rusty_sim
Chemical
- Jun 12, 2017
- 13
Hi
I had a go at the thermal calc due to expansion by Prex (faq378-1339) however I would also like to look at calculating the pressure rise due to nitric acid decomposing into gases (NO2, O2) in a shut in pipe eg. between 2 valves. I do not believe I have the thermodynamics set up correctly. I've also assumed all of the nitric acid reacts and the gases occupy the entire internal volume (this might be a bad assumption).
Q=mΔH
Q=VΔP (I think I have butchered the first law with this assumption)
ΔP=mΔH/V
2HNO3 -> H2O + 2NO2 + ½O2
Formation energy of HNO3 = -207 KJ/mol
Formation energy of H2O(l) = -285.83 KJ/mol
Formation energy NO2 = 33.12 KJ/mol
Formation energy of O2 = 0 KJ/mol (by convention pure compounds have zero formation energy at 273.15 K)
Sum of decomposition products = -285.83 + 2*33.12 = -219.59 KJ/mol which is less than -207 KJ/mol for nitric acid
ΔH 12.59 kJ/mol (219.59 - 207)
Inputs
Pipe ID 15 mm
Pipe length 5 cm
rho (l) 1380 kg/m3 (nitric acid density)
M (HNO3) 63.01 kg/kmol (molar mass nitric acid)
Calculated
D 0.015 m
h 0.05 m
V 8.83573E-06 m3 (volume of a cylinder)
m (HNO3) 0.012193306 kg (volume * density)
n (HNO3) 0.000193514 mol
ΔH 199.809554 kJ/kg (converting kJ/mol to kJ/kg)
Q 2.436339131 kJ
ΔP 275737.1846 kj/m3
ΔP 2.757371846 bar
However as I calculated the mass from the internal pipe volume, there is dependence between m and V so the change in pressure never changes regardless of changing the pipe geometry.
Some guidance would be much appreciated
Kind regards
I had a go at the thermal calc due to expansion by Prex (faq378-1339) however I would also like to look at calculating the pressure rise due to nitric acid decomposing into gases (NO2, O2) in a shut in pipe eg. between 2 valves. I do not believe I have the thermodynamics set up correctly. I've also assumed all of the nitric acid reacts and the gases occupy the entire internal volume (this might be a bad assumption).
Q=mΔH
Q=VΔP (I think I have butchered the first law with this assumption)
ΔP=mΔH/V
2HNO3 -> H2O + 2NO2 + ½O2
Formation energy of HNO3 = -207 KJ/mol
Formation energy of H2O(l) = -285.83 KJ/mol
Formation energy NO2 = 33.12 KJ/mol
Formation energy of O2 = 0 KJ/mol (by convention pure compounds have zero formation energy at 273.15 K)
Sum of decomposition products = -285.83 + 2*33.12 = -219.59 KJ/mol which is less than -207 KJ/mol for nitric acid
ΔH 12.59 kJ/mol (219.59 - 207)
Inputs
Pipe ID 15 mm
Pipe length 5 cm
rho (l) 1380 kg/m3 (nitric acid density)
M (HNO3) 63.01 kg/kmol (molar mass nitric acid)
Calculated
D 0.015 m
h 0.05 m
V 8.83573E-06 m3 (volume of a cylinder)
m (HNO3) 0.012193306 kg (volume * density)
n (HNO3) 0.000193514 mol
ΔH 199.809554 kJ/kg (converting kJ/mol to kJ/kg)
Q 2.436339131 kJ
ΔP 275737.1846 kj/m3
ΔP 2.757371846 bar
However as I calculated the mass from the internal pipe volume, there is dependence between m and V so the change in pressure never changes regardless of changing the pipe geometry.
Some guidance would be much appreciated
Kind regards