# energy balance

## Recommended Posts

I am seeking comment on the method of the following calculations to determine the energy output of combustion of fuel. Any errors in the methodolgy would be appreciated.

The method is using total enthalpies with mole quantities

The ultimate analysis of the fuel is used to obtain weight values of the composition:

C 45 (% weight) 243 (mol) 0 (ΔHf kJ/mol)

H 6 433 (including air, water) 218

N 5 23 473

O 35 155 (including air, water) 249

water 5 -

ash 4 -

CO2 188 (mol/hr) -394 (ΔHf kJ/mol) 37 (Cp J/K/mol)

H2O 227 -286 75

N2 3120 - 29

O 620 - 29

fuel feed 5 kg/hr

air supply 1500 l/min

combustion temperature 1070 K

From dQ/dt=Σ H(outlets)-Σ H(inlets)

C: 0

H: 433*218

N: 23*473

O: 155*249

total enthalpy: 1.44 x10^5

CO2: 188*(-394+((1070*37)-(298*37)))

H20: 227*(-286+((1070*75)-(298*75)))

N2: 3120*(0+((1070*29)-(298*29)))

O2: 620*(0+((1070*29)-(298*29)))

total enthalpy: 1.02 x10^8

energy balance = 1.02 x10^8 J/hr = 28 kW

##### Share on other sites

Your first step is to write down the reaction equation. What goes in, what comes out... and in which phase it is (solid, liquid, gas).

Also, you must write down units and equations, because despite me being a chemical engineer (solving mass and energy balances every day) I do not understand what you wrote down, or what you're trying to calculate.

However, your answer (28 kW) is in the correct range of biomass-type fuels: 5 kg/hr, with a heat of combustion (dHc) of about 20 MJ/kg (typical for wood or grass or other plant materials) -->

P = dHc * flow = 20*10^6 J/kg*5kg/hr = 100*10^6 J/hr = 27.8*10^3 J/s = 27.8 kW.

with P = power in W =J/s

dHc = heat of combustion in J/kg

flow = flow in kg/hr or kg/s

... So it might be correct. I cannot check it though.

Edited by CaptainPanic

##### Share on other sites

i agree with captain panic, you need to be much clearer with your working, sure it takes a few seconds longer to put in the units and write it out in a readable format, but this will be a MASSIVE timesaver if you make a mistake(which you inevitably will, you wouldn't be human if you didn't). especially if it is on a larger balance than a combustion unit.

also, on larger systems your are doing a balance on, draw some diagrams. these are ridiculously useful, i even do it for small unit processes where it isn't strictly necessary out of habit.

##### Share on other sites

The objective is to calculate energy output of fuel when stoichiometry is not known, therefore enthalpies with mole values were used, instead of enthalpies of formation of reactants and products.

The equation is:

biomass (CxHyOz)(s) + xO2(g) -> xCO2(g) + yH2O(l)

so a simple empirical formula could be x=4, y=6, z=3.

The enthalpies units are J/hr, because of the fuel feed rate. The data was tab separated so a quick copy into Calc will show better the units for the values, e.g.

heat capacity Cp (J/K/mol)

CO2(g)=37

H2O(l)=75

N(g)=29

O(g)=29

## Create an account

Register a new account