9/1/2023 0 Comments 20 kw to amp![]() ![]() ![]() What are you going to do, go to the customer when the inevitable failures occur and wave the datasheet saying "But it says here it should have worked"? Use beefier limit switches or split each 5 kW load into two 2.5 kW loads, which will draw only 12 A each and will be fine with the existing limit switches. Having a switch rated for 25 A regularly carry 24 A loads for sustained periods of time is asking for trouble. So in theory, these 25 A limit switches can handle your 24 A load. However, the Amp rating of these thermal limits switches are almost certainly RMS, not peak, check the datasheet. 24 A RMS therefore has peak currents of 34 A. You are right in that the instantaneous peak current is higher, but for sine waves it is higher by the square root of 2 from the RMS, not the square root of 3. Also, the correct units is kW (killo-Watts), not KW (Kelvin-Watts)). (By the way, the extra 1000 in your equations is incorrect. Each element is apparently driven with 208 V accross it and is drawing 5 kW of power. You co-worker is right in calculating the current per element. But intuitively I don't know why it works for Wye but not for delta connections. With a Wye connection my coworker's argument would work because the voltage across each element is 120V. So instantaneous amps would be 24 amps*1.73 = 41.7 amps (to reconcile the two approaches). My thought is that even though on average each element is producing 5 KW of heat, instantaneously it might be producing more than that, as long as all three elements together are producing 15 KW at any given instant. P (kW) 3 x PF x I (A) x V L-L (V) / 1000, which means that the power in Kilowatts is calculated by the square root of three multiplied by the power factor times the phase current in amps by the line to line voltage RMS. Obviously one can't come up with two different answers. So no need to branch for the thermal limit switch. So amps per leg is 5 KW *1000/208V = 24 amps. Treating each leg separately, one can treat it as a 'single phase' circuit with 208V across the element. Rta: // The first thing to do is multiply the Hp by 746, then you must divide the previous result between the multiplication of the voltage line line, the efficiency, the root of three and the power factor, giving as a result: 12.33 Amps. One kilowatt is equal to 1000000 milliwatts: 1kW 1000000mW. One kilowatt is defined as energy consumption of 1000 joules for 1 second: 1kW 1000J / 1s. Since I1 = I2 = I3 = 41.7 amps in each leg, I need to split into two branches to divide the amps into two because of the 25 amp limit switch.Ī coworker is arguing the following: Each element is producing 15KW/3 = 5 KW per element. One kilowatt (kW) is equal to 1000 watts (W): 1kW 1000W. On each leg there is a thermal limit switch with 25 amp rating. My calculation for amps in each leg is 15KW*1000/ = 41.7 amps. Lets say I have a 208V 3-phase heater (delta connection), with a heating element on each leg L1, L2, 元. I am programming an app for ordering 3 phase heaters, and I have an argument to resolve with a coworker. I am a mechanical engineer with basic understanding of electricity. Online electrical calculators for calculations of electricity and electronics. ![]()
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