HT3eChap2_1_69

(1)

2-1 2-1

Chapter 2

Chapter 2 HEAT CONDUCTION EQUATION

HEAT CONDUCTION EQUATION

Introduction Introduction 2-1C

2-1C Heat transfer is aHeat transfer is avector vector quantity since it has direction as well as magnitude. Therefore, we mustquantity since it has direction as well as magnitude. Therefore, we must specify both direction and magnitude in order to describe heat transfer completely at a point. Temperature, specify both direction and magnitude in order to describe heat transfer completely at a point. Temperature, on the other hand, is

on the other hand, is a scalar quantity.a scalar quantity.

2-2C

2-2C The termThe term steady steady impliesimpliesno change with time at any point within the medium whileno change with time at any point within the medium whiletransient transient impliesimplies variation with time

variation with time orortime dependence. Therefore, the temperature or heat flux remains unchanged withtime dependence.Therefore, the temperature or heat flux remains unchanged with time during steady heat transfer through a medium at any location although both quantities may vary from time during steady heat transfer through a medium at any location although both quantities may vary from one location to another.

one location to another. During transient heat tDuring transient heat transfer, the temperaturransfer, the temperature and heat flux mae and heat flux may vary with timey vary with time as well as location. Heat transfer is one-dimensional if it occurs primarily in one direction. It is as well as location. Heat transfer is one-dimensional if it occurs primarily in one direction. It is two-dimensional if heat tranfer in the

dimensional if heat tranfer in the third dimension is negligible.third dimension is negligible.

2-3C

2-3C Heat transfer to a canned drink can Heat transfer to a canned drink can be modeled as two-dimensional since temperature differencesbe modeled as two-dimensional since temperature differences (and thus heat transfer) will exist in the radial and axial directions (but there will be

(and thus heat transfer) will exist in the radial and axial directions (but there will be symmetry about thesymmetry about the center line and no heat transfer in the azimuthal direction. This would

center line and no heat transfer in the azimuthal direction. This would be a transient heat transfer processbe a transient heat transfer process since the temperature at any point within the drink will change

since the temperature at any point within the drink will change with time during heating. Also, we wouldwith time during heating. Also, we would use the cylindrical coordinate system to solve

use the cylindrical coordinate system to solve this problem since a cylinder is this problem since a cylinder is best described in cylindricalbest described in cylindrical coordinates. Also, we would place

coordinates. Also, we would place the origin somewhere on the center line, possibly at the center of the origin somewhere on the center line, possibly at the center of thethe bottom surface.

bottom surface.

2-4C

2-4C Heat transfer to a potato in an Heat transfer to a potato in an oven can be modeled as one-dimensional since temperature differencesoven can be modeled as one-dimensional since temperature differences (and thus heat transfer) will exist in the radial direction only because of symmetry about the center point. (and thus heat transfer) will exist in the radial direction only because of symmetry about the center point. This would be a transient heat transfer process since the

This would be a transient heat transfer process since the temperature at any point within the potato willtemperature at any point within the potato will change with time during cooking. Also, we

change with time during cooking. Also, we would use the spherical coordinate system to solve thiswould use the spherical coordinate system to solve this problem since the entire outer surface of a spherical body can

problem since the entire outer surface of a spherical body can be described by a constant value of the be described by a constant value of the radiusradius in spherical coordinates. We would place the origin at

in spherical coordinates. We would place the origin at the center of the potato.the center of the potato.

2-5C

2-5C Assuming the egg to be round, Assuming the egg to be round, heat transfer to an egg in heat transfer to an egg in boiling water can be modeled as one-boiling water can be modeled as one-dimensional since temperature differences (and thus heat transfer) will

dimensional since temperature differences (and thus heat transfer) will primarily exist in the radialprimarily exist in the radial direction only because of symmet

direction only because of symmetry about the center point. ry about the center point. This would be a transient heat transfer processThis would be a transient heat transfer process since the temperature at any point within the egg will change with

since the temperature at any point within the egg will change with time during cooking. Also, we wtime during cooking. Also, we w ouldould use the spherical coordinate system to solve this problem since the entire outer surface of a spherical body use the spherical coordinate system to solve this problem since the entire outer surface of a spherical body can be described by a constant value

can be described by a constant value of the radius in spherical coordinates. We would of the radius in spherical coordinates. We would place the origin atplace the origin at the center of the egg.

the center of the egg.

