Solution manual heat and mass transfer a practical approach 3rd edition cengel CH02 1

The mathematical formulation, the variation of temperature, and the rate of evaporation of nitrogen are to be determined for steady one-dimensional heat transfer.. The mathematical formu

2-129 A spherical liquid nitrogen container is subjected to specified temperature on the inner surface and

convection on the outer surface The mathematical formulation, the variation of temperature, and the rate of evaporation of nitrogen are to be determined for steady one-dimensional heat transfer

Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there

is thermal symmetry about the midpoint 2 Thermal conductivity is constant 3 There is no heat generation

Properties The thermal conductivity of the tank is given to be k = 18 W/m ⋅°C Also, hfg = 198 kJ/kg for nitrogen

Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical

formulation of this problem can be expressed as

dr

d

and T(r1)= T1=−196°C

])([)(

2 2

C

2

1 2

2 1

Solving for C1 and C2 simultaneously gives

1 2

2 1 2

1 1 1

1 1 2

2 1 2

1 2

1

1

and 1

)(

r r hr

k r r

T T T r

C T C hr

k r r

T T r

C

−

−

−+

=+

1.2)m1.2)(

C W/m25(

C W/m182

1.21

C)20196(

1

11)

(

2

1 2 1 2

2 1 2

1 1 1

1 1

1 1 1

−

r r

T r

r r r

hr

k r r

T T T r r

C r

C T r

C r

T

(c) The rate of heat transfer through the wall and the rate of evaporation of nitrogen are determined from

negative)since

tank the(to

W 261,200)

m1.2)(

C W/m25(

C W/m182

1.21

C)20196(m)1.2()C W/m18(4

1

)(44

)4(

2

2 1 2

1 2 1

2 1 2

π

hr

k r r

T T r k kC

r

C r k dx

dT kA

Q&

kg/s 1.32

=

=

=

J/kg000,198

J/s200,261

2-130 A spherical liquid oxygen container is subjected to specified temperature on the inner surface and

convection on the outer surface The mathematical formulation, the variation of temperature, and the rate

of evaporation of oxygen are to be determined for steady one-dimensional heat transfer

Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there

is thermal symmetry about the midpoint 2 Thermal conductivity is constant 3 There is no heat generation

Properties The thermal conductivity of the tank is given to be k = 18 W/m ⋅°C Also, hfg = 213 kJ/kg for

oxygen

Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical

formulation of this problem can be expressed as

dr

d

and T(r1)= T1=−183°C

C

2

1 2

2 1

Solving for C1 and C2 simultaneously gives

1 2

2 1 2

1 1 1

1 1 2

2 1 2

1 2

1

1

and 1

)(

r r hr

k r r

T T T r

C T C hr

k r r

T T r

C

−

−

−+

=+

1.2)m1.2)(

C W/m25(

C W/m182

1.21

C)20183(

1

11)

(

2

1 2 1 2

2 1 2

1 1 1

1 1

1 1 1

−

r r

T r

r r r

hr

k r r

T T T r r

C r

C T r

C r

T

(c) The rate of heat transfer through the wall and the rate of evaporation of oxygen are determined from

1

)(44

)4(

2

1 2 1

2 1 2

T T r k kC

r

C r k dx

dT kA

Q&

2-131 A large plane wall is subjected to convection, radiation, and specified temperature on the right

surface and no conditions on the left surface The mathematical formulation, the variation of temperature

in the wall, and the left surface temperature are to be determined for steady one-dimensional heat transfer

Assumptions 1 Heat conduction is steady and one-dimensional since the wall is large relative to its

thickness, and the thermal conditions on both sides of the wall are uniform 2 Thermal conductivity is constant 3 There is no heat generation in the wall

Properties The thermal conductivity and emissivity are given to be k = 8.4 W/m⋅°C and ε = 0.7

Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left

surface, and the mathematical formulation of this problem can be expressed as

T(L)= T2 =45°C

45°Cε

T T h C

T T

T T h kC

/]}

)273[(

][

{

])

