




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

Broom Pusher and
Member
Registered: 10/19/00
Posts: 2724
Loc: Anaheim, CA. USA

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).
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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!

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