TH-FL SMS CH-2 Prb- 17

[nirvick]

Can anyone help me on this ? I can't get the answer 33%. what did I wrong here?




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[Slay the P.E.]

Percent excess air means how much above the theoretical air is the actual reaction consuming. So you need to calculate both Air-Fuel ratios; real and theoretical and then compare them.

First, the real reaction air-fuel ratio: There are 22 mol of air for every 3.3 mol of fuel, so AFactual = 22/3.3 = 6.66

Now, the stoichiometric reaction: There are 5 mol of air per mol of fuel, so AFtheoretical = 5/1 = 5

So, how much more (percentage wise) is the real reaction consuming? excess air = (AFactual - AFtheoretical)/AFtheoretical = (6.66 - 5)/5 = 0.33 = 33%

 

[Slay the P.E.]

To further clarify, The amount of excess air is usually expressed in terms of the stoichiometric air as percent excess air or percent theoretical air. For example, 33 percent excess air is equivalent to 133 percent theoretical air, and 200 percent excess air is equivalent to 300 percent theoretical air. Of course, the stoichiometric air can be expressed as 0 percent excess air or 100 percent theoretical air.

Make sure you read the question carefully and be clear if they're asking for percent excess air or percent theoretical air.

Good luck!

 

[Ramnares P.E.]

Slay beat me to it but yes, it's asking for EXCESS air which implies comparison to the actual stoichiometric equation. 

 

[MikeGlass1969]

I saw a couple of these problems on my exam....SKIPPED....and came back to these when/if I had time.

These were supposed to be up my alley being HVAC&R but I was never comfortable doing these analysis.

 

[nirvick]

thank you .I understand now. really appreciate that.

 

[Slay the P.E.]

thank you .I understand now. really appreciate that.

My pleasure!