2-6C

2-6C Heat transfer to a hot dog Heat transfer to a hot dog can be modeled as two-dimensional since temperature differences (and thuscan be modeled as two-dimensional since temperature differences (and thus heat transfer) will exist in the radial and axial directions (but there will be symmetry about the center line heat transfer) will exist in the radial and axial directions (but there will be symmetry about the center line and no heat transfer in the azimuthal direction. This would be

and no heat transfer in the azimuthal direction. This would be a transient heat transfer process since thea transient heat transfer process since the temperature at any point within the hot dog will change w

temperature at any point within the hot dog will change w ith time during cooking. Also, we ith time during cooking. Also, we would use thewould use the cylindrical coordinate system to solve this problem since

cylindrical coordinate system to solve this problem since a cylinder is best a cylinder is best described in cylindricaldescribed in cylindrical coordinates. Also, we would place

coordinates. Also, we would place the origin somewhere on the center line, possibly at the center the origin somewhere on the center line, possibly at the center of the hotof the hot dog. Heat transfer in a

dog. Heat transfer in a very long hot dog could be very long hot dog could be considered to be one-dimensional in preliminaryconsidered to be one-dimensional in preliminary calculations.

calculations.

PROPRIETARY MATERIAL

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. . © 2007 The McGraw-Hill Companies, Inc. Limited distribuLimited distribution permitted only to teachers andtion permitted only to teachers and educators for course preparation.

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2-2 2-2

2-7C

2-7C Heat transfer to a roast beef in an Heat transfer to a roast beef in an oven would be transient since the temperature at any point withinoven would be transient since the temperature at any point within the roast will change with time during cooking. Also,

the roast will change with time during cooking. Also, by approximating the roast as a spherical object, thisby approximating the roast as a spherical object, this heat transfer process can be modeled as one-dimensional since temperature differences (and thus heat heat transfer process can be modeled as one-dimensional since temperature differences (and thus heat transfer) will primarily exist in the radial direction because

transfer) will primarily exist in the radial direction because of symmetry about the center of symmetry about the center point.point.

2-8C

2-8C Heat loss from a hot water tank in Heat loss from a hot water tank in a house to the surrounding medium can be a house to the surrounding medium can be considered to be aconsidered to be a steady heat transfer problem. Also, it can be considered to

steady heat transfer problem. Also, it can be considered to be two-dimensional since temperaturebe two-dimensional since temperature differences (and thus heat transfer) will exist in the radial and axial directions (but there will be

differences (and thus heat transfer) will exist in the radial and axial directions (but there will be symmetrysymmetry about the center line and no heat transfer in

about the center line and no heat transfer in the azimuthal direction.)the azimuthal direction.)

2-9C

2-9C Yes, the heat flux vector at a pointYes, the heat flux vector at a point P P on an isothermal surface of a medium has to be perpendicular toon an isothermal surface of a medium has to be perpendicular to the surface at that point.

the surface at that point.

2-10C

2-10C Isotropic materials have the same properties in all directions, and we do not need to Isotropic materials have the same properties in all directions, and we do not need to be concernedbe concerned about the variation of

about the variation of properties with direction for such properties with direction for such materials. The properties of anisotropic materialsmaterials. The properties of anisotropic materials such as the fibrous or composite materials, however, may change with d

such as the fibrous or composite materials, however, may change with d irection.irection.

2-11C

2-11C In heat conduction analysis, the conversion of electrical, chemical, or nuclear energy into heat (orIn heat conduction analysis, the conversion of electrical, chemical, or nuclear energy into heat (or thermal) energy in solids is called heat generation.

thermal) energy in solids is called heat generation.

2-12C

2-12C The phrase “thermal energy generation” is equivalent to “heat generation,” and they are usedThe phrase “thermal energy generation” is equivalent to “heat generation,” and they are used interchangeably. They imply the conversion of some other form of energy into thermal energy. The ph interchangeably. They imply the conversion of some other form of energy into thermal energy. The ph raserase “energy generation,” however, is vague since the

“energy generation,” however, is vague since the form of energy generated is not clear.form of energy generated is not clear.

2-13

2-13 Heat transfer through the walls, door, and Heat transfer through the walls, door, and the top and bottom sections of an oven the top and bottom sections of an oven is transient in natureis transient in nature since the thermal conditions in the kitchen and the oven, in

since the thermal conditions in the kitchen and the oven, in general, change with time. However, general, change with time. However, we wouldwe would analyze this problem as a steady heat transfer problem under the worst anticipated conditions such as the analyze this problem as a steady heat transfer problem under the worst anticipated conditions such as the highest temperature setting for the oven, and the anticipated lowest temperature in the kitchen (the so highest temperature setting for the oven, and the anticipated lowest temperature in the kitchen (the so called “design” conditions). If the heating element of the oven is large enough to

called “design” conditions). If the heating element of the oven is large enough to keep the oven at thekeep the oven at the desired temperature setting under the presumed worst conditions, then it is large enough to

desired temperature setting under the presumed worst conditions, then it is large enough to do so under alldo so under all conditions by cycling on and off.

conditions by cycling on and off.

