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A potentiometric zirconia oxygen sensor is an electrochemical device that uses a high-temperature stabilized zirconium oxide electrolyte made from ceramics. The sensor consists of a tube which is closed off at one end with electrodes. These are typically platinum, deposited onto the inner and outer surfaces.
The tube is gas tight at the and operates at temperatures commonly in the range 500 to 750ºC. It conducts electricity by means of oxygen ions and the potential difference across the cell is given by the Nernst equation:
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Where:
E is the potential difference / volts
R is the gas constant / 8.314 J mol-1 K-1
F is the Faraday constant / 96484 coulomb mol-1
p1 and p2 are the partial pressures of oxygen outside and inside the zirconia tube, respectively.
The partial pressure of a gas in a mixture is equal to the volume fraction of the gas times the total pressure of the gas mixture. So oxygen found in the air at a standard atmospheric pressure of 1013 mbars means a mole fraction of 20.9/100 = 0.209.
Therefore p1 = 0.209 x 1013 = 211.7 mbars
If the gas sample is vented and exposed to the atmosphere, there will be a pressure drop (∆p) between the sensor inlet and the exit of the outlet tube. The pressure of the gas sample inside the zirconia tube = ∆p + total external pressure. At the recommended flow rate of 150 ml/min the back pressure due to, for example, 5m of 1/8″ tubing is about 20 mbars. If longer runs of tubing are required, ¼” tubing is recommended to minimize back pressure.
We consider the case of the sensor being zeroed and calibrated at low atmospheric pressure (e.g. 980 mbars) and then being used at a higher atmospheric pressure (e.g. 1040 mbars). We calculate the sensor output and the instrument readings for samples of 1 ppm and 1% oxygen.
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Note that E is now in mV. We take the operating temperature as 650 ºC
For 1ppm oxygen
p1 = 0.209 x 980 = 204.8 mbars
p2 = 1 x 10-6 x (980 + 20) = 1 x 10-3 mbars
The sensor potential will be E = 0.0496 x (273.3 + 650) x log10 204.8/1×10-3 = 243.24 mV
For 1% vol oxygen
p1 = 0.209 x 980 = 204.8 mbars
p2 = 1/100 x (980 + 20) = 10.0 mbars
The sensor potential will be E = 0.0496 x (273.3 + 650) x log10 204.8/10.0 = 60.05 mV
For 1ppm oxygen
p1 = 0.209 x 1040 = 217.4 mbars
p2 = 1 x 10-6 x (1040 + 20) = 1.06 x 10-3 mbars
The sensor potential will be E = 0.0496 x (273.3 + 650) x log10 217.4./1.06×10-3 = 243.26 mV
For 1% vol oxygen
p1 = 0.209 x 1040 = 217.4 mbars
p2 = 1/100 x (1040 + 20) = 10.6 mbars
The sensor potential will be E = 0.0496 x (273.3 + 650) x log10 217.4/10.6 = 60.07 mV
For 1% vol oxygen
The percent change in E = [(243.26/243.24) x 100] – 100 = 0.008%
i.e. the sensor potential E is 0.008% greater at 1040 mbars.
For 1% vol oxygen
The percent change in E = [(60.07/60.05) x 100 – 100] = 0.033%
i.e. the sensor potential E is 0.033% greater at 1040 mbars.
As stated, the instrument is calibrated to 980 mbars. Therefore the theoretical readings are normalised to 1.00 ppm and 1.00% vol oxygen for the two gas concentrations at that pressure.
By substituting the sensor potentials from case two, into the equations for case one, we can solve for p2 and determine the theoretical readings.
For 1 ppm oxygen the equation for the sensor potential is:
243.26 = 0.0496 x (273.3 + 650) x log10 204.8/p2
By solving the equation for p2 we find that the oxygen partial pressure would be 9.999E-04 mbars and the instrument reading would be 0.999 ppm.
For 1% vol oxygen the equation for the sensor potential is:
60.07=0.0496 x (273.3+650) x log10 204.8/p2
By solving the equation for p2 we find that the oxygen partial pressure would be 9.999 mbars and the instrument reading would be 0.999% vol.
Under normal operating conditions the effect of changes in atmospheric pressure on the measurement of oxygen concentrations using the zirconia oxygen sensor is negligible. And barometric pressure correction is not required.
August 2012
By Dr Malcolm Taylor PhD, GRSC
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