GATE EE

Dispatch & control

Economic dispatch, voltage and frequency control, power factor correction

This topic covers how real power is shared among generators at least cost, how frequency is held by governor action and supplementary control, and how voltage and reactive power are managed with taps, compensators and power factor correction. In GATE questions you will use the equal incremental cost condition, the penalty-factor form of it, the droop relation for load sharing, the area control error, and the tan-difference formula for the capacitor kVAr.

9 previous-year questions · 9 solved · 6 practice questions

Economic load dispatch without losses

Convention: the fuel cost of unit $i$ is $C_{i} (P_{i})$ in Rs/h with $P_{i}$ in MW, and a quadratic form is assumed unless stated. The incremental cost $IC_{i} = d C_{i} / d P_{i}$ is the extra cost per hour for one more MW. The aim is to minimise the total cost $\sum C_{i}$ while the generation equals the demand $P_{D}$ and each unit stays inside $P_{i, m i n} \leq P_{i} \leq P_{i, m a x}$ .

QUADRATIC COST AND INCREMENTAL COST

C_{i} = a_{i} + b_{i} P_{i} + c_{i} P_{i}^{2}, \frac{d C _{i}}{d P _{i}} = b_{i} + 2 c_{i} P_{i}

Cost in Rs/h, $P_{i}$ in MW; incremental cost is in Rs/MWh and is linear in $P_{i}$ .

EQUAL INCREMENTAL COST CRITERION (NO LOSSES)

\frac{d C _{i}}{d P _{i}} = λ (all units inside limits), i \sum P_{i} = P_{D}

A unit held at $P_{i, m a x}$ has $d C_{i} / d P_{i} \leq λ$ ; a unit held at $P_{i, m i n}$ has $d C_{i} / d P_{i} \geq λ$ . $λ$ is the system incremental cost in Rs/MWh.

CLOSED FORM FOR QUADRATIC COSTS

P_{i} = \frac{λ - b _{i}}{2 c _{i}}, λ = \frac{P _{D} + \sum _{i} \frac{b _{i}}{2 c _{i}}}{\sum _{i} \frac{1}{2 c _{i}}}

Valid while every unit is inside its limits. The fixed term $a_{i}$ does not affect the dispatch.

Procedure with generator limits

Solve the unconstrained problem using the closed form for $λ$ .
If a unit violates a limit, fix it at that limit.
Subtract the fixed generation from $P_{D}$ and re-solve for the remaining units.
Check that the remaining units are inside limits, and that each fixed unit satisfies its inequality on IC against the final $λ$ .

WORKED EXAMPLE

Two units have $C_{1} = 100 + 8 P_{1} + 0.01 P_{1}^{2}$ and $C_{2} = 120 + 6 P_{2} + 0.02 P_{2}^{2}$ (Rs/h, MW). Find the economic dispatch for $P_{D} = 300$ MW with no limits and no losses.

Solution

$IC_{1} = 8 + 0.02 P_{1}$ , $IC_{2} = 6 + 0.04 P_{2}$ , so $P_{1} = 50 λ - 400$ and $P_{2} = 25 λ - 150$ .
Sum: $75 λ - 550 = 300$ , so $λ = 11.333$ Rs/MWh.
$P_{1} = 50 (11.333) - 400 = 166.67$ MW and $P_{2} = 25 (11.333) - 150 = 133.33$ MW.
Check: $IC_{1} = 8 + 0.02 (166.67) = 11.333$ and $IC_{2} = 6 + 0.04 (133.33) = 11.333$ .

Answer: $P_{1} = 166.67$ MW, $P_{2} = 133.33$ MW, $λ = 11.33$ Rs/MWh.

EXAM TIP

Do not equalise the cost $C_{i}$ or the generation $P_{i}$ ; equalise the slopes. Inside limits, $d P_{i} / d λ = 1/ (2 c_{i})$ , so the unit with the smaller $c_{i}$ (the flatter IC line) takes the larger share of any change in load.

Economic dispatch with transmission losses

Power delivered to the load is generation minus the transmission loss $P_{L}$ . A unit whose extra output causes more loss is made to count for less. Convention: all $P_{i}$ and $P_{L}$ in MW and the loss coefficients $B_{ij}$ in MW $^{- 1}$ ; the $B$ matrix is symmetric and treated as constant, and the simplest form has no $B_{0}$ or $B_{00}$ terms.

