Ammonia (Am) mass passing in the absorption of the gas (G) mixture in the absorber per unit time is found from the material balance equation:
M = G(Yin - Yfin) = L(Kfin - Xin) (3.2)
where L, G – consumption respectively liquid absorbent and an inert gas portion, kg/s; Ẍin, Ẍfin – the initial and the final concentration of ammonia in the absorber, kg Am/kg W; Ῡin, Ῡfin – the initial and the final concentration of ammonia in the gas, kg Am/kg G.
Express phase compositions, the load on gas and liquid in the selected dimension to calculate:
Yin = yin/(ρ0y – yin); Ẍin = xin/(100-xin), (3.3)
Yin = ∙
where ρ0y – ammonia average density under normal conditions .
Ῡin = = 0,065 kg Am/kg G;
Ῡfin = 0,065 – 0,0637 = 0,0013 kg Am/kg G;
Ẍin = 0.
Final concentration of ammonia in the absorber Ẍfin determines its consumption (which in turn affects the dimensions as absorber and desorber), as well as a portion of the energy consumption associated with the transfer of fluids and its regeneration. Therefore Ẍfin opt, based on optimal flow absorber. For ammonia production absorb consumption water L take 1,5 times greater than the minimum Lmin . In this case the final concentration of Ẍfin defines a material balance equation using data on equilibrium (im. 3.2 and 3.3):
M = Lmin( = 1,5Lmin(Ẍfin - Ẍin) (3.4)
Ẍfin = ( )/1,5 = (0,0482+0,5∙0)/1,5 = 0,032 kg Am/kg W,
where - the concentration of ammonia in the liquid equilibrium with the initial composition.
Consumption of the inert gas
G = V0(1 - ygen)∙(ρoy – yin) (3.5)
where ygen – volume fraction of ammonia in the gas, which is equal
ygen = yinν0/MAm = 0,1 m3Am/m3G;
MAm – the molar mass of ammonia .
G = 0,5∙(1 – 0,1)∙(1,238 – 0,075) = 0,52 kg/s.
Productivity of absorber sequestered component
M = G(Ῡin - Ῡfin) = 0,52(0,065 – 0,0013) = 0,033 kg/s. (3.6)
Consumption of absorber is:
L = M/(Ẍfin - Ẍin) = 0,033/(0,032 – 0) = 1,031 kg/s.
Then the ratio, or specific consumption, consist:
l = L/G = 1,031/0,52 = 1,98 kg/kg.
M = G(Yin – Yfin) (5.6)
L = M/(Xfin – Xin)
l = L/G
ΔYav = (ΔYb – ΔYl)/[ln(ΔYb/ΔYl)] (5.7)
ΔYb = Yin –
ΔYl = Yfin –
Ky = (5.8)
ρy = ρoy ∙
lg[ ∙ ( )0,16] = A – B( )1/4∙( )1/8 (5.9)
w = 0,5wlim
d = = (5.10)