So how does this data on n and Q apply to a ‘real’ problem? T variation with time (position) Hot...
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Transcript of So how does this data on n and Q apply to a ‘real’ problem? T variation with time (position) Hot...
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