273[(

][

4 surr

4 2 2

1

4 surr

4 2 2

1

−+εσ+

m )4.0(C

W/m4.8

]K)290()K318)[(

K W/m100.7(5.67+

C)2545)(

W/m14(C45

)(])273[(

][)

()()

(

4 4

4 2 8 2

4 surr

4 2 2

2 1 2

1 2 1

x

x

x L k

T T

T T h T C x L T L C T x C

x

T

−+

°

=

−

−+εσ+

−+

=

−+

2-132 The base plate of an iron is subjected to specified heat flux on the left surface and convection and

radiation on the right surface The mathematical formulation, and an expression for the outer surface temperature and its value are to be determined for steady one-dimensional heat transfer

Assumptions 1 Heat conduction is steady and one-dimensional 2 Thermal conductivity is constant 3

There is no heat generation 4 Heat loss through the upper part of the iron is negligible

Properties The thermal conductivity and emissivity are given to be k = 18 W/m⋅°C and ε = 0.7

Analysis (a) Noting that the upper part of the iron is well insulated

and thus the entire heat generated in the resistance wires is transferred

to the base plate, the heat flux through the inner surface is determined

4 base

0

m10150

Taking the direction normal to the surface of the wall to be the x

direction with x = 0 at the left surface, the mathematical formulation

of this problem can be expressed as

][])([])([)

surr

4 2 2

4 surr

L T T

L T h dx

L dT

4 2

4 2 4 2 8 2

2 C)( 26) 0.7(5.67 10 W/m K )[( 273) 295 ] 66,667 W/m W/m

30

2-133 The base plate of an iron is subjected to specified heat flux on the left surface and convection and

radiation on the right surface The mathematical formulation, and an expression for the outer surface temperature and its value are to be determined for steady one-dimensional heat transfer

Assumptions 1 Heat conduction is steady and one-dimensional 2 Thermal conductivity is constant 3

There is no heat generation 4 Heat loss through the upper part of the iron is negligible

Properties The thermal conductivity and emissivity are given to be k = 18 W/m⋅°C and ε = 0.7

Analysis (a) Noting that the upper part of the iron is well insulated

and thus the entire heat generated in the resistance wires is transferred

to the base plate, the heat flux through the inner surface is determined

4 base

0

m10150

Taking the direction normal to the surface of the wall to be the x

direction with x = 0 at the left surface, the mathematical formulation

of this problem can be expressed as

273[(

][

])

([])([

)

surr

4 2 2

4 surr

L T T

L T

4 2 4 2 8 2

2 C)( 26) 0.7(5.67 10 W/m K )[( 273) 295 ] 100,000 W/m W/m

2-134E The concrete slab roof of a house is subjected to specified temperature at the bottom surface and

convection and radiation at the top surface The temperature of the top surface of the roof and the rate of heat transfer are to be determined when steady operating conditions are reached

Assumptions 1 Steady operating conditions are reached 2 Heat transfer is one-dimensional since the roof

area is large relative to its thickness, and the thermal conditions on both sides of the roof are uniform 3 Thermal properties are constant 4 There is no heat generation in the wall

Properties The thermal conductivity and emissivity are given to be k = 1.1 Btu/h⋅ft⋅°F and ε = 0.8

Analysis In steady operation, heat conduction through the roof must be equal to net heat transfer from the

outer surface Therefore, taking the outer surface temperature of the roof to be T2 (in °F),

])

460[(

4 2 8

2 2 2

R]310)460)[(

RftBtu/h10

FftBtu/h2.3(ft0.8

F)62()

Using an equation solver (or the trial and error method), the outer surface temperature is determined to be

T2 = 38°F

Then the rate of heat transfer through the roof becomes

Btu/h 28,875

F)3862()ft3525)(

FftBtu/h1.1

2 1

L

T T kA

Q&

Discussion The positive sign indicates that the direction of heat transfer is from the inside to the outside

