A 3 phase immersion will have 3 (or a multiple of 3) elements which are purely resistive.
Do not use typical power factor values for accurate calculations. The actual current for each resistive element will be 9.23 A not the 11 A you calculated.The current flowing in each line is limited by the wiring for each phase, which is rated at 16 A continuous. NEC 430-150, 10 HP @ 460V, 3 phase I = 14 Amp 3. This article has been viewed 288,833 times. This is a general guide to test any type of heating element for failure using an multimeter to measure the resistance of the element. 0000003035 00000 n
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How can I tell if the switch or the element is bad? 0000039403 00000 n
415 x 14.5 x 3 = 18 kW. 0000062640 00000 n
415 x 14.5 x 3 = 18 kW.
In other words, if your circuit breaker trips for \$R\$ in Y-configuration, it'll trip for \$3R\$ in Δ-configuration.
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Well the fuse is 16A. 0000039328 00000 n
It doesn't depend on your wiring, it can be calculated as a wye or delta but the result is the same: 11kW.If you calculate it as a wye/star, you have 230VAC of voltage and 16A of current times x3. Then remove the heating element using a socket or a wrench and you can now check the element as shown in first method. Don't I need to convert it to Vrms first?
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You are using the formulas for D/C power, but the voltage you are calculating with is Peak to Peak A/C. P = 400V * 16A * sqrt(3) = 11.0kW.These are for purely resistive loads. 0000063305 00000 n
Ask some one who is, such as an electrician or an electrical engineer.
If you really can’t stand to see another ad again, then please We use cookies to make wikiHow great. This limits the total W to 3.693 kW per resistor and 11 kW total. The purpose of these is to maintain a constant load on a generator when no load is present. Leverage Watlow's growing toolkit of calculators, equations, reference data and more to help design your thermal system. Having trouble finding the right product? Amp Draw is now over the 48 per circuit allowed bu UL and NEC; The Element output is now at 1/3 more than factory design limit. If I design the heating elements to connect between phase and null = 230 V RMS * 16 A, I get 3 * 3680 W = 11 kW max Power out of the socket.If I design the heating elements to connect between two phases I get 400 V RMS, but I can draw less than 16 A * 400 V. How much less?In which case I can draw 14,4 kW out of the socket. KW, KVA KW is real consumed power turned into heat, and is the product of volts x current x power factor. say the element is wired to 480V 3Ø, has 48Ω resistance between connection points for 10A draw per phase-- I calculate 4.8kW on each third of the element … 4 0 obj
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Since the power going into the two circuits can't be different, it's impossible that you find one to give you more heating than the other. 0000001096 00000 n
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Please fill out the following form, and we will contact you soon. I have several resistive heating element and I want to draw the maximum power from a 400 V, 3 phase, 16 A socket. 0000039477 00000 n