tadavidson
Senior Member
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- Georgia
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- Electrical Contractor
Related to a recent post in other forum. At 480volts 3 phase into a 15kw heater how much less is the kw output if the voltage is only 455v.
Kw output at lowered voltage calculation.
- Thread starter tadavidson
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tadavidson
Senior Member
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- Georgia
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- Electrical Contractor
Would I be correct in I= P÷Ex1.73 then p=IxE÷1.73?
p = 1.73 x v x i if pf is 1, as it is in this case
v goes down but so does i because R of the heater is the same
p = v^2 / R
take the ratio
15 = 480^2 x R
p? = 455^2 x R
R cancels
p? = 15 x (455/480)^2 = 13.5
tadavidson
Senior Member
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- Georgia
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- Electrical Contractor
Yes thank you.
Electric-Light
Senior Member
(new/old)^2 = n
old watt * n = new watt
only good for resistance load that remains constant. It will be quite a bit off for a light bulb.
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And off in a totally different way for a motor, especially a motor with constant mechanical load.
Your formula is the reduced torque when voltage supplied is lower than the rated voltage.
As you know, Power is a function of torque and speed. The motor will run slower when the voltage is lower so the power drawn by the motor will be lesser than your guesstimate.
Power = wT = 2 x pi x rot. speed x Torque
jumper
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Er, what motor? OP said heater which kinda implies a resistive load to me.
Not that I got anything against motors and your info is great, but,,,,,,:?
Sorry, I'd better get some sleep.
he's talking about a resistance heater?
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