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MCC Compartment Temperature Rise

MCC Compartment Temperature Rise

MCC Compartment Temperature Rise

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

Any help would be appreiated.

RE: MCC Compartment Temperature Rise

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.  

RE: MCC Compartment Temperature Rise

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 --

RE: MCC Compartment Temperature Rise

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

RE: MCC Compartment Temperature Rise

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.

RE: MCC Compartment Temperature Rise

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

RE: MCC Compartment Temperature Rise

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







      




RE: MCC Compartment Temperature Rise

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

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