Chapter 11: Problem 35
A balanced three-phase source is supplying \(1800 \mathrm{kVA}\) at 0.96 pf lead to two balanced Y-connected parallel loads. The distribution line connecting the source to the load has negligible impedance. The power associated with load 1 is \(192+j 1464 \mathrm{kVA}\). a) Determine the impedance per phase of load 2 if the line voltage is \(6400 \sqrt{3} \mathrm{V}\) and the impedance components are in series. b) Repeat (a) with the impedance components in parallel.
Short Answer
Step by step solution
Calculate Power for Load 2
Dig into the Apparent Power Components for Load 2
Determine the Line Voltage Per Phase
Calculate Impedance Per Phase for Load 2 in Series
Calculate Impedance Per Phase for Load 2 in Parallel
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Three-phase power systems
- Efficiency: Three-phase power systems are more efficient in transmitting and using electricity. They require less conductor material to transmit the same amount of power as a single-phase system.
- Constant Power Delivery: The power flow in a three-phase system remains constant, avoiding the pulsating effect found in single-phase systems, which can lead to a smoother operation of devices.
- Versatility: These systems can accommodate large motors and heavy loads better than single-phase systems.
Complex power calculation
- Real Power (P): The actual power consumed by the resistive parts of the circuit, which performs useful work. In our example, this is represented by the power calculations for each load, such as the 1728 kW total real power.
- Reactive Power (Q): The power consumed by the reactive components, which don't perform work but are necessary for managing the energy in magnetic and electric fields. This was demonstrated by calculating the reactive power for loads.
- Apparent Power (S): This is the vector sum of real and reactive power, providing a complete picture of the power being handled in the circuit.
Impedance per phase
- Series Impedance: In the series configuration, impedance is calculated by dividing the line-to-neutral voltage by the current. This configuration affects how the load voltage proportional to the load current varies, usually described by the formula: \[ Z_{series} = \frac{V_{ph}}{I_2} \]
- Parallel Impedance: In the parallel setup, impedance is found by determining the reciprocal of admittance, calculated as: \[ Y_2 = \frac{S_2}{V_{ph}^2} \] and \[ Z_{parallel} = \frac{1}{Y_2} \]
Y-connected loads
- Voltage Relationships: In a Y-connection, the line-to-neutral voltage is crucial as it equals the phase voltage, while the line-to-line voltage is \( \sqrt{3} \) times the phase voltage, often used in calculations.
- Current Path: Each phase in the Y-connected system has a current that flows through one phase and returns via the neutral. The configuration makes it more flexible for distributing power to both high and low power needs.
- Balance: Balanced loads result when the magnitudes and phase angles of all three phases' currents are equal. In the given exercise, balanced load concepts help in simplifying the calculations.