Solution manual heat and mass transfer a practical approach 3rd edition cengel CH04

Transient Heat Conduction in Multidimensional Systems 4-74C The product solution enables us to determine the dimensionless temperature of two- or three-dimensional heat transfer problem

Transient Heat Conduction in Multidimensional Systems

4-74C The product solution enables us to determine the dimensionless temperature of two- or

three-dimensional heat transfer problems as the product of dimensionless temperatures of one-three-dimensional heat transfer problems The dimensionless temperature for a two-dimensional problem is determined by determining the dimensionless temperatures in both directions, and taking their product

4-75C The dimensionless temperature for a three-dimensional heat transfer is determined by determining

the dimensionless temperatures of one-dimensional geometries whose intersection is the three dimensional geometry, and taking their product

4-76C This short cylinder is physically formed by the intersection of a long cylinder and a plane wall The

dimensionless temperatures at the center of plane wall and at the center of the cylinder are determined first Their product yields the dimensionless temperature at the center of the short cylinder

4-77C The heat transfer in this short cylinder is one-dimensional since there is no heat transfer in the axial

direction The temperature will vary in the radial direction only

4-78 A short cylinder is allowed to cool in atmospheric air The temperatures at the centers of the cylinder

and the top surface as well as the total heat transfer from the cylinder for 15 min of cooling are to be determined

Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in

both the axial x- and the radial r- directions 2 The thermal properties of the cylinder are constant 3 The

heat transfer coefficient is constant and uniform over the entire surface 4 The Fourier number is τ > 0.2 so

that the one-term approximate solutions (or the transient temperature charts) are applicable (this

assumption will be verified)

Properties The thermal properties of brass are given to be ρ=8530kg/m3,c p =0.389kJ/kg⋅°C,

Analysis This short cylinder can physically be formed by the intersection of a long cylinder of radius D/2 =

4 cm and a plane wall of thickness 2L = 15 cm We measure x from the midplane

(a) The Biot number is calculated for the plane wall to be

02727.0)

C W/m

110(

)m075.0)(

C W/m40

The constants λ1andA1corresponding to this

Biot number are, from Table 4-2,

0045.1 and 1620

075.0(

s/min)60min/s)(15m1039.3(

2

2 5

0045.1( (0.1620) (5.424)

1 0

,

0

2 2

T T

i wall

τ λ

θ

We repeat the same calculations for the long cylinder,

01455.0)

C W/m

110(

)m04.0)(

C W/m40

04.0(

s)60/s)(15m1039.3(

2

2 5

0036.1( (0.1677 (19.069)

1 ,

2 2

T T

i

o cyl

o

τ λ

θ

Then the center temperature of the short cylinder becomes

C 86.4°

.020150

20),

t T t

T

T T

T t T

cyl o wall o cylinder short i

θθ

(b) The center of the top surface of the cylinder is still at the center of the long cylinder (r = 0), but at the outer surface of the plane wall (x = L) Therefore, we first need to determine the dimensionless temperature

at the surface of the wall

)0045.1()/cos(

),()

T t x T t

.020150

20),

0

,

(

505.0587.0860.0)

,()

L

T

t L T

T

T t L

T

cyl o wall cylinder

short i

θθ

(c) We first need to determine the maximum heat can be transferred from the cylinder

kJ325C)20150)(

CkJ/kg

389.0)(

kg43.6()(

kg43.6m)15.0(m)04.0()kg/m8530(max

2.

3 2

L r m

i p

ρπρV

Then we determine the dimensionless heat transfer ratios for both geometries as

133.01620.0

)1620.0sin(

)871.0(1)sin(

1

1

1 ,

Q

415.01677.0

0835.0)587.0(21)(2

1

1

1 1 , max

The heat transfer ratio for the short cylinder is

max max

max max

=

−+

long wall

plane cylinder

Q Q

Q Q

Q Q

Q

Then the total heat transfer from the short cylinder during the first 15 minutes of cooling becomes

kJ 160

=

=

=0.493Qmax (0.493)(325kJ)

Q

4-79 EES Prob 4-78 is reconsidered The effect of the cooling time on the center temperature of the cylinder, the center temperature of the top surface of the cylinder, and the total heat transfer is to be investigated

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

"From Table 4-2 corresponding to this Bi number, we read"

lambda_1_w=0.1620 "w stands for wall"

A_1_w=1.0045

tau_w=(alpha*time*Convert(min, s))/L^2

theta_o_w=A_1_w*exp(-lambda_1_w^2*tau_w) "theta_o_w=(T_o_w-T_infinity)/(T_i-T_infinity)"

"For long cylinder"

Bi_c=(h*r_o)/k "c stands for cylinder"

