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

9-7C The greater the volume expansion coefficient, the greater the change in density with temperature, the greater the buoyancy force, and thus the greater the natural convection curren

Chapter 9 NATURAL CONVECTION Physical Mechanisms of Natural Convection

9-1C Natural convection is the mode of heat transfer that occurs between a solid and a fluid which moves

under the influence of natural means Natural convection differs from forced convection in that fluid motion in natural convection is caused by natural effects such as buoyancy

9-2C The convection heat transfer coefficient is usually higher in forced convection because of the higher

fluid velocities involved

9-3C The hot boiled egg in a spacecraft will cool faster when the spacecraft is on the ground since there is

no gravity in space, and thus there will be no natural convection currents which is due to the buoyancy force

9-4C The upward force exerted by a fluid on a body completely or partially immersed in it is called the

buoyancy or “lifting” force The buoyancy force is proportional to the density of the medium Therefore, the buoyancy force is the largest in mercury, followed by in water, air, and the evacuated chamber Note that in an evacuated chamber there will be no buoyancy force because of absence of any fluid in the medium

9-5C The buoyancy force is proportional to the density of the medium, and thus is larger in sea water than

it is in fresh water Therefore, the hull of a ship will sink deeper in fresh water because of the smaller buoyancy force acting upwards

9-6C A spring scale measures the “weight” force acting on it, and the person will weigh less in water

because of the upward buoyancy force acting on the person’s body

9-7C The greater the volume expansion coefficient, the greater the change in density with temperature, the

greater the buoyancy force, and thus the greater the natural convection currents

9-8C There cannot be any natural convection heat transfer in a medium that experiences no change in

volume with temperature

9-9C The lines on an interferometer photograph represent isotherms (constant temperature lines) for a gas,

which correspond to the lines of constant density Closely packed lines on a photograph represent a large temperature gradient

9-10C The Grashof number represents the ratio of the buoyancy force to the viscous force acting on a fluid

The inertial forces in Reynolds number is replaced by the buoyancy forces in Grashof number

9-11 The volume expansion coefficient is defined as

β 1 For an ideal gas, P=ρRT or

P T RT

P T

RT P

P

11

11

/1

ρ

∂

∂ρβ

Natural Convection over Surfaces

9-12C Rayleigh number is the product of the Grashof and Prandtl numbers

9-13C A vertical cylinder can be treated as a vertical plate when

4 / 1

35

Gr

L

9-14C No, a hot surface will cool slower when facing down since the warmer air in this position cannot rise

and escape easily

9-15C The heat flux will be higher at the bottom of the plate since the thickness of the boundary layer

which is a measure of thermal resistance is the lowest there

9-16 Heat is generated in a horizontal plate while heat is lost from it by convection and radiation The

temperature of the plate when steady operating conditions are reached 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 We assume the surface temperature to be 50°C Then the properties of air at 1 atm and the film

temperature of (Ts+T∞)/2 = (50+20)/2 = 35°C are (Table A-15)

1 -

2 5

K003247.0K)27335(

11

1

C W/m

02625

0

=+

m20.0()m16.0[(

2

)m20.0)(

m16.0(

=+

/sm10655.1(

)m04444.0)(

K2050)(

K003247.0)(

m/s81.9(Pr)(

2 2 5

3 -1

593,222(54.054

C W/m

02625

)20)(

m20.016.0)(

C W/m928.6()

C W/m

02625

)20)(

m20.016.0)(

C W/m464.3()

4 2 8 2

4 surr 4 rad

)K27317()K273(

102659.3

)K27317()K273(

.K W/m1067.5)(

m20.016.0)(

2)(

9.0(

)(

2

+

−+

×

=

+

−+

T

T

T T A

=

+

−+

×+

−+

−

=

++

=

−

s

s s

s T

T T

T

Q Q

Q Q

4 4

9 rad

bottom top

total

)K27317()K273(102659.3)20(0.1108)

20(0.2217 W

9-17 Flue gases are released to atmosphere using a cylindrical stack 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

2 5

K003356.0K)27325(

11

1

C W/m

02551

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 stack, L c = L=10m.Then,

12 2

2 5

3 -1

2

2

3

10953.2)7296.0()

