- Joined Aug 21, 2019
Hi, no I don't really know where the square root of 3 comes from in the conversion of a phase voltage/current to a line voltage/current.Hi,
Do you know where the square root of 3 factor comes from?
That is part of the key to solving this problem.
Hint: the spec's are in line to neutral voltage but the motor runs on line to line voltages.
So given is: V(line to line)=440 V and V(an)= V(phase)<0°. But the motor is connected in delta and in delta V(line to line) = V(delta) therefore: Van = 440<0° ?? I don't understand why the factor square root of 3 is necessary here (since the motor is connected in delta) ?Hello again,
Well for now i think as long as you know that the square root of 3 relates the line to neutral voltage to the line to line voltage you're doing just fine. So can you see where one of the calculations comes from now?
The next thing then is to take the line to neutral voltage and convert it to line to line voltage because that's what the motor is running off of.
Yes, thank you. In respect to where the factor sqrroot of 3 comes from it helps.hi N,
Does this link help, ref root 3.?
PS: I don' think the solution from the text is correct, as the solution manual is known to be full of mistakes.But is my method for calculation correct? and if not, what would be the correct method ? Because I've found different methods on the internet with al different solutions for this problem. So I'm very skeptical.
yes, that I understand. And is it because there is a neutral line that you can assume that the voltage source is wye(Y) connected? and therefore: Vp= (440/sqrt(3))= 254V ? and Van=Vp<0° = 254<0° ?Hello again,
Let's take one at a time.
Do you understand how they got:
Hi,yes, that I understand. The first calculation that I don't understand is the one for the lighting: Vp= (440/sqrt(3)), I would say that the voltage across the lighting load is Vcn and since Van=Vp<0° ==> Vbn=Vp<-120° and Vcn=Vp<120°. And because motor is delta connected Vl=Vp=440V, therefore: Van=440<0° , Vbn=440<-120° and Vcn=440<120°, no ?
Because I looked at it from the perspective of the motor load which is delta connected. So I thought for delta: Vl=Vp and given is that Van=Vp<0° so therefore Van=Vl<0°. But I realise now that I can assume that the source is wye(Y) connected (because of the presence of a neutral line) and work from thereHi,
Why on earth would you think that the line to neutral voltage would be the same as the line to line voltage?
Did you look at the phasor drawing for three phase? You can see right away that the distance between the tips of the line to neutral arrows are farther apart than the length of one of the phase arrows. That means the line to line voltage is always greater than the line to neutral voltage. Since the relationship is VLL/sqrt(3)=VLN the line to neutral voltage is equal to the line to line voltage divided by the square root of 3.
Since the lighting load is connected line to neutral, the voltage across that must be less than the line to line voltage of 440vac.
If this does not make sense, then look at the phasor diagram and note also that the line to line resultants form a perfect equilateral triangle the sides of which will always be longer than the distance from the center to any vertex, and are related by the square root of three.
Oh yes the source is wye. That means the voltage line to neutral is 440/sqrt(3).Because I looked at it from the perspective of the motor load which is delta connected. So I thought for delta: Vl=Vp and given is that Van=Vp<0° so therefore Van=Vl<0°. But I realise now that I can assume that the source is wye(Y) connected (because of the presence of a neutral line) and work from there
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