"From Table 4-2 corresponding to this Bi number, we read"

"theta_L_w=(T_L_w-(T_L_o-T_infinity)/(T_i-T_infinity)=theta_L_w*theta_o_c "center temperature of the top surface"

J_1=0.0835 "From Table 4-3, at lambda_1_c"

Q/Q_max=Q_w+Q_c*(1-Q_w) "total heat transfer"

time [min] T o,o [C] T L,o [C] Q [kJ]

4-80 A semi-infinite aluminum cylinder is cooled by water The temperature at the center of the cylinder 5

cm from the end surface is to be determined

Assumptions 1 Heat conduction in the semi-infinite cylinder is two-dimensional, and thus the temperature

varies in both the axial x- and the radial r- directions 2 The thermal properties of the cylinder are constant

3 The heat transfer coefficient is constant and uniform over the entire surface 4 The Fourier number is τ >

0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified)

Properties The thermal properties of aluminum are given to be k = 237 W/m.°C and α = 9.71×10-5

m2/s

Analysis This semi-infinite cylinder can physically be formed by the intersection of a long cylinder of

radius r o = D/2 = 7.5 cm and a semi-infinite medium The dimensionless temperature 5 cm from the surface

of a semi-infinite medium is first determined from

8951.01049.01)

,(

1049.0)7308.0)(

0468.1(8699.0)2433.0()0458.0exp(

)1158.0(

237

)608)(

1071.9()140()608)(

1071.9(2

05.0

)237(

)608)(

1071.9()140(237

)05.0)(

140(exp)608)(

1071.9(2

05.0

2exp

5

2 2

T t x T

erfc erfc

erfc erfc

k

t h t

x erfc k

t h k

hx t

x erfc T

i

i

θ

αα

αα

D 0 = 15 cm

z

Semi-infinite cylinder

237

)m075.0)(

C W/m140

The constants λ1andA1corresponding to this

Biot number are, from Table 4-2,

λ1 = 0.2948 and A1 = 1.0110

The Fourier number is

2.0286.8m)

075.0(

s)60/s)(8m1071.9(

2

2 5

0110.1( (0.2948) (8.286)

1 ,

2 2

T T

i

o cyl

o

τ λ

θ

The center temperature of the semi-infinite cylinder then becomes

C 56.3°

.010

115

10),0

,

(

4405.04921.08951.0)

,()

, inf

semi cylinder

infinite semi

t x T t

x

T

t x T

T

T t x

T

cyl o i

θθ

4-81E A hot dog is dropped into boiling water The center temperature of the hot dog is do be determined

by treating hot dog as a finite cylinder and also as an infinitely long cylinder

Assumptions 1 When treating hot dog as a finite cylinder, heat conduction in the hot dog is

two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions When treating

hot dog as an infinitely long cylinder, heat conduction is one-dimensional in the radial r- direction 2 The

thermal properties of the hot dog are constant 3 The heat transfer coefficient is constant and uniform over the entire surface 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the

transient temperature charts) are applicable (this assumption will be verified)

Properties The thermal properties of the hot dog are given to be k = 0.44 Btu/h.ft.°F, ρ = 61.2 lbm/ft3

cp = 0.93 Btu/lbm.°F, and α = 0.0077 ft2

/h

Analysis (a) This hot dog can physically be formed by the intersection of a long cylinder of radius r o = D/2

= (0.4/12) ft and a plane wall of thickness 2L = (5/12) ft The distance x is measured from the midplane

44.0(

)ft12/5.2)(

F.Btu/h.ft120

The constants λ1andA1corresponding to this

Biot number are, from Table 4-2,

2728.1 and 5421

12/5.2(

h)/h)(5/60ft

0077.0(

2 2

2728.1( (1.5421) (0.015)

1 0

,

0

2 2

T T

i wall

τ λ

θ

We repeat the same calculations for the long cylinder,

1.9)

FBtu/h.ft

44.0(

)ft12/4.0)(

F.Btu/h.ft120

12/4.0(

h)/h)(5/60ft

0077.0(

2 2

5618.1( (2.1589 (0.578)

1 ,

2 2

T T

i

o cyl

.021240

212),

t T t

T

T T

T t T

cyl o wall o i

θθ

After 10 minutes

2.003.0ft)

12/5.2(

h)/h)(10/60ft

0077.0(

2 2

2728.1( (1.5421) (0.03)

1 0

,

0

2 2

T T

i wall

τ λ

θ

2.0156.1ft)

12/4.0(

h)/h)(10/60ft

0077.0(

2 2

5618.1( (2.1589 (1.156)

