Three-Phase Power Calculations

Introduction

In the last lesson, we established the fundamentals of three-phase systems. You learned to describe star (wye) and delta connections and, crucially, to distinguish between phase and line quantities, understanding the $\sqrt{3}$ relationship that connects them.

• Recap: Star Connection: $V_L=\sqrt{3}\times V_p$ and $I_L=I_p$
• Recap: Delta Connection: $V_L=V_p$ and $I_L=\sqrt{3}\times I_p$

Today, we will use this foundation to perform one of the most common tasks in power system analysis: calculating power.

Learning Outcome: By the end of this lesson, you will be able to calculate real power ($P$), reactive power ($Q$), and apparent power ($S$) in balanced three-phase loads.

We will cover:

• How to determine total power from the power of a single phase.
• The standard formulas for calculating three-phase power using line voltage and line current.
• Step-by-step examples for both star and delta connected loads.

For this lesson, we will focus on balanced systems, where the load impedance is identical for all three phases. This is a very common and practical assumption for analysing large motors or entire substations.

1. From Single-Phase to Three-Phase Power

Remember the power triangle from our lessons on single-phase circuits? The same concepts of real power ($P$), reactive power ($Q$), and apparent power ($S$) apply to each phase of a three-phase system.

In a balanced system, each of the three phases has an identical load. This means each phase draws the same amount of real power and the same amount of reactive power. Therefore, the simplest way to find the total power is to calculate it for one phase and multiply by three.

• $S_{Total}=3\times S_{Phase}$
• $P_{Total}=3\times P_{Phase}$
• $Q_{Total}=3\times Q_{Phase}$

2. The Universal Three-Phase Power Formulas

While calculating power on a per-phase basis is always correct, it's often more convenient to work directly with line quantities ($V_L$ and $I_L$). These are the values most commonly specified for equipment and measured in the field (e.g., a "400V, three-phase motor" refers to the line voltage).

Through a mathematical derivation, we can create a set of formulas that use line quantities. A key insight is that these formulas are the same for both star and delta connections.

The resulting formulas are fundamental to all three-phase power calculations:

• Apparent Power (S): $S_{Total}=\sqrt{3}\times V_L\times I_L$ (in VA)
• Real Power (P): $P_{Total}=\sqrt{3}\times V_L\times I_L\times \cos (\varphi )$ (in W)
• Reactive Power (Q): $Q_{Total}=\sqrt{3}\times V_L\times I_L\times \sin (\varphi )$ (in VAR)

Here, $\varphi$ is the phase angle of the load impedance, and $\cos (\varphi )$ is the power factor.

For a detailed text summary of these formulas and how they apply to both star and delta connections, you can refer to the following resource.

3. Calculation Walkthrough

Let's apply these formulas. The process usually involves two main stages:

1. Determine the phase and line voltages and currents.
2. Use these values to calculate the total power.

First, let's refresh our understanding of voltage and current calculations with some worked examples.

Now, let's extend those examples to calculate power.

Example 1: Star-Connected Resistive Load

From the first video example, we have a star-connected load with:

• Line Voltage ($V_L$): 415 V
• Load Resistance per phase: 25 $\Omega$
• Calculated Line Current ($I_L$): 9.58 A

Since the load is purely resistive, the power factor is 1 ($\cos (\varphi )=1$) and the phase angle $\varphi$ is 0°.

1. Calculate Total Apparent Power ($S_{Total}$):
   $S_{Total}=\sqrt{3}\times V_L\times I_L=\sqrt{3}\times 415\text{V}\times 9.58\text{A}=6891\text{VA}\approx 6.9\text{kVA}$

2. Calculate Total Real Power ($P_{Total}$):
   $P_{Total}=S_{Total}\times \cos (\varphi )=6891\text{VA}\times 1=6891\text{W}\approx 6.9\text{kW}$

3. Calculate Total Reactive Power ($Q_{Total}$):
   $Q_{Total}=S_{Total}\times \sin (\varphi )=6891\text{VA}\times \sin (0^{\circ })=0\text{VAR}$
   As expected for a resistive load, there is no reactive power.