B-COEFFICIENT LOSS FORMULA

P_{L} = i \sum j \sum P_{i} B_{ij} P_{j}, \frac{\partial P _{L}}{\partial P _{i}} = 2 j \sum B_{ij} P_{j}

Symmetric $B_{ij}$ . For two units: $P_{L} = B_{11} P_{1}^{2} + 2 B_{12} P_{1} P_{2} + B_{22} P_{2}^{2}$ . $\partial P_{L} / \partial P_{i}$ is the incremental transmission loss of unit $i$ .

COORDINATION EQUATIONS

\frac{d C _{i}}{d P _{i}} + λ \frac{\partial P _{L}}{\partial P _{i}} = λ, i \sum P_{i} = P_{D} + P_{L}

Holds for units inside limits. Here $λ$ is the incremental cost of power delivered at the load.

PENALTY FACTOR FORM

L_{i} \frac{d C _{i}}{d P _{i}} = λ, L_{i} = \frac{1}{1 - \frac{\partial P _{L}}{\partial P _{i}}} \approx 1 + \frac{\partial P _{L}}{\partial P _{i}}

Every unit inside limits has the same product of penalty factor and incremental cost. The approximation holds when $\partial P_{L} / \partial P_{i}$ is small. A unit with larger $\partial P_{L} / \partial P_{i}$ has a larger $L_{i}$ and so a lower optimal IC.

Iterative solution

Guess $P_{i}$ (for example the lossless dispatch) and compute $P_{L}$ and each $\partial P_{L} / \partial P_{i}$ from the $B$ matrix.
Compute the penalty factors $L_{i}$ .
Solve $L_{i} IC_{i} = λ$ together with $\sum P_{i} = P_{D} + P_{L}$ for new $P_{i}$ .
Repeat until the $P_{i}$ stop changing.

WORKED EXAMPLE

Two units have $IC_{1} = 16 + 0.02 P_{1}$ and $IC_{2} = 20 + 0.04 P_{2}$ (Rs/MWh, MW). The loss is $P_{L} = 0.001 P_{2}^{2}$ MW (unit 1 is at the load). For $λ = 26$ Rs/MWh, find $P_{1}$ , $P_{2}$ , $P_{L}$ and the load $P_{D}$ .

Solution

Unit 1: $\partial P_{L} / \partial P_{1} = 0$ , so $L_{1} = 1$ and $16 + 0.02 P_{1} = 26$ , giving $P_{1} = 500$ MW.
Unit 2: $\partial P_{L} / \partial P_{2} = 0.002 P_{2}$ . From the coordination equation, $20 + 0.04 P_{2} = 26 (1 - 0.002 P_{2})$ , so $0.092 P_{2} = 6$ and $P_{2} = 65.22$ MW.
$P_{L} = 0.001 (65.22)^{2} = 4.25$ MW, so $P_{D} = 500 + 65.22 - 4.25 = 560.96$ MW.
Check: $L_{2} = 1/ (1 - 0.1304) = 1.15$ and $L_{2} IC_{2} = 1.15 \times 22.61 = 26.0$ .

Answer: $P_{1} = 500$ MW, $P_{2} = 65.22$ MW, $P_{L} = 4.25$ MW, $P_{D} \approx 561$ MW.

EXAM TIP

With losses the units do not run at equal incremental cost; they run at equal $L_{i} IC_{i}$ . If $B_{ij} = 0$ for all $i, j$ , every $L_{i} = 1$ and the problem reduces to the lossless case.

Frequency falls when generation is less than load. The turbine governor senses speed and changes the valve opening. A governor with speed regulation (droop) $R$ lowers output as speed rises. Convention: $R$ is a per-unit number on the unit's own rating, $R = \frac{Δ f / f _{0}}{Δ P / P _{r a t e d}}$ with a negative sign understood, so 5% droop means $R = 0.05$ and a 5% frequency change takes the unit from no load to full load. $Δ P_{D}$ denotes a load increase in MW.

GOVERNOR DROOP

Δ P_{i} = - \frac{P _{r a t e d, i}}{R _{i}} \frac{Δ f}{f _{0}}

$R_{i}$ in per unit on the unit's own rating; $Δ P_{i}$ is the change in generator output in steady state with the speed changer setting fixed.

LOAD SHARING OF PARALLEL UNITS

Δ P_{i} \propto \frac{P _{r a t e d, i}}{R _{i}}

All units settle at one common frequency. A smaller $R$ or a larger rating takes a larger share of a load change, as long as no unit reaches its limit.

