiation

9-64C The effective thermal conductivity of an enclosure represents the enhancement on heat transfer as result of convection currents relative to conduction The ratio of the effective thermal conductivity to the ordinary thermal conductivity yields Nusselt number Nu=k eff /k

9-65 Conduction thermal resistance of a medium

is expressed as Thermal resistance

of a rectangular enclosure can be expressed by

replacing L with characteristic length of

c , and thermal conductivity k with

effective thermal conductivity k eff to give L c

A

Q&

)/(

)/(k A L kNuA L

R= c eff = c

9-66 The U-factors for the center-of-glass section of a double-pane window and a triple-pane window are

to be determined Also, the percentage decrease in total heat transfer when triple-pane window is used is to

be estimated

Assumptions 1 Steady operating conditions exist 2 Heat transfer through the window is one-dimensional

3 The thermal resistance of glass sheets is negligible

Properties The thermal conductivity of air space is given to be 0.025 W/m⋅ºC

Analysis The convection heat transfer coefficient of the

air space is determined from

Air space

i h

15 mm

m015.0

C W/m

025

Noting that the radiation across the air space is of the

same magnitude as the convection, the combined heat

transfer coefficient of the space is

C W/m4)C W/m2(2

rad conv

h

Disregarding the thermal resistance of glass sheets, which

are small, the U-factor for the center region of a double

pane window is determined from

C W/m

=

⎯→

⎯++

=++

space

14

16

11111

U h

h h

Noting that there are two air spaces, the U-factor for triple-pane window is

C W/m

=

⎯→

⎯+++

=+++

space space

14

14

16

111111

U h

h h h

Considering that about 70 percent of total heat transfer through a window is due to center-of-glass section, the percentage decrease in total heat transfer when triple-pane window is used in place of double-pane window is

415.1190.2)70.0()

70.0(Decrease

%

double

triple double

U

U U

That is, triple-pane window decreases the heat transfer through the center region by 35.4 percent while the decrease for the entire window is 24.8 percent The use of triple-pane window is usually not justified economically except for extremely cold regions

9-67 Two glasses of a double pane window are maintained at specified temperatures The fraction of heat transferred through the enclosure by radiation is to be determined

Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas

with constant properties 3 The air pressure in the enclosure is 1 atm

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

of (T1+T2)/2 = (280+336)/2 = 308 K = 35°C are (Table A-15E)

1 -

2 5

K003247.0K308

11

1

C W/m

Analysis The characteristic length in this case is the distance

between the two glasses, L c = L = 0.4 m Then,

3 -1

2 2

3 2 1

10029.3)7268.0()

/sm10655.1(

)m4.0)(

K280336)(

K003247.0)(

m/s81.9(Pr)

.0

5.1)10029.3(7268.02.0

7268.022.0Pr

2.0

Pr22

0

4 / 1 28 0 8 4

/ 1 28 0

Then,

2m5.4m)3(m)5.1

4.0

K)280336()ft5.4)(

00.35)(

C W/m

02625.0

2 1

L

T T kNuA

Q& s

The effective emissivity is

1475.0778

.6190.0

115.0

11111

ff 2

1 ff

=

−+

e

εε

εε

The rate of heat transfer by radiation is

W4.248])K280(K)336)[(

.K W/m1067.5)(

m5.4)(

1475.0(

)(

4 4

4 2 8 2

4 2 4 1 eff rad

Q& ε sσ

Then the fraction of heat transferred through the enclosure by radiation becomes

0.30

=+

=+

=

4.2489.578

4.248rad

conv

rad rad

Q Q

Q f

&

&

&

9-68E Two glasses of a double pane window are maintained at specified temperatures The rate of heat transfer through the window by natural convection and radiation, and the R-value of insulation are to be determined

Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas

with constant properties 3 The air pressure in the enclosure is 1 atm

40°F 65°F

L =1 in

H = 4 ft

Q&

Air

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

of (T1+T2)/2 = (65+40)/2 = 52.5°F are (Table A-15E)

1 -

2 3

R001951.0R)4605.52(

11

0

FBtu/h.ft

01415

0

=+

Analysis (a) The characteristic length in this case is the distance

between the two glasses, L c = L = 1 in Then,

824,27)7332.0()

/sft101548.0(

)ft12/1)(

R4065)(

R001951.0)(

ft/s2.32(Pr)(

2 2 3

3 -1

2 2

3 2

The aspect ratio of the geometry is H/L = 4×12/1 = 48 (which is a little over 40, but still close enough for

an approximate analysis) For these values of H/L and Ra L, the Nusselt number can be determined from

692.1ft

12/1

ft4)

7332.0()824,27(42.0Pr

42

0

3 0 012

0 4

/ 1 3

0 012 0 4 /

Nu

Then,

2ft24ft)6(ft)4

F)4065()ft24)(

692.1)(

FBtu/h.ft

01415.0

2 1

L

T T kNuA

Q& s

(b) The rate of heat transfer by radiation is

Btu/h 454.3

=+

−+

.RBtu/h.ft10

1714.0)(

ft24)(

4 4

4 2 8

2

4 2 4

T A

Q&rad ε sσ

Then the total rate of heat transfer is

Btu/h7.6263.4544

=+

= convection rad

Q& & &

Then the effective thermal conductivity of the air, which also accounts for the radiation effect and the value become

R-FBtu/h.ft.0.08704

F)4065)(

ft24(

)ft12/1)(

Btu/h7.626()

