solucionario ccc 1

8/3/2019 solucionario ccc 1

1/15

1. INTRODUCTION

1.1 A composite wall consist of alternative layers of fir ( 5 cm thick ) , aluminum ( 1

cm thick ), lead ( 1 cm thick ), and corkboard ( 6 cm thick ). The temperature is 60

C of the outside of the fir and 10 C on the outside of the corkboard. Plot thetemperature gradient through the wall. Does the temperature profile suggest any

simplifying assumptions that might be made in subsequent analysis of the wall?

Solution:

Thermal Conductivities:

kfir= 0.12 W/m.K (Table A.2, Appendix A)

kalu = 237 W/m.K (Table A.1, Appendix A)

kld= 35 W/m.K (Table A.1, Appendix A)

kcb = 0.04 W/m.K (Table A.2, Appendix A).

Question No. 1: Plot the temperature gradient through the wall.

Answer:

Question No. 2: Does the temperature profile suggest any simplifying assumptions

that might be made in subsequent analysis of the wall?

Answer:

Yes, since the thermal conductivity of aluminum and lead are very high than fir and

corkboard, they are considered isothermal. Therefore consider only fir and corkboard.

fir+ Tcb = 60 C 10 C = 50 K

cbfir L

Tk

L

Tkq

=

=

Lfir= 5 cm = 0.05 m

Lcb = 6 cm = 0.06 m


2/15

1. INTRODUCTION

Then,

( )( )

( )( )( )m

TKmWm

TKmWq cbfir

06.0/04.0

05.0/12.0 ==

Tcb = 3.6Tfir

Then,

fir+ 3.6fir= 50 K

fir= 10.87 K

( )

=

=

m

KKmW

L

Tkq

fir 05.0

87.10./12.0 = 26.09 W/m

2

Considering all walls:

Tfir+ Talu + Tld+ Tcb = 60 C 10 C = 50 K

cbldalufir L

Tk

L

Tk

L

Tk

L

Tkq

=

=

=

=

Lfir= 5 cm = 0.05 m

Lcb = 6 cm = 0.06 m

Lalu = 1 cm = 0.01 m

Lld= 1 cm = 0.01 m

=

alu

fir

firalu

L

k

L

k

TT

=

ld

fir

firld

L

k

L

k

TT

=

cb

fir

fircb

L

k

L

k

TT

Then


3/15

1. INTRODUCTION

+

+

+

cbldalu

fir

fir

L

k

L

k

L

kL

k

T

111

1 = 50 K

+

+

+

06.0

04.0

1

01.0

35

1

01.0

237

1

05.0

12.01firT = 50

Tfir= 10.87 K

( )

=

=

m

KKmW

L

Tkq

fir 05.0

87.10./12.0 = 26.09 W/m

2

There it is equal to simplified solution.

1.2 Verify Equation (1.15).

Solution:

Equation (1.15)

TT

dt

dTbody

body

For verification only

Equation (1.3)

dt

dTmc

dt

dUQ ==

Equation (1.16)

TTQ body

Then

TT

dt

dTmc body

TT

dtdT

body

Then

TT

dt

dTbody

bodywhere mc is constant.


4/15

1. INTRODUCTION

1.3 q = 5000 W/m2

in a 1 cm slab and T= 140 C on the cold side. Tabulate the

temperature drop through the slab if it is made of

Silver

Aluminum

Mild steel (0.5 % carbon)

Ice

Spruce

Insulation (85 % magnesia)

Silica aerogel

Indicate which situations would be unreasonable and why.