STEADY-STATE FREQUENCY CHANGE FOR A LOAD CHANGE

Δ f = - \frac{Δ P _{D}}{i \sum \frac{P _{r a t e d, i}}{R _{i} f _{0}} + D}

Free governor action, no supplementary control. $Δ P_{D}$ is the load increase in MW and $D$ is load damping in MW/Hz (set $D = 0$ if not given).

WORKED EXAMPLE

Two units share a 50 Hz system: 100 MW with $R = 0.04$ and 200 MW with $R = 0.05$ , both on their own ratings. A load increase of 30 MW occurs. Neglect load damping; both units have spare capacity. Find the new frequency and each unit's share.

Solution

$\sum P_{r a t e d} / R = 100/0.04 + 200/0.05 = 2500 + 4000 = 6500$ MW per unit of $Δ f / f_{0}$ .
$Δ f / f_{0} = - 30/6500 = - 0.004615$ , so $Δ f = - 0.2308$ Hz.
Unit 1: $Δ P_{1} = 2500 \times 0.004615 = 11.54$ MW. Unit 2: $Δ P_{2} = 4000 \times 0.004615 = 18.46$ MW.
Check: $11.54 + 18.46 = 30$ MW.

Answer: $f = 49.77$ Hz; unit 1 takes 11.54 MW and unit 2 takes 18.46 MW.

EXAM TIP

Droop percentages are on each unit's own MW rating. Convert to a common base, or use $P_{r a t e d} / R$ as above, before combining units of different size.

Load-frequency control basics and area control error

Governor droop alone leaves a steady frequency error. Load-frequency control (LFC) adds a supplementary integral loop that moves the speed changer setting of selected units until the error is removed. In an interconnected system each area also has to hold its scheduled exchange on the tie line. Convention in this section: $R_{i}$ in Hz/MW, so $R_{i} (Hz/MW) = R_{i, pu} f_{0} / P_{r a t e d, i}$ ; $D$ in MW/Hz; tie-line flow positive out of the area.

AREA FREQUENCY RESPONSE CHARACTERISTIC

β = D + i \sum \frac{1}{R _{i}} (MW/Hz), Δ f = - \frac{Δ P _{D}}{β}

For one isolated area with governor action only (no integral control). $1/ R_{i}$ is the governor response of unit $i$ and $D$ the load damping.

TWO AREAS, GOVERNOR ACTION ONLY

Δ f = - \frac{Δ P _{D 1} + Δ P _{D 2}}{β _{1} + β _{2}}, Δ P_{12} = \frac{β _{1} Δ P _{D 2} - β _{2} Δ P _{D 1}}{β _{1} + β _{2}}

Steady state with no supplementary control; $Δ P_{12}$ is the change in flow from area 1 to area 2. For a load step in area 1 only, $Δ P_{12} = - β_{2} Δ P_{D 1} / (β_{1} + β_{2})$ : area 2 sends part of the step over the tie line.

TIE-LINE POWER AND SYNCHRONISING COEFFICIENT

P_{12} = \frac{∣ V _{1} ∣∣ V _{2} ∣}{X _{12}} sin δ_{12}, Δ P_{12} = T Δ δ_{12}, T = \frac{∣ V _{1} ∣∣ V _{2} ∣}{X _{12}} cos δ_{12}

Lossless tie line of reactance $X_{12}$ , small deviations about the operating angle $δ_{12}$ ; per phase or per unit consistently.

AREA CONTROL ERROR

ACE = Δ P_{t i e} + B_{f} Δ f

$Δ P_{t i e}$ is the tie-line flow deviation (positive out of the area) and $B_{f} > 0$ is the frequency bias in MW/Hz, commonly set equal to the area's $β$ . The integral controller changes the set point as $Δ P_{r e f} = - K \int ACE d t$ : negative ACE raises generation.

What the supplementary integral loop achieves

Condition: each area has integral control acting on its own ACE.
In steady state the integral controllers make $ACE_{1} = ACE_{2} = 0$ . Since $Δ P_{t i e, 1} = - Δ P_{t i e, 2}$ , this forces $Δ f = 0$ and $Δ P_{t i e} = 0$ .
So after a load step in one area the frequency returns to its scheduled value and the tie-line flow returns to its schedule: the area that had the load change supplies it from its own units.
Primary control (droop) acts first; supplementary control (LFC) is slower and removes the remaining error.