2 1

L Q k

L

T T A k Q

s eff s

eff total

0.08704

)ft12/1

eff value

k L R

9-69E EES Prob 9-68E is reconsidered The effect of the air gap thickness on the rates of heat transfer by

natural convection and radiation, and the R-value of insulation is to be investigated

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

9-70 Two surfaces of a spherical enclosure are maintained at specified temperatures The rate of heat transfer through the enclosure is to be determined

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

pressure in the enclusure is 1 atm

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

of (T1+T2)/2 = (350+275)/2 = 312.5 K = 39.5°C are (Table A-15)

1 -

2 5

K003200.0K5.312

11

1

C W/m

15252

2 2

3 2 1

10415.7)7256.0()

/sm10697.1(

)m05.0)(

K275350)(

K003200.0)(

m/s81.9(Pr)(

m05.0)

()

+

=+

o i

o

i

c

D D

D

D

L F

[(0.00590)(7.415 10 )] 0.1315 W/m C7256

.0861.0

7256.0C) W/m

Pr74

0

4 / 1 5 4

/ 1

4 / 1 4

/ 1 eff

)m25.0)(

m15.0()C W/m

1315.0()(

c

o i

T T L

D D k

Q&

9-71 EES Prob 9-70 is reconsidered The rate of natural convection heat transfer as a function of the hot surface temperature of the sphere is to be plotted

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

9-72 The absorber plate and the glass cover of a flat-plate solar collector are maintained at specified temperatures The rate of heat loss from the absorber plate by natural convection is to be determined

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

by radiation is negligible 4 The air pressure in the enclusure is 1 atm

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

of (T1+T2)/2 = (80+40)/2 = 60°C are (Table A-15)

Solar radiation

θ Insulation

AbsorberPlate 80°C

Glass Cover, 40°C

1.5 m

L = 2.5

1 -

2 5

K003003.0K)27360(

11

1

C W/m

02808

0

=+

Analysis For θ =0°, we have horizontal

rectangular enclosure The characteristic length

in this case is the distance between the two

glasses L c = L = 0.025 m Then,

4 2

2 5

3 -1

2 2

3 2 1

10689.3)7202.0()

/sm10896.1(

)m025.0)(

K4080)(

K003003.0)(

m/s81.9(Pr)(

)10689.3(10

689.3

17081

44.11

118

RaRa

1708144.11

Nu

3 / 1 4 4

3 / 1

⎥⎦

⎤

⎢⎣

⎡ −+

=

+ +

+ +

Then

2m5.4m)3(m)5.1

C)4080()m5.4)(

223.3)(

C W/m

02808.0

2 1

L

T T kNuA

)30cos(

)10689.3()30cos(

)10689.3(

)308.1sin(

17081)30cos(

)10689.3(

17081

)cosRa(cos

Ra

)8.1(sin17081cosRa

17081

4

6 1 4

3 / 1 6

1

+

θ

θθ

W 621

C)4080()m5.4)(

074.3)(

C W/m

02808.0

2 1

L

T T kNuA

025.0

m2)

7202.0()10689.3(42.0Pr

42

0

3 0 012

0 4

/ 1 4 3

0 012 0 4 /

Nu

W 315

C)4080()m5.4)(

557.1)(

C W/m

02808.0

2 1

L

T T kNuA

Q& s

Discussion Caution is advised for the vertical case since the condition H/L < 40 is not satisfied

9-73 A simple solar collector is built by placing a clear plastic tube around a garden hose The rate of heat loss from the water in the hose per meter of its length by natural convection is to be determined

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

by radiation is negligible 3 The air pressure in the enclosure is 1 atm

Properties The properties of air at 1 atm and the anticipated average temperature of (Ti+ o)/2 = (65+35)/2

= 50°C are (Table A-15)

1 -

2 5

K003096.0K)27350(

11

1

C W/m

02735

0

=+

Analysis We assume the plastic tube temperature to be 35°C

We will check this assumption later, and repeat calculations,

if necessary The characteristic length in this case is

cm7.12

6.15

= o i

c

D D

L

Then,

000,10)7228.0()

/sm10798.1(

)m017.0)(

K3565)(

K003096.0)(

m/s81.9(Pr)(

2 2 5

3 -1

2 2

)016.0/05.0ln(

)(

)/ln(

5 3/5 - 3/5

3

-4 5

5 / 3 5 / 3 3

4

+

=+

o i

c

i o

D D

L

D D F

[(0.1821)(10,000)] 0.05670 W/m.C7228

.0861.0

7228.0)C W/m

02735.0(386.0

)Ra(Pr861.0

Pr386

.0

4 / 1 4

/ 1

4 / 1 cyl

4 / 1 eff

k

Then the rate of heat transfer between the cylinders becomes

)65()016.0/05.0ln(

)C W/m

05670.0(2)()/ln(

2

0 eff

T T

T D D

k

i o

are C5.302/)2635(2/)

= T T∞

T avg s

1 -

2 5

K003295.0K)2735.30/(

1/1and ,7281

0

Pr

/s,m10613.1 C, W/m

02592

0

=+

k

βν

The characteristic length in this case is the outer diameter of the solar collector L c = D o = 0.05 m Then,

3 -1

2 2

3

10018.1)7281.0()

/sm10613.1(

)m05.0)(

K2635)(

K003295.0)(

m/s81.9(Pr)

( )

7281.0/559.01

)10018.1(387.06.0Pr

/559.01

387.06

.0

2

27 / 8 16 / 9

6 / 1 5 2

27 / 8 16 / 9

6 / 1

C W/m

02592

(Eq 2) C

)26)(

m1571.0)(

C W/m063.4()

C,8

W 8.22

=

°

T o 39.0C, &