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Problem d: Use de nition (8.16) to rewrite the expression above as the conservation law

of kinetic energy:

@EK + (vEK ) = (v F) : (8.18)

@t r

This equation states that the kinetic energy current is given by J = vEK , this term de-

scribes how kinetic energy is transported by the ow. The term (v F) on the right hand

side denotes the source of kinetic energy. This term is relatively easy to understand. Sup-

pose the force F acts on the uid over a distance r, the work carried out by the force is

given by ( r F). If it takes the uid a time t to move over the distance r the work per

unit time is given by ( r= t F). However, r= t is simply the velocity v, and hence the

term (v F) denotes the work performed by the force per unit time. Since work per unit

time is called the power, equation (8.18) states that the power produced by the force F is

the source of kinetic energy.

In order to invoke the potential energy as well we assume for the moment that the

force F is the gravitational force. Suppose there is a gravitational potential V (r), then

the gravitational force is given by

F= (8.19)

; rV

and the potential energy EP is given by

EP = V : (8.20)

CHAPTER 8. CONSERVATION LAWS

78

Problem e: Take the (partial) time derivative of (8.20), use the continuity equation (8.6)

to eliminate @ =@t, use that the potential V does not depend explicitly on time and

employ expressions (8.19) and (8.20) to derive the conservation law of potential

energy:

@EP + (vE ) = F) : (8.21)

P

@t r ;(v

Note that this conservation law is very similar to the conservation law (8.18) for kinetic

energy. The meaning of the second term on the left hand side will be clear to you by

now, it denotes the divergence of the current vEP of potential energy. Note that the right

hand side of (8.20) has the opposite sign of the right hand side of (8.18). This re ects the

fact that when the force F acts as a source of kinetic energy, it acts as a sink of potential

energy the opposite signs imply that kinetic and potential energy are converted into each

other. However, the total energy E = EK + EP should have no source or sink.

Problem f: Show that the total energy is source-free:

@E + (vE) = 0 : (8.22)

@t r

8.4 The heat equation

In the previous section the momentum and energy current could be derived from Newton's

law. Such a rigorous derivation is not always possible. In this section the transport of heat

is treated, and we will see that the law for heat transport cannot be derived rigorously.

Consider the general conservation equation (8.3) where T is the temperature. (Strictly

speaking we should derive the heat equation using a conservation law for the heat content

rather than the temperature. The heat content is given by CT, with C the heat capacity.

When the speci c heat is constant the distinction between heat and temperature implies

multiplication with a constant, for simplicity this multiplication is left out.)

The source term in the conservation equation is simply the amount of heat (normalized

by the heat capacity) supplied to the medium. For example, the decay of radioactive

isotopes is a major source of the heat budget of the earth. The transport of heat is a ected

by the heat current J. In the earth, heat can be transported by two mechanisms, heat

conduction and heat advection. The rst process is similar to the process of di usion, it

accounts for the fact that heat ows from warm regions to colder regions. The second

process accounts for the heat that is transported by the ow eld v is the medium.

Therefore, the current J can be written as a sum of two components:

J = Jconduction+Jadvection : (8.23)

The heat advection is simply given by

Jadvection = vT (8.24)

which re ects that heat is simply carried around by the ow. This viewpoint of the process

of heat transport is in fact too simplistic in many situation. Fletcher 24] describes how

the human body during outdoor activities looses heat through four processes conduction,

8.4. THE HEAT EQUATION 79

advection, evaporation and radiation. He describes in detail the conditions under which

each of these processes dominate, and how the associated heat loss can be reduced. In

the physics of the atmosphere, energy transport by radiation and by evaporation (or

condensation) also plays a crucial role.

Thigh Thigh

T T

J

J

Tlow Tlow

Figure 8.1: Heat ow and temperature gradient in an isotropic medium (left panel) and

in a medium consisting of alternating layers of copper and styrofoam (right panel).

For the moment we will focus on the heat conduction.This quantity cannot be derived

from rst principles. In general, heat ows from warm regions to cold regions. The

vector points from cold regions to warm regions. It therefore is logical that the heat

rT

conduction points in the opposite direction from the temperature gradient:

Jconduction = (8.25)

; rT

see the left panel of gure (8.1). The constant is the heat conductivity. (For a given

value of the heat conduction increases when increases, hence it measures indeed the

rT

conductivity.) However, the simple law (8.25) does not hold for every medium. Consider

a medium consisting of alternating layers of a good heat conductor (such as copper)

and a poor heat conductor (such as styrofoam). In such a medium the heat will be

preferentially transported along the planes of the good heat conductor and the conductive

heat ow Jconduction and the temperature gradient are not antiparallel, see the right panel

in gure (8.1). In that case there is a matrix operator that relates Jconduction and : rT

Jiconduction = ij @j T, with ij the heat conductivity tensor. In this section we will restrict

;

ourselves to the simple conduction law (8.25). Combining this law with the expressions

(8.23), (8.24) and the conservation law (8.3) for heat gives:

@T + )=S :

(vT (8.26)

@t r ; rT

As a rst example we will consider a solid in which there is no ow (v = 0). For a

constant heat conductivity , expression (8.26) reduces to:

@T = 2T + S : (8.27)

@t r

The expression is called the \heat equation", despite the fact that it holds only under

special conditions. This expression is identical to Fick's law that accounts for di usion

CHAPTER 8. CONSERVATION LAWS

80

processes. This is not surprising since heat is transported by a di usive process in the

absence of advection.

We now consider heat transport in a one-dimensional medium (such as a bar) when

there is no source of heat. In that case the heat equation reduces to

@T = @ 2 T : (8.28)

@t @x2

If we know the temperature throughout the medium at some initial time (i.e. T(x t = 0)

is known), then (8.28) can be used to compute the temperature at later times. As a special

case we consider a Gaussian shaped temperature distribution at t = 0:

2!

x

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