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Model 4: Heat flow in an electrical conductor

 A copper conductor is sheathed in an insulator

material. The insulator also stops heat from escaping. Imagine that the wire is hot after

conducting electricity. We will model heat losses after the power is switched off.

2003/ 1

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Step 1: Model and equationWe construct a simple heat flow model of the copper

conductor of length L:

Copper

Insulation

Lfigure 17: copper conductor

To describe the temperature from 0 to L we need:

•an equation

•The temperature at the boundaries.

•The initial temperature.

•Properties of the conductor material.

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In any body heat will flow in the direction of decreasing temperature. The rate of flow is

proportional to the gradient of the temperature:

In one-dimension we can say:  Rate of heat flow

dx

duKA

where u = temperature, K= thermal conductivity and A = cross-sectional area of the conductor. Because the wire is insulated heat only flows in the x-direction. We apply conservation of heat to a segment of the wire [x, x+dx]: Net change of heat in [x, x+dx] = Net flux of heat across boundaries + total heat generated in [x, x+dx]. Note: We assume that there is no heat generated within the body for this problem.

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Flux of heat across the boundaries

Copper

Insulation

L

x x + Δxu(0,t)=0 u(L,t)=0

Extract the element between x and x + Δx:

x x + Δx

xx

uKA

Rate of heat flowing into (or out of) back end

xxx

uKA

Rate of heat flowing into (or out of) front end

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The net heat transported into or out of this segment is the difference between the heat flux at front and

back:

xxx x

u

x

uKAheatflux

Heat is produced as a consequence of electrical resistance. Imagine that we have left the power on and the conductor has heated up. When switched off the conductor will cool as heat is conducted out of the ends.

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The total quantity of heat in the segment is: σρΔxAu, where σ = specific heat

ρ = density As it cools down we can say that the change in heat energy is:

x x + Δx

xx

uKA

Rate of heat flowing into (or out of) back end

xxx

uKA

Rate of heat flowing into (or out of) front end

t

uxAheatenergy

t

uxAheatenergy

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Energy cannot be created so the sum of the heat leaving or entering through the ends added to the

change in heat energy within the segment = 0We can write:

Δheat energy + Δheat flux = 0

We can therefore write Δheat energy = -Δheat flux

xxx x

u

x

uKA

t

uxA

Rearrange:

xx

u

x

uK

t

u xxx

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If we let Δx→ 0, we get the 1 dimensional heat equation:

xx

u

x

uK

t

u xxx

x0

2

2

x

uK

t

u

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The expression K/σρ is called the diffusivity and is sometimes expressed as α2

The 1 dim. heat equation is written:

2

22

x

u

t

u

(27)

(i.e. α2 = K/σρ)

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Step 2: General solutionWe can now produce a formula for the temperature along the wire if we can find a general form for the solution to the 1-

dimensional equation.

Instead of guessing we again use the method of “Separation of Variables”, making the assumption that the solution will be some function of x only multiplied by some function of t only:

u = X(x)T(t)

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Substitute the solution u=X(x)T(t) into the PDE: 2

22

x

u

t

u

You get: Tx

X

t

TX

2

22

Now separate variables (get everything that’s a function of x on one side and everything that’s a function of t on the other):

Xx

X

Tt

T2

2

2

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Both sides are equal to a constant. Call it k: k

Xx

X

Tt

T

2

2

2

We get the two equations:

02 Tkdt

dT

02

2

kXdx

Xd

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Choose a sign for the constant that will give you sensible results: k= -2

You now get 2 differential equations:

022 Tdt

dT

022

2

Xdx

Xd

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Solutions to T and X are:

tAeT22

)cos(sin xCxBX

(Demonstrate to yourself that these two expressions are solutions to the equations.)

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The solution for u(x,t) equals T times X:

)cos()sin(,22

xCxBAetxu t

or

))cos()sin((,22

xExDetxu t

This looks OK. There is an exponential ‘decay’ term for describing heat loss and the sinusoidal terms will be useful for fitting a solution within a fixed length as we will see.  What form of solution would you get if you chose k= +2 ?

(28)

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Step 3: Apply the boundary conditions Both ends were fixed at 0ºC.

At x=0

))0cos()0sin((,022

EDetu t

0 1

0,022

Eetu t

To satisfy this b.c. we set E=0

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At x=L

0)sin(,22

LDetLu t

We could make D=0 but this would not give us a useful solution.

We therefore make sin(L) = 0

This will be zero for L= nπ, n=1,2,3…

All the terms in the solution are given below:

)sin(,2

2

L

xneDtxu

tL

n

nn

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The solution is the sum of the terms: )sin(,

22

L

xneDtxu

tL

n

nn

)sin(,2

2

1 L

xneDtxu

tL

n

nn

(29)

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Step 4: Satisfy the initial conditions If we know the initial temperature, we can find values for the constants Dn. Lets assume that the

wire temperature was the same across the entire length of wire, say u(x,0)=T0.

i.e. )sin(0,

10 L

xnDTxu

nn

(30)

T0

x=0x=L

Temperature at t=0:

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To determine the Dn coefficients we again use a

property of the sine function (Recall the guitar string model):

L L

dxL

xm

L

xn

0 2sinsin

if n=m

0 if nm

(13)

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We multiply both sides of equation 30 by sin (mπx) and then integrate between the limits 0 and L.

(30) )sin(0,1

0 L

xnDTxu

nn

2

sin0,0

LDdx

L

xmxu m

L

dxL

xm

L

xnDdx

L

xmxu

L

nn

L

sinsinsin0,

0 10

All terms are equal to zero except when n =m.

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Rearrange to find the value of the m’th term of the following equation:

2

sin0,0

LDdx

L

xmxu m

L

dxL

xmxu

LD

L

m

sin0,

2

0

(15)

m

mTdx

L

xmT

LD

L

m cos12

sin2

0

0

0

Now we can evaluate Dm :

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We now have an expression for our mth constant:

mm

TDm cos12

0

Put this into the solution:

)sin(cos12

,2

2

1

0

L

xnen

n

Ttxu

tL

n

n

(31)

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I will show the solution to this problem in our class/tutorial. The following constants will be used for calculating the diffusivity of copper:Specific heat = 386 J/(kgC)Density = 8.96*103 kg/m3 Thermal conductivity K = 385 J/(sec*meter*C).

Therefore 2 = 1.11E-4 What are the units?

Remember: α2 = K/σρ

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Assume L=0.1m and T0 = 100°C. I have used Microsoft Excel for this. The black line represents the sum of the 14 non-zero terms I have calculated. The blue line is the first term. Altogether I have plotted 5 of the non-zero terms.

u(x,t)

-50-40-30-20-10

0102030405060708090

100110120130140

0 0.2 0.4 0.6 0.8 1

Distance x

Te

mp

era

ture

Solution at t=0

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Solutions:

u(x,t)

-50-40-30-20-10

0102030405060708090

100110120130140

0 0.2 0.4 0.6 0.8 1

Distance x

Te

mp

era

ture

t= 1 sec

u(x,t)

-50-40-30-20-10

0102030405060708090

100110120130140

0 0.2 0.4 0.6 0.8 1

Distance x

Te

mp

era

ture

t= 10 sec