Recent content by GeorgeP1

my course work supplied me with a table for gas enthalpy at various temperatures. I didn't do ant equations here. I just simply used the data from the table. then used the theory of heat in = heat out. My heat in I calculated to be 2581kj. Then i tabulated the output estimates which were within...

N2 = 66.1 kj / mol is the enthalpy at 2000°C This is how I have derived the maximum flame temp. to be 1987°C. Then from this obtained the heat capacity. what do you make it?

Butane = -2660 kJ / mol Propane = -2046 kJ / mol Butane = -2540 kJ / mol Heat released by combustion of 1 mole of the fuel : = (0.75x2660) + (0.1x2046) + (0.15x2540) = 1995 + 204.6 + 381 = 2581 kJ To burn 1 mole of fuel mix we require 6.3 mole of O2 and produce in our flue gas 3.9 mole of...

Hi chestermiller. Its the flame temperature. I calculated the flame temperature to be 1987°C. (which might also be incorrect). what do you make it? Regards.

Homework Statement A fuel gas consists of 75% butane (C4H10), 10% propane (C3H8) and 15% butene (C4H8) by volume. It is to be fed to the combustion chamber in 10% excess air at 25°C, where it is completely burnt to carbon dioxide and water. The flue gases produced are to be used to generate 5...

how does this look?

Flue gas = H2O = 4.75 moles CO2 = 3.9 moles O2 = 0.63 moles N2 = 26.06 moles Total = 35.34 moles assuming a flame temperature of 2000°C Heat content = enthalpy x no. of molesFor N2, 66.10 x 26.06 = 1722.6 kJ CO2, 108.32 x 3.9 = 422.5 kJ O2, 69.65 x 0.63 = 43.9 kJ H2O...

butane = -2660 kj/mol propane = -2046 kj/mol butene = -2540 kj/mol heat released by 1 mole of fuel = (0.75x2660)+(0.1x2046)+(0.15x2540) =2581 kj (heat in) is this correct so far?

Homework Statement A fuel gas consists of 75% butane (C4H10), 10% propane (C3H8) and 15% butene (C4H8) by volume. It is to be fed to the combustion chamber in 10% excess air at 25°C, where it is completely burnt to carbon dioxide and water. The flue gases produced are to be used to generate 5...

:-)

so per kmol is 40.89 mols / 3.1804 = 0.013 Mj / Kmol does this make sense??

Excellent. Thanks chestermiller My previous attempt was a country mile out then. what about the net calorific value (CV) per kmol of the fuel/air mix at 25°C ? Any ideas?

propane = 0.003 m^3 = 0.2574 MJ /M^3 Butene = 0.0044 M^3 = 0.463 Mj / M^3 Total = 3.1804 Mj /M^3

I make it 0.02206 M^3 for butane does that make the net CV for butane 2.46 Mj / M^3

great... yes the net cv is for 1 m^3 of each gas.