EXAM TIP

Watch the signs. After a load rise in area 1, area 1 has $Δ f < 0$ and $Δ P_{t i e} < 0$ (it draws extra power in), so both ACE terms are negative and its units raise generation. Area 2 has $Δ f < 0$ but $Δ P_{t i e} > 0$ ; with $B_{f} = β_{2}$ the two terms cancel at the governor-only steady state, so area 2 does not change its setting.

Voltage control and reactive power

In a mainly inductive network, bus voltage is tied to reactive power: drawing lagging reactive power lowers the bus voltage and supplying it raises the voltage. Voltage is controlled locally by tap changers, shunt and series compensation and generator excitation, while real power and frequency are system-wide quantities.

APPROXIMATE VOLTAGE DROP ON A LINE

Δ V \approx \frac{R P + X Q}{V}

Per phase or per unit consistently; $P$ and $Q$ are the real and lagging reactive power received, $V$ the receiving-end voltage; valid for a small angle across the line. In a mainly inductive line ( $X ≫ R$ ) the drop depends mostly on $Q$ .

VOLTAGE RISE FROM REACTIVE INJECTION

Δ V \approx \frac{X Δ Q _{inj}}{V}, \frac{Δ V}{V} \approx \frac{Δ Q _{inj}}{S _{sc}}

Reactive power $Δ Q_{inj}$ injected (for example by a shunt capacitor) at a bus whose Thevenin source reactance is $X$ ; $S_{sc}$ is the three-phase short-circuit level at that bus (per-unit form, $X$ dominant).

OFF-NOMINAL TAP SETTING FOR A LINE

t_{s}^{2} = \frac{V _{2} / V _{1}}{1 - \frac{R P + X Q}{V _{1} V _{2}}}, t_{s} t_{r} = 1

Tap-changing transformers at both ends of a line of impedance $R + j X$ . $V_{1}$ and $V_{2}$ are the nominal voltages at the two ends referred to the line side, $P$ and $Q$ are the power received at end 2, and the angle across the line is small. With $V_{1} = V_{2} = V$ : $t_{s}^{2} = 1/ (1 - (R P + X Q) / V^{2})$ .

Methods of voltage control

Method	How it acts	Remark
Tap-changing transformer	Changes the turns ratio, shifting the secondary voltage by steps	Generates no reactive power; it only redistributes the reactive flow
Shunt capacitor	Supplies reactive power (draws leading current)	Raises voltage; its output varies as $V^{2}$ , so it falls when voltage is low
Shunt reactor	Absorbs reactive power	Lowers voltage at light load or on long lines
Series capacitor	Cancels part of the line's series reactance	Voltage rise grows with load current, so it adjusts itself to the load
Synchronous condenser	Over-excited it supplies $Q$ , under-excited it absorbs $Q$	Continuously variable reactive output
Generator excitation (AVR)	Changes the generator's reactive output	Over-excitation supplies $Q$ , under-excitation absorbs $Q$
SVC / STATCOM	Power-electronic control of reactive power	Fast, continuously variable supply or absorption of $Q$

EXAM TIP

A tap changer supplies no reactive power. Raising the load-side voltage with a tap makes the load draw more reactive power through the transformer, which lowers the supply-side voltage. When the system is short of $Q$ , this can make the upstream voltage worse.

Power factor correction with shunt capacitors

A shunt capacitor at the load supplies part of the lagging reactive power the load needs, so the source supplies less. The load's own real power $P$ and reactive demand do not change, and the load voltage is taken as constant (the small voltage rise from the capacitor is neglected). Convention: lagging power factor, $Q$ lagging positive, $ϕ_{1}$ before and $ϕ_{2}$ after correction.

CAPACITOR RATING

Q_{c} = P (tan ϕ_{1} - tan ϕ_{2}), ϕ = cos^{- 1} (pf)

Total three-phase kVAr for total real power $P$ in kW. The rating is the reactive power the bank delivers at its rated voltage.

CAPACITANCE FROM THE KVAR RATING

C_{Y} = \frac{Q _{c}}{ω V _{L}^{2}} (star, per phase), C_{Δ} = \frac{Q _{c}}{3 ω V _{L}^{2}} (delta, per phase)

$Q_{c}$ is the three-phase VAr, $V_{L}$ the line voltage in V, $ω = 2 π f$ , and $C$ comes out in F. Note $C_{Δ} = C_{Y} /3$ .