Heat transfer from the oven is three-dimensional in nature since heat will be entering through all Heat transfer from the oven is three-dimensional in nature since heat will be entering through all six sides of the oven.

six sides of the oven. However, heat transfer through any wall or However, heat transfer through any wall or floor takes place in the direction normalfloor takes place in the direction normal to the surface, and thus it can

to the surface, and thus it can be analyzed as being one-dimensional. Therefore, this problem can bebe analyzed as being one-dimensional. Therefore, this problem can be simplified greatly by considering the heat transfer as being one- dimensional at each of the four sides as simplified greatly by considering the heat transfer as being one- dimensional at each of the four sides as well as the top and bottom sections, and

well as the top and bottom sections, and then by adding the calculated values of heat transfers at eachthen by adding the calculated values of heat transfers at each surface.

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2-3 2-3

2-14E

2-14E The power consumed by the resistance wire of The power consumed by the resistance wire of an iron is given. The an iron is given. The heat generation and the heatheat generation and the heat flux are to be determined.

flux are to be determined. Assumptions

Assumptions Heat is generated uniformly in the resistance wire.Heat is generated uniformly in the resistance wire. _q_{q = 1000 W}_{= 1000 W}

L

L = 15 in= 15 in D

D = 0.08 in= 0.08 in Analysis

Analysis A 1000 W iron A 1000 W iron will convert electrical energy intowill convert electrical energy into heat in the wire at a rate of 1000 W. Therefore, the rate of heat heat in the wire at a rate of 1000 W. Therefore, the rate of heat generation in a resistance wire is simply equal to the power generation in a resistance wire is simply equal to the power rating of a resistance heater. Then the rate of

rating of a resistance heater. Then the rate of heat generation inheat generation in the wire per unit volume is determined by dividing the total the wire per unit volume is determined by dividing the total rate of heat generation by the volume of the

rate of heat generation by the volume of the wire to bewire to be

3 3 7 7 _Btu/h_Btu/h _ft_ft 10 10 7.820 7.820

××

⋅⋅

==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

==

W W 1 1 Btu/h Btu/h 412 412 .. 3 3 ft) ft) 12 12 / / 15 15 ]( ]( 4 4 / / ft) ft) 12 12 / / 08 08 .. 0 0 (( [[ W W 1000 1000 )) 4 4 / / (( 22 22 gen gen wire wire gen gen gen gen π π π π D D L L E E E E ee

&&

V V

Similarly, heat flux on the outer surface of the wire

Similarly, heat flux on the outer surface of the wire as a result of this heat generation is determined byas a result of this heat generation is determined by dividing the total rate of heat generation by the surface area of

dividing the total rate of heat generation by the surface area of the wire to bethe wire to be

2 2 5 5 ft ft Btu/h Btu/h 10 10 1.303 1.303

××

⋅⋅

==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

==

W W 1 1 Btu/h Btu/h 412 412 .. 3 3 ft) ft) 12 12 / / 15 15 (( ft) ft) 12 12 / / 08 08 .. 0 0 (( W W 1000 1000 gen gen wire wire gen gen π π π π DL DL E E A A E E q q

&&

Discussion

Discussion Note that heat generation is expressed per unit volume in Btu/hNote that heat generation is expressed per unit volume in Btu/h

⋅⋅

ftft33whereas heat flux iswhereas heat flux is expressed per unit surface area in Btu/h

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2-4 2-4

2-15E EES

2-15E EES Prob. 2-14E is reconsidered. The Prob. 2-14E is reconsidered. The surface heat flux as a function of wsurface heat flux as a function of w ire diameter is to beire diameter is to be plotted.

plotted. Analysis

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

"GIVEN" "GIVEN" E_dot=1000 [W] E_dot=1000 [W] L=15 [in] L=15 [in] D=0.08 [in] D=0.08 [in] "ANALYSIS" "ANALYSIS" g_dot=E_dot/V_wire*Convert(W, Btu/h) g_dot=E_dot/V_wire*Convert(W, Btu/h) V_wire=pi*D^2/4*L*Convert(in^3, ft^3) V_wire=pi*D^2/4*L*Convert(in^3, ft^3) q_dot=E_dot/A_wire*Convert(W, Btu/h) q_dot=E_dot/A_wire*Convert(W, Btu/h) A_wire=pi*D*L*Convert(in^2, ft^2) A_wire=pi*D*L*Convert(in^2, ft^2) D D [in] [in] qq [Btu/h.ft [Btu/h.ft22]] 0.02 521370 0.02 521370 0.04 260685 0.04 260685 0.06 173790 0.06 173790 0.08 130342 0.08 130342 0.1 104274 0.1 104274 0.12 86895 0.12 86895 0.14 74481 0.14 74481 0.16 65171 0.16 65171 0.18 57930 0.18 57930 0.2 52137 0.2 52137 0 0..0022 00..0044 00..0066 00..0088 00..11 00..1122 00..1144 00..1166 00..1188 00..22 0 0 50000 50000 100000 100000 150000 150000 200000 200000 250000 250000 300000 300000 350000 350000 400000 400000 450000 450000 500000 500000 550000 550000