Solution:

L = 1 cm = 0.01 m

(a) Silver Slab

SiL

Tkq

= = 5000 W/m

2

Thermal conductivity of silver at 140 C, 99.99+ % Pure, Table A.1, Appendix A

ksi = 420 W/m.K

( )

=

m

TKmWq Si

01.0

/420 = 5000 W/m2

TSi = 0.12 K

(b) Alumium Slab

aluL

Tkq

= = 5000 W/m

2

Thermal conductivity of aluminum at 140 C, 99.99+ % Pure, Table A.1, App. A

Kalu = 237.6 W/m.K

( )

=

m

TKmWq alu

01.0/6.237 = 5000 W/m

2

Talu = 0.21 K

(c) Mild Steel Slab

msL

Tkq

= = 5000 W/m

2

Thermal conductivity of mild steel at 140 C, Table A.1, Appendix AKms = 50.4 W/m.K


5/15

1. INTRODUCTION

( )

=

m

TKmWq ms

01.0/4.50 = 5000 W/m

2

Tms = 0.992 K

(d) Ice Slab

iceL

Tkq

= = 5000 W/m

2

Thermal conductivity of ice at 140 C, Table A.1, Appendix A

ice at 0 C, kice = 2.215 W/m.K

Note: there is no ice at 140 C, but continue calculation at 0 C.

( )

=

m

TKmWq ice

01.0

/215.2 = 5000 W/m2

Tice = 22.57 K

(e) Spruce Slab

SiL

Tkq

= = 5000 W/m

2

Thermal conductivity of spruce at 140 C, Table A.1, Appendix A

Ksp = 0.11 W/m.K @ 20 C (available)

( )

=

m

TKmWq

Sp

01.0/11.0 = 5000 W/m

2

TSp = 454.55 K

(f) Insulation (85 % Magnesia)

SiL

Tkq

= = 5000 W/m

2

Thermal conductivity of insulation at 140 C, Table A.1, Appendix A

Kin = 0.074 W/m.K @ 150 C (available)

( )

=

m

TKmWq in

01.0/074.0 = 5000 W/m

2

TSi = 675.8 K

(g) Silica Aerogel Slab

SiL

Tkq

= = 5000 W/m

2

Thermal conductivity of silica aerogel at 140 C, Table A.1, Appendix A

ksa = 0.022 W/m.K @ 120 C


6/15

1. INTRODUCTION

( )

=

m

TKmWq sa

01.0/022.0 = 5000 W/m

2

Tsa = 2,273 K

Tabulation:

Slab Temperature Drop, K

Silver 0.12

Aluminum 0.21

Mild Steel (0.5 % Carbon) 0.992

Ice 22.57

Spruce 454.55

Insulation (85 % Magnesia) 675.8Silica Aerogel 2273

The situation which is unreasonable here is the use of ice as slab at 140 C, since ice

will melt at temperature of 0 C and above. Thats it.

1.4 Explain in words why the heat diffusion equation, eq. no. (1.13), shows that in

transient conduction the temperature depends on the thermal diffusitivity, , butwe can solve steady conduction problems using just k(as in Example 1.1).

Solution:

Equation (1.13)

xdt

dTcAx

dt

TTdcA

dt

dUQ

ref

net =

==

Answer: The application of heat diffusion equation eq. no. (1.13) depends on the

thermal diffusivity as the value oft

T

is not equal to zero as it I s under unsteady

state conduction. While in steady conduction depends only on kbecause the value of

t

T

= 0 for steady state conduction giving 2

2

x

T

= 0 , so dx

dT

kq = .

1.5 A 1-m rod of pure copper 1 cm2 in cross section connects a 200 C thermal

reservoir with a 0 C thermal reservoir. The system has already reached steady

state. What are the rates of change of entropy of (a) the first reservoir, (b) the

second reservoir, (c) the rod, and (d) the whole universe, as a result of the


7/15

1. INTRODUCTION

process? Explain whether or not your answer satisfies the Second Law of

Thermodynamics.

Solution:

Equation (1.9)

L

Tkq

=

Thermal conductivity of copper at 100 C, Table A.1, Appendix A

k= 391 W/m.K

L = 1 m

T= 200 C 0 C = 200 K

( )

=

m

KKmWq

1

200/391 = 78,200 W/m2.K

Q = qA

A = 1 cm2

= 1 x 10-4

m2

Q = (78,200 W/m2.K)(1 x 10

-4m

2) = 7.82 W

(a)( )K

W

T

QS rev

273200

82.7

1

1+

=

=

& = - 0.01654 W/K

(b)( )K

W

T

QS rev

2730

82.7

2

2+

+==

& = + 0.02864 W/K

(c) =rS& = 0.0 W/K (see Eq. 1.5, steady state)

(d) =+= 21 SSSUn&&& = - 0.01654 W/K + 0.02864 W/K = + 0.0121 W/K

Since 0UnS& , therefore it satisfied Second Law of Thermodynamics.