/sm10562.1(

)m10)(

K1040)(

K003356.0)(

m/s81.9(Pr)(

/ 1 12

4

/

1

35 thusand 0.6

<

246.0)7296.0/10953.2(

)10(3535

Gr

L D Gr

)10953.2(1.01

C W/m

02551

(b) When the stack 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:

400,213/s

m10562.1

m)m/s)(0.63600

/100020(

7296.0()400,213(027.0PrRe027

0

Nu= 0.805 1/3 = 0.805 1/3= (from Table 7-1)

C W/m15.20)9.473(m6.0

C W/m

02551.0

9-18 Heat generated by the electrical resistance of a bare cable is dissipated to the surrounding air The surface temperature of the cable 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 The temperature of the surface of the cable is constant

Properties We assume the surface temperature to be 100°C Then the properties of air at 1 atm and the film

temperature of (Ts+T∞)/2 = (100+20)/2 = 60°C are (Table A-15)

1 -

2 5

K003003.0K)27360(

11

1

CW/m

02808

0

=+

/sm10896.1(

)m005.0)(

K20100)(

K003003.0)(

m/s81.9(Pr)(

2 2 5

3 -1

)2.590(387.06

.0Pr

/559.01

387.06

.0

2

27 / 8 16 / 9

6 / 1 2

27 / 8 16 / 9

6 / 1

m005.0(

C W/m17.13)346.2(m005.0

C W/m

02808.0

k

h

s

C128.8

C)20)(

m06283.0)(

C W/m17.13()AV)(1.5

0

(

)(

2 2

T

T

T T hA Q&

which is not close to the assumed value of 100°C Repeating calculations for an assumed surface

temperature of 120°C, [Tf = (Ts+T∞)/2 = (120+20)/2 = 70°C]

1 -

2 5

K002915.0K)27370(

11

1

CW/m

02881

0

=+

/sm10995.1(

)m005.0)(

K20120)(

K002915.0)(

m/s81.9(Pr)(

2 2 5

3 -1

)6.644(387.06

.0Pr

/559.01

387.06

.0

2

27 / 8 16 / 9

6 / 1 2

27 / 8 16 / 9

6 / 1

Nu

m005.0

C W/m

02881

T

T A

T T hA Q

C)20)(

m06283.0)(

C W/m76.13()5.1

2 2

&

which is sufficiently close to the assumed value of 120°C

9-19 A horizontal hot water pipe passes through a large room The rate of heat loss from the pipe by natural convection and radiation 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 The temperature of the outer surface of the pipe is constant

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

of (Ts+T∞)/2 = (73+27)/2 = 50°C are (Table A-15)

2 5

K003096.0K)27350(

11

1

C W/m

02735

0

=+

2 5

3 -1

2

2

3

10747.6)7228.0()

/sm10798.1(

)m06.0)(

K2773)(

K003096.0)(

m/s81.9(Pr)(

)10747.6(387.06.0Pr

/559.01

387.06

.0

2

27 / 8 16 / 9

6 / 1 5 2

27 / 8 16 / 9

6 / 1

m06.0(

C W/m950.5)05.13(m06.0

C W/m

02735.0

k

h

s

W 516

)K27327()K27373(.K W/m1067.5)(

m885.1)(

8.0(

)( s surr

s

Q& ε σ

9-20 A power transistor mounted on the wall dissipates 0.18 W The surface temperature of the transistor is

to be determined

Assumptions 1 Steady operating conditions exist 2 Air is

an ideal gas with constant properties 3 Any heat transfer

from the base surface is disregarded 4 The local

atmospheric pressure is 1 atm 5 Air properties are

evaluated at 100°C

Air 35°C

Power transistor, 0.18 W

D = 0.4 cm

ε = 0.1

Properties The properties of air at 1 atm and the given

film temperature of 100°C are (Table A-15)

1 -

2 5

K00268.0K)273100(

11

2

C W/m

03095

0

=+

Analysis The solution of this problem requires a trial-and-error approach since the determination of the

