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jnxPsuOutletPowerFactorValue

OBJECT-TYPE
1.3.6.1.4.1.2636.3.58.1.2.4.1.10
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read-only
Status
current
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0
Power factor percentage of each PSU (2k/3k). Algorithm for calculation of Power Factor is below. For PowerOut values that fall in between 618.93W and 915.24W, say 700W, the appropriate PF ranges from 0.910191 & 0.917994. Following linear equation could help deduce a fairly accurate input power value. Linear equation y = mx + b (where m is the slope and b is the Y intercept) Slope m = (y2 - y1) / (x2 - x1) Y intercept b = y - mx Plugging it all together for our example: m = (915.24 - 618.93) / (0.917994 - 0.910191) = 37973.86 b = 915.24 - (37973.86 * 0.917994) = -33944.5 for 700W (y), our efficiency (x) would then be: x = (700 - (-33944.5)) / 37973.86 = 0.912326 = 91% PowerIn = 700W /0.912326 = 767.26W
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