heater in the cabinet.
How big the power of the heater should be?
Two factors:
1. delta temperature between inner cabinet and environment.
2. the surface area of cabinet
Use the formula bellow:
Q=k*A*dT
k: materials heat loss index, for steel and stainless steel, k=5.5
A: surface area of cabinet
dT: delta temperature between env and inner cabinet.(`C)
So for example, when we use cabinet RITTAL WM483614 with stainless steel
surface and size 1.22x0.91x0.41m, used in -40`C lowest.
Q=5.5*2*(1.22*0.91+0.91*0.41+0.41*1.22)*40=11*(1.1102+0.3731+0.5002)*40=
872.74W
Oh, so big power...note that we ignore the heat power of the electric
components in the cabinet.
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