8.3 Three-Phase Generation, Star-Delta & UK Distribution

Three-Phase Generation, Star-Delta & UK Distribution

While single-phase alternating current supplies ordinary domestic appliances and lighting, public utility distribution and industrial power systems operate almost universally on three-phase alternating supplies. Developed in the late 19th century through the pioneering work of Nikola Tesla, Mikhail Dolivo-Dobrovolsky, and Charles Proteus Steinmetz, three-phase systems represent one of the most elegant and efficient electrical engineering designs in human history.

Compared to single-phase systems, three-phase networks deliver constant, non-pulsating electrical power, require approximately $25\%$ less conductor material (copper or aluminium) to transmit the exact same quantity of electrical energy, provide two distinct operating voltages from a single transformer, and naturally establish a rotating magnetic field that enables robust, self-starting industrial electric motors without external starting switches.

---

Generation of a Three-Phase Alternating Supply

A three-phase AC generator (alternator) operates on the exact same principles of electromagnetic induction as a single-phase machine, but its stationary housing (stator) contains three separate, identical windings (coils) designated as Phase 1 (L1), Phase 2 (L2), and Phase 3 (L3).

Stator Geometry and Phase Displacement

• The three stator windings are spaced around the circular periphery of the machine, separated physically by an angle of $120^\circ$ ($\frac{2\pi}{3}$ radians, or one-third of a full $360^\circ$ revolution).
• A central rotor, magnetized by direct current to create alternating North and South magnetic poles, is driven by an external mechanical prime mover (such as a steam, gas, or hydro turbine).
• As the rotor magnetic field sweeps across the stator windings at constant speed, it cuts each coil at identical velocity but displaced in time by $120^\circ$.

Consequently, the alternator generates three identical sinusoidal electromotive forces having equal amplitude and equal frequency ($50\text{ Hz}$ in the UK), but displaced from each other in phase by $120^\circ$:

$e_{L1} = E_{pk} \sin(\omega t)$ $e_{L2} = E_{pk} \sin(\omega t - 120^\circ)$ $e_{L3} = E_{pk} \sin(\omega t - 240^\circ) = E_{pk} \sin(\omega t + 120^\circ)$

UK Phase Sequence and Harmonised Conductor Identification

The order in which the three phase voltages reach their positive maximum peaks is known as the phase sequence (standard sequence is L1 $\rightarrow$ L2 $\rightarrow$ L3). Phase sequence is vital because it determines the rotational direction of three-phase induction motors.

Under BS 7671 (Table 51) and European Harmonisation Standards (introduced in the UK in 2004 via Amendment 2 to BS 7671:2001), phase conductors in three-phase fixed wiring installations are strictly identified by colour:

---

Winding Configurations: Star (Y) versus Delta ($\Delta$)

An alternator or transformer secondary winding contains six terminal ends (two per phase: $U_1$-$U_2$, $V_1$-$V_2$, $W_1$-$W_2$). Rather than transmitting six independent wires, the windings are interconnected into one of two configurations: Star or Delta.

1. Star (Wye, Y) Connection

In a Star connection, one end of each of the three phase windings ($U_2, V_2, W_2$) is joined together at a central common junction known as the star point or neutral point. The remaining three ends ($U_1, V_1, W_1$) connect directly to the three external line conductors (L1, L2, L3).

• Phase Voltage ($V_{ph}$): The voltage measured across an individual phase winding, or between any single line conductor and the neutral point (Line-to-Neutral: L1-N, L2-N, L3-N). In the UK, nominal $V_{ph} = 230\text{ V}$.
• Line Voltage ($V_L$): The voltage measured between any two line conductors (Line-to-Line: L1-L2, L2-L3, L3-L1).
• Voltage Relationship in Star: Because the line voltage represents the vector difference between two phase voltages displaced by $120^\circ$: $V_L = 2 \times V_{ph} \times \cos(30^\circ) = 2 \times V_{ph} \times \frac{\sqrt{3}}{2} = \sqrt{3} \times V_{ph} \approx 1.732 \times V_{ph}$ In the UK low-voltage network, with a nominal single-phase voltage of $230\text{ V}$: $V_L = \sqrt{3} \times 230\text{ V} \approx 1.73205 \times 230\text{ V} \approx 398.37\text{ V} \approx 400\text{ V}$ (Prior to European harmonisation in 1995, UK nominal voltages were $415\text{ V} / 240\text{ V}$. Today, the statutory rating is $400\text{ V} / 230\text{ V} +10\% / -6\%$, allowing seamless compatibility).
• Current Relationship in Star: Because each line conductor connects directly in series with its corresponding phase winding, the current flowing in the line conductor is identical to the current flowing through the phase winding: $I_L = I_{ph}$

The Neutral Conductor in Star Systems: Balanced vs Unbalanced Loads

The star connection allows the star point to be brought out as a fourth conductor: the neutral conductor (creating a 3-phase 4-wire system).

