Robert Cinca - Physics
I then worked out the net energy flow into the wire:
This lead to me finding out the temperature increase after each time increment,
equation:
, where
is the
I using
then the
worked
out the net energy
flow into
the mass
wire: of the filament (as worked out
10
I earlier)
then worked
out
the
net energy
into of
the Tungsten
wire:
.
and is
the
specific
heat flow
capacity
This lead to me finding out the temperature increase after each time increment,
This
lead
to
me finding out the
temperature
each time
, where
is the increase
mass
of after
the filament
(as increment,
worked out
using
the out
equation:
To
work
the current at each
time increment,
I just
used
the equation:
10
using
the
equation:
,
where
is
the
mass
of
the
filament
(as
worked out
earlier) and is the specific heat capacity of Tungsten .
10
I earlier)
continued
until heat
I reached
a final
steady current
(please see Figure 15
.
and this
is process
the specific
capacity
of Tungsten
To
out the of
current
at each time increment, I just used the equation:
for work
appearance
the model).
To work out the current at each time increment, I just used the equation:
I continued this process until I reached a final steady current (please see Figure 15
I continued
this process
until I reached a final steady current (please see Figure 15
for
appearance
of the model).
for appearance of the model).
Figure 15: Hot Body Radiator Model
Excel Model Results
Figure 15: Hot Body Radiator Model
Figure 15: Hot Body Radiator Model
Running the excel model with various final currents, adjusted by altering the potential
divider Model
resistance,
yielded a proportional relationship between surge current time
Excel
Results
and the
inverse
square of the final steady current. The equation of the trend line that
Excel
Model
Results
Running
the
excel
model
with
final
currents,
ΔT = (7.3
± 2.1)
I -2 + (15 adjusted
± 37). by altering the potential
has resulted from the
graph
is: various
Running
the excel model
various final relationship
currents, adjusted
altering
the potential
divider resistance,
yielded with
a proportional
between by surge
current
time
divider
yielded
a proportional
time that
and
the resistance,
inverse square
of the
final steady relationship
current. The between
equation surge
of the current
trend line
-2
and resulted
the inverse
square
of the is: final
of the trend line that
has
from
the graph
ΔT = steady
(7.3 ± current.
2.1) I + The
(15 ± equation
37).
-2
has resulted from the graph is: ΔT = (7.3 ± 2.1) I + (15 ± 37).
10”
Metals-Specific Heats”, The Engineering Toolbox [Online][Date Accessed 14/07/13] URL:
http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html (The Engineering Toolbox)
10”
Metals-Specific Heats”, The Engineering Toolbox
39 [Online][Date Accessed 14/07/13] URL:
10”
http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html
(The Engineering
Toolbox)
Metals-Specific Heats”, The Engineering Toolbox [Online][Date Accessed
14/07/13]
URL:
http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html (The Engineering Toolbox)
21