Three Phase AC Circuits

Perhaps the simplest approach to this problem is to calculate the power dissipation of each resistor inside of each three-resistor array. Since power is a scalar quantity (i.e. it adds directly, not trigonometrically), the 15 kW total heat output of each array means each resistor inside of each array must dissipate 5 kW of power.

In the delta-connected heater, each resistor sees full line voltage (480 VAC), therefore the resistance may be calculated as such:

$$R = {V^2 \over P} = {480^2 \over 5000} = 46.08~\Omega$$

In the wye-connected heater, each resistor sees $1 \over \sqrt{3}$ of the full line voltage (480 VAC), which is 277.1 VAC. Therefore the resistance may be calculated as such:

$$R = {V^2 \over P} = {277.1^2 \over 5000} = 15.36~\Omega$$

Apparent power is the product of the voltage and current, as with any standard electrical power calculation, and also multiplied by $\sqrt{3}$

$$S=V \cdot I \cdot \sqrt{3}$$

$S$ = 198.15 kVA

Converting from VA to HP divides the total VA by 746 (watts per HP).

$$P_{mech} = {S \over 746}$$

$P_{mech}$ = 265.6 HP

With 100:5 ratios at each CT, the line current to this motor is twenty times the amount of current through each ammeter:

$$(2.81) \left({100 \over 5}\right) = 56.2 ~A$$

At a line voltage of 480 VAC and a line current of 56.2 amps, the total electrical power in this 3-phase system may be calculated as follows:

$$P_{total} = (\sqrt{3}) (I_{line}) (V_{line})$$

$$P_{total} = (\sqrt{3}) (56.2) (480) = 46.724~kW$$

At an efficiency of 88%, the total power would be reduced from this theoretical maximum.

$$P_{actual}=P_{total} \cdot 0.88$$

$P_{actual}=41.117~kW$

Since we know there are 746 watts to every horsepower, we may convert this kW figure into HP as follows:

$$41.117~kW=41117~W$$

$$\left({41117~W \over 1}\right) \left({1~HP \over 746~W}\right) = 55.12~HP$$