So how does this data on n and Q apply to a ‘real’ problem? T variation with time (position) Hot drawing of wire Modify the JMA equation so that we add.

So how does this data on n and Q apply to a ‘real’ problem?

T variation with time (position)

Hot drawing of wire

Modify the JMA equation so that we add up the amount of recrystallization occurring at each place (at each temperature)on the

wire.

In order to determine the temperature variation along the length of the wire, we need to describethe heat transport processes. We know from every day experiences that heat travels from hotregions to cold regions. When the heat energy transfers by atomic vibration transferring frommolecule to molecule, the process is known as conduction. When the heat energy transferoccurs by material flow away from hot regions, carrying the heat with it, the process is known asconvection. Heat conduction was described as a flux of heat energy,q, by Fourier using theequation

q x kxTd

d

where T is the temperature, xis is the distance along a specific direction and k is the thermalconductivity. Convection at a solid/fluid interface is treated as a flux of heat energy using theequation q h Ts Tf

where h is the heat transfer coefficient, a function of the fluid velocity, density and heat capacity.The temperatures, Ts and Tf, refer to the solid surface temperature and the fluid temperature farfrom the surface, respectively. For forced convectio perpindicular to a cylinder (wire).

NuD C ReDm Pr

1

3

Where C~1, m~1/3 for laminar flow. For fully turbulent flow 4000 ReD 40000 , C=0.193 and

m=0.805

In addtion to heat conduction and heat convection, this particular problem requires that we alsoaccount for the heat energy transported by the moving wire. Using the heat capacity/unit mass forthe wire, Cp, and the radius of the wire, R, the heat flux from the moving wire is q Cp T Vo

To use the heat transport equations to determine the steady state temperature variation, weequate the heat flowing into and out of the imaginary volume of length, x, and cross sectional

area, R2 .

In order to determine the temperature variation along the length of the wire, we need to describethe heat transport processes. We know from every day experiences that heat travels from hotregions to cold regions. When the heat energy transfers by atomic vibration transferring frommolecule to molecule, the process is known as conduction. When the heat energy transferoccurs by material flow away from hot regions, carrying the heat with it, the process is known asconvection. Heat conduction was described as a flux of heat energy,q, by Fourier using theequation

qx kxTd

d

where T is the temperature, xis is the distance along a specific direction and k is the thermalconductivity. Convection at a solid/fluid interface is treated as a flux of heat energy using theequation q h T s T f

where h is the heat transfer coefficient, a function of the fluid velocity, density and heat capacity.The temperatures, Ts and Tf, refer to the solid surface temperature and the fluid temperature farfrom the surface, respectively. For forced convectio perpindicular to a cylinder (wire).

NuD C ReDm Pr

1

3

Where C~1, m~1/3 for laminar flow. For fully turbulent flow 4000 ReD 40000 , C=0.193 and

m=0.805

In addtion to heat conduction and heat convection, this particular problem requires that we alsoaccount for the heat energy transported by the moving wire. Using the heat capacity/unit mass forthe wire, Cp, and the radius of the wire, R, the heat flux from the moving wire is q Cp T Vo

To use the heat transport equations to determine the steady state temperature variation, weequate the heat flowing into and out of the imaginary volume of length, x, and cross sectional

area, R2 .

q C p T Vo

Dividing both sides by x R2

[1]

kxT x x d

d k

xT x( )d

d

x

Cp Vo T x x T x( )

x 2

h

R

T x x T x( ) 2

Tf

0

Taking the limit as x goes to zero leads to

[2] 2

xTd

d

2 C p Vo

k xTd

d

2 hk R

T T f 0

A solution to equation [2] may be found by setting T Tf , allowing C1 exp C2 x and substituting into

equation [2] gives

[3a] C2a

Cp Vo

k

Cp Vo

k

28 h

k R

1

2

or

[3b] C2b

Cp Vo

k

Cp Vo

k

28 h

k R

1

2

Conservation of energy at steady state gives

Need to know how T varies with x, where x=Vo*t

A solution to equation [2] may be found by setting T Tf , allowing C1 exp C2 x and

substituting into

Dividing both sides by x R2

[1]

kxT x x d

d k

xT x( )d

d

x

Cp Vo T x x T x( )

x 2

h

R

T x x T x( ) 2

Tf

0

Taking the limit as x goes to zero leads to

[2] 2

xTd

d

2 C p Vo

k xTd

d

2 hk R

T T f 0

A solution to equation [2] may be found by setting T Tf , allowing C1 exp C2 x and substituting into

equation [2] gives

[3a] C2a

Cp Vo

k

Cp Vo

k

28 h

k R

1

2

or

[3b] C2b

Cp Vo

k

Cp Vo

k

28 h

k R

1

2

2xd

d

2 C p Vo

k xd

d

2 hk R

0

Make homogeneous

nonhomogeneous

[2]

[2]

[3a] C2a

Cp Vo

k

Cp Vo

k

28 h

k R

1

2

or

[3b] C2b

Cp Vo

k

Cp Vo

k

28 h

k R

1

2

Solve for C2

Thus C 1a exp C 2a x C 1b exp C 2b x . The coefficients can be found by examining the

boundary conditions, x 0( ) T1 Tf and x L( ) T2 Tf .

Thus C1a exp C2a x C1b exp C2b x . The coefficients can be found by examining the

boundary conditions, x 0( ) T1 Tf and x L( ) T2 Tf .

Thus C1a exp C2a x C1b exp C2b x . The coefficients can be found by examining the

boundary conditions, x 0( ) T1 Tf and x L( ) T2 Tf .

o C 1a C 1b

L C 1a exp C 2a L C 1b exp C 2b L

solveC 1a

C 1b

o exp C 2b L L

exp C 2a L exp C 2b L

L exp C 2a L o

exp C 2a L exp C 2b L

[3a] C2a

Cp Vo

k

Cp Vo

k

28 h

k R

1

2

or

[3b] C2b

Cp Vo

k

Cp Vo

k

28 h

k R

1

2

Find the heat transfer coefficient

Cpair 1051J

kg K air 3.3510

5kg

m s kair 0.0513

J

s m K air 0.5

kg

m3

Dwire .02 mTf 300K

PrD

air Cpair

kair

ReD v( ) air v Dwire

air

h v( ) 0.193ReD vm

s

.805

PrD

1

3kair

Dwire

at an air velocity of 1m/s h 1( ) 42.904kg

s3K

7801kg

m3

Cp 473

J

kg K

R .01 mL 0 Kk Cp 2 10

5m

2

s

Vo .01m

s

h 43W

m2

K

k 73.797kgm

s3K

o 800K 300KL 0.5 m L 300K 300K

Other parameters

Check Bi, just to be sure

hD wire

4

k2.913 10

3

No gradients across the wire

0 0.2 0.40

200

400

600

x( )

x0.47 0.48 0.49 0.50

200

400

600

x( )

x

Tf=300 K

X t( ) 1 exp

0

x

Vo

x

Vo

3

d

d

d

f x( )

0

x

Vo

1 103

5000K

300 2d exp

5000 K

300

x

Vo

3

exp5000 K

300

x

Vo

3

3

x

Vo

d

0 0.2 0.40

0.1

0.2

1 exp 1 f x( )( )

x

Temperature variation along wire

Recrystallized volume fraction as a function of position along wire