Hi, I have a 150 HP rated, 150 kVA

[mech5]

Hi,
I have a 150 HP rated, 150 kVA, 180A @460v VFD drive that we are adding to a room. How much heat will be dissipated to the room?
Thanks

 

[MintJulep]

As a SWAG, 15% of the power being used by the driven equipment.

Actual power, not nameplate rating.

 

[DRWeig]

It depends on the motor load.

My best answer: ask the VFD manufacturer, they give this info out often. It's often in the data sheet.

If that phone of yours is just too heavy to pick up, consult the attached document. It's from the DOE EERE. Say that three time real fast... The difference between actual drive efficiency and 100 percent equals heat released.

For those who don't want to bother reading,

DOE = US Dept of Energy
EERE = Energy Efficiency and Renewable Energy division of DOE


Best to you,

Goober Dave

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[willard3]

You should read forum policies, no 9 of which reads as follows:

>>>>>>>Academic related posts are discouraged. If you post topics related to undergraduate course-work or projects, your posts will be red flagged and marked for deletion. Students generally have vast resources available to them (libraries, professors, etc), and need to learn how to use those sources instead of relying on the Eng-Tips.com fora (it is similar to having someone else do your homework for you; not a good idea.)

 

[willard3]

I meant to post this elsewhere

 

[jkauwale]

Hi Mech5,

Typically 2% to 6% of the operating horsepower will be dissipated as heat to the space. During the preliminary design stages, I typically use 5% of the horsepower rating of the VFD equipment.

In your case, 150 HP * 5% = 381,818 Btu/hr or 1.6 tons. [Assumes operating horsepower of 150 HP]

For more accurate results, as DrWeig stated, please check the manufacturer's product data for the efficiency of the VFD. Also use the operating horsepower of the equipment.

For example, check out the product data for this VFD:

http://www05.abb.com/global/scot/sc...773e007661d6/$file/ACH550_PHPB01U_EN_REVC.pdf

This product data indicates an efficiency of 98%.



Justin K, P.E.

http://www.engproguides.com