Therefore, the house is losing heat as expected

2-135 The surface and interface temperatures of a resistance wire covered with a plastic layer are to be

determined

Assumptions 1 Heat transfer is steady since there is no change with time 2 Heat transfer is

one-dimensional since this two-layer heat transfer problem possesses symmetry about the center line and

involves no change in the axial direction, and thus T = T(r) 3 Thermal conductivities are constant 4 Heat

generation in the wire is uniform

Properties It is given that kwire=18 W/m⋅°C and kplastic=1.8 W/m⋅°C

Analysis Letting denote the unknown interface temperature, the mathematical formulation of the heat transfer problem in the wire can be expressed as

dT r

Multiplying both sides of the differential

equation by r, rearranging, and integrating give

k

e dr

r k

e dr

dT

r =−& + (a) Applying the boundary condition at the center (r = 0) gives

2

)0(

0× =− gen × +C1 → C1=

k

e dr

dT &

Dividing both sides of Eq (a) by r to bring it to a readily integrable form and integrating,

r k

e dr

k

e r

2 2 1 gen

4

e T C C

r k

−

=Substituting this C2 relation into Eq (b) and rearranging give

)(

4)

The solution of the differential equation is determined by integration to be

1

C dr

dT

r = →

r

C dr

dT = 1 → T(r)=C1lnr+C2where C1 and C2 are arbitrary constants Applying the boundary conditions give

r = r1: C1lnr1+C2 =T I → C2 =T I −C1lnr1

ln

hr

k r r T T

plastic 1

2 1

1 1

ln

lnln

)(

r r

hr

k r r

T T T r C T r C r

+

−+

=

−+

We have already utilized the first interface condition by setting the wire and plastic layer temperatures equal to at the interface The interface temperature is determined from the second interface condition that the heat flux in the wire and the plastic layer at

plastic 1

2 plastic 1

gen 1

plastic plastic 1

wire wire

1ln

2 )()

(

r hr

k r r

T T k

r e dr

r dT k dr

r dT

°+

=C25m)C)(0.007 W/m

(14

C W/m1.8m

0.003

m007.0lnC)

W/m2(1.8

m)003.0)(

W/m105

1

(

ln2

2

2 3

6

2

plastic 1

2 plastic

2 1 gen

T hr

k r

r k

r e

=

C) W/m(184

m)003.0)(

W/m105.1(+C1.974

)

0

(

2 3

6

wire

2 1 gen wire

k

r e T

Thus the temperature of the centerline will be slightly above the interface temperature

2-136 A cylindrical shell with variable conductivity is

subjected to specified temperatures on both sides The rate of

heat transfer through the shell is to be determined

Assumptions 1 Heat transfer is given to be steady and

one-dimensional 2 Thermal conductivity varies quadratically 3

There is no heat generation

Properties The thermal conductivity is given to be

)1

(

)

(T k0 T2

Analysis When the variation of thermal conductivity with

temperature k(T) is known, the average value of the thermal

conductivity in the temperature range between is

determined from

2

1 and T T

2 2 0

1 2

3 1

3 2 1

2 0

1 2

3 0

1 2

2 0

1 2

avg

31

33

)1()

(

2

1 2

1 2

1

T T T T k

T T

T T T

T k T

T

T T k T

T

dT T k T

T

dT T k

k

T

T T

T T

T

β

ββ

2

1 2

2 1 2 1 2 1

2 2 0

1 2

2 1 avg cylinder

r r

T T L T T T T k

r r

T T L k

Discussion We would obtain the same result if we substituted the given k(T) relation into the second part

of Eq 2-77, and performed the indicated integration

2-137 Heat is generated uniformly in a cylindrical uranium fuel rod The temperature difference between

the center and the surface of the fuel rod is to be determined

Assumptions 1 Heat transfer is steady since there is no indication of any change with time 2 Heat transfer

is one-dimensional since there is thermal symmetry about the center line and no change in the axial

direction 3 Thermal conductivity is constant 4 Heat generation is uniform

Properties The thermal conductivity of uranium at room temperature is k = 27.6 W/m⋅°C (Table A-3)

