Calculating Demand for Four Wire Delta Service

[AHarb]

Hi,

I'm trying to figure out how to calculate the maximum possible demand through a 9S meter served by a four-wire delta bank. I know the meter is capable of a maximum 20 amps per phase. If we assume there are 20 amps per phase and we have a unity power factor (cos (theta) = 1), which of the 3 voltages should we use in the power calculation? Below is a diagram showing the four-wire delta bank and the voltages.



Any help would be appreciated.

Thanks in advance.

Adam

 

[Julius Right]

The maximum permitted through the each transformer phase winding is 20/sqrt(3).If the receiver is
three-phase rated it is not important where the neutral is. The permissible apparent power will be:
S=sqrt(3)*240*20=8.313 kVA.
If you have only single-phase consumers then you have to limit the current to the maximum permitted
through the transformer windings. A part of winding AC serves both A-N and B-N circuit [the C-N and other part of B-N circuit].
Let's take both IA-N and IB-N current through A-N part of transformer winding. If p.f.=1 for all phases then the current will be in phase with the voltage. Then IB-N will be 90 dgr. lagging IA-N so the common current module will be sqrt*(IA_N^2 +IB-N^2/4) and should be not more than 20/sqrt(3).
If, at the end, IA_N=IB-N then the total maximum current flowing through A-N will be sqrt (IA-N^2+IA-N^2/4) =IA-N*sqrt (1.25) =20/sqrt (3) IA-N=IC-N=IB-N=10.328 A.
You may change this pattern as you wish but you have to keep the common current at 20/sqrt(3) level [or less].

 

[Julius Right]

Sorry, I forgot you question .Then you have to multiply each circuit supply voltage by the current:
S=2*120*10.328+208*10.328=4.627 kVA:weeping:

 

[AHarb]

Sorry, I forgot you question .Then you have to multiply each circuit supply voltage by the current:
S=2*120*10.328+208*10.328=4.627 kVA:weeping:

Thanks for the reply and the help. Where are you getting that the maximum permitted through each transformer phase winding is 20/sqrt(3)? I'm confused how you came up with that. Are you somehow working off the fact that the line to line voltage is the same for all 3 phases (240 V)?

I was thinking about this over the weekend and can't we just substitute the line to line (V_LL) voltage in the P = 3*(V_LN*I_L) equation? That would then give us P = ((V_LL*20)/sqrt(3))*3. Since we know V_LL is 240V then we come up with P = ((240*20)/sqrt(3))*3 = 8.31 kVA. Is that incorrect?

 

[GoldDigger]

Remember that the individual line current will be the vector sum of the line-to-line currents, which is one of the places the square root of three comes into the picture.
For the transformer, the currents in the windings are what is important. For a delta wound transformer, that will be the line to line current, for a wye wound secondary, that will be the line to neutral current.
Make sure you use the current value that corresponds to the voltage value.

Tapatalk!

 

[Julius Right]

Thank you, GoldDigger.The attached sketch could enlighten more.

 

[AHarb]

Thank you, GoldDigger.The attached sketch could enlighten more. View attachment 9854

I'm still not getting where the sqrt(3) comes in to the equation. The drawing helps and I understand the IA = IAB - IAC, but then I don't see what you're plugging in to introduce the sqrt(3). Sorry if this is a dumb question. It's been a while since I've used vector math so it may be something with vectors that I've forgotten.

 

[GoldDigger]

In a 60 degree right triangle the ratio of the opposite side to the hypotenuse (the sine of the angle) is (sqrt3)/2.
That gives you the ratio of the length of the vectors in the diagram.

Tapatalk!

 

[Julius Right]