Question

A group of small appliances on a \(60 \mathrm{Hz}\) system requires \(25 \mathrm{kVA}\) at 0.96 pf lagging when operated a \(125 \mathrm{V}(\mathrm{rms}) .\) The impedance of the feeder supplying the appliances is \(0.006+j 0.048 \Omega\). The voltage at the load end of the feeder is \(125 \mathrm{V}(\mathrm{rms})\). a) What is the rms magnitude of the voltage at the source end of the feeder? b) What is the average power loss in the feeder? c) What size capacitor (in microfarads) at the load end of the feeder is needed to improve the load power factor to unity? d) After the capacitor is installed, what is the rms magnitude of the voltage at the source end of the feeder if the load voltage is maintained at \(125 \mathrm{V}(\mathrm{rms}) ?\) e) What is the average power loss in the feeder for \((\mathrm{d}) ?\)

Short answer

a) 134.645 V b) 240 W c) 94.375 µF d) 126.152 V e) 221.184 W

Step-by-step solution

Calculate Apparent Power Components

The apparent power is given as 25 kVA with a power factor of 0.96 lagging. Calculate real power (P) and reactive power (Q) using:\[ P = S \times \text{power factor} = 25,000 \times 0.96 = 24,000 \text{ W} \]\[ Q = S \times \sqrt{1 - \text{power factor}^2} = 25,000 \times \sqrt{1 - 0.96^2} = 7,000 \text{ VAR} \]

Determine Load Current

Calculate the load current using the formula for apparent power:\[ S = V \times I^* \] Where \( V = 125 \text{ V (rms)} \) and \( S = 25,000 \text{ VA}\), so:\[ I = \frac{S}{V} = \frac{25,000}{125} = 200 \text{ A}\]

Calculate Voltage Drop Across Feeder

The impedance of the feeder is given as \( Z = 0.006 + j 0.048 \text{ } \Omega \). The current is 200 A. The voltage drop is given by:\[ \Delta V = I \cdot Z = 200 \cdot (0.006 + j0.048) \]Calculate:\[ \Delta V = 200 \cdot 0.006 + j200 \cdot 0.048 = 1.2 + j9.6 \text{ V} \]The magnitude of this voltage drop is:\[ |\Delta V| = \sqrt{1.2^2 + 9.6^2} = 9.645 \text{ V} \]

Calculate Source Voltage

The source voltage is the sum of the load voltage and the voltage drop:\[ V_{source} = V_{load} + |\Delta V| \]Given \( V_{load} = 125 \text{ V} \), we have:\[ V_{source} = 125 + 9.645 = 134.645 \text{ V (rms)} \]

Compute Average Power Loss in Feeder

The power loss in the feeder is calculated as:\[ P_{loss} = I^2 \cdot R = 200^2 \cdot 0.006 \]\[ P_{loss} = 240 \text{ W} \]

Determine Capacitor Size for Unity Power Factor

To achieve a unity power factor, the reactive power supplied by the capacitor, \( Q_c \), must counteract the original reactive power at the load:\[ Q_c = Q = 7,000 \text{ VAR} \] Using the formula \( Q_c = \frac{V^2}{X_c} \) where \( X_c = \frac{1}{\omega C} \), solve for C:\[ C = \frac{Q_c}{\omega V^2} = \frac{7,000}{2 \pi \times 60 \times (125^2)} \]\[ C \approx 94.375 \text{ µF} \]

Find Source Voltage After Capacitor Installation

After power factor correction to unity, reactive components contribute no voltage drop, thus only real power causes voltage drop:Calculate new current since reactive part is canceled:\[ I_{new} = \frac{P}{V} = \frac{24,000}{125} = 192 \text{ A} \]\[ \Delta V_{new} = 192 \times 0.006 = 1.152 \text{ V}\]Thus, the new source voltage is:\[ V_{source,new} = V_{load} + \Delta V_{new} = 125 + 1.152 = 126.152 \text{ V (rms)} \]

Compute Average Power Loss After Correction

With updated current, the new power loss is:\[ P_{loss,new} = I_{new}^2 \cdot R = 192^2 \times 0.006 \]\[ P_{loss,new} = 221.184 \text{ W}\]

Key concepts

Apparent Power

Apparent power is an essential concept in electrical engineering. It represents the total power in an electrical system. Unlike real power which only takes into account the actual energy consumed, apparent power includes both real and reactive power.
Apparent power is measured in Volt-Amps (VA) and is utilized to gauge the capacity of an electrical system. The formula for apparent power (\(S\)) is:

• The product of the root-mean-square (RMS) voltage (\(V\)) and the root-mean-square current (\(I\))
• Given by \(S = V \times I^*\)
• In our example, the apparent power is \(25 \text{kVA}\)

Apparent power offers a broader understanding by encompassing all power components in a circuit, providing a fuller picture of power flow.

Reactive Power

While real power does the useful work, reactive power happens due to energy storage in the form of magnetic and electric fields. It is essential because it helps transfer real power in AC systems while not performing any work.Reactive power is measured in VAR (Volt-Amps Reactive). In scenarios with power factor issues, compensating for reactive power, for example with capacitors, can improve the efficiency of the system.For calculation, use the formula:

• Reactive power (\(Q\)) is calculated with \(Q = S \times \sqrt{1 - \text{power factor}^2}\)
• In this context, \(Q = 7,000 \text{VAR}\)

Understanding reactive power and its correct management is crucial for ensuring electrical systems operate efficiently and are not overburdened.

Voltage Drop Calculation

Voltage drop is a reduction in voltage in an electrical circuit between its source and the load. It occurs mainly due to the impedance of components in the circuit.
The formula used to calculate this loss of voltage is:

• Voltage drop (\(\Delta V\)) is the product of the current (\(I\)) and impedance (\(Z\))
• Given by \(\Delta V = I \cdot Z\)
• For this exercise, calculate \(\Delta V = 200 \cdot (0.006 + j0.048)\)

is \(9.645\text{ V}\) in magnitude.Managing voltage drop is vital for maintaining efficient power delivery and ensuring that electrical devices operate correctly at the desired voltage level.

Power Loss in Electrical Systems

Power loss in electrical systems usually happens due to the resistance of the wires and components. Such losses can lead to inefficiencies and cost increases if not properly managed.To calculate power loss:

• Use the formula \(P_{loss} = I^2 \cdot R\)
• In this situation, with current \(200 \text{A}\) and resistance \(0.006 \Omega\), \(P_{loss} = 240 \text{ W}\)

Continuously managing and monitoring power loss in systems is necessary for optimal thermal management and can help in reducing costs by improving efficiency and extending the life of electrical components.