1 ,

2 2

T T

i

o cyl

.021240

212),0,0(

0084.00071.0185.1)

t T t

T

T T

T t T

cyl o wall o i

θθ

After 15 minutes

2.0045.0ft)

12/5.2(

h)/h)(15/60ft

0077.0(

2 2

2728.1( (1.5421) (0.045)

1 0

,

0

2 2

T T

i wall

τ λ

θ

2.0734.1ft)

12/4.0(

h)/h)(15/60ft

0077.0(

2 2

5618.1( (2.1589) (1.734)

1 0

,

0

2 2

T T

.021240

212),

t T t

T

T T

T t T

cyl o wall o i

θθ

(b) Treating the hot dog as an infinitely long cylinder will not change the results obtained in the part (a)

since dimensionless temperatures for the plane wall is 1 for all cases

4-82E A hot dog is dropped into boiling water The center temperature of the hot dog is do be determined

by treating hot dog as a finite cylinder and an infinitely long cylinder

Assumptions 1 When treating hot dog as a finite cylinder, heat conduction in the hot dog is

two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions When treating

hot dog as an infinitely long cylinder, heat conduction is one-dimensional in the radial r- direction 2 The

thermal properties of the hot dog are constant 3 The heat transfer coefficient is constant and uniform over the entire surface 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the

transient temperature charts) are applicable (this assumption will be verified)

Properties The thermal properties of the hot dog are given to be k = 0.44 Btu/h.ft.°F, ρ = 61.2 lbm/ft3

cp = 0.93 Btu/lbm.°F, and α = 0.0077 ft2

/h

Analysis (a) This hot dog can physically be formed by the intersection of a long cylinder of radius r o = D/2

= (0.4/12) ft and a plane wall of thickness 2L = (5/12) ft The distance x is measured from the midplane

44.0(

)ft12/5.2)(

F.Btu/h.ft120

The constants λ1andA1corresponding to this

Biot number are, from Table 4-2,

2728.1 and 5421

12/5.2(

h)/h)(5/60ft

0077.0(

2 2

2728.1( (1.5421) (0.015)

1 0

,

0

2 2

T T

i wall

τ λ

θ

We repeat the same calculations for the long cylinder,

1.9)

FBtu/h.ft

44.0(

)ft12/4.0)(

F.Btu/h.ft120

12/4.0(

h)/h)(5/60ft

0077.0(

2 2

5618.1( (2.1589) (0.578)

1 0

,

0

2 2

T T

.020240

202),

t T t

T

T T

T t T

cyl o wall o i

θθ

After 10 minutes

2.003.0ft)

12/5.2(

h)/h)(10/60ft

0077.0(

2 2

2728.1( (1.5421) (0.03)

1 0

,

0

2 2

T T

i wall

τ λ

θ

2.0156.1ft)

12/4.0(

h)/h)(10/60ft

0077.0(

2 2

5618.1( (2.1589) (1.156)

1 0

,

0

2 2

T T

.020240

202),0,0(

0084.00071.0185.1)

t T t

T

T T

T t T

cyl o wall o i

θθ

After 15 minutes

2.0045.0ft)

12/5.2(

h)/h)(15/60ft

0077.0(

2 2

2728.1( (1.5421) (0.045)

1 0

,

0

2 2

T T

i wall

τ λ

θ

2.0734.1ft)

12/4.0(

h)/h)(15/60ft

0077.0(

2 2

5618.1( (2.1589) (1.734)

1 0

,

0

2 2

T T

.020240

202),

t T t

T

T T

T t T

cyl o wall o i

θθ

(b) Treating the hot dog as an infinitely long cylinder will not change the results obtained in the part (a) since dimensionless temperatures for the plane wall is 1 for all cases

4-83 A rectangular ice block is placed on a table The time the ice block starts melting is to be determined

Assumptions 1 Heat conduction in the ice block is two-dimensional, and thus the temperature varies in

both x- and y- directions 2 The thermal properties of the ice block are constant 3 The heat transfer

coefficient is constant and uniform over the entire surface 4 The Fourier number is τ > 0.2 so that the

one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified)

Properties The thermal properties of the ice are given to be k = 2.22 W/m.°C and α = 0.124×10-7

m2/s

Analysis This rectangular ice block can be treated as a short

rectangular block that can physically be formed by the intersection

of two infinite plane wall of thickness 2L = 4 cm and an infinite

plane wall of thickness 2L = 10 cm We measure x from the

bottom surface of the block since this surface represents the

adiabatic center surface of the plane wall of thickness 2L = 10 cm

Since the melting starts at the corner of the top surface, we need to

determine the time required to melt ice block which will happen

when the temperature drops below 0°C at this location The Biot

numbers and the corresponding constants are first determined to

be

Ice block -20 °C

Air 18°C

1081.0)C W/m

22.2(

)m02.0)(

C W/m12

22.2(

)m05.0)(

C W/m12

The ice will start melting at the corners because of the maximum exposed surface area there Noting that

and assuming that τ > 0.2 in all dimensions so that the one-term approximate solution for transient heat conduction is applicable, the product solution method can be written for this problem as

