ups load

[ceb58]

I need some help explaining to a co-worker the draw on a ups. We have 240v ups systems that contain 4 modules, each module is 3.5kw for a total of 14kw. The ups has a power factor of .8 which when you beat the numbers out it's 72.9 OCP 80. The load it serves is only 3360 watts. Now correct me if I am off base. He thinks that the ups being 14kw it is going to pull 14kw no matter what the load. I am trying to make him understand that 14kw is the max load. Since the load is only pulling 3360 watts the ups should be pulling around 4200 watts (counting in the .8pf). Now am I completely wrong or do I need to beat him in the head to understand?

 

[Buck Parrish]

Perhaps he is confuseing it with the type of working method as a transformer.

 

[cadpoint]

I need some help explaining to a co-worker the draw on a ups. We have 240v ups systems that contain 4 modules, each module is 3.5kw for a total of 14kw. The ups has a power factor of .8 which when you beat the numbers out it's 72.9 OCP 80. The load it serves is only 3360 watts. Now correct me if I am off base. He thinks that the ups being 14kw it is going to pull 14kw no matter what the load. I am trying to make him understand that 14kw is the max load. Since the load is only pulling 3360 watts the ups should be pulling around 4200 watts (counting in the .8pf). Now am I completely wrong or do I need to beat him in the head to understand?

M O

P = IE or , W = AV one in the same ...

You need to show the numbers were the numbers drop in.

I think you need to explain it mathwise not verbally, if you implied that I missed it.
It easiest to forget that it's a simple ratio equation but words don't explain it correctly at times.
So like others things were all aren't wired the same... how they see the light well, you know

The Voltage is contant your wattage gets add up, that amps goes up to make up its side of the ratio (equation). your correct on the PF side its a percentage not a ratio like before (math application), but required.

Ask them to get back with you when the bulb blooms...

 

[gar]

090812-2233 EST

Take any "black box" that has power input to it and power output from it, then "power input in watts" = "power losses in the black box in watts" + "power output in watts". This says nothing about power factor, that is not part of the equation. However, on a short time basis this could be modified by electrical energy being stored or released inside the black box.

Assuming the internal battery is fully charged and that any limitation on input power does not require energy from the battery, then above equation holds.

Operating at the partial load indicated I might expect the UPS to be 60% to 70% efficient. Thus, on this assumption, the 3360 W output load might require 3360/0.6 = 5600 W input. As the output loading increases the efficiency should go up and probably reach a maximum efficiency before 100% output load.

.

 

[broadgage]

A UPS rated at 14 KW but loaded to only about 3.5 KW certainly wont normaly draw 14 KW from the supply, where would all this power go ?
It certainly wont produce over 10KW of heat !

Under normal operating conditions, I would expect it to draw about 4 or 5 kw. Exact figures can only be obtained by accurate measurement or enquiry of the manufacturer.

Care must however be taken when sizing wire and breakers for large but lightly loaded UPS units.
Remember that following an outage, the UPS will draw extra power from the line in order to charge the internal batteries.
This extra load can be greater than expected in the case of large but lightly loaded UPS units.

As a rough "rule of thumb", in the absence of detailed instructions, circuits supplying fully loaded UPS units should be sized at 150% of the rated output, to allow for losses, and for charging the battery after an outage.

Therefore for a 3.5KW UPS, the branch circuit should be able to supply about 5.3 KW.

For a 14 KW UPS , the branch circuit should be able to supply about 21 KW.

Now consider a 14KW UPS but only loaded to about 3.5 KW as in the O/P.
Although in normal use, it will only draw about 4/5 KW this does NOT mean that a branch circuit designed for 4/5 KW will suffice.
Following an outage, the UPS will require up to 7KW (50 % of its rating) to charge the battery, AND 4/5KW to supply the load.
Therefore a branch circuit suitable for about 11/12 KW will be required.

Quite a difference between a fully loaded 3.5KW UPS and a 14 KW one loaded to only 3.5 KW !

The input wattage under normal conditions of a 3.5 KW UPS and a 14 KW UPS loaded to only 3.5 KW, will probably be broadly similar.
However the bigger unit will require a much larger circuit ampacity on account of the extra battery charging demand following an outage.