1.6 Two thermal energy reservoirs at temperatures of 27 C and 43 C, respectively,

are separated by a slab of material 10 cm thick and 930 cm2

in cross-sectional

area. The slab has a thermal conductivity of 0.14 W/m.K. The system is operating

at steady-state conditions. What are the rates of change of entropy of (a) the

higher temperature reservoir, (b) the lower temperature reservoir, (c) the slab, and

(d) the whole universe as a result of this process? (e) Does your answer satisfy the

Second Law of Thermodynamics?

Solution:

Equation (1.9)

L

Tkq

=

Thermal conductivity , k= 0.14 W/m.K


8/15

1. INTRODUCTION

A = 930 cm2

= 0.093 m2

L = 10 cm = 0.10 mT= 27 C (- 43 C) = 70 KT1 = 27 + 273 = 300 K

T2 = -43 + 273 = 230 K

( )

=

m

KKmWq

10.0

70./14.0 = 98 W/m

2

Q = qA = (98 W/m2)(0.093 m

2) = 9.114 W

(a)( )K

W

T

QS rev

300

114.9

1

1

=

=& = - 0.03038 W/K

(b)( )K

W

T

QS rev

230

114.9

2

2

+==& = + 0.03963 W/K

(c) =rS& = 0.0 W/K (see Eq. 1.5, steady state)

(d) =+= 21 SSSUn&&& = - 0.03038 W/K + 0.03963 W/K = + 0.00925 W/K

Since 0UnS& , therefore it satisfied Second Law of Thermodynamics.

1.7 (a) If the thermal energy reservoirs in Problem 1.6 are suddenly replaced with

adiabatic walls, determine the final equilibrium temperature of the slab. (b) What

is the entropy change for the slab for this process? (c) Does your answer satisfythe Second Law of Thermodynamics in this instance? Explain. The density of the

slab is 26 lb/ft3

and the specific heat 0.65 Btu/lb-F.

Solution:

( )

=

3

33

/1

/018.16/26

ftlb

mkgftlb = 416.468 kg/m

3

( )

=

FlbBtu

KkgJFlbBtuc

./1

./8.4186./65.0 = 2721.42 J/kg.K

k= 0.14 W/m.KT= 27 C (-43 C) = 70 CT1 = 27 C + 273 = 300 K

T2 = - 43 C + 273 = 230 K

A = 0.093 m2

L = 0.10 m

(a) =2

1

T

TT

dQ

T

Q


9/15

1. INTRODUCTION

=2

1

T

TTcVdT

TQ

( )

=

1

212 lnT

TcV

T

TTcV

( )

=

1

212 lnT

T

T

TT

( ) ( )

=

=

300

230ln

300230

ln1

2

12

T

T

TTT = 263.45 K

(b)( ) ( )

T

TTcAL

T

TTcV

T

QS 1212

=

==

( )( )( )( )( )

45.263

30023010.0093.042.2721468.416 =S = - 2801 J/K

(c) This will not satisfy the Second Law of Thermodynamic since this is not a rate of

entropy of production of the universe.

1.8 A copper sphere 2.5 cm in diameter has a uniform temperature of 40 C. The

sphere is suspended in a slow-moving air stream at 0 C. The air stream produces a

convection heat transfer coefficient of 15 W/m2.K. Radiation can be neglected.

