How did you use that calculation? No where are you using the power at all. What is "sq" mean? "square", "square root"? Square or square root of what?This is the question. What is the normal operating current of a balanced three phase 15kw heater (delta connected) when applied to balanced voltage 440/3/60
I got 19.7 A
Calc I used-- volts x amps x 3 sq/1000 = 19.7
Now who is getting sloppy?P = V*I*sqrt(3)
I = P/V*sqrt(3)
You are absolutely correct!Now who is getting sloppy?
I believe it's:
I = P/(V*sqrt(3))
This is the symbol I meant (3√)How did you use that calculation? No where are you using the power at all. What is "sq" mean? "square", "square root"? Square or square root of what?
Look at your units.
On the right hand side you have 19.7, which you apparently mean 19.7A.
Does the left hand side yield units of amps?
Since you did get the correct answer, I'll show you how you might consider presenting your work:
Power in a balanced three-phase circuit:
P = V*I*sqrt(3)
where V is the phase-to-phase voltage and I is the line current. Thus
I = P/(V*sqrt(3)) // EDIT: Parens in denominator added to fix the mistake that LDC3 caught.
I = 15kW/(440Vrms*sqrt(3)) = 19.7Arms
It would be 48So what would 16√9 be? would it be 4*9=36 or 16*3=48?
The convention I've always seen (could be different elsewhere, I realize) is that the thing being taken the square root of goes under the radical and the radical has that check-mark looking part on the left side. Thus, if you only have that character is a text only situation, the thing that you are taking the square root of goes to the right. This is also consistent with the usual convention used with infix notation in which unary operators, like operate on a single operand that is located to the right of the operator. So √9 would be 3 and 9√ would be undefinted, just like 9- would be undefined.
I think its on the wrong side of the numberOkay, so then if 16√9 means 16 multiplied by the square root of 9, what does 16√ mean?
25kw/√3 *440=19.7AShifting gears a bit, if you haven't already done so, it might be good for you to see if you can derive that equation that you used, namely
P = V*I*sqrt(3)
for a balanced three phase load and show that it applies for both wye and delta connected generators and loads.
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Units, units, UNITS!25kw/√3 *440=19.7A
by Lisa Boneta
by Luke James
by Gary Elinoff
by Gary Elinoff