L’àmbit de la regulació

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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 Δ

 z  z and thicknessand thickness

Δ 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 Δ

t t can be expressed ascan be expressed as t  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 ))

Dividing the equation above by

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

Δ Δ

r r ))

Δ Δ

 z  z givesgives

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 ,,

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,

 ⎠ ⎟⎟

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 

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

1

∂∂ ⎛  ⎛  ∂∂  ⎞  ⎞ ∂∂

²²

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2-12 2-12

2-31

2-31Consider a thin disk element of thicknessConsider a thin disk element of thickness

Δ Δ

 z  z and diameterand diameter D D in a long cylinder (Fig. P2-31). Thein a long cylinder (Fig. P2-31). The density of the cylinder is

density of the cylinder is

ρρ

, the specific heat is, the specific heat is c,c, and the area of the cylinder normal to the direction of heatand the area of the cylinder normal to the direction of heat transfer

transfer is is , , which which is is constant. constant. AnAn energy balanceenergy balance on this thin element of thicknesson this thin element of thickness

Δ Δ

 z  z during aduring a small time interval

small time interval

Δ Δ

t t can be expressed ascan be expressed as 4

−− ^&&

_++ΔΔ

^&&

_{elementelement} ^{elementelement}

&&

But the change in the energy content of the element and the rate of heat generation within the element can But the change in the energy content of the element and the rate of heat generation within the element can be expressed as

be expressed as

))

 E 

&&

_{elementelement}

== &&

_{gengenV  V  elementelement}

== &&

_gengen

Δ Δ

Substituting,

Dividing by A AΔΔ z  z givesgives

T 

1

&& && &&

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2-13 2-13

2-32

2-32 For a medium in which the heat conduction equation is given byFor a medium in which the heat conduction equation is given by

t 

((aa) Heat transfer is transient, () Heat transfer is transient, (bb) it is two-dimensional, () it is two-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.

2-33

2-33 For a medium in which the heat conduction equation is given byFor a medium in which the heat conduction equation is given by 0

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

conductivity is variable.

2-34

2-34 For a medium in which the heat conduction equation is given byFor a medium in which the heat conduction equation is given by

t 

((aa) Heat transfer is transient, () Heat transfer is transient, (bb) it is two-dimensional, () it is two-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.

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2-14 2-14

2-38C

2-38C The boundary condition at a perfectly insulated surface (atThe boundary condition at a perfectly insulated surface (at  x x = 0, for example) can be expressed as= 0, for example) can be expressed as 0

)) 0 ,, 0 0 or ((

or 0

)) 0 ,, 0 0

((

==

∂∂

== ∂∂

∂∂

−− ∂∂

 x  x

t  t  T  T   x

 x t  t  T  k  T 

k  which indicates zero heat flux.which indicates zero heat flux.

2-39C

2-39CYes, the temperature profile in a medium must be perpendicular to an insulated surface since theYes, the temperature profile in a medium must be perpendicular to an insulated surface since the slope

slope

∂∂

T T  /  / 

∂∂

 x x

==

00 at at that that surface.surface.

2-40C

2-40CWe try to avoid the radiation boundary condition in heat transfer analysis because it is a non-linearWe try to avoid the radiation boundary condition in heat transfer analysis because it is a non-linear expression that causes mathematical difficulties while solving the problem; often making it impossible to expression that causes mathematical difficulties while solving the problem; often making it impossible to obtain analytical solutions.

obtain analytical solutions.

2-412-41 A A spherical spherical container container of of inner inner radius radius , , outer outer radius radius , , and and thermalthermal conductivity

conductivity k k is given. The boundary condition on the inner surface of theis given. The boundary condition on the inner surface of the container for steady one-dimensional conduction is to be expressed for the container for steady one-dimensional conduction is to be expressed for the following cases:

following cases:

r 

r ₁₁ r r ₂₂

r 

r 

11

r  r 

22

((aa) Specified temperature of 50) Specified temperature of 50

°°

C:C: T T ((r r ₁₁))

==

5050

°°

CC

((bb) Specified heat flux of 30 W/m) Specified heat flux of 30 W/m²²towards the center:towards the center: (( ¹¹)) 3030W/mW/m²² dr 

==

dr  r  r  dT  k dT  k 

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2-15 2-15

2-432-43 A long pipe of inner radiusA long pipe of inner radius r r 11, outer radius, outer radius r r 22, and thermal conductivity, and thermal conductivity k k is considered. The outer surface of the pipe is subjected to convection to ais considered. The outer surface of the pipe is subjected to convection to a medium

medium at at with with a a heat heat transfer transfer coefficient coefficient of of hh.. Assuming steady one-Assuming steady one-dimensional conduction in the radial direction, the convection boundary dimensional conduction in the radial direction, the convection boundary condition on the outer surface of the pipe can be expressed as

condition on the outer surface of the pipe can be expressed as

∞ T ∞ T 

r  r ₂₂ hh,, T T _∞∞

r  r ₁₁

]]

