MvBurgh
Chemical
- Dec 8, 2008
- 1
Hi All,
Im trying to calculate the heater sizing for an intertec cabinet that will be placed in a freezing location.
However im getting weird results based on:
k.A
--- * dT
L
I'm getting for a small window (0.36 m2) a heat-loss of 1440W.
Thermal conductivity of the glass is 0.8 W/m.K
Can someone indicate where im going wrong?
Best Regards,
Martin
Im trying to calculate the heater sizing for an intertec cabinet that will be placed in a freezing location.
However im getting weird results based on:
k.A
--- * dT
L
I'm getting for a small window (0.36 m2) a heat-loss of 1440W.
Thermal conductivity of the glass is 0.8 W/m.K
Can someone indicate where im going wrong?
Best Regards,
Martin