)4951.0cos(

)05.0(

)10124.0()4951.0(exp)0408.1(

)3208.0cos(

)02.0(

)10124.0()3208.0(exp)0173.1(4737.0

)/cos(

)/cos(

1820

180

),(),()

,,

,

(

2

7 2

2

2

7 2

3 3 1 1

2 1 1 1 1

wall,2 3 2 wall,1 1 3

2

1

2 2

t

t t

L L e

A L L e

A

t L t

L t

L

L

λλ

θθ

θ

τ λ τ

λ

Therefore, the ice will start melting in about 21 hours

Discussion Note that

2.0384.0m)

05.0(

s/h)/s)(77,500m

10124.0(

2

2 7

4-84 EES Prob 4-83 is reconsidered The effect of the initial temperature of the ice block on the time period before the ice block starts melting is to be investigated

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

"From Table 4-2 corresponding to this Bi number, we read"

lambda_1_w1=0.3208 "w stands for wall"

4-85 A cylindrical ice block is placed on a table The initial temperature of the ice block to avoid melting for 2 h is to be determined

Assumptions 1 Heat conduction in the ice block is two-dimensional, and thus the temperature varies in

both x- and r- directions 2 Heat transfer from the base of the ice block to the table is negligible 3 The

thermal properties of the ice block are constant 4 The heat transfer coefficient is constant and uniform over the entire surface 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the

transient temperature charts) are applicable (this assumption will be verified)

Air

T∞ = 24°C

x

r (r o , L)

Analysis This cylindrical ice block can be treated as a short

cylinder that can physically be formed by the intersection of a

long cylinder of diameter D = 2 cm and an infinite plane wall

of thickness 2L = 4 cm We measure x from the bottom surface

of the block since this surface represents the adiabatic center

surface of the plane wall of thickness 2L = 4 cm The melting

starts at the outer surfaces of the top surface when the

temperature drops below 0°C at this location The Biot

numbers, the corresponding constants, and the Fourier numbers

are

1171.0)C W/m

22.2(

)m02.0)(

C W/m13

22.2(

)m01.0)(

C W/m13

02.0(

s/h)6003h/s)(3m10124.0(

2

2 7 2

L t

ατ

2.03392.1m)

01.0(

s/h)6003h/s)(3m10124.0(

2

2 7 2

o r

t

α

τ

Note that τ > 0.2 in all dimensions and thus the one-term approximate solution for transient

heat conduction is applicable The product solution for this problem can be written as

24

240

)/()

/cos(

24

240

),(),()

,

,

(

) 3392 1 ( ) 3393 0 ( )

3348 0 ( 3319 0 (

1 0 1 1

1

cyl wall

block

2 2

2 2

T

r r J e A L L e

A T

t r t L t

r

L

i

o o i

o o

λλ

]

θθ

θ

τ λ τ

λ

which gives T i =−6.6°C

Therefore, the ice will not start melting for at least 3 hours if its initial temperature is -6.6°C or below

4-86 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces The center temperatures of each geometry in 10, 20, and 60 min are to be determined

Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in

all x-, y, and z- directions 2 Heat conduction in the cylindrical block is two-dimensional, and thus the temperature varies in both axial x- and radial r- directions 3 The thermal properties of the granite are

constant 4 The heat transfer coefficient is constant and uniform over the entire surface 5 The Fourier

number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified)

Properties The thermal properties of the granite are given to be k = 2.5 W/m.°C and α = 1.15×10-6

5.2(

)m025.0)(

C W/m40

s/min)60min/s)(10m1015

1

(

2

2 6

369.0)

0580.1(50020

500),0

,0,0

,

0

(

3 ) 104 1 ( ) 5932 0 (

3 1

3 wall block

2 2

t T

e t

T

e A T

T

T t T

t t

i

τ λ

θθ

After 20 minutes

2.0208.2m)

025.0(

s/min)60min/s)(20m1015

.0)

0580.1(500

t T e

t

T

After 60 minutes

2.0624.6m)

025.0(

s/min)60min/s)(60m1015.1(

2

2 6

.0)

0580.1(50020

500),0

e t

T

Note that τ > 0.2 in all dimensions and thus the one-term approximate solution for transient heat

conduction is applicable