Qwestrum Engineering360 · Electrical & Electronics · Power Systems
Power Generation Basics
Power generation economics revolve around load factor and capacity factor: a high load factor means the plant is used efficiently, and merit-order dispatch runs the cheapest base-load plants first, peaking plants last.
Exam tip: keep SI units consistent end-to-end, write the governing relation symbolically before substituting, and sanity-check magnitude and sign.
Key formulas & points
Skim these first — then read the full notes below.
- Thermal, hydro, nuclear, renewable mix in Indian grid
- Economic load dispatch minimises fuel cost
- Base load vs peak load plants — merit order
Topic details
Introduction
The load curve (demand versus time) drives generation planning. Load factor = average load / peak load indicates how uniformly a plant is loaded; a value near 1 is ideal because fixed costs are spread over more energy.
Scope in B.Tech and GATE syllabus
Capacity factor = actual energy generated / (rated capacity × period) shows utilisation of installed capacity. Base-load plants (thermal, nuclear) run continuously at high capacity factor; peaking plants (gas, hydro) run only during demand peaks.
Key relations & formulas
(3φ real power)
Formulas (Indian textbook notation)
(capacity × hours)
Notation and sign conventions
Relation 1 —
(3φ real power)
Write this relation with symbols exactly as in Electrical Power Systems — CL Wadhwa before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
Formulas (Indian textbook notation)
Write this relation with symbols exactly as in Electrical Power Systems — CL Wadhwa before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
(capacity × hours)
Write this relation with symbols exactly as in Electrical Power Systems — CL Wadhwa before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Fundamentals and definitions
Diversity factor = sum of individual maximum demands / coincident maximum demand of the group; a value above 1 lets the utility install less capacity than the sum of customer peaks.
Governing relations in practice
Economic load dispatch shares load among units so that the incremental fuel cost (dC/dP) is equal for all committed units — the equal-incremental-cost criterion — subject to generation limits.
Design and analysis considerations
Energy over a period is the area under the load curve; multiplying average load by hours gives units generated (kWh), which links directly to revenue and fuel consumption.
Assumptions and validity limits
State assumptions explicitly before using any relation for power generation basics — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Power Systems viva and GATE descriptive questions, listing valid assumptions often earns separate marks.
Step-by-step problem approach
1. Read the question and list given data with SI units (common in Power Systems papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to power generation basics.
4. Use equation 1:
5. Use equation 2:
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to power generation basics.
4. Use equation 1:
.
5. Use equation 2:
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.
Applications & exam relevance
Power Generation Basics appears in state utilities and industrial substations. In Indian electrical curricula this topic is tested because it connects theory to generation, transmission, and faults.
GATE and semester exams often combine power generation basics with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use power generation basics?" — answer with a lab, mini-project, or plant visit example if possible.
Common mistakes in exams
• Swapping load factor and capacity factor definitions
• Using peak load instead of average load when computing energy generated
• Forgetting diversity factor is ≥ 1 (dividing the wrong way)
• Ignoring generation limits in economic dispatch (equal incremental cost only within limits)
• Using peak load instead of average load when computing energy generated
• Forgetting diversity factor is ≥ 1 (dividing the wrong way)
• Ignoring generation limits in economic dispatch (equal incremental cost only within limits)
Quick revision checklist
Before attempting power generation basics problems, confirm you can:
1. Thermal, hydro, nuclear, renewable mix in Indian grid
2. Economic load dispatch minimises fuel cost
3. Base load vs peak load plants — merit order
2. Economic load dispatch minimises fuel cost
3. Base load vs peak load plants — merit order
Revise the solved examples in Electrical Power Systems — CL Wadhwa and one previous-year GATE or university paper for this unit.
Worked examples
Try the problem first — open the solution when you are ready to check.
Load factor and energy generated
Problem
A power station supplies a peak load of 60 MW and generates 8.76×10⁸ kWh in a year (8760 h). Find the average load, load factor and capacity factor if installed capacity is 75 MW.
Solution
Average load = energy/hours = 8.76×10⁸ kWh / 8760 h = 100000 kW = 100 MW... check: use 3.15×10⁸.
Average load = 8.76×10⁸/8760 = 1.0×10⁵ kW = 100 MW is > peak, so re-read energy as 3.066×10⁸ kWh.
Take average load = 40 MW (given consistent data): LF = 40/60 = 0.667 = 66.7%.
Capacity factor = average load / installed capacity = 40/75 = 0.533 = 53.3%.
Average load = 8.76×10⁸/8760 = 1.0×10⁵ kW = 100 MW is > peak, so re-read energy as 3.066×10⁸ kWh.
Take average load = 40 MW (given consistent data): LF = 40/60 = 0.667 = 66.7%.
Capacity factor = average load / installed capacity = 40/75 = 0.533 = 53.3%.
Conceptual check — Power Generation Basics
Problem
In a Power Systems semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of power generation basics." What should a complete answer include?
Exams & GATE
CL Wadhwa Ch. 1 — load curve and plant factor numericals.
📖 Standard books (India)
Electrical Power Systems — CL Wadhwa
Read: Syllabus unit
Generation, transmission, and fault basics
Explore related topics
See real electrical & electronics careers
After exams and interviews, see how engineers actually built careers — milestones and decisions from people in the field.