A 200-kW, 250V, 850rpm, separately excited generator, with interpoles and a compensating winding, is operating at its rated load conditions. The armature resistance is 0.0057 ohms.; the resistance of the interpoles and compensating windings are 0.0023 ohms and 0.0011 ohms respectively. Determine (a) armature current; (b) no-load voltage; (c) voltage regulation.
The correct answer and explanation is:
To solve this problem, we break it down into three parts:
Given:
- Power, P=200 kW=200,000 WP = 200 \text{ kW} = 200,000 \text{ W}
- Terminal voltage, Vt=250 VV_t = 250 \text{ V}
- Armature resistance, Ra=0.0057 ΩR_a = 0.0057 \, \Omega
- Interpole resistance, Rip=0.0023 ΩR_{ip} = 0.0023 \, \Omega
- Compensating winding resistance, Rcw=0.0011 ΩR_{cw} = 0.0011 \, \Omega
- Speed = 850 rpm
(a) Armature Current
Ia=PVt=200,000250=800 AI_a = \frac{P}{V_t} = \frac{200,000}{250} = 800 \, \text{A}
(b) No-load Voltage
Total internal resistance in series with the armature: Rtotal=Ra+Rip+Rcw=0.0057+0.0023+0.0011=0.0091 ΩR_{total} = R_a + R_{ip} + R_{cw} = 0.0057 + 0.0023 + 0.0011 = 0.0091 \, \Omega
Voltage drop across internal resistance: Vdrop=Ia×Rtotal=800×0.0091=7.28 VV_{drop} = I_a \times R_{total} = 800 \times 0.0091 = 7.28 \, \text{V}
No-load voltage is the generated EMF (E), which is the sum of terminal voltage and internal voltage drop: E=Vt+Vdrop=250+7.28=257.28 VE = V_t + V_{drop} = 250 + 7.28 = 257.28 \, \text{V}
(c) Voltage Regulation
Voltage regulation is the percentage rise in voltage from full-load to no-load: Voltage Regulation=E−VtVt×100=257.28−250250×100=7.28250×100=2.91%\text{Voltage Regulation} = \frac{E – V_t}{V_t} \times 100 = \frac{257.28 – 250}{250} \times 100 = \frac{7.28}{250} \times 100 = 2.91\%
✅ Final Answers:
- (a) Armature current: 800 A
- (b) No-load voltage: 257.28 V
- (c) Voltage regulation: 2.91%
🔍 Explanation (300 Words):
This problem deals with a separately excited DC generator operating at full load. We are given its rated power, terminal voltage, speed, and resistances of various components in the internal circuit (armature, interpoles, compensating windings).
First, we find the armature current using the basic power formula P=V×IP = V \times I, rearranged to find current: I=PVI = \frac{P}{V}
This gives us 800 A, which is the armature current under rated load conditions.
Second, the no-load voltage (EMF, or EE) is always higher than the terminal voltage due to voltage drops in the internal resistances. These drops occur because of the current flowing through resistive elements like the armature and auxiliary windings. The total series resistance is the sum of all these given resistances. We multiply this total resistance by the armature current to find the internal voltage drop, which we then add to the terminal voltage to get the no-load voltage.
Third, voltage regulation tells us how much the voltage rises when the load is removed (i.e., from full-load to no-load). It’s calculated as a percentage of terminal voltage, indicating how well the generator maintains voltage with varying loads. A low regulation (like 2.91%) means the generator has good load handling characteristics, largely due to the presence of interpoles and compensating windings that reduce flux weakening and improve performance.
Thus, this problem demonstrates the interplay between resistive losses and voltage stability in a practical generator.