- #1
marsten
- 15
- 1
Hi
1. Homework Statement
I'm trying to calculate the enthalphy of flue gas leaving a gas turbine, in kJ/kg. I just want to confirm that I'm using a correct method.
hfg = (VCO2⋅ hCO2 + VH2O⋅ hH2O + VN2⋅ hN2 + VO2⋅ hO2) / (VCO2 + VH2O + VN2 + VO2)
ρfg = (p⋅ M) / (R⋅ Tfg)
I have already calculated the total flue gas volume and the volumes of the flue gas components, and I'm fairly sure that I've got that right.
The next step is to calculate the flue gas enthalphy using the following equation (the enthalpies are taken from a table at Tfg = 497 °C):
hfg = (VCO2⋅ hCO2 + VH2O⋅ hH2O + VN2⋅ hN2 + VO2⋅ hO2) / (VCO2 + VH2O + VN2 + VO2)
hfg = (0,59⋅ 991 + 0,55⋅ 776 + 5,09⋅ 653 + 0,81⋅ 688) / (0,59 + 0,55 + 5,09 + 0,81) = 695 kJ/mn3
Then, from the ideal gas law, I get the density of the flue gas (M is an average that I've calculated, and p = 1 bar):
ρfg = (p⋅ M) / (R⋅ Tfg)
ρfg = (1⋅ 105⋅ 29⋅ 10-3) / (8,314⋅ 770) = 0,45 kg/m3
And finally, to convert from kJ/mn3 to kJ/kg, I divide the calculated enthalphy with the density:
hfg = 695 kJ/mn3 / 0,45 kg/m3 = 1532 kJ/kg
Is my method correct? My biggest concern is whether or not m3 really cancel off mn3.
Thanks!
Marcus
1. Homework Statement
I'm trying to calculate the enthalphy of flue gas leaving a gas turbine, in kJ/kg. I just want to confirm that I'm using a correct method.
Homework Equations
hfg = (VCO2⋅ hCO2 + VH2O⋅ hH2O + VN2⋅ hN2 + VO2⋅ hO2) / (VCO2 + VH2O + VN2 + VO2)
ρfg = (p⋅ M) / (R⋅ Tfg)
The Attempt at a Solution
I have already calculated the total flue gas volume and the volumes of the flue gas components, and I'm fairly sure that I've got that right.
The next step is to calculate the flue gas enthalphy using the following equation (the enthalpies are taken from a table at Tfg = 497 °C):
hfg = (VCO2⋅ hCO2 + VH2O⋅ hH2O + VN2⋅ hN2 + VO2⋅ hO2) / (VCO2 + VH2O + VN2 + VO2)
hfg = (0,59⋅ 991 + 0,55⋅ 776 + 5,09⋅ 653 + 0,81⋅ 688) / (0,59 + 0,55 + 5,09 + 0,81) = 695 kJ/mn3
Then, from the ideal gas law, I get the density of the flue gas (M is an average that I've calculated, and p = 1 bar):
ρfg = (p⋅ M) / (R⋅ Tfg)
ρfg = (1⋅ 105⋅ 29⋅ 10-3) / (8,314⋅ 770) = 0,45 kg/m3
And finally, to convert from kJ/mn3 to kJ/kg, I divide the calculated enthalphy with the density:
hfg = 695 kJ/mn3 / 0,45 kg/m3 = 1532 kJ/kg
Is my method correct? My biggest concern is whether or not m3 really cancel off mn3.
Thanks!
Marcus
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