Synrad FH FLYER - Page 231

appendix a
SYNRAD FH Series Flyer Operator’s Manual Version 3.4
229
Cooling
Example – the cycle time of the mark is 0.75 seconds and there are 1.25 seconds between ‘start mark’
signals. 0.75 / 1.25 equals a DC
MARK
Duty Cycle of 0.6.
DC
MARK
Duty cycle is ______ .
Solve the Heat Load equation
Flyer’s Heat Load equation is:
T
INT
= T
AMB
+ 14 °C + (T
GRAPH
× DC
MARK
)
Enter the values you have determined for T
AMB
, T
GRAPH
, and DC
MARK
. Solve for T
INT
, which is the maxi-
mum internal temperature that Flyer will reach. Using the values shown in our examples, we can solve the
Heat Load equation to determine the maximum internal temperature. Our example values are:
T
AMB
= 30 °C;
T
GRAPH
= 9 °C;
DC
MARK
= 0.6.
T
INT
= T
AMB
+ 14 °C + (T
GRAPH
× DC
MARK
)
T
INT
= 30 + 14 + (9 × 0.6)
T
INT
= 30 + 14 + (5.4)
T
INT
= 49.4 °C.
Our example calculated temperature is below 50 °C, which is the maximum internal temperature that
Flyer should reach, so no external cooling is required in this case. Remember that this Heat Load calcula-
tion is based on “worse-case” parameters including still air temperatures—even a small amount of airow
over Flyer’s heat sink will reduce the maximum internal air temperature signicantly.