)) ((

)) [[

((

2 2 2

2

== −−

∞∞

−−

hhT T r r  T T  dr 

dr  r  r  dT  k dT  k 

2-44

2-44 A spherical shell of inner radiusA spherical shell of inner radius r r ₁₁, outer radius, outer radius r r ₂₂, and thermal, and thermal conductivity

conductivity k k is considered. The outer surface of the shell isis considered. The outer surface of the shell is subjected

subjected to to radiation radiation to to surrounding surrounding surfaces surfaces at at . . Assuming Assuming nono convection and steady one-dimensional conduction in the radial convection and steady one-dimensional conduction in the radial direction, the radiation boundary condition on the outer surface of the direction, the radiation boundary condition on the outer surface of the shell can be expressed as

shell can be expressed as

surr

T surr

T 

r 

r ₁₁ r r ₂₂ T T _surrsurr k 

k 

ε  ε 

[ [

surrsurr⁴⁴

]]

4 4 2 2 2

2)) (( ))

(( T T r r  T T  dr 

dr  r  r  dT  k dT 

k 

== −−

−−

εσ εσ 

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2-16 2-16

2-462-46 Heat conduction through the bottom section of a steel pan that is used to boil water on top of anHeat conduction through the bottom section of a steel pan that is used to boil water on top of an electric range is considered. Assuming constant thermal conductivity and one-dimensional heat transfer, electric range is considered. Assuming constant thermal conductivity and one-dimensional heat transfer, the mathematical formulation (the differential equation and the boundary conditions) of this heat

the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem is to be obtained for steady operation.

conduction problem is to be obtained for steady operation.

 Assumptions

 Assumptions 11Heat transfer is given to be steady and one-dimensional.Heat transfer is given to be steady and one-dimensional. 22Thermal conductivity is given toThermal conductivity is given to be constant.

be constant. 33 There is no heat generation in the medium.There is no heat generation in the medium. 44 The top surface atThe top surface at x x == L L is subjected tois subjected to convection and the bottom surface at

convection and the bottom surface at  x x = 0 is subjected to uniform heat flux.= 0 is subjected to uniform heat flux.

 Analysis

 Analysis The heat flux at the bottom of the pan isThe heat flux at the bottom of the pan is

2

Then the differential equation and the boundary conditions for this heat conduction problem can be Then the differential equation and the boundary conditions for this heat conduction problem can be expressed as

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2-17 2-17

2-482-48 Heat conduction through the bottom section of an aluminum pan that is used to cook stew on top of anHeat conduction through the bottom section of an aluminum pan that is used to cook stew on top of an electric range is considered (Fig. P2-48). Assuming variable thermal conductivity and one-dimensional electric range is considered (Fig. P2-48). Assuming variable thermal conductivity and one-dimensional heat transfer, the mathematical formulation (the differential equation and the boundary conditions) of this heat transfer, the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem is to be obtained for steady operation.

heat conduction problem is to be obtained for steady operation.

 Assumptions

 Assumptions 11Heat transfer is given to be steady and one-dimensional.Heat transfer is given to be steady and one-dimensional. 22Thermal conductivity is given toThermal conductivity is given to be variable.

be variable. 33 There is no heat generation in the medium.There is no heat generation in the medium. 44 The top surface atThe top surface at x x == L L is subjected tois subjected to specified temperature and the bottom surface at

specified temperature and the bottom surface at x x = 0 is subjected to uniform heat flux.= 0 is subjected to uniform heat flux.

 Analysis

 Analysis The heat flux at the bottom of the pan isThe heat flux at the bottom of the pan is

2

Then the differential equation and the boundary conditions for this heat conduction problem can be Then the differential equation and the boundary conditions for this heat conduction problem can be expressed as

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2-18 2-18

2-502-50 A spherical metal ball A spherical metal ball that is heated in an that is heated in an oven to a temperature of oven to a temperature of T T _iithroughout is dropped into athroughout is dropped into a large body of water at

large body of water at T T _∞∞where it is cooled by convection. Assuming constant thermal conductivity andwhere it is cooled by convection. Assuming constant thermal conductivity and transient one-dimensional heat transfer, the mathematical formulation (the differential equation and the transient one-dimensional heat transfer, the mathematical formulation (the differential equation and the boundary

boundary and initial conditions) of this and initial conditions) of this heat conduction problem is heat conduction problem is to be to be obtained.obtained.

 Assumptions

 Assumptions 11Heat transfer is given to be transient and one-dimensional.Heat transfer is given to be transient and one-dimensional. 22Thermal conductivity is givenThermal conductivity is given to be constant.

to be constant. 33 There is no heat generation in the medium.There is no heat generation in the medium. 44 The outer surface atThe outer surface at r r == r r ₀₀is subjected tois subjected to convection.

convection.

 Analysis

 Analysis Noting that there is thermal symmetry about the midpoint and convection at the outer surface,Noting that there is thermal symmetry about the midpoint and convection at the outer surface, the differential equation and the boundary conditions for this heat conduction problem can be expressed as the differential equation and the boundary conditions for this heat conduction problem can be expressed as

t 

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In document La protecció de la part feble del contracte en el dret civil català i en el dret civil europeu (página 79-82)