EFFECT ON LINE CURRENT AND LOSS

\frac{I _{2}}{I _{1}} = \frac{cos ϕ _{1}}{cos ϕ _{2}}, \frac{loss _{2}}{loss _{1}} = (\frac{cos ϕ _{1}}{cos ϕ _{2}})^{2}

At constant $P$ and constant load voltage, the source-side current falls when $cos ϕ_{2} > cos ϕ_{1}$ , and feeder $I^{2} R$ loss falls with its square.

MOST ECONOMICAL POWER FACTOR

sin ϕ_{2} = \frac{y}{x}

Constant $P$ ; demand charge $x$ per kVA of maximum demand per year and capacitor annual cost $y$ per kVAr per year (both linear). It minimises $x P / cos ϕ_{2} + y P (tan ϕ_{1} - tan ϕ_{2})$ .

WORKED EXAMPLE

A three-phase load takes 400 kW at 0.8 lagging. Find the capacitor kVAr needed to raise the power factor to 0.95 lagging, and the new apparent power.

Solution

$tan ϕ_{1} = 0.75$ , so $Q_{1} = 400 \times 0.75 = 300$ kVAr.
$sin ϕ_{2} = 1 - 0.9 5^{2} = 0.3122$ , so $tan ϕ_{2} = 0.3287$ and $Q_{2} = 131.5$ kVAr.
$Q_{c} = 300 - 131.5 = 168.5$ kVAr.
New apparent power $S_{2} = 400/0.95 = 421.1$ kVA (down from $500$ kVA).

Answer: $Q_{c} \approx 168.5$ kVAr; $S_{2} \approx 421$ kVA.

EXAM TIP

Use the tan difference, not a difference of angles or of power factors. Correcting past unity pf makes the load leading, and then the line current rises again.

PREVIOUS-YEAR QUESTIONS

9 questions from the EE papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE EE 2026 · GATE EE 2025 · GATE EE 2024 · GATE EE 2023 · GATE EE 2022 · GATE EE 2020

GATE 2026 EE Q.58

GATE 2026 · EEQ.58 · 2 marks · NAT

A system with two generators G1 and G2 (without generator limits) is shown. The total load on the system is 1184 MW. The expressions for the cost of generation (C_1 and C_2) and real power loss (P_Loss) are as follows: C_1(P_G1) = 1000 + 50 P_G1 + 0.01 (P_G1)^2 Rs/MWh; C_2(P_G2) = 2000 + 50 P_G2 + 0.001 (P_G2)^2 Rs/MWh; P_Loss = 0.001 (P_G2 - 50)^2 MW. When the generators are operating at their optimal generation, meeting the total load requirement, the real power loss in the system is ________ MW (Round off to one decimal place). Consider the Lagrange multiplier lambda = 70.25 for optimal generation.

GATE 2026 EE question 58: Single-line diagram: generator G1 on the left feeds a bus with power P_G1 flowing in (arrow to the right) and a load of… — Figure from the GATE 2026 EE paper. It is part of the question.

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GATE 2026 EE Q.58: answer and solution

GATE 2025 EE Q.63

GATE 2025 · EEQ.63 · 2 marks · NAT

Using shunt capacitors, the power factor of a 3-phase, 4 kV induction motor (drawing 390 kVA at 0.77 pf lag) is to be corrected to 0.85 pf lag. The line current of the capacitor bank, in A, is ________ (round off to one decimal place).

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GATE 2025 EE Q.63: answer and solution

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GATE 2025 EE Q.64

GATE 2025 · EEQ.64 · 2 marks · NAT

Two units, rated at 100 MW and 150 MW, are enabled for economic load dispatch. When the overall incremental cost is 10,000 Rs./MWh, the units are dispatched to 50 MW and 80 MW respectively. At an overall incremental cost of 10,600 Rs./MWh, the power output of the units are 80 MW and 92 MW, respectively. The total plant MW-output (without overloading any unit) at an overall incremental cost of 11,800 Rs./MWh is ________ (round off to the nearest integer).

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GATE 2025 EE Q.64: answer and solution

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GATE 2024 EE Q.34

GATE 2024 · EEQ.34 · 1 mark · NAT

The incremental cost curves of two generators (Gen A and Gen B) in a plant supplying a common load are shown in the figure. If the incremental cost of supplying the common load is Rs. 7400 per MWh, then the common load in MW is ______ (rounded off to the nearest integer).

