RTUEE / EC / EEEYr 2024 · Sem 52024

Q5Power System - I

Question

10 marks

Q.5. (a) Draw and explain the structure of electrical power system indicating the voltage level in each transmission level. [4]

(b) A 3-phase, 4-wire system is used for lighting. Compare the amount of copper required with that needed for a 2-wire DC system with the same lamp voltage. The cross-sectional area (A) of neutral is same as outer conductors. [6]

Answer

The electrical power system structure consists of generation (11-25 kV), step-up transmission (up to 765 kV EHV), sub-transmission (33-132 kV) and distribution (11 kV primary, 415V/230V secondary) levels in a hierarchical chain; comparing a 3-phase 4-wire lighting system (same lamp voltage, same total power and loss, neutral csa = phase csa) against an equivalent 2-wire DC system shows the 3-phase 4-wire system requires only one-third (33.3%) of the copper volume of the 2-wire DC system, a 66.7% saving.

(a) Structure of the Electrical Power System

The electrical power system is structured as a hierarchical chain of progressively stepped-down (or up) voltage levels, from generation through to final consumer utilization:

Structure of Electrical Power SystemGeneration 11-25kVStep-up Xfmr →220-765kVEHV TransmissionSub-transmission 33-132kVPrimary Dist. 11kVDistribution XfmrSecondary 415V/230VConsumers (households, industry, commercial)

Generation level (11-25 kV): electrical power is generated at moderate voltage (typically 11-25 kV) at power stations, since generator insulation design becomes progressively more difficult and costly at higher voltages. Step-up transformation and EHV transmission (220 kV up to 765 kV or higher): a step-up transformer at the generating station raises the voltage to the transmission level, and bulk power is carried over long distances at extra-high voltage to minimize transmission losses and maximize power-transfer capacity per corridor. Sub-transmission (33-132 kV): at regional grid/receiving substations, voltage is stepped down from the EHV transmission level to sub-transmission voltage, feeding a network of substations serving cities, towns or large industrial consumers directly. Primary distribution (11 kV): further step-down at distribution substations feeds primary distribution feeders (commonly 11 kV in India) running through urban and rural areas, from which distribution transformers tap off. Secondary distribution (415 V three-phase / 230 V single-phase): the final distribution transformers step down to utilization voltage, directly supplying individual households, shops and small industrial/commercial consumers through the final low-voltage distribution network and service mains.

(b) Comparison of 3-Phase 4-Wire and 2-Wire DC Copper Requirement

Given: a 3-phase, 4-wire system supplies lighting load; the neutral conductor's cross-sectional area is the same as each of the three phase (outer) conductors; compare against a 2-wire DC system delivering the same total power at the same lamp voltage V, same distance, and same percentage power loss.

2-wire DC system: for total power P at voltage V, DC current Idc = P/V; loss = 2Idc²Rdc (2 conductors).

3-phase, 4-wire system: since the lamp (utilization) voltage equals V (same as the DC case, taken as the phase voltage of the 3-phase system), each phase carries P/3 of the total load: Iph = (P/3)/V = P/(3V) = Idc/3. In a perfectly balanced 3-phase load, the neutral carries no net current — but per the problem statement, the neutral conductor is still sized with the same cross-section as the outer (phase) conductors, so the 4-wire system uses 4 total conductors (3 phase + 1 neutral), each of cross-section aph.

Equal-loss condition: the total power loss in the 3-phase system (in the 3 phase conductors; the neutral, carrying no current in the balanced case, contributes no loss) is 3Iph²Rph. Setting this equal to the DC system's loss 2Idc²Rdc:

Since R = ρl/a (same ρ, l for both systems), Rph=6Rdc implies aph = adc/6.

Copper volume comparison: DC system uses 2 conductors of area adc: Voldc = 2×adc×l. The 3-phase 4-wire system uses 4 conductors (3 phase + 1 neutral, all of area aph = adc/6, since the neutral is specified to be the same size as the phase conductors): Vol3ph = 4×aph×l = 4×(adc/6)×l = (2/3)×adc×l.

Result: the 3-phase, 4-wire system requires only one-third (33.3%) of the copper volume needed by the equivalent 2-wire DC system for the same lamp voltage, same total power, same distance and same percentage loss — equivalently, a copper saving of 66.7% is achieved by using the 3-phase 4-wire system instead of an equivalent 2-wire DC distribution scheme. This substantial material saving (in addition to the well-known advantages of AC for easy voltage transformation) is a further practical reason why 3-phase 4-wire distribution is the standard choice for supplying combined lighting and general power loads in modern electrical distribution systems, rather than a 2-wire DC scheme.

Back to Paper