Calculate the load current.
Determine the load apparent power
Given:
- Load voltage = 375 V
- Load power = 64 kW
- Load power factor = 0.85 (lagging)
The apparent power can be calculated as:
S = P / cos(φ) = 64 kW / 0.85 = 75.29 kVA
Calculate the load current
I_load = S / V = 75.29 × 10³ VA / 375 V = 200.78 A
The load current is 200.78 A at a power factor of 0.85 lagging.
\(\boxed{I_{load} = 200.78 \text{ A at PF } = 0.85 \text{ lagging}}\)
Calculate the transformer secondary and primary voltage and current.
Calculate the transformer secondary current
The transformer secondary current is equal to the load current:
I_secondary = I_load = 200.78 A
Calculate the transformer secondary voltage
The secondary voltage at the transformer terminals will be higher than the load voltage due to the voltage drop in the feeder.
However, since the hint states to place the feeder on the primary side of the transformer, the secondary voltage of the transformer is equal to the load voltage:
V_secondary = V_load = 375 V
Calculate the transformer primary current
The transformer turns ratio is:
n = V_primary / V_secondary = 3000 V / 380 V = 7.895
The primary current (neglecting magnetizing current for now) is:
I_primary = I_secondary / n = 200.78 A / 7.895 = 25.43 A
Calculate the transformer primary voltage
Since the feeder is on the primary side, the primary voltage at the transformer terminals will be lower than the supply voltage due to voltage drop in the feeder. We'll calculate the exact primary voltage in part c when we determine the supply voltage.
For now, the nominal primary voltage is 3 kV.
\(\boxed{\begin{array}{l}
V_{secondary} = 375 \text{ V} \\
I_{secondary} = 200.78 \text{ A} \\
V_{primary} = 3000 \text{ V (nominal)} \\
I_{primary} \approx 25.43 \text{ A (excluding magnetizing current)}
\end{array}}\)
Calculate the transformer magnetizing current and network supply current and voltage.
Calculate the magnetizing current
The magnetizing current has two components:
- Current through magnetizing resistance (R_m = 6.2 kΩ)
- Current through magnetizing reactance (X_m = 3 kΩ)
At primary voltage of 3 kV:
I_R = V_primary / R_m = 3000 V / 6200 Ω = 0.484 A (in phase with voltage)
I_X = V_primary / X_m = 3000 V / 3000 Ω = 1 A (90° leading the voltage)
The magnetizing current magnitude is:
I_mag = √(I_R² + I_X²) = √(0.484² + 1²) = 1.11 A
The angle is:
θ = tan⁻¹(I_X/I_R) = tan⁻¹(1/0.484) = 64.2°
So the magnetizing current is 1.11 A at 64.2° leading.
Calculate the total primary current
The total primary current is the phasor sum of the load current referred to the primary and the magnetizing current.
The load current referred to the primary is 25.43 A at power factor 0.85 lagging, which means an angle of -31.79° (cos⁻¹(0.85) = 31.79°).
Converting to rectangular form:
- Load current (primary): 25.43∠-31.79° = 21.62 - j13.4 A
- Magnetizing current: 1.11∠64.2° = 0.484 + j1 A
Total primary current:
I_primary_total = (21.62 - j13.4) + (0.484 + j1) = 22.1 - j12.4 A
In polar form:
I_primary_total = √(22.1² + 12.4²)∠tan⁻¹(-12.4/22.1) = 25.35∠-29.3° A
Calculate the voltage drop in the feeder
Feeder impedance = (0.2 + j0.58) Ω/mi × 8 mi = 1.6 + j4.64 Ω
Voltage drop in feeder:
ΔV = I_primary_total × Z_feeder = 25.35∠-29.3° × (1.6 + j4.64)
= 25.35∠-29.3° × 4.91∠71° = 124.5∠41.7° V
Calculate the supply voltage
The supply voltage is the sum of the transformer primary voltage and the feeder voltage drop:
V_supply = V_primary + ΔV = 3000∠0° + 124.5∠41.7° V
Converting the voltage drop to rectangular form:
ΔV = 124.5∠41.7° = 93.1 + j82.9 V
Therefore:
V_supply = 3000 + 93.1 + j82.9 = 3093.1 + j82.9 V
In polar form:
V_supply = √(3093.1² + 82.9²)∠tan⁻¹(82.9/3093.1) = 3094.1∠1.53° V
\(\boxed{\begin{array}{l}
I_{magnetizing} = 1.11 \text{ A at } 64.2° \text{ leading} \\
I_{primary\_total} = 25.35 \text{ A at PF } = 0.87 \text{ lagging} \\
V_{supply} = 3094.1 \text{ V}
\end{array}}\)
Determine the voltage regulation. Is it acceptable?
Calculate the voltage regulation
Voltage regulation is defined as:
VR = (V_no_load - V_full_load) / V_full_load × 100%
For the transformer:
- V_no_load (secondary) = 380 V (rated secondary voltage)
- V_full_load (secondary) = 375 V (load voltage)
Voltage regulation = (380 - 375) / 375 × 100% = 1.33%
Evaluate if the voltage regulation is acceptable
For power distribution transformers, a voltage regulation of 1-3% is typically considered acceptable. The calculated voltage regulation of 1.33% falls within this range, indicating good performance.
The voltage regulation is low enough to ensure that voltage variations between no-load and full-load conditions are minimal, which is desirable for stable operation of connected equipment.
\(\boxed{\text{Voltage regulation} = 1.33\% \text{, which is acceptable}}\)
Load current = 200.78 A at power factor 0.85 lagging
Transformer secondary voltage = 375 V
Transformer secondary current = 200.78 A
Transformer primary voltage = 3000 V (nominal)
Transformer primary current ≈ 25.43 A (excluding magnetizing current)
Magnetizing current = 1.11 A at 64.2° leading
Total primary current = 25.35 A at power factor 0.87 lagging
Supply voltage = 3094.1 V
Voltage regulation = 1.33%, which is acceptable
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