GATE 2024 EE question 34: Graph of incremental cost (Rs./MWh, vertical axis, 5000 to 13000 in steps of 1000) against generation (MW, horizontal… — Figure from the GATE 2024 EE paper. It is part of the question.

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GATE 2024 EE Q.34: answer and solution

GATE 2024 EE Q.47

GATE 2024 · EEQ.47 · 2 marks · MSQ

Which of the following options is/are correct for the Automatic Generation Control (AGC) and Automatic Voltage Regulator (AVR) installed with synchronous generators?

(A)
AGC response has a local effect on frequency while AVR response has a global effect on voltage.
(B)
AGC response has a global effect on frequency while AVR response has a local effect on voltage.
(C)
AGC regulates the field current of the synchronous generator while AVR regulates the generator's mechanical power input.
(D)
AGC regulates the generator's mechanical power input while AVR regulates the field current of the synchronous generator.

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GATE 2023 EE Q.21

GATE 2023 · EEQ.21 · 1 mark · MCQ

The expressions of fuel cost of two thermal generating units as a function of the respective power generation P_G1 and P_G2 are given as
F_1(P_G1) = 0.1 a P_G1^2 + 40 P_G1 + 120 Rs/hour, 0 MW ≤ P_G1 ≤ 350 MW
F_2(P_G2) = 0.2 P_G2^2 + 30 P_G2 + 100 Rs/hour, 0 MW ≤ P_G2 ≤ 300 MW
where a is a constant. For a given value of a, optimal dispatch requires the total load of 290 MW to be shared as P_G1 = 175 MW and P_G2 = 115 MW. With the load remaining unchanged, the value of a is increased by 10% and optimal dispatch is carried out. The changes in P_G1 and the total cost of generation, F (= F_1 + F_2) in Rs/hour will be as follows

(A)
P_G1 will decrease and F will increase
(B)
Both P_G1 and F will increase
(C)
P_G1 will increase and F will decrease
(D)
Both P_G1 and F will decrease

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GATE 2022 EE Q.43

GATE 2022 · EEQ.43 · 2 marks · MCQ

The fuel cost functions in rupees/hour for two 600 MW thermal power plants are given by Plant 1: $C_{1} = 350 + 6 P_{1} + 0.004 P_{1}^{2}$ ; Plant 2: $C_{2} = 450 + a P_{2} + 0.003 P_{2}^{2}$ , where $P_{1}$ and $P_{2}$ are power generated by plant 1 and plant 2, respectively, in MW and $a$ is constant. The incremental cost of power ( $λ$ ) is 8 rupees per MWh. The two thermal power plants together meet a total power demand of 550 MW. The optimal generation of plant 1 and plant 2 in MW, respectively, are

(A)
200, 350
(B)
250, 300
(C)
325, 225
(D)
350, 200

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GATE 2022 EE Q.56

GATE 2022 · EEQ.56 · 2 marks · NAT

Two generating units rated for 250 MW and 400 MW have governor speed regulations of 6% and 6.4%, respectively, from no load to full load. Both the generating units are operating in parallel to share a load of 500 MW. Assuming free governor action, the load shared in MW, by the 250 MW generating unit is __________. (round off to nearest integer)

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GATE 2022 EE Q.56: answer and solution

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GATE 2020 EE Q.23

GATE 2020 · EEQ.23 · 1 mark · NAT

A single 50 Hz synchronous generator on droop control was delivering 100 MW power to a system. Due to increase in load, generator power had to be increased by 10 MW, as a result of which, system frequency dropped to 49.75 Hz. Further increase in load in the system resulted in a frequency of 49.25 Hz. At this condition, the power in MW supplied by the generator is ______ (rounded off to 2 decimal places).

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GATE 2020 EE Q.23: answer and solution

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PRACTICE

6 questions to test yourself

PRACTICE 12 marks · NAT

Two units have incremental costs $IC_{1} = 10 + 0.04 P_{1}$ and $IC_{2} = 8 + 0.06 P_{2}$ (Rs/MWh, $P$ in MW). Neglect losses and limits. For a total demand of 250 MW, the economic output of unit 1 in MW is ____.

