The forum discussion centers on calculating the amount of steam generated per hour from flue gases produced by burning a fuel gas mixture consisting of 75% butane, 10% propane, and 15% butene. The calculation involves determining the heat capacity of the flue gas, which is found to be 1.34 kJ/kg after accounting for a 5% heat loss. The final calculation yields a steam production rate of 38,300 kg/h based on the available heat and the enthalpy change required to convert water at 90°C to steam at 5 bar.
PREREQUISITESChemical engineers, process engineers, and anyone involved in energy production or thermodynamic calculations related to steam generation and combustion processes.
Chestermiller said:Check the units
Rogue said:Right, I think I should be adding enthalpy into the system into my qmc equation.
But I'm not sure I understand why.
Why would you divide the heat load by the heat capacity of water x dT + system enthalpy?
Rogue said:I noticed it was small, especially with my steam produced being so small.
But as I haven't done any mass and energy balance modules etc or much in the way conversion kmol to kg/h, I'm just trying best I can.
I don't confirm these mole percents or your molecular weight. I getRogue said:Flue gas:
CO2 = 3.9. (11%)
H2O = 4.75. (13.5%)
O2 = 0.6. (0.6%)
N2 = 25.96. (25.96%)
Total 35.21
Rogue said:Sorry, enthalpy of steam at 5bar.
I guess this tells you the energy required to produce the steam at 5bar?
Don't think I've factored this in before on other heat transfer equations though.
Chestermiller said:I don't confirm these mole percents or your molecular weight. I get
11.06%
13.48%
1.78%
73.68%
And a molecular weight of 28.5 kg/kmole
What about the molecular weight?Rogue said:Sorry Chester, my mistake. I did have these correct but it would appear I have transferred across onto here incorrectly.
Chestermiller said:Check out my post #12 from a year ago. This has the correct results.
Chestermiller said:What about the molecular weight?
What do you get for the heat load now?
I got less than that. With the 5% heat loss, I get 90 MJ./hrRogue said:Yeh, got that. Too many pages with scribbles.
My heat load is 101276973 after 5% heat loss.
The difference coming from my flame temp difference.
Thanks Chester
Chestermiller said:I got less than that. With the 5% heat loss, I get 90 MJ./hr
Where did the 2050 come from?Rogue said:My delta T was 1750 (2050-300), which is different to your original answer.
This is determined by a previous question where flame temperature needs to be calculated.Chestermiller said:Where did the 2050 come from?
If this is the case, the heat load could have been determined much more accurately by neglecting the heating to the adiabatic flame temperature all together and, instead, simply using Hess' Law to determine the enthalpy change between reactants at 25 C and products at 300 C. In this way, one would only need to know the average heat capacities of the products over the range from 25 C to 300 C.Rogue said:This is determined by a previous question where flame temperature needs to be calculated.
There will always be some differences in answer to this question as it has to be interpolated from a graph drawn by the student.
Accuracy will obviously depend on each student to an extent.
Any rounding errors calculating enthalpy in the fuel gas will also affect this.
Chestermiller said:If this is the case, the heat load could have been determined much more accurately by neglecting the heating to the adiabatic flame temperature all together and, instead, simply using Hess' Law to determine the enthalpy change between reactants at 25 C and products at 300 C. In this way, one would only need to know the average heat capacities of the products over the range from 25 C to 300 C.