0 registered members (),
65
guests, and
9
spiders. 
Key:
Admin,
Global Mod,
Mod



#209935  05/15/13 11:19 PM
Re: Electrical demand from multiple breakers.
[Re: Woodiyaki]


I think you have a safe bet that the total demand is <= the summation. You would have to take a look at the process for more clues. Consider if Bkr A feeds a conveyor lifting a heavy load to a compactor hopper. Bkr B feeds the compactor motor. Unless you had a continuous feed of identical materials, your maximum demand peaks would not likely be concurrent. Then you would probably have to look into any net power factor concerns. Joe



#209936  05/15/13 11:31 PM
Re: Electrical demand from multiple breakers.
[Re: Woodiyaki]


Woodiyaki: Welcome to ECN forums from one of the 'Jersey Guys'.
IMHO, your math is OK.
Scott35 may have a detailed comment.
John



#209938  05/16/13 07:49 AM
Re: Electrical demand from multiple breakers.
[Re: Woodiyaki]


As a design issue you might want to use 125% of the larger and the other at full load. (basically the motor rule)
Greg Fretwell



#209940  05/16/13 12:04 PM
Re: Electrical demand from multiple breakers.
[Re: Woodiyaki]

New Member

Joined: May 2013
Posts: 2
OK


Joe: I see your point, that makes good sense to know which breaker is feeding what part of the process, I'll have to see what I can determine on that front at a later time. What I'm interested in knowing right now is simply the peak demand for the overall two breaker system, the two data points given were the peak demand of each breaker over the same time interval. This is what lead me to my summation assumption. I'm glad to know that others seem to concur that this is OK to do. The two breaker system is the system we are looking to move the process off of. The single breaker system that we would like to move the process onto already runs a similar process (minimal changeover would be required), and can do it with greater capability at a lower cycle time. The similar process has a peak demand of 58 kW over the same time frame for a net reduction of 5 kW in peak demand. So now I think I can begin some calculations that could turn into dollar signs for my boss (which hopefully in turn will be enough to appease his boss ) Thanks to everyone for the welcome and the input, I greatly appreciate it.



#210016  05/22/13 08:08 AM
Re: Electrical demand from multiple breakers.
[Re: Woodiyaki]


Woodiyaki;
Hope I am not too late, but nevertheless, here is my 2¢...
*** First off, some Engineering Jargon regarding the Calculations... ***
To perform the Calculations for this Process Equipment, it would be best to record the Peak Load Amperes of each Circuit, or the Peak KVA of each Circuit.
Reading Peak KW Consumption without knowing the Power Factor at that Peak Period will not allow for your calculated Load values to be converted in to KVA  and eventually in to Peak Load Amperes.
For example:
480V 3 Phase Machine records a Peak True Power Draw of 25,000W (25 KW);
a. If the Peak Power Factor is 70% (0.7 PF), the Apparent Power Drawn from the Supply will be 35,725 VA (35.73 KVA), and the resulting Peak Load Amperes will be 42.99A. (25000 x 1÷0.7 = 35725 va; 35725 ÷ 480 x 1.732 = 42.99A)
b. If the Peak Power Factor is 90% (0.9 PF), the Apparent Power Drawn from the Supply will be 27,750 VA (27.75 KVA), and the resulting Peak Load Amperes will be 33.39A.
c. If the Peak Power Factor is 100% (1.0 PF, or Unity PF), the Apparent Power Drawn from the Supply will be 25,000 VA (25.00 KVA), and the resulting Peak Load Amperes will be 30.08A.
...........................................................
Unity Power Factor would be found with 100% Pure Resistance type Loads. All other Load types will rank below Unity P.F.; Where the connected Load's Reactive Power is highest in Inductive Reactance [XL], the Power Factor is "Lagging"; with higher Capacitive Reactance [XC], the Power Factor is "Leading".
In either case (XL or XC), the Power Factor will be less than Unity, so the Apparent Power Drawn from the Supply will be a combination of "TRUE POWER" (Kilo Watts), and "REACTIVE POWER" (Kilo Voltamps Reactive).
....................................................................................................................................................................................
With the above being said, let's apply numbers to your query:
 Placing the Equipment on a Single Branch Circuit 
The Equipment appears to draw a Peak True Power Load of 63.00 KW (30.1 KW + 32.9 KW). Not sure of the System Voltage this Equipment is connected to, so I will use 480V 3 Phase for these examples.
If the Peak Power Factor is 0.8, the Equipment will have an Apparent Power Draw of 78.75 KVA. 78.75 KVA at 480V 3 Phase = 94.77 Amps.
If the entire FLA (Full Load Amperes) was Continuous, the minimum Ampacity of the Branch Circuit will be 118.46 Amps (98.44 KVA).
A single 125 Amp 3 Phase 3 Wire Branch Circuit could be used to feed the Equipment in this example.
Things to include for the total Calculations:
1.: Total Coincidental Peak KVA Load Values, 2.: System Voltage and Phase, 3.: FLA of the Largest Motor, 4.: Load Amperes / KVA running for more than 179 Minutes, 5.: Locked Rotor Amperes for Motors > 2 HP (especially for Motors connected to hard starting shaft loads).
Good luck!!!
Scott (EE)
Scott " 35 " Thompson Just Say NO To Green Eggs And Ham!