D [in]

q q [ [ B B t t u u / / h h - - f f t t 2 2 ] ] 2-16

2-16 A certain thermopile used for heat flux meters is considered. The minimum heat flux this meter canA certain thermopile used for heat flux meters is considered. The minimum heat flux this meter can detect is to be determined.

detect is to be determined. Assumptions

Assumptions 11 Steady operating conditions exist.Steady operating conditions exist. Properties

Properties The thermal conductivity of kapton is given to be The thermal conductivity of kapton is given to be 0.345 W/m0.345 W/m

⋅⋅

K.K. Analysis

Analysis The minimum heat flux can be determined fromThe minimum heat flux can be determined from

2 2 W/m W/m 17.3 17.3

°°

Δ

00..11 CC )) C C W/m W/m 345 345 0 0 (( t t k k

&&

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2-5 2-5

2-17

2-17 The rate of heat generation per unit volume in the The rate of heat generation per unit volume in the uranium rods is given. The total rate of heaturanium rods is given. The total rate of heat generation in each rod is to be

generation in each rod is to be determined.determined. Assumptions

Assumptions Heat is generated uniformly in the uranium rods.Heat is generated uniformly in the uranium rods. g g = 7= 7

××

101077W/mW/m33

L L = 1 m= 1 m

D

D = 5 cm= 5 cm Analysis

Analysis The total rate of heat generation in the rod The total rate of heat generation in the rod isis

determined by multiplying the rate of heat generation per unit determined by multiplying the rate of heat generation per unit volume by the volume of the rod

volume by the volume of the rod

kW kW 137 137 = = W W 10 10 1.374 1.374 m) m) 1 1 ]( ]( 4 4 / / m) m) 05 05 .. 0 0 (( [[ )) W/m W/m 10 10 7 7 (( )) 4 4 / / (( 22 77 33 22 55 gen gen rod rod gen gen gen gen

==

ee

==

ee π π D D L L

==

××

π π

==

××

E

&&

V V

&&

2-18

2-18 The variation of the absorption of solar energy The variation of the absorption of solar energy in a solar pond with depth in a solar pond with depth is given. A relation for theis given. A relation for the total rate of heat generation in a water layer at the

total rate of heat generation in a water layer at the top of the pond is top of the pond is to be determined.to be determined. Assumptions

Assumptions Absorption of solar radiation by water is modeled as heat Absorption of solar radiation by water is modeled as heat generation.generation. Analysis

Analysis The total rate of heat generation in a water layer The total rate of heat generation in a water layer of surface areaof surface area A A and thicknessand thickness L L at the top of theat the top of the pond

pond is determined is determined by integration by integration to beto be

b b )) ee (1 (1 ee A A ₀₀ −−bLbL −− == −−

₌₌

−−

==

∫ ∫

&&

∫ ∫

&&

L L bx bx L L x x bx bx b b ee ee A A Adx Adx ee ee d d ee E E 0 0 0 0 0 0 00 gen gen gen gen (( )) V V V V 2-19

2-19The rate of heat generation per unit volume in a stainless steel plate is given. The heat flux on theThe rate of heat generation per unit volume in a stainless steel plate is given. The heat flux on the surface of the plate is to be

surface of the plate is to be determineddetermined.. Assumptions

Assumptions Heat is generated uniformly in steel plate.Heat is generated uniformly in steel plate.

ee

L

Analysis

AnalysisWe consider a unit surface area of 1 We consider a unit surface area of 1 mm22. The total rate of heat. The total rate of heat generation in this section of the plate is

generation in this section of the plate is

W W 10 10 1.5 1.5 m) m) )(0.03 )(0.03 m m 1 1 )( )( W/m W/m 10 10 5 5 (( )) (( 66 33 22 55 gen gen plate plate gen gen gen

gen

==

ee

==

ee A A

××

L L

==

××

==

××

E

&&

V V

&&

Noting that this heat will be dissipated from both sides of the plate, Noting that this heat will be dissipated from both sides of the plate, the heat flux on either surface of the plate becomes

the heat flux on either surface of the plate becomes

2 2 kW/m kW/m 75 75

==

××

==

22 2 2 5 5 plate plate gen gen W/m W/m 000 000 ,, 75 75 m m 1 1 2 2 W W 10 10 5 5 .. 1 1 A A E E q q

