3-phase alternator is used, consider 500KvA source having resistances of 0.3ohm and 0.01 ohm respectively and the corresponding leakage resistances are 0,7ohm and 0.05ohm. The primary and secondary voltages are 100V and 500V and the core loss 3.5Kw. Assume power factor is 0.5. Calculate the efficiency on full load and half load.

To calculate the efficiency of a 3-phase alternator at both full load and half load, we need to consider several factors, including the input power, output power, and losses in the system. Let's break down the calculations step by step.

Understanding the Parameters

We have the following parameters for the alternator:

• Rated Power (S): 500 kVA
• Primary Voltage (V1): 100 V
• Secondary Voltage (V2): 500 V
• Power Factor (PF): 0.5
• Core Loss: 3.5 kW
• Resistance (R1): 0.3 ohm
• Leakage Resistance (R2): 0.01 ohm
• Leakage Resistance (R3): 0.7 ohm
• Leakage Resistance (R4): 0.05 ohm

Calculating Output Power

The output power (P_out) can be calculated using the formula:

P_out = S × PF

For full load:

P_out_full = 500 kVA × 0.5 = 250 kW

For half load:

P_out_half = (500 kVA / 2) × 0.5 = 125 kW

Calculating Total Losses

Total losses in the alternator consist of core losses and copper losses. The core loss is given as 3.5 kW. We need to calculate the copper losses at full load and half load.

Copper Loss Calculation

Copper losses can be calculated using the formula:

P_copper = I² × R

Where I is the current and R is the total resistance. First, we need to find the current:

For full load:

I_full = S / (√3 × V2) = 500,000 VA / (√3 × 500 V) ≈ 577.35 A

Now, let's calculate the total resistance:

Total Resistance (R_total) = R1 + R2 + R3 + R4 = 0.3 + 0.01 + 0.7 + 0.05 = 1.06 ohm

Now, we can calculate the copper losses at full load:

P_copper_full = I_full² × R_total = (577.35)² × 1.06 ≈ 348.54 kW

For half load:

I_half = (500 kVA / 2) / (√3 × 500 V) ≈ 288.68 A

P_copper_half = I_half² × R_total = (288.68)² × 1.06 ≈ 86.78 kW

Calculating Total Power Losses

Now, we can find the total losses for both full load and half load:

Total Losses_full = Core Loss + Copper Loss_full = 3.5 kW + 348.54 kW ≈ 352.04 kW

Total Losses_half = Core Loss + Copper Loss_half = 3.5 kW + 86.78 kW ≈ 90.28 kW

Efficiency Calculation

Efficiency (η) can be calculated using the formula:

η = (Output Power) / (Output Power + Total Losses)

For full load:

η_full = P_out_full / (P_out_full + Total Losses_full) = 250 kW / (250 kW + 352.04 kW) ≈ 41.52%

For half load:

η_half = P_out_half / (P_out_half + Total Losses_half) = 125 kW / (125 kW + 90.28 kW) ≈ 58.05%

Final Results

In summary, the efficiency of the 3-phase alternator is:

• Full Load Efficiency: Approximately 41.52%
• Half Load Efficiency: Approximately 58.05%

This analysis shows how the efficiency of an alternator can vary significantly between full load and half load conditions, primarily due to the impact of copper losses. Understanding these concepts is crucial for optimizing the performance of electrical machines.