Calculating Power with Given Values

[BatmanisWatching1987]

A balanced delta load consist of three 20∠25° impedance. The 60 Hz line voltage is 208 V (rms).

I got the following values using the information above
Phase Current = 10.4∠-25° A
Line Current = 18∠-55° A
Phase Voltage = Line Voltage = 208V

What is the correct Total Power
Since

P_3ø = (3)*V_p*I_p*cosθ = 3*208*10.4*cos(-25°) = 5882 W

P_3ø = √3*V_l*I_l*cosθ = √3*208*18*cos(-55°) = 3720 W.

Why I'm I getting 2 completely different answers?

 

[charlie b]

I can't solve the whole problem for you. But I will observe that you are treating the phase angle of all four voltages (phase voltages A, B, and C and line voltage) as zero. That is not correct. Phase voltages A, B, and C will differ from each other by 120 degrees, and I don't know what the angle of line voltage to phase voltage would be.

 

[BatmanisWatching1987]

I can't solve the whole problem for you. But I will observe that you are treating the phase angle of all four voltages (phase voltages A, B, and C and line voltage) as zero. That is not correct. Phase voltages A, B, and C will differ from each other by 120 degrees, and I don't know what the angle of line voltage to phase voltage would be.

I thought for a Delta configuration, Line Voltage = Phase Voltage

 

[gar]

190404-2243 EDT

BatmanisWatching1987:

You are blindly applying equations without understanding their derivation.

A solution to the problem is:

1. Line-to-line voltage is 208.

2. Current thru one line-to-line load is 208/20 = 10.4 A, that is voltage divided by impedance.

3. Need to know the series resistance of one line-to-line load. This is determined from the solution of a right triangle with one included angle of 25 deg, and a hypotenuse (impedance) of 20 ohms. The result is 20*cos 25 = 20*0.9063 = 18.13 ohms. This results from how phase angle is defined.

4. Power in one line-to-line load is I^2*R = 10.4^2*18.13 = 108.16*18.13 = 1960.9 watts. There are three of these so total power is 3*1960.9 = 5883 total watts.

There are some underlying assumptions. The waveforms are sinusoidal, and steady state in value, and loads are linear.

You can work back from this to determine a single equation to use, and then qualify that equation by defining all of the assumptions.

.

.

 

[gar]

190404-2450 EDT

BatmanisWatching1987:

Your second approach is really one relative to assuming you know the impedance from line to neutral in a balanced wye load, but you don't. However, there is a transform equation that would allow you to determine this.

Fundamental in this second calculation for power in your second method (not the transform) is a determination of the line-to-neutral voltage which is based on cos 30 which happens to equal 1/2 sq-root of 3, and in turn how thru gyrations ends up as a sq-root of 3 in the numerator.

.

 

[Julius Right]

P3ø = (3)*Vp*Ip*cosθp = 3*208*10.4*cos(-25°) = 5882 W
θp=angle between EAB and IAB
P3ø = √3*Vl*Il*cosθL = √3*208*18*cos(-25°) = 5882 W.
θL=angle between IA and VA[VA it is a virtual voltage between terminal A and a virtual N]