Question

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

For series: Calculate \( Z_{series} = \frac{V_{ph}}{I_2} \); for parallel: Calculate \( Z_{parallel} = \frac{1}{Y_2} \).

Step-by-step solution

Calculate Power for Load 2

Determine the total power provided by the source: \[ S_{total} = 1800 \text{ kVA} \]. Then, calculate the apparent power for load 2 by subtracting the power for load 1 from the total power: \[ S_2 = S_{total} - S_1 = 1800 \angle \cos^{-1}(0.96) - (192 + j 1464) \text{ kVA} \].

Dig into the Apparent Power Components for Load 2

Calculate the real and reactive components for the total power: \[ P_{total} = 1800 \times 0.96 = 1728 \text{ kW} \] and \[ Q_{total} = \sqrt{1800^2 - 1728^2} \]. Subtract the real and reactive components of load 1 to find the components of load 2: \( P_2 = P_{total} - 192 \text{ kW} \) and \( Q_2 = Q_{total} - 1464 \text{ kVAR} \).

Determine the Line Voltage Per Phase

Calculate the line-to-neutral voltage for load 2. Given line voltage \[ V_L = 6400 \sqrt{3} \], convert to line-to-neutral voltage using: \[ V_{ph} = \frac{V_L}{\sqrt{3}} = 6400 \text{ V} \].

Calculate Impedance Per Phase for Load 2 in Series

Determine the phase current using: \[ I_2 = \frac{S_2}{\sqrt{3} \times V_{ph}} \]. Then calculate the impedance per phase by dividing the line-to-neutral voltage by the phase current: \[ Z_{series} = \frac{V_{ph}}{I_2} \].

Calculate Impedance Per Phase for Load 2 in Parallel

For parallel configuration, calculate the admittance: \[ Y_2 = \frac{S_2}{V_{ph}^2} \]. Determine the equivalent impedance by inverting the admittance: \[ Z_{parallel} = \frac{1}{Y_2} \].

Key concepts

Three-phase power systems

Three-phase power systems are an essential part of electrical engineering, commonly used for power generation, transmission, and distribution. Unlike single-phase systems, which rely on a single electrical wave, three-phase systems use three separate but synchronized waves, each phase offset by 120 degrees. This configuration provides several benefits:

• 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.

Understanding three-phase power is crucial because it forms the backbone of modern power distribution networks, enabling us to handle vast amounts of electricity efficiently even in complex systems.

Complex power calculation

In three-phase power systems, complex power is a crucial concept that combines real power (measured in kilowatts, kW) and reactive power (measured in kilovolt-amperes reactive, kVAR). Together, they form what is known as apparent power (measured in kilovolt-amperes, kVA). The relationships are represented in the power triangle:

• 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.

Calculating complex power involves using the formula: \[ S = P + jQ \]Here, complex power equations allow us to evaluate each component's contribution, helping us manage systems efficiently.

Impedance per phase

Impedance is a fundamental concept when dealing with electrical circuits, particularly in the context of three-phase systems. Impedance per phase measures how each phase of the circuit resists the flow of alternating current (AC) and consists of two components: resistance (R) and reactance (X).

• 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 \[ Z_{parallel} = \frac{1}{Y_2} \]

Understanding impedance per phase helps electrical engineers design circuits that optimize power delivery by analyzing the total effect of resistive and reactive parts in each phase.

Y-connected loads

Y-connected loads, or star-connected loads, are one of the common configurations used in three-phase systems. Each component, or load in this configuration, is connected to a common center point, forming a shape similar to the letter 'Y'.

• 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.

Y-connected configurations are vital because they enable balanced electrical distribution, streamline computation, and reduce system complexities. They are favored in many industrial applications due to these advantages.