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Output bigger than Input???

Output bigger than Input???

Output bigger than Input???

(OP)
Hi,

I am new in this field but have been trying to calculate the energy content of flue gas from an incinerator.

I am using the stochiometric method and assuming 10% excess air for the diesel and carcass combustion while knowing the hourly inputs.

But I am getting about 45,000 Btu/ hr more being outputted that the gross input into the incinerator.

Anyone any ideas on where I'm going wrong?

I assume this method gives the LHV and no deductions for latent heat and losses have to taken from it. Is this correct?

Does the pressure in the flue chamber have a significant effect on the energy content?
Is 10% excess air assumption valid? I also dont have an exact composition for the animal carcass waste.

Is measuring the actual composition of the flue gas the most accurate way of doing this? Is this difficult and does it require alot of equipment?

Thanks for any help at all.

RE: Output bigger than Input???


I'm not an expert on incineration, just thinking aloud. What is the NCV of the carcasses ? Depending on the ratio of meat-to-bone this value could be much higher than assumed.

From what I've read on the subject, the needed % of xs air depends on the firing conditions, refuse being generally incinerated with about 50% xs air.

Others would surely add much more useful information.

RE: Output bigger than Input???

boole:

When dealing with ideal gas, pressure has no impact on the enthalpy/energy values.

As 25362 pointed out, the fuel (diesel + carcasse) composition may need some adjustment in order to get the energy balance correct.

boole, use the spreadsheet mentioned in your other post to do your energy and combustion calculations -- you should be able to estimate the flame temperature and combustion gas composition with the spreadsheet ...

Thanks,

Gordan
http://engware.i-dentity.com

RE: Output bigger than Input???


On the change of the molar heat capacity of the gases, Cp, change with pressure, P, Feric is right.

(∂ Cp/∂ P)T = -T (∂ 2V/∂T2)P

When speaking of an ideal gas: PV=RT. And (∂ 2V/∂ T2)P = 0
 

RE: Output bigger than Input???


Gordan, you're welcome !

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