Weight and absorbs the substance flow absorber

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(Y_in - Y_fin) = L(K_fin - X_in) (3.2)

where L, G – consumption respectively liquid absorbent and an inert gas portion, kg/s; Ẍ_in, Ẍ_fin – the initial and the final concentration of ammonia in the absorber, kg Am/kg W; Ῡ_in, Ῡ_fin – 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:

Y_in = y_in/(ρ_0y – y_in); Ẍ_in = x_in/(100-x_in), (3.3)

Y_in = ∙

where ρ_0y – ammonia average density under normal conditions [2].

We get:

Ῡ_in = = 0,065 kg Am/kg G;

Ῡ_fin = 0,065 – 0,0637 = 0,0013 kg Am/kg G;

Ẍ_in = 0.

Final concentration of ammonia in the absorber Ẍ_fin 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 Ẍ_fin opt, based on optimal flow absorber[3]. For ammonia production absorb consumption water L take 1,5 times greater than the minimum L_min [4]. In this case the final concentration of Ẍ_fin defines a material balance equation using data on equilibrium (im. 3.2 and 3.3):

M = L_min( = 1,5L_min(Ẍ_fin - Ẍ_in) (3.4)

Thence

Ẍ_fin = ( )/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 - y_gen)∙(ρ_oy – y_in) (3.5)

where y_gen – volume fraction of ammonia in the gas, which is equal

y_gen = y_inν₀/M_Am = 0,1 m³Am/m³G;

M_Am – the molar mass of ammonia [4].

Then

G = 0,5∙(1 – 0,1)∙(1,238 – 0,075) = 0,52 kg/s.

Productivity of absorber sequestered component

M = G(Ῡ_in - Ῡ_fin) = 0,52(0,065 – 0,0013) = 0,033 kg/s. (3.6)

Consumption of absorber is:

L = M/(Ẍ_fin - Ẍ_in) = 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(Y_in – Y_fin) (5.6)

L = M/(X_fin – X_in)

l = L/G

ΔY_av = (ΔY_b – ΔY_l)/[ln(ΔY_b/ΔY_l)] (5.7)

ΔY_b = Y_in –

ΔY_l = Y_fin –

K_y = (5.8)

ρ_y = ρ_oy ∙

lg[ ∙ ()^0,16] = A – B()^1/4∙()^1/8 (5.9)

w = 0,5 w _lim

d = = (5.10)

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