Juniper Psu Outlet Power Factor Value

1.3.6.1.4.1.2636.3.58.1.2.4.1.10

Algorithm for calculation of Power Factor is below. For PowerOut values that fall in between 618.93W and 915.24W. let.s 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 PPowerIn = 700W /0.912326 = 767.26W