Since copper is highly conductive, temperature gradients in the sphere willsmooth out rapidly, and its temperature can be taken as uniform throughout the

cooling process (i.e., Bi


10/15

1. INTRODUCTION

( )( )

=

TTcV

Ah

dt

TTd

( ) C

Ah

cV

tTT +

=

ln

at T(t= 0) Ti,

( )

= TTC ln

( ) ( )

+

= TT

Ah

cV

tTT

ilnln

xi T

t

Ah

cV

t

TT

TT=

=

ln

=

Ah

cVTx

xT

t

i

eTT

TT

=

T = 0 C + 273 = 273 K

iT = 40 C + 273 = 313 K

=

Ah

cVTx

3

3

4rV =

r= (1/2)(2.5 cm) = 1.25 cm = 0.0125 m24 rA =

( ) hcr

rh

rc

Ah

cVTx

34

3

4

2

3

=

==

h = 15 W/m2.K

Properties of copper, Table A.1, App. A

= 8954 kg/m3

cp = 384 J/kg.K

= 11.57 x 10-5 m2/s

2

( )( )( )( )KmW

mKkgJmkgTx

./153

0125.0./3843/8954= = 955 sec

Then:


11/15

1. INTRODUCTION

( ) xTt

ieTTTT

=

( )

+= TeTTT x

T

t

i

( ) KeTt

273273313 955 +=

KeT

t

27340 955 +=

95540

t

eT

= C

where tin seconds

Tabulation:

Time, t, seconds Temperature, T, C

0 40

10 39.6

20 39.2

40 38.4

60 37.6

80 36.9

100 36.2

200 32.7

300 29.6

400 26.8

600 22

800 18

1000 14.75000 0.3

10000 0.0

100000 0.0

1000000 0.0

0.0

Plot:


12/15

1. INTRODUCTION

1.9 Determine the total heat transfer in Problem 1.8 as the sphere cools from 40 C to

0 C. Plot the net entropy increase resulting from the cooling process above, S vs

T(K).

Solution:

T = 0 C + 273 = 273 K

24 rA = ,3

3

4rV =

r= 0.0125 m

= 8954 kg/m3

cp = 384 J/kg.K

= 11.57 x 10-5 m2/s

2

T= 40 C 0 C = 40 K

Total Heat Transfer:

Q = cVT = (8954 kg/m3)(384 J/kg.K)(4/3)()(0.0125 m)

3(40 K)

Q = 1125 J - - - - Answer.

Plotting the net-entropy increase:

Equation (1.24)


13/15

1. INTRODUCTION

b

T

Tb

dTTT

cVS

b

b

=

0

11

( )( ) ( )b

T

Tb

dTTT

S

b

b

=

0

110125.0

3

43848954

3

=

0

0lnln13.28

b

b

b

bT

T

TT

T

TS

=

0

0ln13.28

b

bbb

T

T

T

TTS

Tb0 = 40 C = 313 K

=313

ln273

31313.28

bbTT

S

Tb, C Tb, K S

40 313 0

35 308 0.0622

30 303 0.117

25 298 0.1642

20 293 0.2034

15 288 0.2344

10 283 0.2569

5 278 0.2707

0 273 0.2754

Plot:


14/15


15/15

1. INTRODUCTION

T2 = 6 C

dxdTkAQ

=

x

DD

L

DD =

121

D1 = 15 cm = 0.15 m

D2 = 7.5 cm = 0.075 m

L = 30 cm = 0.30 m

x

Dm

m

mm =

15.0

30.0

075.015.0

D = 0.15 m 0.25x

2

4DA =

dx

dTDkQ

=

2

4

( )dx

dTxmkQ

225.015.0

4

=

( ) dTkdxxQ

=

425.015.0

2

( ) dTkdxxQ

m

=

425.015.0

3.0

0

2

Thermal Conductivity of Portland Cement, Table A.2, Appendix A.

k= 0.70 W/m.K

( ) ( )[ ] ( ) ( )6404

70.025.015.025.0

11

3.0

0

1

=

xQ

( ) ( )( ) ( )[ ] ( ) ( )344

70.015.03.025.015.0411

=

Q

( ) ( ) ( )344

70.015.0

1

075.0

14

=

Q

Q = -0.70 W Ans.

solucionario ccc 1

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