Answer and solution

Answer: $130$ (accepted range $\pm 0.5$ )

Equal IC: $10 + 0.04 P_{1} = 8 + 0.06 P_{2}$ with $P_{2} = 250 - P_{1}$ .
$10 + 0.04 P_{1} = 23 - 0.06 P_{1}$ , so $0.10 P_{1} = 13$ .
$P_{1} = 130$ MW; then $P_{2} = 120$ MW and both ICs equal 15.2 Rs/MWh.

PRACTICE 22 marks · NAT

A 415 V, 50 Hz, three-phase load takes 100 kW at 0.8 power factor lagging. A delta-connected capacitor bank raises the power factor to 0.95 lagging. The capacitance required per phase, in $μ$ F, is ____ (round to one decimal place).

Answer and solution

Answer: $259.6$ (accepted range $\pm 1$ )

$Q_{c} = 100 (tan ϕ_{1} - tan ϕ_{2}) = 100 (0.75 - 0.3287) = 42.13$ kVAr.
Delta: $C_{Δ} = Q_{c} / (3 ω V_{L}^{2}) = 42132/ (3 \times 314.16 \times 41 5^{2})$ .
$C_{Δ} = 42132/ (1.6232 \times 1 0^{8}) = 2.596 \times 1 0^{- 4}$ F $\approx 259.6 μ$ F.

PRACTICE 31 mark · MCQ

Two identical 100 MW generators run in parallel at the same no-load frequency. Their governor speed regulations are 4% and 6% on their own ratings. With a common load of 100 MW in total and free governor action, the shares of unit 1 and unit 2 are respectively

(A)
50 MW and 50 MW
(B)
60 MW and 40 MW
(C)
40 MW and 60 MW
(D)
66.7 MW and 33.3 MW

Answer and solution

Answer: (B)

Share is proportional to $P_{r a t e d} / R$ , which is $1/0.04 = 25$ and $1/0.06 = 16.67$ for equal ratings.
Unit 1 fraction: $25/ (25 + 16.67) = 0.6$ .
So 60 MW and 40 MW.

PRACTICE 41 mark · NAT

A load draws constant real power at constant voltage through a feeder of fixed resistance. A shunt capacitor at the load improves its power factor from 0.8 lagging to unity. The feeder $I^{2} R$ loss after correction, as a fraction of the loss before correction, is ____ (round to two decimal places).

Answer and solution

Answer: $0.64$ (accepted range $\pm 0.005$ )

At constant $P$ and $V$ , $I \propto 1/ cos ϕ$ .
$I_{2} / I_{1} = 0.8/1 = 0.8$ .
Loss ratio $= 0. 8^{2} = 0.64$ .

PRACTICE 51 mark · NAT

The transmission loss of a system is $P_{L} = 0.0002 P_{1}^{2}$ MW, with $P_{1}$ in MW. When unit 1 generates 500 MW, its penalty factor is ____ (round to two decimal places).

Answer and solution

Answer: $1.25$ (accepted range $\pm 0.005$ )

$\partial P_{L} / \partial P_{1} = 0.0004 P_{1} = 0.0004 \times 500 = 0.2$ .
$L_{1} = 1/ (1 - 0.2) = 1.25$ .

PRACTICE 62 marks · MSQ

Which of the following statements are correct? (Losses described by $P_{L} = \sum\sum P_{i} B_{ij} P_{j}$ .)

(A)
The penalty factor of unit $i$ is $1/ (1 - \partial P_{L} / \partial P_{i})$ .
(B)
For units inside their limits the optimum has $L_{i} IC_{i}$ equal for all units.
(C)
In the lossless problem, a unit held at its upper limit must have incremental cost equal to $λ$ .
(D)
For two parallel units of equal rating under free governor action, the unit with the smaller speed regulation $R$ takes the larger share of a load change.

Answer and solution

Answer: (A), (B), (D)

A is the definition of the penalty factor.
B is the coordination equation written with penalty factors.
C is false: a unit at its upper limit has $IC \leq λ$ ; it is not required to equal $λ$ .
D is true because the share is proportional to $P_{r a t e d} / R$ .

OFFICIAL GATE 2027 SYLLABUS WORDING

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GATE EE

Economic Load Dispatch (with and without considering transmission losses) … Voltage and Frequency control, Power factor correction

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Economic load dispatch without losses

Economic dispatch with transmission losses

Frequency control: governor droop and load sharing

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Voltage control and reactive power

Power factor correction with shunt capacitors

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