&&

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2-6 2-6

Heat Conduction Equation Heat Conduction Equation 2-20

2-20 The one-dimensional transient heat conduction equation for a plane wall with constant thermalThe one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and heat generation is

conductivity and heat generation is

t t T T α α k k ee x x T T

∂∂

==

++

∂∂

gengen 11 2 2 2 2

_&&

. Here

. Here T T is the is the temperaturtemperature,e, x x is the space variable,is the space variable, is the heat generation per unit volume,

is the heat generation per unit volume, k k is the thermal conductivity,is the thermal conductivity, α α is the thermal diffusivity, andis the thermal diffusivity, and t t is the time. is the time. gen gen ee

&&

2-21

2-21 The one-dimensional transient heat conduction equation for a plane wall with constant thermalThe one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and heat generation is

conductivity and heat generation is

t t T T k k ee r r T T r r r r r r

∂∂

α

==

++

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

11 1

1

&&

gengen

. Here

. Here T T is the is the temperaturtemperature,e, r r is the spaceis the space variable,

variable, g g is the heat generation per unit volume,is the heat generation per unit volume, k k is the thermal conductivity,is the thermal conductivity,

α

is the thermal diffusivity,is the thermal diffusivity, and

and t t is the time.is the time.

2-22

2-22We consider a thin element of thicknessWe consider a thin element of thickness

Δ

x x in a large plane wall (see Fig. 2-13 in the text). Thein a large plane wall (see Fig. 2-13 in the text). The density of the wall is

density of the wall is ρ ρ , the specific heat is, the specific heat is c,c, and the area of the wall normal to the direction of heatand the area of the wall normal to the direction of heat transfer is

transfer is A A. In the absence of any . In the absence of any heat generation, anheat generation, an energy balanceenergy balance on this thin element of thicknesson this thin element of thickness

Δ

x x during a small time interval

during a small time interval

Δ

t t can be expressed ascan be expressed as

t t E E Q Q Q Q_x_x _x_x _x_x

Δ

==

−−

_&&

₊₊_Δ_Δ elementelement

&&

where where )) (( )) (( element

element E E t t t t E E t t mcmcT T t t t t T T t t cAcA x x T T t t t t T T t t

E E

==

−−

==

−−

==

Δ

−−

Δ

₊₊_Δ_Δ ₊₊_Δ_Δ ρ ρ ₊₊_Δ_Δ Substituting, Substituting, t t T T T T x x cA cA Q Q Q Q_x_x _x_x _x_x t t t t t t

Δ

−−

Δ

==

−−

&&

₊₊_Δ_Δ ρ ρ ++ΔΔ

&&

Dividing by

Dividing by A AΔΔ x x givesgives

t t T T T T cc x x Q Q Q Q A A t t t t t t x x x x x x

Δ

−−

==

Δ

−−

11

&&

++ΔΔ

&&

ρ ρ ++ΔΔ Taking

Taking the the limit limit as as

Δ

x x

→

00 and and

Δ

t t

→

00yieldsyields

t t T T ρ ρcc x x T T kA kA x x A A

∂∂

==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

1 1

since from the definition of the derivative and Fourier’s law of heat conduction, since from the definition of the derivative and Fourier’s law of heat conduction,

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

−−

∂∂

==

∂∂

==

Δ

−−

Δ Δ ++ → → Δ Δ _x_x T T kA kA x x x x Q Q x x Q Q Q Q_x_x _x_x _x_x x x

&&

0 0 lim lim Noting that the area

Noting that the area A A of a plane wall is constant, the one-dimensional transient heat conduction equationof a plane wall is constant, the one-dimensional transient heat conduction equation in a plane wall with constant thermal conductivity

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2-7 2-7

2-23

2-23We consider a thin cylindrical shell element of thicknessWe consider a thin cylindrical shell element of thickness

Δ

r r in a long cylinder (see Fig. 2-15 in thein a long cylinder (see Fig. 2-15 in the text). The density of the cylinder is

text). The density of the cylinder is ρ ρ , the specific heat is, the specific heat is c,c, and the length isand the length is L. L. The area of the cylinderThe area of the cylinder normal to the direction of heat transfer at

normal to the direction of heat transfer at any locationany location isis A A

==

22π π rLrL where r wherer is the value of the radius at thatis the value of the radius at that location

location.. Note that the heat transfer areaNote that the heat transfer area A A depends ondepends on r r in this casein this case , , and thus and thus it varies with it varies with location. location. AnAn energy balance

energy balance on this thin cylindrical shell element of thicknesson this thin cylindrical shell element of thickness

Δ

r r during a small time intervalduring a small time interval

Δ

t can bet can be expressed as expressed as t t E E E E Q Q Q Q_r_r _r_r _r_r

Δ

==

++

−−

₊₊_Δ_Δ elementelement element element

&&

element E E t t t t E E t t mcmcT T t t t t T T t t cAcA r r T T t t t t T T t t

E E

==

−−

==

−−

==

Δ

−−

Δ

₊₊_Δ_Δ ₊₊_Δ_Δ ρ ρ ₊₊_Δ_Δ r r A A ee ee E

E

&&

_element_element

==

&&

_gen_genV V _element_element

==

&&

_gen_gen

Δ

Substituting, Substituting, t t T T T T r r cA cA r r A A ee Q Q Q Q_r_r _r_r _r_r t t t t t t

