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AN #20211:

Back to Basics

Energy Savings - PF Correction

Improving system power factor reduces apparent power [volt-amps] for same amount of true power [watts].

Example:         Find capacitor KVAR to bring 100KW load from 80% to 95% Power Factor.   Factor is 0.421, so 0.421 x 100 = 42.1KVAR needed.

AN #30505:

Watts Transducer - Applications

Example:     Consider a standard 480/277V, 3-phase, 4-wire system. KW meter full scale = 100KW (selected),     PT ratio = 2 (PTs or voltage input 1/2 tap, 480/240 = 2), CT ratio = 30 (selected 150/5 CTs, 150/5 =30).

Find Cal-Watts and Single-phase Test-Watts for this watts transducer.

1. Calibration Watts (Cal-Watts) Short Method Calculations [1]:

Formula: Cal-Watts = Full Scale Rating / (PT x CT ratios)

Therefore,     Cal-Watts = 100,000 / (2 x 30) = 1,666.66 Watts

1,666.66

2. Calibration Watts (Cal-Watts) Full Method Calculations:

Find full scale Amps in terms of secondary current.

I [Amps] = (KW x 1000) / (V[Volts] x 1.732)

Therefore,     I [Amps] = 100,000 / (480 x 1.732) = 120.281 Amps

Isec. = 120.281 / 30 = 4.00937 Amps

Find Cal-Watts in terms of the current and voltage actually seen by the device.

Cal-Watts = Isec. x Vsec. x 1.732

Therefore,     Cal-Watts = 4.00937 x 240 x 1.732 = 1,666.66 Watts

1,666.66

3. Single-phase Test-Watts [1]:

Formula: Single-phase Test-Watts = Cal-Watts / K

where

K=1 for 1-phase, 2-wire (1 element)

K=2 for 3-phase, 3-wire (2 element)

K=4 for 3-phase, 4-wire (2 1/2 element)

Therefore,     Single-phase Test-Watts = 1,666.66 / 4 = 416.66 Watts

Note: Transducer's voltage coils shall be rated for 139 Volts nominal, because of Line-Neutral input (240/1.732 = 138.56). In this case the current required for single-phase scaling is 3.00Amps (416.66/138.56 = 3.00).

416.66

References: [1] Westinghouse Application Data 43-860 A WE A

LMM

AN #20211: Back to Basics	Energy Savings - PF Correction Improving system power factor reduces apparent power [volt-amps] for same amount of true power [watts]. Example: Find capacitor KVAR to bring 100KW load from 80% to 95% Power Factor. Factor is 0.421, so 0.421 x 100 = 42.1KVAR needed.


AN #30505:	Watts Transducer - Applications Example: Consider a standard 480/277V, 3-phase, 4-wire system. KW meter full scale = 100KW (selected), PT ratio = 2 (PTs or voltage input 1/2 tap, 480/240 = 2), CT ratio = 30 (selected 150/5 CTs, 150/5 =30). Find Cal-Watts and Single-phase Test-Watts for this watts transducer.
	1. Calibration Watts (Cal-Watts) Short Method Calculations [1]: Formula: Cal-Watts = Full Scale Rating / (PT x CT ratios) Therefore, Cal-Watts = 100,000 / (2 x 30) = 1,666.66 Watts	1,666.66
	2. Calibration Watts (Cal-Watts) Full Method Calculations: Find full scale Amps in terms of secondary current. I [Amps] = (KW x 1000) / (V[Volts] x 1.732) Therefore, I [Amps] = 100,000 / (480 x 1.732) = 120.281 Amps Isec. = 120.281 / 30 = 4.00937 Amps Find Cal-Watts in terms of the current and voltage actually seen by the device. *Cal-Watts = Isec. x Vsec. x 1.732* Therefore, Cal-Watts = 4.00937 x 240 x 1.732 = 1,666.66 Watts	1,666.66
	3. Single-phase Test-Watts [1]: Formula: Single-phase Test-Watts = Cal-Watts / K where K=1 for 1-phase, 2-wire (1 element) K=2 for 3-phase, 3-wire (2 element) K=4 for 3-phase, 4-wire (2 1/2 element) Therefore, Single-phase Test-Watts = 1,666.66 / 4 = 416.66 Watts Note: Transducer's voltage coils shall be rated for 139 Volts nominal, because of Line-Neutral input (240/1.732 = 138.56). In this case the current required for single-phase scaling is 3.00Amps (416.66/138.56 = 3.00).	416.66

	References: [1] Westinghouse Application Data 43-860 A WE A	LMM