Question 2.38: A 400 V, three-phase, 50 Hz power supply is applied across t...
A 400 V, three-phase, 50 Hz power supply is applied across the three terminals of a delta-connected three-phase load. The resistance and reactance of each phase is 6 Ω and 8 Ω, respectively.
Calculate the line current, phase current, active power, reactive power, and apparent power of the circuit.
The load is delta connected. Hence
V_{Ph} = V_{L} = 400 V
Z_{Ph} =R + jX = 6 + j8 = \sqrt{6²+8²} ⌊\tan^{-1} \frac{8}{6 } = 10 ⌊53° Ω
I_{Ph} =\frac{V_{Ph}}{Z_{Ph}} = \frac{400 ⌊0°}{10⌊53°}=40 ⌊-53° A
| Star Connection | Delta Connection |
| 1. Line current is the same as phase current,i.e.,
I_{L}=I_{Ph} |
1. Line current is \sqrt {3} × the phase current, i.e.,
I_{L}=\sqrt {3} I_{Ph} |
| 2. Line voltage is \sqrt {3} the phase voltage, i.e.,
V_{L}=\sqrt {3} V_{Ph} |
2. Line voltage is the same as phase voltage, i.e.,
V_{L} = V_{Ph} |
| 3. Total power =\sqrt {3} V_{L} I_{L} \cos Φ | 3. Total power =\sqrt {3} V_{L} I_{L} \cos Φ |
| 4. Per phase power = V_{Ph} I_{Ph} \cos Φ | 4. Per phase power = V_{Ph} I_{Ph} \cos Φ |
| 5. Three-phase three-wire and three-phase four-wire systems are possible | 5. Three-phase three-wire system is possible |
| 6. Line voltages lead the respective phase voltages by 30° | 6. Line currents lag the respective phase currents by 30° |
Power factor, \cosΦ = \cos 53° = 0.6 lagging
I_{L}=\sqrt {3} I_{Ph}= 1.732 × 40 =69.28 A
Power factor, \cosΦ = \frac{R}{Z}=\frac{6}{10}=0.6 lagging
\sin Φ = 0.8
Active Power =\sqrt {3} V_{L} I_{L} \cosΦ = 1.732 × 400 × 69.28 × 0.6
= 28798 W = 28.798 kW
\simeq 28.8 kW
Reactive Power = \sqrt {3} V_{L} I_{L} \sinΦ=1.732 × 400 × 69.28 × 0.8
= 38397 VAR = 38.397 kVAR
\simeq 38.4 kVAR
Apparent Power = 3 V_{Ph} I_{Ph} =\sqrt {3} V_{L} I_{L}
= 1.732 × 400 × 69.28
= 47997 VA = 47.997 kVA
\simeq 48 kVA
The power triangle is shown in Fig. 2.85.
Apparent Power in
kVA =\sqrt {(Active Power)^{2} +(Rective Power)^{2} }=\sqrt {(28.8)^{2} +(38.4)^{2} }
=\sqrt {829.44 + 1474.56}
=\sqrt{2304}=48