Δ

−−

Δ

==

Δ

++

−−

&&

₊₊_Δ_Δ

&&

_gen_gen ρ ρ ++ΔΔ

&&

where

where A A

==

22π π rLrL. . Dividing the Dividing the equation above equation above byby A A

Δ

r r givesgives

t t T T T T cc ee r r Q Q Q Q A A t t t t t t r r r r r r

Δ

−−

==

++

Δ

−−

11

&&

++ΔΔ

&&

_gen_gen ρ ρ ++ΔΔ Taking

Δ

r r

→

00 and and

Δ

t t

→

00yieldsyields

t t T T cc ee r r T T kA kA r r A A

∂∂

ρρ

==

++

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

gen gen 1 1

&&

since, from the definition of the derivative and Fourier’s law of heat conduction, since, from the definition of the derivative and Fourier’s law of heat conduction,

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

−−

∂∂

==

∂∂

==

Δ

−−

Δ Δ ++ → → Δ Δ _r_r T T kA kA r r r r Q Q r r Q Q Q Qr r r r r r r r

&&

0 0 lim lim

Noting that the heat transfer area in this case is

Noting that the heat transfer area in this case is A A

==

22π π rLrL and the thermal conductivity is constant, the one-and the thermal conductivity is constant, the one-dimensional transient heat conduction equation in a cylinder becomes

dimensional transient heat conduction equation in a cylinder becomes

t t T T ee r r T T r r r r r r

∂∂

α

==

++

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

11 1 1 gen gen

&&

where

(14)

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2-8 2-8

2-24

2-24We consider a thin spherical shell element of thicknessWe consider a thin spherical shell element of thickness

Δ

r r in a sphere (see Fig. 2-17 in the text).. Thein a sphere (see Fig. 2-17 in the text).. The density of the sphere is

density of the sphere is ρ ρ , the specific heat is, the specific heat is c,c, and the length isand the length is L. L. The area of the sphere normal to theThe area of the sphere normal to the direction of heat transfer at any

direction of heat transfer at any locationlocation is is wherewhere r r is the value of the radius at that locationis the value of the radius at that location .. Note that the heat transfer area

Note that the heat transfer area A A depends ondepends on r r in this casein this case , , and thus and thus it varies with it varies with location. location. When there isWhen there is no heat generation, an

no heat generation, an energy balanceenergy balance on this thin spherical shell element of thicknesson this thin spherical shell element of thickness

Δ

r r during a smallduring a small time interval

time interval

Δ

2 2 4 4 r r A A

==

π π t t E E Q Q Q Q_r_r _r_r _r_r

Δ

==

−−

_&&

&&

element E E t t t t E E t t mcmcT T t t t t T T t t cAcA r r T T t t t t T T t t

E E

==

−−

==

−−

==

Δ

−−

Δ

₊₊_Δ_Δ ₊₊_Δ_Δ ρ ρ ₊₊_Δ_Δ Substituting, Substituting, t t T T T T r r cA cA Q Q Q Q_r_r _r_r _r_r t t t t t t

Δ

−−

Δ

==

−−

++ΔΔ Δ Δ ++ ρ ρ

&&

where

where A A

==

44 r π π r 22. Dividing . Dividing the the equation equation above above byby A AΔΔr r givesgives

t t T T T T cc r r Q Q Q Q A A t t t t t t r r r r r r

Δ

−−

==

Δ

−−

11

&&

++ΔΔ

&&

ρ ρ ++ΔΔ Taking

Δ

r r

→

00 and and

Δ

t t

→

00yieldsyields

t t T T ρ ρcc r r T T kA kA r r A A

∂∂

==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

1 1

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

−−

∂∂

==

∂∂

==

Δ

−−

Δ Δ ++ → → Δ Δ _r_r T T kA kA r r r r Q Q r r Q Q Q Q_r_r _r_r _r_r r r

&&

0 0 lim lim Noting

Noting that that the the heat heat transfer transfer area area in in this this case case is is and and the the thermal thermal conductivityconductivity k k is constant, theis constant, the one-dimensional transient heat conduction equation in a sphere becomes

one-dimensional transient heat conduction equation in a sphere becomes

2 2 4 4 r r A A

==

π π t t T T α α r r T T r r r r r r

∂∂

==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

11 1 1 ₂₂ 2 2 where

(16)

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2-9 2-9

2-26

2-26 For a medium in which the heat conduction equation is given in its simplest byFor a medium in which the heat conduction equation is given in its simplest by 0 0 1 1 gen gen

==

++

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

ee dr dr dT dT rk rk dr dr d d r r

&&

::

((aa) Heat transfer is steady, () Heat transfer is steady, (bb) it is one-dimensional, () it is one-dimensional, (cc) there is heat ) there is heat generationgeneration, and (, and (d d ) the thermal) the thermal conductivity is variable.

conductivity is variable.

