GATE Electrical Engineering Power Systems Syllabus

What is a key reason AC won the early 'war of currents' over DC for transmission?

AC voltage can be easily stepped up and down using transformers, allowing transmission at high voltage (low current, low $I^{2}R$ loss) and distribution at safe low voltage. Early DC had no economical voltage-conversion device, limiting it to short distances.

Why is power transmitted at high voltage in AC systems? Quantify the loss dependence.

For a fixed power $P=VI$, raising $V$ lowers the current $I$. Line loss is $P_{loss}=I^{2}R=\left(\dfrac{P}{V}\right)^{2}R$, so doubling the voltage cuts the line loss to one-quarter. High voltage therefore minimises copper loss and conductor cross-section.

State the fundamental principle of HVDC (DC) transmission for moving bulk power.

In HVDC, AC is rectified to DC at the sending end (converter station), transmitted as DC at high voltage over the line/cable, and inverted back to AC at the receiving end. Power flow and direction are controlled by adjusting the converter firing angles and DC voltage.

List three major technical advantages of HVDC over HVAC transmission.

(1) No reactive power / charging-current limit, so very long overhead lines and especially underground/submarine cables are feasible. (2) Asynchronous interconnection of two AC systems possible (different frequencies). (3) Full, fast control of power flow; no stability angle limit and lower line losses (no skin effect, fewer conductors).

What is the 'break-even distance' in the HVDC vs HVAC economic comparison?

It is the line length at which the total cost of HVDC equals that of HVAC. Below it AC is cheaper; above it DC is cheaper. DC has higher terminal (converter) cost but lower per-km line cost, so the two cost-vs-distance lines cross — typically ~600–800 km for overhead lines and much shorter (~50 km) for cables.

Why is DC preferred over AC for long submarine/underground cable transmission?

Cables have very high shunt capacitance. Under AC, the large charging current $I_c=\omega C V$ consumes most of the conductor's capacity within a few tens of km. DC has no charging current in steady state ($\omega=0$), so DC cables can carry full load over long distances.

Name the two main converter configurations/types used in modern HVDC and the device each is based on.

(1) Line-Commutated Converter (LCC / CSC) based on thyristors, which require an AC voltage for commutation and absorb reactive power. (2) Voltage-Source Converter (VSC) based on IGBTs with PWM, which can independently control real and reactive power and can feed weak/passive networks.

Why does an AC system have no DC counterpart to the skin effect, and how does this help DC lines?

Skin effect arises from time-varying flux ($f\neq0$). In steady DC, $f=0$, so current distributes uniformly over the full cross-section and the line resistance equals the DC resistance. This lowers effective resistance and conductor losses compared with AC for the same conductor.

Compare the insulation requirement of a line for the same peak voltage in AC vs DC.

For the same insulation level (peak voltage $V_m$), DC transmits $V_m$ continuously, whereas AC's RMS value is $V_m/\sqrt{2}$. Thus a line/cable insulated for a given peak can carry a higher continuous (DC) voltage, giving DC better utilisation of insulation.

How many conductors does a bipolar HVDC link use, and what happens if one pole fails?

A bipolar link uses two conductors, one at $+V$ and one at $-V$, with a neutral/ground return. If one pole fails, the system can continue at reduced (about half) power as a monopolar link using ground or the healthy pole's return, improving reliability.

State a major disadvantage of HVDC transmission related to power-system topology.

DC circuit breakers and tapping are difficult and expensive, so building multi-terminal DC grids is hard — HVDC is best suited to point-to-point links. Converter stations are also costly and generate harmonics and absorb reactive power (LCC), requiring filters and compensation.

For a thyristor (LCC) HVDC rectifier, give the average DC output voltage in terms of firing angle $\alpha$.

$$V_d=V_{d0}\cos\alpha$$ where $V_{d0}$ is the maximum (uncontrolled, $\alpha=0$) average DC voltage. For a 6-pulse bridge $V_{d0}=\dfrac{3\sqrt{2}}{\pi}V_{LL}$, with $V_{LL}$ the line-to-line RMS AC input voltage.

In economic dispatch, define the incremental fuel cost (IC) of a generating unit.

The incremental fuel cost is the rate of change of fuel cost with output power, $IC=\dfrac{dC_i}{dP_i}$ (Rs/MWh). For a quadratic cost $C_i=a_iP_i^{2}+b_iP_i+c_i$, the incremental cost is $\dfrac{dC_i}{dP_i}=2a_iP_i+b_i$, a straight line in $P_i$.

State the optimal economic-dispatch condition for N units NEGLECTING transmission losses.

All units operate at equal incremental cost: $$\frac{dC_1}{dP_1}=\frac{dC_2}{dP_2}=\cdots=\frac{dC_N}{dP_N}=\lambda$$ subject to $\sum_i P_i=P_D$ (total demand). $\lambda$ is the system incremental cost in Rs/MWh.

Without transmission losses, how is the system $\lambda$ found for quadratic cost curves with demand $P_D$?

Set each $2a_iP_i+b_i=\lambda\Rightarrow P_i=\dfrac{\lambda-b_i}{2a_i}$. Substitute into $\sum_iP_i=P_D$: $$\lambda=\frac{P_D+\sum_i \dfrac{b_i}{2a_i}}{\sum_i \dfrac{1}{2a_i}}$$ then back-substitute to get each $P_i$ (respecting generator limits).

