Calculating energy usage of a compressor pack (Rack)?

[buddy]

Hi Guys,

I didnt know if this was the right forum to post on energy issues?

Anyway, I just wish to check which is the best way to calculate a Multi compressor rack energy useage?

The measurements are required for comparison purposes of a "before and after" energy saving initiative.

e.g -

A Low Temperature compressor rack operating with 3 x Bitzer 5 Horsepower compressors is planned to have glass sliding doors fitted to the runs of Island freezers.

Calculation method??

5 hp compressor x .746 kw x 4.0 (price of electrity per Kw/Hr) x 24 hrs running x 365 days = per year savings per compressor? then x 3?

Before and after readings will be recorded on a current logger but wish to confirm if the calculatiuon method being used is correct so we can convert it into actual money saved for the customer?

Also will there be liquid floodback, or any other detrimental effects with fitting glass sliding doors to island freezers?

I have heard energy savings of up to 40% are acheavable...is this correct?

best regards.

[mad fridgie]

Are you the store owner, a consultant, an engineer or something else

[buddy]

Thanks for the quick responce.

Strange situation this one, sort of got 1 foot in both camps on this one.

Working for a Supermarket chain but also the refrigeration contractor has hired me to assist.

Before anybody comments, strick integrity and no conflicts of interest.

best regards

[buddy]

Correct calculation anybody?

[mad fridgie]

Under a fixed set of conditions you are measuring change, between before and after. If you are measuring current, then you need to measure power factor, if the client is paying for Kwhrs, or volts as well is client is paying for KVA. The you need to bring time into your calculations to give averages.

[buddy]

Thanks Mad Fridgie,

I am asking your opinion on what you consider is the correct calculation method taking into account all the factors you have mentioned?

best regards

[buddy]

Sorry,
Kw/Hrs the client pays for.

[Magoo]

Buddy.
3 phase 400vac
Kw = volts x amps x 1.73 x Power factor x efficiency / 1000
better to use a Kw/Hr check meter, better still a system power analyser from Techrentals.
And install on system for a period of time to get better trends and results.
Commercial users are usually pinged with peak load penalties as a multiplier. Power factor correction systems can save large amounts as well.

[buddy]

Thanks for your contribution Magoo,

We shall be using a Kw/Hr check meter.

So for example the Store has plans to fit Glass sliding doors to its run of open Island freezers.

We fit the Kw/hr check meter to the incoming 3 phase on the Low Temperature compressor rack.

We collect the data over 24 hrs with out the glass sliding doors fitted, lets say for arguements sake to make it easy a 100Kw/Hr reading is the data.

Then we gather Kw/hr data for another 24 hours after the glass sliding doors have been fitted and again for arguements sake to make it easy a 80 Kw/hr reading is the data ( a 20% reduction).

The calculation would be what in your opinion.

I am aware there are variables but can you please expalin the 1.73 that is in your calculation and power factor we dont know or efficiency.

Need something simple to show the food retailer the payback converted into a monetary value.

best regards

[Magoo]

Hi Buddy,
1.73 is applied to calc for 3 phase power. [ square root of 3 ], power factor can range between .9 and .99, higher the better or less power consumption. Efficiency for semi hermetics is high being gas cooled, around 0.9.
The check meter needs to be on system for a period of time [ say a month ] to get a better trend.
Your example of 20% reduction as in 20KwHrs x unit cost for power, saving can vary due to store traffic and weather conditions.

[buddy]

Hi Buddy,
1.73 is applied to calc for 3 phase power. [ square root of 3 ], power factor can range between .9 and .99, higher the better or less power consumption. Efficiency for semi hermetics is high being gas cooled, around 0.9.
The check meter needs to be on system for a period of time [ say a month ] to get a better trend.
Your example of 20% reduction as in 20KwHrs x unit cost for power, saving can vary due to store traffic and weather conditions.

Hi Magoo,

Thank you again for your input.

I understand about all stores being different.

Example calculation.

5 hp compressor x .746 kw x 4.0 (price of electrity per Kw/Hr) x 24 hrs running x 365 days = per year savings per compressor? then x 3?


Question?

Is the above calculation method good enough for measuring the power consumption of a Compressor rack?

What calculation would you recommend for measuring Compressor rack energy consumption and converting it into a monetary value?

best regards

[Magoo]

Hi Buddy.
A 5 HP compressor does not run at max capacity during normal operation, and probably doesnt run 24/7.
The only accurate way to get power consumption is to temporarily fit a system analyser. Install to the mains incomer to the motor control panel so as you include the whole system, fans etc., as well as compressors. the analyser will print out peak loads, accumulative KwHrs, power factor and miles of other stuff.
Because racks cycle compressos etc., with varying conditions. The formula earlier gives a drawn power for that moment in time the readings are taken. You need an accumultive value over a period of time to be half confident before quoting savings to a client.
A check KwHr meter is the simplist way.

[Segei]

One thing is calculation, another thing is measurement. What do you want? To measure or to calculate.

[buddy]

Thanks guys for your valued input.

Sergei,

I know how to measure and want confirmation on the calculation.

best regards

Buddy.

[Magoo]

Hi Buddy.
so where are you at. You have formulas, you have recommendations and power monitoring consumption.

[buddy]

Hi Buddy.
so where are you at. You have formulas, you have recommendations and power monitoring consumption.

Hi Magoo,

We will fit a Kwh analyser to an MT compressor rack overnight, gather data and then install the energy saving features and do a comparison of the "before and after" data.

The food retailer just wants to know if it will reduce his energy usage.

Not exact sceience I know because of the variables, but unless any one else has any better suggestions that is the way we are going to do it.

best regards.

[mad fridgie]

OVER NIGHT? at least monitor indoor and out door temps. Changes in these conditions on a single over night test, could negate or over estimate the savings made.
You must have some base data from the cabinet manufactures. Calculate the losses of the cabinet fixed sides, sub tract these losses from the design rating, what is left is a combination of air infiltration and thermal losses through the air curtain. Do the same but now calculate based upon the new "false front fitted". You will still get a bit of air infiltration, make an educated guess. The difference in what is saved ref Kw. Look at the COP of your rack, devide the Ref Kw by the rack COP, this will give the electrical saving. Then calculate the daily saving, multiply the hourly saving by the number of hours, the false front is on.

[Segei]

You should compare apple and apple. Ambient conditions and loads should be similar. Do several tests to get average result.

[buddy]

Thanks Guys for all your replies and apologies for my late reply.

Well I did get jumped on as expected.

The food retailers are harsh and just wont understand but your are right apples for apples, near enough same ambients overnight on different days, a before and and after comparison and reduce your energy saving values to take into account all the variables.

No laboratory conditions on site, 8 year old equipment etc etc.

Thanks for your input.

[Magoo]

Hi Buddy,
thanks for update, so what is the next step. Did you note the power factor on the site?.

[buddy]

Hi Magoo,

I tried to upload a data chart for you to look at but it keeps on asking me for the URL???

How do I upload a file or picture?