2-27

2-27 For a medium in which the heat conduction equation is given byFor a medium in which the heat conduction equation is given by

t t T T α α r r T T r r r r r r

∂∂

==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

11 1 1 22 2 2

((aa) Heat transfer is transient, () Heat transfer is transient, (bb) it is one-dimensional, () it is one-dimensional, ( cc) there is no heat generation, and () there is no heat generation, and ( d d ) the thermal) the thermal conductivity is constant.

conductivity is constant.

2-28

2-28 For a medium in which the heat conduction equation is given in its simplest byFor a medium in which the heat conduction equation is given in its simplest by ₂₂ 00

2 2

==

++

dr dr dT dT dr dr T T d d r r ::

((aa) Heat transfer is steady, () Heat transfer is steady, (bb) it is one-dimensional, () it is one-dimensional, (cc) there is no heat generation, and () there is no heat generation, and ( d d ) the thermal) the thermal conductivity is constant.

(18)

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2-10 2-10

2-29

2-29We consider a small rectangular element of lengthWe consider a small rectangular element of length

Δ

x x, width, width

Δ

y y, and height, and height

Δ

z z = 1 (similar to the one= 1 (similar to the one in Fig. 2-21). The density of the body is

in Fig. 2-21). The density of the body is

ρρ

and the specific heat isand the specific heat is c.c. Noting that heat conduction is two-Noting that heat conduction is two-dimensional and assuming no heat generation, an

dimensional and assuming no heat generation, an energy balanceenergy balance on this element during a small timeon this element during a small time interval

interval

Δ

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

Δ

++

Δ

−−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

element element the the of of content content energy energy the the of of change change of of Rate Rate and and + + at at surfaces surfaces at the at the conduction conduction heat heat of of Rate Rate and and at at surfaces surfaces at the at the conduction conduction heat heat of of Rate Rate y y y y x x x x y y x x or or t t E E Q Q Q Q Q Q Q

Q_x_x _y_y _x_x _x_x _y_y _y_y

Δ

==

−−

++

_&&

₊₊_Δ_Δ

_&&

&&

Noting that the volume of the element is

Noting that the volume of the element is V V _element_element

==

Δ

x x

Δ

y y

Δ

z z

==

Δ

x x

Δ

y y

××

11, the change in the energy content of , the change in the energy content of the element can be expressed as

the element can be expressed as

)) ((

element

element E E t t t t E E t t mcmcT T t t t t T T t t cc x x y y T T t t t t T T t t

E E

==

−−

==

−−

==

Δ

−−

Δ

₊₊_Δ_Δ ₊₊_Δ_Δ ρ ρ ₊₊_Δ_Δ Substituting, Substituting, t t T T T T y y x x cc Q Q Q Q Q Q Q

Q_x_x _y_y _x_x _x_x _y_y _y_y t t t t t t

Δ

−−

Δ

==

−−

++

&&

₊₊_Δ_Δ

&&

₊₊_Δ_Δ ρ ρ ++ΔΔ

&&

Dividing by

Dividing by ΔΔ x xΔΔ y y givesgives

t t T T T T cc y y Q Q Q Q x x x x Q Q Q Q y y t t t t t t y y y y y y x x x x x x

Δ

−−

==

Δ

−−

Δ

−−

Δ

−−

Δ

−−

++ΔΔ ++ΔΔ ρ ρ ++ΔΔ

&&

₁₁ 1 1

Taking the thermal conductivity

Taking the thermal conductivity k k to be constant and noting that the heat transfer surface areas of theto be constant and noting that the heat transfer surface areas of the element for heat conduction in the

element for heat conduction in the x x andand yy directions aredirections are A A_x_x

==

Δ

y y

××

11andand A A_y_y

==

Δ

x x

××

11,, respectivelyrespectively, , andand taking

taking the the limit limit as as

Δ

x x,,

Δ

y y,,andand

Δ

t t

→

00 yieldsyields

t t T T α α y y T T x x T T

∂∂

==

∂∂

++

∂∂

11 2 2 2 2 2 2 2 2

2 2 2 2 0 0 1 1 1 1 1 1 lim lim x x T T k k x x T T k k x x x x T T z z y y k k x x z z y y x x Q Q z z y y x x Q Q Q Q z z y y x x x x x x x x x x

