Shunt Capacitor Bank Savings Calculator

How Shunt Capacitor Banks Save Money

Inductive loads like motors draw lagging reactive power, resulting in a poor power factor. This causes:

• Higher current flow through transformers and cables
• Increased I²R heating losses in the electrical system
• Higher electricity bills due to these losses

Shunt capacitor banks provide leading reactive power that offsets the lagging reactive power, improving the power factor. This results in:

• Reduced current flow through the system
• Lower I²R heating losses in transformers and cables
• Cost savings that can pay back the capacitor investment

Transformer Impedance

The transformer's impedance is typically specified as a percentage on the nameplate. To convert this to actual ohmic values:

• Base impedance: Z_base = V² / S_VA
• Actual impedance: Z = Z_base × (Z% / 100)
• Resistance component: R = Z × cos(θ)
• Reactance component: X = Z × sin(θ)

Where θ is the impedance angle (typically 85-88° for power transformers).

Power Factor Correction

Power factor (PF) is the ratio of real power (P) to apparent power (S):

PF = P/S = P/√(P² + Q²)

Where:

• P = Real power (kW)
• Q = Reactive power (kVAR)
• S = Apparent power (kVA)

Required Capacitor Size

To improve the power factor from PF₁ to PF₂, the required capacitive reactive power is:

Q_C = P × (tan(cos⁻¹(PF₁)) - tan(cos⁻¹(PF₂)))

Current Reduction

The current before power factor correction:

I₁ = P / (√3 × V × PF₁) for three-phase systems

The current after power factor correction:

I₂ = P / (√3 × V × PF₂) for three-phase systems

Power Loss Calculation

The power loss in a resistive element is:

P_loss = I² × R

The reduction in power loss after power factor correction is:

ΔP_loss = R × (I₁² - I₂²)

Annual Cost Savings

Annual cost savings = ΔP_loss × Operating hours × Electricity rate

Payback Period

Payback period = Capacitor bank cost / Annual cost savings