A gas stream produced as a by-product of a petroleum processing operation is to

Question

A gas stream produced as a by-product of a petroleum processing operation is to be used as a fuel in a small electric generation facility. The gas composition, in volume %, is:
CH4 (methane) 82%, C2H6 (ethane) 18%
The gas flow rate is 200 standard liters per minute (0oC definition).
1. Determine the air feed rate for a stoichiometric feed ratio.
2. The actual combustion is to take place with 15% excess air. Determine the composition (vol%) and flowrate (std L /min) of the gases leaving the burner if 95% of the fuel is completely consumed and the product gas contains no CO (all carbon in the fuel which reacts forms CO2). Apply the 95% to each component of the fuel.
3. If the electrical generation process is 30% efficient, determine the maximum electrical current and power that can be generated at a voltage of 240 volts.
Fuel Energy Values: Methane: 900 kJ/mol, Ethane: 1560 kJ/mol

Explanation / Answer

We want to write the balanced chemical equations. We do this by fiddling with the coefficients until there are the same number of each atom on both sides. CH4 + 2O2 --> CO2 + 2H2O 2C2H6 + 7O2 --> 4CO2 + 6H2O Equal volumes contain equal moles. Air is 20.95% oxygen In one minute we produce 200*.82 l of CH4 = 164 l In one minute we produce 200*.18 of C2H6 = 36 l We need twice as much O2 for the methane = 164l*2 = 328l We need 7/2 as much O2 for the ethane = 36l*7/2 = 126l Total O2 needed is 328l + 126l = 454 Air*0.2095 = oxygen Air = 454l/0.2095 = 2167l or 2200l This is the stoichiometric feed ratio 2200 l air answer 2. We add 15% more air 164*.95 = 155.8l of the CH4 reacts 36*.95 = 34l of the C2H6 reacts. This produces 3*155.8l and (10/2)*34l of CO2 and H2O The 10/2 is from adding up the 4 from CO2 and the 6 from H20 and dividing by the 2 from C2H6 we also have the excess air excess air is found from excess O2 155.8l*2 = 311.6l O2 34l*7/2 = 119.7l O2 O2 = 431.3 l consumed air = 2167l put in more air = 2167l*1.15 = 2492 we take away the oxygen consumed 2492 - 431 = 2061l flowing out CO2 and H2O coming out = 467.4l + 171l CH4 and C2H6 coming out = 10l all together = 2061l + 638.4l + 10l = 2709l/min or about 2700l/min out of time part c to follow

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