Feed Water Conductivity Equivalent (FCE)
The concept of EDI feed water conductivity equivalent arose from the need for a simple field method of estimating the ionic load on a EDI device. Certainly the best way to determine the ionic load is to perform a complete water analysis and determine the concentration of all ionized and ionizable constituents, but in some cases this is not practical. What had frequently been substituted for a complete water analysis was a simple measurement of the conductivity of the EDI feed water. This could introduce considerable error, as a conductivity measurement does not detect the full amount of weakly ionized species such as carbon dioxide (CO_{2}) and silica (SiO_{2}). For example, 10 µS/cm water could contain 4 ppm of NaCl or 60 ppm CO_{2}. Just as this would have a huge impact on the service cycle of a conventional demineralizer, it can have a major impact on sizing a EDI system. For this reason we have developed the concept of EDI feed conductivity equivalent, which attempts to take into account weak ions such as CO_{2} and silica.
Calculating FCE
 Measure EDI feed water conductivity (µS/cm)
 Measure the ppm CO_{2}
 ppm as CO_{2} x 2.79 = µS/cm
 Measure the ppm SiO_{2}
 ppm SiO_{2} x 1.94 = µS/cm
 Add measured conductivity to CO_{2 }& SiO_{2 }µS/cm
In the field, the CO_{2} concentration in the EDI feed water can be measured using Hach test kit Model CA23 (#143601). The smallest increment for this test kit is 1.25 mg/l. The EDI feed water conductivity can be measured with a handheld conductivity meter such as the Myron L Model 4P or with the permeate conductivity meter on the RO system.
Example FCE Calculation
 Measured conductivity = 6 µS/cm
 Measured CO_{2} = 5 ppm as CO_{2}
 Measured SiO_{2} = 0.5 ppm as SiO_{2}
 FCE = 6 + (5 x 2.79) + (0.5 x 1.94) = 20.9 µS/cm
DC Volts and Amps
 Voltage (potential) causes current to flow
 Current causes transfer of salt, regeneration of resin
 Amount of current required proportional to product water flow, amount of salt being removed
 Use Faraday’s law to calculate current required
Calculating Amps Required
 Faraday’s Law
 I = 1.31 (Q)(FCE)/(# cells)(eff)
 I = DC current, amps (per module)
 Q = product flow rate, liters/min/module
 FCE = feed conductivity equivalent, µS/cm
 eff = current efficiency, % (assume 10% current efficiency)
Example DC Current Calculation
 Product flow = 50 lpm
 Feed = 5 µS/cm + 3.75 ppm CO_{2}
 FCE = 5 + (3.75 x 2.79) = 15.5 µS/cm
 I = 1.31(50 lpm)(15 mS/cm)/(24 cells)(10%)
 = 4.2 amps per module
Setting DC Amperage
 Use constant current power supply
 Estimate amps required per module
 Set amperage to calculated value
 Power supply adjusts voltage to maintain current
 Important to record (trend) voltage & amperage
Definition of EDI Recovery
 % Recovery (R) = (QP)(100)/(QP + QR)
 QP = Product (Dilute) flow rate
 QR = Reject (Concentrate) flow rate
Calculating Reject Flow
 Reject = ((100  R)/R) x product flow
 where R is the percent recovery
 Example: Product flow = 50 m^{3}/h
 Recovery = 90%
 Reject = ((10090)/90) x 50
 = 5.6 m^{3}/h
