×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Contact US

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Combustion Products Calculation
2

Combustion Products Calculation

Combustion Products Calculation

(OP)

A friend asked me to calculate the exhaust products that would be produced by a propane powered IC engine generating 5kw electrical at a fuel efficiency of 33%.

This is my first cut at the solution … but engines are not my expertise at all and my chemistry skills are inexperienced at best.

I am seeking someone at a higher pay grade than myself to review this calc for any corrections or sanity mods needed.

All comments welcome!
Best regards,
B
———————————————————
C3H8 MASS FLOW CALC

5 Kilowatts (Electrical) Delivered for 1 hour
33% Fuel Efficiency => 15kWh fuel input

1kg of propane = 14.019 kWh
1.07 kg propane mass required/hr

1kg propane = 22.678 moles
Propane fuel mass input = 24.265 moles

Chemistry:
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)

24.265 moles propane + 121.33 moles O2
produces:

72.795 moles CO2 + 97.06 moles H2O

Mass of nitrogen ingested for Stoich:

21% O2 in the atmospheric => 4.762 N2 ratio

N2 ingested = 122.33 molesO2 x 4.762N2 ratio
= 577.75 moles N2

Hourly Exhaust Mass @ 5 kWh(e):

72.795 moles CO2. (@44.01 g/mol)
97.06 moles H2O. (@18.015 g/mol)
577.75 moles N2. (@28.013 g/mol)

Makes:
3204 g CO2
1752 g H2O
16185 g N2

21.149 kg/hr total combust mass flow @
15.155% CO2, 8.287% H2O & 76.56% N2

VOLUMETRIC FLOW
STP Molar Gas Volume = 22.4 liters

747.60 total moles

747.60 mol x 22.4 liters = 16,746 liters/hr

.0353 ft3 / liter

591.1 cubic feet / hour @STP

RE: Combustion Products Calculation

You have the big-picture gist of it although I'm not about to double-check your numbers. But, beware of one thing that could be a "gotcha": Incomplete combustion.

You have made an implicit assumption of stoichiometric air/fuel ratio and 100% complete combustion. For an engine with modern emission controls and a warmed-up 3-way catalyst with lambda-sensor feedback control of the air/fuel ratio, and with the exhaust emission control systems considered to be an integral part of the engine, this is a reasonable approximation; any products of incomplete combustion will be in the parts-per-million range.

If this engine is not one with modern emission controls, or during a cold-start warm-up phase in which the catalyst has not "lit-up" yet, there will be significant products of incomplete combustion: carbon monoxide, hydrocarbons (unburned or partially burned fuel and trace amounts of lubricating oil), and oxides of nitrogen.

"Lean" combustion - excess air, but within the bounds of a sparkplug still being able to ignite it - tends to increase NOx, and to some extent HC if lambda is pushing the lean-combustion limits.

"Rich" combustion - insufficient air - tends to increase CO and HC.

RE: Combustion Products Calculation

(OP)
Brian,
Def no need to rerun the numbers. The process was my only concern.

Your points about operating conditions, transient behavior are spot on. Those might come into play if there is further opportunity to develop this rabbit hole further.

Appreciate you taking the time to comment!

Best regards,
B

RE: Combustion Products Calculation

The engine has to put out more than 5 Kw of power to generate 5 Kw, I think the average 5 kw generator uses a 10HP engine.

RE: Combustion Products Calculation

But only about 5.5 kW crankshaft when generating 5 kW electrical. The 10 hp (7.5 kW) engine is only needed to handle transients when the 5 kW alternator briefly generates far more than 5 kW.

je suis charlie

RE: Combustion Products Calculation

The presumption is that the stated thermal efficiency of the engine is at whatever operating point (i.e. less than 100% of rated torque) that it takes for the generator to generate that much electricity. Given the assumptions, the maximum rated torque/power output of the engine is irrelevant. It's still 33% efficient as defined in the original problem definition.

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members! Already a Member? Login



News


Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close