What role do generator inequality constraints play in lossless economic dispatch?

Each unit must satisfy $P_{i,\min}\le P_i\le P_{i,\max}$. If equal-$\lambda$ allocation drives a unit beyond a limit, that unit is fixed (clamped) at the violated limit and removed from the equal-incremental-cost set; the remaining units are redispatched to meet demand at a common $\lambda$.

State the coordination equation for economic dispatch WHEN transmission losses are considered.

$$\frac{dC_i}{dP_i}+\lambda\frac{\partial P_L}{\partial P_i}=\lambda\quad\Longleftrightarrow\quad \frac{dC_i}{dP_i}=\lambda\left(1-\frac{\partial P_L}{\partial P_i}\right)$$ where $P_L$ is the total transmission loss and $\dfrac{\partial P_L}{\partial P_i}$ is the incremental transmission loss of unit $i$.

Define the penalty factor $L_i$ in loss-included economic dispatch and write the optimality condition using it.

The penalty factor is $$L_i=\frac{1}{1-\dfrac{\partial P_L}{\partial P_i}}$$ The optimal condition becomes $L_i\dfrac{dC_i}{dP_i}=\lambda$ for all units; i.e. the penalised incremental costs are equal.

How does considering transmission losses change the optimal loading of a generator far from the load?

A remote unit has a large incremental loss $\dfrac{\partial P_L}{\partial P_i}>0$, giving a penalty factor $L_i>1$. Its penalised incremental cost $L_i\dfrac{dC_i}{dP_i}$ rises, so it is loaded LESS than under the lossless equal-$\lambda$ rule, even if its fuel cost is low.

Write George's (Kron's) loss formula using B-coefficients for transmission losses.

$$P_L=\sum_{i}\sum_{j}P_i\,B_{ij}\,P_j$$ (simplest form $P_L=\sum_i\sum_j P_iB_{ij}P_j$). The $B_{ij}$ are the loss (B) coefficients; they are symmetric ($B_{ij}=B_{ji}$) and assumed constant for a given operating condition.

For a two-plant system with loss $P_L=B_{11}P_1^{2}$, what is the incremental transmission loss for plant 1?

$$\frac{\partial P_L}{\partial P_1}=2B_{11}P_1$$ and the incremental loss for plant 2 (which contributes no loss term here) is $\dfrac{\partial P_L}{\partial P_2}=0$, giving a penalty factor $L_2=1$ for plant 2.

State the power-balance (equality) constraint for economic dispatch WITH losses.

$$\sum_{i=1}^{N}P_i = P_D + P_L$$ Total generation equals demand plus transmission losses. (Without losses this reduces to $\sum_i P_i=P_D$.)

What is the physical meaning of the Lagrange multiplier $\lambda$ in economic dispatch?

$\lambda$ is the system incremental cost — the additional fuel cost (Rs/h) to supply one extra MW of load. It equals each unit's penalised incremental cost at the optimum and is the marginal price of power for the system.

Derive the optimality condition for loss-included dispatch using the Lagrangian.

Minimise $\mathcal{L}=\sum_iC_i(P_i)-\lambda\!\left(\sum_iP_i-P_D-P_L\right)$. Setting $\dfrac{\partial\mathcal{L}}{\partial P_i}=0$ gives $$\frac{dC_i}{dP_i}-\lambda\left(1-\frac{\partial P_L}{\partial P_i}\right)=0$$ i.e. the coordination equation $\dfrac{dC_i}{dP_i}=\lambda\big(1-\partial P_L/\partial P_i\big)$.

In the lossless case, why are unit incremental costs equal rather than the fuel costs themselves?

Total cost is minimised when shifting 1 MW from any unit to any other produces no net cost change. That equality of marginal (incremental) costs — not total costs — characterises the minimum; differing total costs are irrelevant to the optimal marginal allocation.

For a 6-pulse LCC HVDC bridge, give the no-load maximum DC voltage in terms of the AC line voltage.

$$V_{d0}=\frac{3\sqrt{2}}{\pi}\,V_{LL}\approx1.35\,V_{LL}$$ where $V_{LL}$ is the RMS line-to-line voltage of the converter transformer secondary; the controlled output is $V_d=V_{d0}\cos\alpha$.

Compare corona and radio-interference behaviour of DC vs AC lines at the same voltage.

DC lines generally exhibit lower corona loss and radio interference than AC lines of comparable voltage, because corona loss depends on the peak gradient; for the same insulation, DC's continuous voltage equals AC's peak, but DC avoids the repetitive AC peaks and the foul-weather loss increase is less severe.

What is a back-to-back HVDC link and what is its main purpose?

A back-to-back HVDC link has the rectifier and inverter in the same station with essentially no DC line (zero/short length). Its purpose is to interconnect two AC systems that are asynchronous (different frequency or uncontrolled phase), allowing controlled power exchange between them.

Summarise when to choose AC versus DC for a new transmission project (key decision factors).

Choose AC for shorter distances, where intermediate tapping/networks are needed, and where transformer voltage conversion suffices. Choose DC for very long overhead lines beyond the break-even distance, long submarine/underground cables (charging-current limited), asynchronous ties, and where precise, fast, lossless-angle power-flow control is required.