Cathodic Protection Rectifier Sizing
Sizes an impressed-current rectifier from design current, bed and cable resistance, back EMF, and margin.
Example
You enter
- Design current (A) 10
- Anode bed resistance (ohms) 1.64
- Header cable length (ft) 500
- Negative cable length (ft) 300
- Cable resistance (ohms per 1,000 ft) 0.2485
- Back EMF allowance (V) 2
- Design margin (%) 50
- Rectifier efficiency (%) 60
- Energy rate ($/kWh) 0.12
You get
- Cable resistance (ohm) 0.1988
- Circuit resistance 1.8388 ohms, of which 0.1988 is cable
- Required DC voltage 20.39 V at 10.0 A
- With the design margin 30.582
- What the cable costs 1.988
- DC output 204 W
- AC input 340 W at 60% efficiency
- Every year, continuously 2977 kWh, $357
- Annual cost 357.198
Details, formula, and sources
Sizes an impressed-current rectifier from design current, bed and cable resistance, back EMF, and margin, then the AC input and annual energy. The worked 10 A system needs 20.4 V, a tenth of it cable, and costs $357 a year continuously -- forever, since a CP system switched off is not one.
required voltage = design current x (bed + cable resistance) + back EMF; design voltage adds the margin; AC input = DC output / rectifier efficiency.
NACE SP0169 (now AMPP) and the CP designer govern. The rectifier manufacturer's rating governs the selection.
Ohm's law; rectifier ratings are published by manufacturers.
Estimate. AHJ and licensed professional govern.
Field names used by the API: design_current_a, bed_resistance_ohm, header_length_ft, negative_length_ft, cable_ohm_per_kft, back_emf_v, design_margin_pct, rectifier_efficiency_pct, energy_rate_per_kwh, cable_resistance_ohm, total_resistance_ohm, required_voltage_v, design_voltage_v, cable_drop_v, dc_output_w, ac_input_w, annual_kwh, annual_cost
- Back EMF entered as an allowance, commonly about 2 VCP practice
- Selection the next standard rectifier above the design voltage, tapped down on commissioningmanufacturer