• Balanced Load: When identical loads are connected across each of the three phases (such as a 3-phase electric heater or 3-phase motor), the three line currents are identical in magnitude ($I_{L1} = I_{L2} = I_{L3}$) and spaced symmetrically by $120^\circ$. The instantaneous sum of these three sinusoidal currents is zero at every instant: $\vec{I}_N = \vec{I}_{L1} + \vec{I}_{L2} + \vec{I}_{L3} = 0\text{ A}$ In a perfectly balanced three-phase star system, zero current flows in the neutral conductor.
• Unbalanced Load: In commercial or domestic distribution, single-phase loads (computers, lighting, domestic sockets) are connected between individual lines and neutral. If phase loading is uneven (e.g., $I_{L1} = 40\text{ A}, I_{L2} = 25\text{ A}, I_{L3} = 15\text{ A}$), the currents no longer cancel completely. The neutral conductor carries the vector sum of this imbalance back to the transformer star point.

The Danger of a Disconnected / Open Neutral (Floating Neutral)

In a three-phase 4-wire installation supplying single-phase equipment, the integrity of the neutral conductor is critical to life and property.

---

2. Delta (Mesh, $\Delta$) Connection

In a Delta connection, the three phase windings are connected end-to-end to form a closed triangle or mesh ($U_2$ connects to $V_1$, $V_2$ connects to $W_1$, and $W_2$ connects to $U_1$). External line conductors connect to the three corners of the delta mesh.

• No Neutral Point: A delta system possesses no star point; it is strictly a 3-phase 3-wire system.
• Voltage Relationship in Delta: Because each phase winding is connected directly across two line conductors, the phase voltage is identical to the line voltage: $V_L = V_{ph} = 400\text{ V}$
• Current Relationship in Delta: Each line conductor connects to a junction feeding two separate phase windings. The line current is the vector difference between the two phase currents separated by $120^\circ$: $I_L = \sqrt{3} \times I_{ph} \approx 1.732 \times I_{ph}$ Rearranging to find phase winding current: $I_{ph} = \frac{I_L}{\sqrt{3}} \approx \frac{I_L}{1.732} \approx 0.577 \times I_L$

Delta connections are widely used for heavy industrial loads (such as three-phase motors and large three-phase heating banks) and for high-voltage transmission lines where no neutral is required.

---

Three-Phase Power Calculations

To calculate total power in a balanced three-phase system, we consider the sum of the power dissipated in all three individual phase windings:

$P_{\text{total}} = 3 \times P_{ph} = 3 \times (V_{ph} \times I_{ph} \times \cos\phi)$

Deriving the Universal Line Formula

Electricians in the field typically measure line quantities ($V_L$ and $I_L$) with multimeters and clamp meters. Substituting line quantities into the equation:

• In a Star System: Substitute $V_{ph} = \frac{V_L}{\sqrt{3}}$ and $I_{ph} = I_L$: $P = 3 \times \left(\frac{V_L}{\sqrt{3}}\right) \times I_L \times \cos\phi = \left(\frac{3}{\sqrt{3}}\right) \times V_L \times I_L \times \cos\phi = \sqrt{3} \times V_L \times I_L \times \cos\phi$
• In a Delta System: Substitute $V_{ph} = V_L$ and $I_{ph} = \frac{I_L}{\sqrt{3}}$: $P = 3 \times V_L \times \left(\frac{I_L}{\sqrt{3}}\right) \times \cos\phi = \sqrt{3} \times V_L \times I_L \times \cos\phi$

Similarly, for total three-phase apparent power ($S$) and reactive power ($Q$):

$S = \sqrt{3} \times V_L \times I_L \quad (\text{VA or kVA})$ $Q = \sqrt{3} \times V_L \times I_L \times \sin\phi \quad (\text{VAR or kVAR})$

---

Star-Delta Motor Starting

When a standard three-phase squirrel-cage induction motor is started Direct-On-Line (DOL) in delta, it draws an initial starting inrush current between $6\times$ and $8\times$ its rated full-load current. On large motors, this severe current spike causes severe voltage dips across the local supply network, dimming lights and disturbing electronic equipment.