Rayleigh number and thus the Nusselt number depends on the surface temperature which is unknown We start the solution process by “guessing” the surface temperature to be 165°C for the evaluation of h This is the surface temperature that will give a film temperature of 100°C We will check the accuracy of this guess later and repeat the calculations if necessary

The transistor loses heat through its cylindrical surface as well as its top surface For convenience,

we take the heat transfer coefficient at the top surface of the transistor to be the same as that of its side surface (The alternative is to treat the top surface as a vertical plate, but this will double the amount of calculations without providing much improvement in accuracy since the area of the top surface is much smaller and it is circular in shape instead of being rectangular) The characteristic length in this case is the outer diameter of the transistor, L c = D=0.004m Then,

6.292)7111.0()

/sm10306.2(

)m004.0)(

K35165)(

K00268.0)(

m/s81.9(Pr)(

2 2 5

3 -1

)6.292(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

m0000691

04/m)004.0()m0045.0)(

m004.0(4/

C W/m78.15)039.2(m004.0

C W/m

03095.0

=+

=+

π

πDL D

A

Nu D

)K27325()273(

1067.5)(

m0000691

0)(

1.0(

C)35)(

m0000691

0)(

C W/m8.15( W

18

0

)(

)(

4 4

8 2

2 2

4 4

×+

surr s s s

s

T

T T

T T A T T hA

9-21 EES Prob 9-20 is reconsidered The effect of ambient temperature on the surface temperature of the

9-22E A hot plate with an insulated back is considered The rate of heat loss by natural convection is to be

determined for different orientations

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 = (130+75)/2 = 102.5°F are (Table A-15)

1 -

2 3

R001778.0R)4605.102(

11

0

FBtu/h.ft

01535

0

=+

Analysis (a) When the plate is vertical, the characteristic length is

the height of the plate L c = L=2ft Then,

8 2

2 3

3 -1

2

2

3

10503.5)7256.0()

/sft101823.0(

)ft2)(

R75130)(

R001778.0)(

ft/s2.32(Pr)(

.0

492.01

)10503.5(387.0825.0Pr

492.01

Ra387.0825

.0

2

27 / 8 16 / 9

6 / 1 8 2

27 / 8 16 / 9

6 / 1

×+

+

=

Nu

2 2 2

2

ft4)ft2(

F.Btu/h.ft7869.0)6.102(ft

2

FBtu/h.ft

01535.0

k

h

s

and Q&=hA s(T s−T∞)=(0.7869Btu/h.ft2.°F)(4ft2)(130−75)°C=173.1 Btu/h

(b) When the plate is horizontal with hot surface facing up, the characteristic length is determined from

ft5.04

ft244

A

Then,

6 2

2 3

3 -1

2

2

3

10598.8)7256.0()

/sft101823.0(

)ft5.0)(

R75130)(

R001778.0)(

ft/s2.32(Pr)(

5.0

FBtu/h.ft

01535

and Q&=hA s(T s−T∞)=(0.8975Btu/h.ft2.°F)(4ft2)(130−75)°C=197.4 Btu/h

(c) When the plate is horizontal with hot surface facing down, the characteristic length is again

and the Rayleigh number is Then,

δ = 0 5 ft

6

10598

=

Ra

62.14)10598.8(27.027

5.0

FBtu/h.ft

01535

9-23E EES Prob 9-22E is reconsidered The rate of natural convection heat transfer for different

orientations of the plate as a function of the plate temperature is to be plotted

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

T s [F] Q a [Btu/h] Q b [Btu/h] Q c [Btu/h]

9-24 A cylindrical resistance heater is placed horizontally in a fluid The outer surface temperature of the

resistance wire is to be determined for two different fluids

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

atmospheric pressure is 1 atm 4 Any heat transfer by radiation is ignored 5 Properties are evaluated at

500°C for air and 40°C for water

Properties The properties of air at 1 atm and 500°C are (Table A-15)

1 -

2 5

K001294.0K)273500(

11

,6986

0

Pr

/sm10804

7

C W/m

05572

0

=+

2 6K

000377.0 ,32

4

Pr

/sm106582.0/ C, W/m

k

Analysis (a) The solution of this problem requires a trial-and-error approach since the determination of the