Example 2: Delta-Connected Resistive Load

From the second video example, the same 25 $\Omega$ resistors are now in a delta connection:

• Line Voltage ($V_L$): 415 V
• Calculated Line Current ($I_L$): 28.75 A
• Power Factor: 1

1. Calculate Total Apparent Power ($S_{Total}$):
   $S_{Total}=\sqrt{3}\times V_L\times I_L=\sqrt{3}\times 415\text{V}\times 28.75\text{A}=20678\text{VA}\approx 20.7\text{kVA}$

2. Calculate Total Real Power ($P_{Total}$):
   $P_{Total}=S_{Total}\times 1=20678\text{W}\approx 20.7\text{kW}$

3. Calculate Total Reactive Power ($Q_{Total}$):
   $Q_{Total}=0\text{VAR}$

Notice that connecting the same loads in delta draws three times more power than in star ($20.7\text{kW}/6.9\text{kW}=3$). This is because in delta, each resistor has the full line voltage across it, whereas in star it only has the phase voltage ($415V/\sqrt{3}\approx 240V$).

Example 3: Star-Connected Inductive Load (Motor)

Let's take a more realistic example of a three-phase motor connected in star to a 400V supply. Each phase winding has an impedance of $Z_p=8+j6\Omega$.

1. Find Impedance Magnitude and Power Factor:

   • Magnitude: $|Z_p|=\sqrt{8^2+6^2}=\sqrt{100}=10\Omega$
   • Phase Angle: $\varphi =\arctan (6/8)=36.87^{\circ }$
   • Power Factor: $\cos (36.87^{\circ })=0.8$ lagging (since the reactive part 'j6' is positive, it's an inductive load).

2. Find Voltages and Currents:

   • The supply is 400V, so $V_L=400\text{V}$.
   • It's a star connection, so phase voltage is $V_p=V_L/\sqrt{3}=400/\sqrt{3}\approx 230.9\text{V}$.
   • Phase current is $I_p=V_p/|Z_p|=230.9\text{V}/10\Omega =23.09\text{A}$.
   • For a star connection, line current equals phase current: $I_L=I_p=23.09\text{A}$.

3. Calculate Total Powers:

   • Apparent Power:
     $S_{Total}=\sqrt{3}\times V_L\times I_L=\sqrt{3}\times 400\text{V}\times 23.09\text{A}=16000\text{VA}=16\text{kVA}$
   • Real Power:
     $P_{Total}=S_{Total}\times \text{PF}=16000\text{VA}\times 0.8=12800\text{W}=12.8\text{kW}$
   • Reactive Power:
     $Q_{Total}=S_{Total}\times \sin (\varphi )=16000\text{VA}\times \sin (36.87^{\circ })=16000\times 0.6=9600\text{VAR}=9.6\text{kVAR}$

This final example represents a complete calculation for a typical three-phase load.

Conclusion

Key Takeaways:

• In a balanced three-phase system, the total power ($P,Q,S$) is three times the power of a single phase.
• A single set of formulas using line quantities can be used to calculate power for both balanced star and delta connected loads.
• The key formulas are:
  • $S_{Total}=\sqrt{3}{V}_L{I}_L$
  • $P_{Total}=\sqrt{3}{V}_L{I}_L\cos (\varphi )$
  • $Q_{Total}=\sqrt{3}{V}_L{I}_L\sin (\varphi )$
• Calculating three-phase power is a two-step process: first find the line voltage and current, then apply the power formulas.

Next Lesson Preview

We have now covered the fundamentals of three-phase systems, from their structure to power calculations. In the next lesson, we will step back and look at the bigger picture, discussing the practical and economic advantages of using three-phase power transmission over single-phase systems for the national grid. The concepts of power and current you've learned today will be key to understanding why.