Estimating power requirements for industrial complex 2

[stevenyoung]

As a non-electrical engineering manager I am looking for a starting point to learn how to estimate the electric power requirements for an industrial complex using refrigeration compressors/motors, battery charging for forklifts, general lighting, air conditioning system, general office use.

Management's power requirement tend to be rounded up at each level then rounded up again until it is really excessive and the incoming cable and transformers too large.

 

[jbartos]

Suggestions:
1. First you have to have some idea about sizes of those items you would like to have electrically powered.
2. Then obtain power requirements from their technical specifications or nameplates, e.g.
Horsepower(s) (~ 1HP=0.746kW)
Kilowatts or Watts, kW or W
Kilovoltamperes or voltamperes, kVA or VA, kVA = kW / cos(fi) = kW / (power factor)
Kilovoltamperereactive of voltamperereactive, kVAR or VAR = kVA * sin(fi) = kVA * (1 - cos(ph)**2)**0.5
3. Then add all kWs in one column and all kVARs second column
4. Then obtain the total: kVA = (sum(kWs)**2 + sum(kVAR)**2)**0.5
5. Then multiply total kVA by some number similar to 1.25 or higher to have 25% or more spare capacity in the power supply service. It is not unusual to have as much as 100% spare capacity of the power source, if you are not certain about a potential growth of the electrical load in the future.

 

[stevenyoung]

Thanks for the helpful response.

I think I get the general approach.

I guess lighting and heating doesn't have a power factor but motors do.

Sorry to be thick but:

But what is the difference between ph and fi and where do you find it (or them)?

What is an average default value for this (these)?

Is it to be measured in degrees or radians or doesn't it matter?

 

[jbartos]

Clarifications:
1. fi=ph meant to be phi. They are often imbedded in P.F.=cos(phi)=0.8, 0.85, 0.9 etc.
2. Fluorescent fixture or high intensity discharge light (HID) ballast may have a power factor specified or on its nameplate
3. Heating elements do not have power factor; however, the heating units with the motor do, because of the motor load.
4. Power factor angle, if shown, is usually in degrees. Often, you will find P.F. = 0.8, 0.9, etc. which shall be interpreted as p.f. = cos(phi) = 0.8, from which phi = 36.9°
5. Motor power factors vary from about 0.8 (smaller motors) to about 0.9 (larger motors) at full (rated) load. If the load is lower, the power is lower according to curves.

 

[hor1349]

with reference to jbartos response this is to inform you in adition of his mentioned procedure you must consider the simultaneos koeficient(=k) of loads .It means if all loads in one time consumed nominal electrical power we assume (k=1).however we often find that k<1.for further explanation for example in one house we have many electrical devices such as refrigerator,lamps,electrical heater,radio,television ,...but all of these equipments don't consume electrical power in one time.

 

[stevenyoung]

On the one hand we know that not every piece of equipment will be going at the same time, on the other - what about start-up loads?

In a typical refrigeration plant (for fruit and vegetable cool store), what &quot;overload&quot; will a motor cause when starting up, and is it necessary to allow for it?

I suppose there are ways of forcing such motors to stagger their start times.

Do I need to go to this level of detail to estimate transformer size and cable capacity?

Thanks for responses to date.

 

[jbartos]

Suggestion: Please notice that the original posting calls for &quot;estimate of power requirements.&quot; The design stage is supposed to refine the power requirements and provide the higher accuracy. It is not unusual, if the estimate is 15 to 20% higher or lower.
For Stevenyoung Jan 16, 2001:
1. The motor starting inrushes (&quot;overloads&quot are normally imbedded in the power distribution system in terms of larger sizes of electrical equipment including associated appropriate protective devices. If properly engineered and designed, there is no need to pay a special attention to those issues.
2. Loads may be started in load blocks, if needed and justified. Some applications allow that others do not. Load starting in load blocks does reduce sizes of electrical power distribution equipment. Consequently, the cost of the associated electrical installation is lower. However, if the power distribution has its adequate spare capacity, then the load starting in load blocks is not considered.

 

[stevenyoung]

In the dim past as a civil student doing some basic electrical, I seem to recall that there was something called &quot;power factor correction&quot; which presumably adjusts the collective PHIs to reduce their root mean squares. I guess this is not going to be relevant to light/medium duty refrigeration for a fruit and vegetable store. Anyway what does PHI measure?

 

[peterb]

Power factor is probably the most difficult electrical area for non-electrical people to get a handle on.
Electrical equipment which runs on straight thermal energy conversion, such as an incandescent lamp or resistance heater, utilizes all of the electrical power which it consumes to produce &quot;useful&quot; power (KW).
On the other hand, equipment such as induction motors which utilize a magnetic field in the process of producing their output require a certain amount of the input power to produce the magnetic field (reactive power - KVAR)- this power is not converted into &quot;useful&quot; power output (real power - KW).
We draw the power triangle consisting of the real power(KW), the reactive power (KVAR) at a right angle to the real power and the total apparent power (KVA) as the hypoteneuse. PHI is the power factor angle between the KW and KVA and the power factor is (KW/KVA)
Power factor correction can help if the utility charges you for KVA (apparent power). Capacitors provide negative KVARs to supply the reactive power requirements of the load. This cuts down on the amount of reactive power that the utility must supply and reduces their system losses - some rate structures encourage you to do this. An added benefit is an improvement in voltage conditions in your facility, as the operating voltage drop from the utility is reduced.
Hope this helps clear up some of the mystery.