∂∂

−−

==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

−−

==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

Δ

−−

∂∂

Δ

==

∂∂

Δ

==

Δ

−−

Δ

Δ Δ ++ → → Δ Δ

&&

2 2 2 2 0 0 1 1 1 1 1 1 lim lim y y T T k k y y T T k k y y y y T T z z x x k k y y z z x x y y Q Q z z x x y y Q Q Q Q z z x x y y y y y y y y y y

∂∂

−−

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

∂∂

−−

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

∂∂

Δ

−−

∂∂

Δ

==

∂∂

Δ

==

Δ

−−

Δ

Δ Δ ++ → → Δ Δ

&&

Here the property

(20)

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2-11 2-11

2-30

2-30We consider a thin ring shaped volume element of widthWe consider a thin ring shaped volume element of width

Δ

z z and thicknessand thickness

Δ

r r in a cylinder. Thein a cylinder. The density of the cylinder is

density of the cylinder is ρ ρ and the specific heat isand the specific heat is c.c. In general, anIn general, an energy balanceenergy balance on this ring elementon this ring element during a small time interval

during a small time interval

Δ

t t E E Q Q Q Q Q Q Q Q_r_r _r_r _r_r _z_z _z_z _z_z

Δ

==

−−

++

−−

₊₊_Δ_Δ ₎₎ ₍₍ ₊₊_Δ_Δ ₎₎ elementelement

((

&&

Δ

z z

r+ r+

Δ

rr r r r r But the change in the energy content of the element can be expressed as

But the change in the energy content of the element can be expressed as )) (( )) 2 2 (( )) (( element

element E E t t t t E E t t mcmcT T t t t t T T t t cc r r r r z z T T t t t t T T t t

E E

==

−−

==

−−

==

ρρ

ππ

Δ

−−

Δ

₊₊_Δ_Δ ₊₊_Δ_Δ ₊₊_Δ_Δ Substituting, Substituting, t t T T T T z z r r r r cc Q Q Q Q Q Q Q Q_r_r _r_r _r_r _z_z _z_z _z_z t t t t t t

Δ

−−

Δ

==

−−

++

−−

₊₊_Δ_Δ )) (( ₊₊_Δ_Δ )) ((22 )) ++ΔΔ ((

&&

ρ ρ π π

Dividing the equation above by

Dividing the equation above by (( π 22π r r

Δ

r r ))

Δ

z z givesgives

t t T T T T cc z z Q Q Q Q r r r r r r Q Q Q Q z z r r t t t t t t z z z z z z r r r r r r

Δ

−−

==

Δ

−−

Δ

−−

Δ

−−

Δ

−−

++ΔΔ ++ΔΔ _ρ_ρ ++ΔΔ π π π π

&&

2 2 1 1 2 2 1 1

Noting that the heat transfer surface areas of the element for heat conduction in the

Noting that the heat transfer surface areas of the element for heat conduction in the r r andand z z directions aredirections are ,, 2 2 and and 2 2 r r z z A A r r r r A

A_r_r

==

π π

Δ

_z_z

==

π π

Δ

respectively, and taking the limit asrespectively, and taking the limit as

Δ

r r ,,

Δ

z z andand

Δ

t t

→

00 yieldsyields

t t T T cc z z T T k k z z T T k k r r r r T T kr kr r r r r

∂∂

ρρ

==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

++

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

∂φ

∂∂

∂φ

∂∂

++

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

2 2 1 1 1 1

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

−−

==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

Δ

ππ

−−

∂∂

Δ

ππ

==

∂∂

Δ

ππ

==

Δ

−−

Δ

ππ

Δ Δ ++ → → Δ Δ _r_r T T kr kr r r r r r r T T z z r r k k r r z z r r r r Q Q z z r r r r Q Q Q Q z z r r r r r r r r r r 1 1 )) 2 2 (( 2 2 1 1 2 2 1 1 2 2 1 1 lim lim 0 0

&&

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

−−

==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

Δ

ππ

−−

∂∂

Δ

ππ

==

∂∂

Δ

ππ

==

Δ

−−

Δ

ππ

Δ Δ ++ → → Δ Δ _z_z T T k k z z z z T T r r r r k k z z r r r r z z Q Q r r r r z z Q Q Q Q r r r r z z z z z z z z z z ₂₂ ((22 )) 1 1 2 2 1 1 2 2 1 1 lim lim 0 0 && &&

For the case of constant thermal conductivity the equation above reduces to For the case of constant thermal conductivity the equation above reduces to

t t T T z z T T r r T T r r r r r r

∂∂

α

==

∂∂

++

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

11 1 1 2 2 2 2 where

whereα α

==

k k ρ / / ρ cc is the thermal diffusivityis the thermal diffusivity of the material. For the case of steady heat conduction with noof the material. For the case of steady heat conduction with no heat generation it reduces to

heat generation it reduces to 1

(22)

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