Tariff and Power Factor Improvement

Improving power factor with capacitors reduces the reactive kVAr the supply must carry; the capacitor rating to move from φ₁ to φ₂ is Q_c = P(tanφ₁ − tanφ₂), cutting kVA demand charges.

Key formulas & points

Skim these first — then read the full notes below.

  • Most utilities bill on kVA demand — low pf increases cost
  • Capacitors at load vs at substation
  • Avoid over-compensation leading pf

Topic details

Introduction

A low, lagging power factor means the same real power draws more current and kVA, raising I²R losses and the utility’s kVA-based demand charge. Shunt capacitors supply the lagging reactive power locally, reducing the kVAr drawn from the supply.

Scope in B.Tech and GATE syllabus

The capacitor bank rating to improve from an initial angle φ₁ to a target φ₂ (at the same real power P) is Q_c = P(tanφ₁ − tanφ₂). Tariffs often penalise pf below 0.9.

Key relations & formulas

Formulas (Indian textbook notation)

  • kVA=kWcosϕ;kVAr=kWtanϕkVA = \frac{kW}{cos} \phi; kVAr = kW tan \phi
CapacitorkVArQc=P(tanϕ1tanϕ2)Capacitor kVAr Q_{c} = P (tan \phi_{1} - tan \phi_{2})
(improvement from φ₁ to φ₂)

Formulas (Indian textbook notation)

  • Twoparttariff:fixedcharge+energycharge;pfpenaltyifcosϕ<0.9Two-part tariff: fixed charge + energy charge; pf penalty if cos \phi < 0.9

Notation and sign conventions

Relation 1 —
kVA=kWcosϕ;kVAr=kWtanϕkVA = \frac{kW}{cos} \phi; kVAr = kW tan \phi

Formulas (Indian textbook notation)

  • kVA=kWcosϕ;kVAr=kWtanϕkVA = \frac{kW}{cos} \phi; kVAr = kW tan \phi
Write this relation with symbols exactly as in Art & Science of Utilisation of Electrical Energy — H. Partab before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
CapacitorkVArQc=PCapacitor kVAr Q_{c} = P
CapacitorkVArQc=P(tanϕ1tanϕ2)Capacitor kVAr Q_{c} = P (tan \phi_{1} - tan \phi_{2})
(improvement from φ₁ to φ₂)
Write this relation with symbols exactly as in Art & Science of Utilisation of Electrical Energy — H. Partab before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
Twoparttariff:fixedcharge+energycharge;pfpenaltyifcosϕ<0.9Two-part tariff: fixed charge + energy charge; pf penalty if cos \phi < 0.9

Formulas (Indian textbook notation)

  • Twoparttariff:fixedcharge+energycharge;pfpenaltyifcosϕ<0.9Two-part tariff: fixed charge + energy charge; pf penalty if cos \phi < 0.9
Write this relation with symbols exactly as in Art & Science of Utilisation of Electrical Energy — H. Partab before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Fundamentals and definitions

After correction the new kVA = P/cosφ₂ is lower, so the maximum-demand charge falls; the payback of the capacitor bank is the annual saving in demand and energy-loss charges.

Governing relations in practice

Place capacitors close to the inductive load to relieve the whole feeder, or at the substation for bulk correction. Automatic switched banks track the varying load to avoid over-compensation.

Design and analysis considerations

Over-correction produces a leading power factor, which can raise voltage and is also penalised; the target is usually just below unity (e.g. 0.95–0.98).

Assumptions and validity limits

State assumptions explicitly before using any relation for tariff and power factor improvement — 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 Electrical Utilization 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 Electrical Utilization 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 tariff and power factor improvement.
4. Use equation 1:
kVA=kWcosϕ;kVAr=kWtanϕkVA = \frac{kW}{cos} \phi; kVAr = kW tan \phi
.
5. Use equation 2:
CapacitorkVArQc=PCapacitor kVAr Q_{c} = P
.
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

Tariff and Power Factor Improvement appears in industry and railways. In Indian electrical curricula this topic is tested because it connects theory to traction, illumination, and heating.
GATE and semester exams often combine tariff and power factor improvement with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use tariff and power factor improvement?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Using cosφ instead of tanφ in the capacitor kVAr formula
• Forgetting real power P stays constant during correction (only reactive changes)
• Over-correcting into a leading power factor
• Confusing kVA (apparent) with kW (real) demand

Quick revision checklist

Before attempting tariff and power factor improvement problems, confirm you can:
1. Most utilities bill on kVA demand — low pf increases cost
2. Capacitors at load vs at substation
3. Avoid over-compensation leading pf
Revise the solved examples in Art & Science of Utilisation of Electrical Energy — H. Partab 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.

Capacitor rating for pf correction

Problem

A load of 100 kW operates at 0.7 pf lagging. Find the capacitor kVAr needed to raise the power factor to 0.95 lagging.

Solution

cosφ₁ = 0.7 → φ₁ = 45.57°, tanφ₁ = 1.020.
cosφ₂ = 0.95 → φ₂ = 18.19°, tanφ₂ = 0.329.
Q_c = P(tanφ₁ − tanφ₂) = 100 × (1.020 − 0.329).
Q_c = 100 × 0.691 = 69.1 kVAr.

Conceptual check — Tariff and Power Factor Improvement

Problem

In a Electrical Utilization semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of tariff and power factor improvement." What should a complete answer include?

Exams & GATE

H Partab — size capacitor bank for target power factor.

📖 Standard books (India)

  • Art & Science of Utilisation of Electrical EnergyH. Partab

    Read: Syllabus unit

    Traction, illumination, and drives