A Star-Delta Starter resolves this issue mechanically or via contactors:

1. Starting in Star: The motor stator windings are initially connected in Star. The voltage across each winding is reduced from $400\text{ V}$ to $230\text{ V}$ (a factor of $1/\sqrt{3} \approx 0.577$).
2. Current Reduction: Because winding voltage is reduced to $57.7\%$, winding current drops to $57.7\%$. Furthermore, in Star $I_L = I_{ph}$, whereas in Delta $I_L = \sqrt{3} I_{ph}$. As a result, the total starting line current in Star is reduced to exactly one-third ($1/3$ or $33.3\%$) of the DOL Delta starting current: $I_{\text{start(Star)}} = \frac{1}{3} \times I_{\text{start(Delta)}}$
3. Running in Delta: Once the motor accelerates to approximately $75\% - 85\%$ of its rated synchronous speed, the starter switches the winding configuration to Delta, applying full $400\text{ V}$ across each phase winding for normal continuous running torque.

---

UK Low-Voltage Distribution Architecture

In the UK, electrical power is transmitted nationally by National Grid at $400\text{ kV}$ and $275\text{ kV}$, stepped down to $132\text{ kV}$ and $33\text{ kV}$ for regional distribution, and supplied to local town substations at $11\text{ kV}$.

The Secondary Distribution Substation

At the local neighborhood substation or industrial intake room, a distribution transformer steps down the high-voltage supply to low voltage:

• Primary Winding: Connected in $11\text{ kV}$ Delta (3-wire, no neutral needed for transmission).
• Secondary Winding: Connected in $400\text{ V} / 230\text{ V}$ Star (3-phase 4-wire).
• The Substation Earth: The secondary star point is solidly connected to an earth electrode at the substation, establishing the system neutral and reference earth (forming TN-S or TN-C-S / PME earthing systems).
• Consumer Distribution: The 4-wire output (L1, L2, L3, and N) feeds underground distribution mains beneath streets. Domestic properties are allocated a single phase line (Brown) and neutral (Blue) at $230\text{ V}$ RMS, balanced sequentially along the street (House 1 on L1, House 2 on L2, House 3 on L3) to maintain balanced transformer loading. Commercial and industrial facilities receive all three lines plus neutral to operate both $400\text{ V}$ plant and $230\text{ V}$ equipment.

---

Step-by-Step Worked Calculations

Worked Example 1: Star-Connected Commercial Heater

Problem: A commercial water heating unit consists of three identical resistive heating elements, each having a resistance of $23\ \Omega$, connected in Star across a standard UK $400\text{ V}$, $50\text{ Hz}$ three-phase supply. Calculate:

1. Phase voltage ($V_{ph}$).
2. Phase current ($I_{ph}$).
3. Line current ($I_L$).
4. Total power dissipated ($P$).

Solution:

1. Phase voltage: $V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{400\text{ V}}{1.732} = 230.94\text{ V} \approx 230\text{ V}$
2. Phase current: $I_{ph} = \frac{V_{ph}}{R_{ph}} = \frac{230.94\text{ V}}{23\ \Omega} = 10.04\text{ A}$
3. Line current in Star ($I_L = I_{ph}$): $I_L = 10.04\text{ A}$
4. Total power (purely resistive, so $\cos\phi = 1.0$): $P = \sqrt{3} \times V_L \times I_L \times \cos\phi = 1.73205 \times 400 \times 10.04 \times 1.0 = 6,956\text{ W} \approx 6.96\text{ kW}$ (Verification: $P = 3 \times I_{ph}^2 R_{ph} = 3 \times (10.04)^2 \times 23 = 3 \times 100.8 \times 23 = 6,955\text{ W}$)

Worked Example 2: Delta-Connected Industrial Induction Motor

Problem: A three-phase induction motor connected in Delta operates from a $400\text{ V}$, $50\text{ Hz}$ supply. When running under full load, it draws a line current of $28.0\text{ A}$ at a lagging power factor of $0.85$. Calculate:

1. Phase voltage ($V_{ph}$).
2. Phase current ($I_{ph}$) flowing in each stator winding.
3. Total active electrical power input ($P$).
4. Total apparent power ($S$).

Solution:

1. Phase voltage in Delta ($V_L = V_{ph}$): $V_{ph} = 400\text{ V}$
2. Phase current in Delta: $I_{ph} = \frac{I_L}{\sqrt{3}} = \frac{28.0\text{ A}}{1.732} = 16.17\text{ A}$
3. Total active power: $P = \sqrt{3} \times V_L \times I_L \times \cos\phi = 1.73205 \times 400 \times 28.0 \times 0.85 = 16,489\text{ W} \approx 16.49\text{ kW}$
4. Total apparent power: $S = \sqrt{3} \times V_L \times I_L = 1.73205 \times 400 \times 28.0 = 19,399\text{ VA} \approx 19.40\text{ kVA}$