Rayleigh number and thus the Nusselt number depends on the surface temperature which is unknown We start the solution process by “guessing” the surface temperature to be 1200°C for the calculation of h We will check the accuracy of this guess later and repeat the calculations if necessary The characteristic length

in this case is the outer diameter of the wire, L c = D=0.005m Then,

)/sm10804.7(

)m005.0(C)201200)(

K001294.0)(

m/s81.9(Pr)(

2 2 5

3 -1

)7.214(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

m01178.0)m75.0)(

m005.0(

C W/m38.21)919.1(m005.0

C W/m

05572.0

(b) For the case of water, we “guess” the surface temperature to be 40°C The characteristic length in this

case is the outer diameter of the wire, L c = D=0.005m Then,

197,92)32.4()

/sm106582.0(

)m005.0)(

K2040)(

K000377.0)(

m/s81.9(Pr)(

2 2 6

3 -1

)197,92(387.06.0Pr

/559.01

387.06

.0

2

27 / 8 16 / 9

6 / 1 2

27 / 8 16 / 9

6 / 1

Nu

C W/m1134)986.8(m005.0

C W/m

631

m01178.0)(

C W/m1134( W300)

hA

which is sufficiently close to the assumed value of 40°C in the evaluation of the properties and h The film

temperature in this case is (Ts+T∞)/2 = (42.5+20)/2 =31.3°C, which is close to the value of 40°C used in the evaluation of the properties

9-25 Water is boiling in a pan that is placed on top of a stove The rate of heat loss from the cylindrical side

surface of the pan by natural convection and radiation and the ratio of heat lost from the side surfaces of the pan to that by the evaporation of water 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 = (98+25)/2 = 61.5°C are (Table A-15)

Vapor

2 kg/h

Water 100°C

-2 5

K00299.0K)2735.61(

11

1

C W/m

02819

0

=+

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

height of the pan, L c = L=0.12m.Then

6 2

2 5

3 -1

2

2

3

10299.7)7198.0()

/sm10910.1(

)m12.0)(

K2598)(

K00299.0)(

m/s81.9(Pr)(

/ 1 6

4

/

1

35 thus

and 0.25

<

07443.0)7198.0/10299.7(

)12.0(3535

Gr

L D Gr

.0

492.01

)10299.7(387.0825.0Pr

492.01

Ra387.0825

.0

2

27 / 8 16 / 9

6 / 1 6 2

27 / 8 16 / 9

6 / 1

×+

+

=

Nu

2 2

m09425.0)m12.0)(

m25.0(

C W/m720.6)60.28(m12.0

C W/m

02819.0

)K27325()K27398(.K W/m1067.5)(

m09425.0)(

80.0(

)( s surr

kg/s3600/5.1

940

3.472

46

f

9-26 Water is boiling in a pan that is placed on top of a stove The rate of heat loss from the cylindrical side

surface of the pan by natural convection and radiation and the ratio of heat lost from the side surfaces of the pan to that by the evaporation of water 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 = (98+25)/2 = 61.5°C are (Table A-15)

Vapor

2 kg/h

Water 100°C

-2 5

K00299.0K)2735.61(

11

1

C W/m

02819

0

=+

Analysis (a) The characteristic length in this case is

the height of the pan, L c = L=0.12m Then,

6 2

2 5

3 -1

2

2

3

10299.7)7198.0()

/sm10910.1(

)m12.0)(

K2598)(

K00299.0)(

m/s81.9(Pr)(

/ 1 6

4

/

1

35 thus

and 0.25

<

07443.0)7198.0/10299.7(

)12.0(3535

Gr

L D Gr

.0

492.01

)10299.7(387.0825.0Pr

492.01

Ra387.0825

.0

2

27 / 8 16 / 9

6 / 1 6 2

27 / 8 16 / 9

6 / 1

×+

+

=

Nu

2 2

m09425.0)m12.0)(

m25.0(

C W/m720.6)60.28(m12.0

C W/m

02819.0

)K27325()K27398(.K W/m1067.5)(

m09425.0)(

10.0(

)( s surr

kg/s3600/5.1

940

9.52

46

f