MCC Compartment Temperature Rise

[mutttt]

Can someone tell me how to go about working out the potntial temperature rise in a MCC compartment.

Any help would be appreiated.

 

[dpc]

Talk with the MCC manufacturer. There are way too many variables to make a meaningful calculation.

UL may have some standards as well. They don't use 90 deg C wire in there for nothing.

 

[pablo02]

What are your needs? Do a key word search using "MCC heat" for a list of threads (in this forum and others) -- possibly there are some ideas there (you might try some other key words to see if other threads show up..)

There are also a number of resources for estimating the heat gain for electrical equipment: I use a book, "HVAC Equations, Data, and Rules of Thumb" by Arthur A. Bell, Jr. for estimating A/C needs for MCC rooms --

 

[panelman]

rittal have a section of their catalogue devoted to this...I think they have a bit of software available too, on the electronic version of the catalogue if my memory serves

 

[jbartos]

Suggestion: It is necessary to know watt dissipation of various items located in the MCC compartments, e.g. contactor and relays coils wattages, thermal overloads heat dissipation, control transformer wattage, contact resistances, etc.

 

[mutttt]

Sorry guys its actually the temperature rise in a field mounted junction box which I have been asked to calculate.

 

[bklauba]

The electrical enclosure mfrs. have these formulas and discussions in their catalogs (Hoffman & Hammond, amongst others). You will need to know the wattage dissipated inside the JB, and its surface area. The formula will be able to give a temperature rise above ambient (TRAA). Looking at the Hoffman, they supply a linear relation depicted in a chart, that can be summarized as:

TRAA(degs F) = 4.375 degs. F / sq.ft. (1)

So if you have a 48" X 36" X 16" enclosure that has 300 watts being dissipated within it, this would mean that:


2[(48x36)+(48x16)+36x16)] / 144 = 42 sq.ft. (2)

300 watts/42 sq.ft. =~ 7.1 watts/sq.ft. (3)

from (1) we get: 7.1 X 4.375 =~ 31 degs. F. rise

Hope this helps.

bklauba@airmatic.com

 

[jbartos]

Suggestion to the previous posting: Please, would you clarify dimensions in:
""from (1) we get: 7.1 X 4.375 =~ 31 degs. F. rise""