Oxy-Fuel Technology Quiz

Limited to less than 400 tonnes of O2 per day (correct)

Limited to less than 600 tonnes of O2 per day

Limited to less than 800 tonnes of O2 per day

Limited to less than 200 tonnes of O2 per day

$Minimum Power = \left( \frac{\ln\left(\frac{\frac{y_2}{X}}{\frac{y_{O2}}{2}}\right)}{\ln\left(\frac{p_1}{p}\right)} - \frac{\ln y_{O2} + \frac{y_2}{X}}{y_{O2}}\right) \times \left(\frac{RT}{F_{CO2}}\right)$ (correct)

$Minimum Power = \left( \frac{\ln\left(\frac{\frac{y_{O2}}{2}}{\frac{y_{2}}{X}}\right)}{\ln\left(\frac{p}{p_1}\right)} - \frac{\ln y_{O2} + \frac{y_{2}}{X}}{y_{O2}}\right) \times \left(\frac{RT}{F_{CO2}}\right)$

$Minimum Power = \left( \frac{\ln\left(\frac{\frac{y_{O2}}{X}}{\frac{y_{2}}{2}}\right)}{\ln\left(\frac{p}{p_1}\right)} - \frac{\ln y_{O2} + \frac{y_{2}}{X}}{y_{O2}}\right) \times \left(\frac{RT}{F_{CO2}}\right)$

$Minimum Power = \left( \frac{\ln\left(\frac{\frac{y_{2}}{X}}{\frac{y_{O2}}{2}}\right)}{\ln\left(\frac{p}{p_1}\right)} - \frac{\ln y_{O2} + \frac{y_{2}}{X}}{y_{O2}}\right) \times \left(\frac{RT}{F_{CO2}}\right)$

2,470 kmol/s (correct)

2,370 kmol/s

2,570 kmol/s

2,670 kmol/s

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<p>Significant reduction in NOx.</p> Signup and view all the answers