Analysis The temperature difference between the center

6.27(4

m)016.0)(

W/m104(4

2 3

7 2

2-138 A large plane wall is subjected to convection on the inner and outer surfaces The mathematical

formulation, the variation of temperature, and the temperatures at the inner and outer surfaces to be determined for steady one-dimensional heat transfer

Assumptions 1 Heat conduction is steady and one-dimensional 2 Thermal conductivity is constant 3

There is no heat generation

Properties The thermal conductivity is given to be k = 0.77 W/m⋅°C

Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the

inner surface, the mathematical formulation of this problem can be expressed as

T

)]

0([ 1

− ( ) =h2[T(L)−T∞2]

dx

L dT

(

)82

.0)(

12(77

T( )=20−45.45

(c) The temperatures at the inner and outer surfaces are

C 10.9

C 20

(

045.4520

2-139 A hollow pipe is subjected to specified temperatures at the inner and outer surfaces There is also

heat generation in the pipe The variation of temperature in the pipe and the center surface temperature of the pipe are to be determined for steady one-dimensional heat transfer

Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its

thickness, and there is thermal symmetry about the centerline 2 Thermal conductivity is constant

Properties The thermal conductivity is given to be k = 14 W/m⋅°C

Analysis The rate of heat generation is determined from

2 2

2 1

2 2

4/)m17(m)3.0(m)4.0(

W000,254

/)

& V

Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this

problem can be expressed as

dT r dr

r e dr

r e dr

2 gen

ln4

)

k

r e r

where C1 and C2 are arbitrary constants Applying the boundary conditions give

2 1 gen

4)

k

r e r

2 2 gen

4)

k

r e r

Substituting the given values, these equations can be written as

2 1

2

)15.0ln(

)14(4

)15.0)(

750,26(

2 1

2

)20.0ln(

)14(4

)20.0)(

750,26(

Solving for C1 and C2 simultaneously gives

8.257

58

r r

r

)14(4

750,26)

The temperature at the center surface of the pipe is determined by setting radius r to be 17.5 cm, which is

the average of the inner radius and outer radius

C 71.3°

=+

2-140 Heat is generated in a plane wall Heat is supplied from one side which is insulated while the other

side is subjected to convection with water The convection coefficient, the variation of temperature in the wall, and the location and the value of the maximum temperature in the wall are to be determined

Assumptions 1 Heat transfer is steady since there is no indication of any change with time 2 Heat transfer

is one-dimensional since the wall is large relative to its thickness 3 Thermal conductivity is constant 4

Heat generation is uniform

Analysis (a) Noting that the heat flux and the heat

generated will be transferred to the water, the heat

transfer coefficient is determined from the Newton’s

m))(0.04 W/m(10) W/m

T d k e

e T b

x k

e dx

dT

++

−

=+

−

2

&

→

2 3

5 gen

C/m2500)

C W/m20(2

W/m10

)

(x =− x2+ x+

T

This part could also be answered to without any information about the nature of the T(x) function, using qualitative arguments only At steady state, heat cannot go from right to left at any location There is no way out through the left surface because of the adiabatic insulation, so it would accumulate somewhere, contradicting the steady state assumption Therefore, the temperature must continually decrease from left to

right, and Tmax is at x = L

2-141 Heat is generated in a plane wall The temperature distribution in the wall is given The surface

temperature, the heat generation rate, the surface heat fluxes, and the relationship between these heat fluxes, the heat generation rate, and the geometry of the wall are to be determined

Assumptions 1 Heat transfer is steady since there is no indication of any change with time 2 Heat transfer

is one-dimensional since the wall is large relative to its thickness 3 Thermal conductivity is constant 4

Heat generation is uniform

Analysis (a) The variation of temperature is symmetric about x = 0 The surface temperature is

C 72°

W/m 4000

C/m102(C) W/m5(2)(2()

(

m)02.0)(

C/m102(C) W/m5(2)2()

(

2 4

2 4

L b k dx

dT k L

q

bL k dx

dT k L

q

L s

L s