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Absorber design: The volume flow rate of the mixed gas is Vs = 100 m3/h, which is equivalent to 4.464 kmol/h. Average molecular weight = 29.92 wv = 133.57 kg/h. The molar flow rate of air entering the tower is V=4.464×(1-0.12)=3.93 (kmol/h). The amount of CO2 absorbed: Assuming a gas absorption rate of 95%, then: y1=0.12; Y1=0.12÷(1-0.12)=0.1364. Y2=Y1(1-φA)=0.12×0.05=0.00682, and X2=0.001. The amount of CO2 absorbed during the absorption process in the packed tower is: V(Y1-Y2)=3.93×0.12958=0.5092 kmol/h. According to the chemical equilibrium equation, 1 mole of CO2 reacts with 1 kmol of R2CH3N; therefore, the amount is 0.5092 kmol/h. Then, using 20% R2CH3N: (0.5092×119.16)/0.2=303.4(kg/h). The average molecular weight of 20% alkali is: 100÷(20÷119.16+80÷18)=21.68 (kg/kmol). Therefore, L theory = 303.4 ÷ 21.68 = 13.99 (kmol/h). If the amount of base used is 1.8 times the amount calculated theoretically, then: L=13.99×1.8=25.182 (kmol/h). Find the minimum liquid-to-gas ratio. Given that (L/V)min = 13.99 ÷ 3.93 = 3.56, and (L/V)min = (Y1 – Y2) / (X*1 – X2); thus, (0.1364 – 0.00682) / (X*1 – 0.001) = 3.56. From this, it can be determined that X*1 = 0.0374. The concentration of the liquor in the tower was determined. Given that V(Y1–Y2) = L(X1–X2), and X1–X2 = 0.5092/25.182, it follows that X1 = 0.0212. Thus, based on material balance calculations, the liquid and gas compositions at the inlet and outlet of the tower are: X1 = 0.0212, X2 = 0.0001; Y1 = 0.1364, Y2 = 0.00682. The values V = 3.93 kmol/h and L = 25.182 kmol/s were also obtained. Find the height of the packing layer. Determine the total number of gas-phase mass transfer units, NOG. On the m-n diagram, find the equation relating the operating line Y and X: Y = XL/V + (Y2 – X2)/V = 25.182/3.93X + (0.00682 – 6.41×0.001) = 6.41X + 0.00041. To find the phase equilibrium constant m: the intersection point of Y*=0.0374m+b and Y=7.013X+0.00612 is (X2, Y2), namely (0.001, 0.00682). Thus, m=3.56 and b=-0.00326, giving the relationship Y*=3.56X-0.00326. Y*1=3.56×0.0212-0.00326=0.072212, and Y*2=3.56×0.001-0.00326=0.0003. ΔY1=Y1-Y*1=0.1364-0.072212=0.0642, and ΔY2=Y2-Y*2=0.00682-0.0003=0.00652. ΔYm=(0.0642-0.00652)/ln (0.0642/0.00652)=0.0252. NOG=(Y1-Y2)/ΔYm=5.14. To find the total gas-phase mass transfer unit height: KY = 7.95 kg/(m2•h), and HOG = V/(KYαΩ) = 0.1026/(0.3924×200×3.14×0.22) = 0.4 m. The height of the packing layer is Z=HOG×NOG=0.4×5.14=2.6 (m). Find the tower diameter of the packed tower: D = 4VS/πu. It is known that VS = 100 m3/h, 0.0278 m3/s; the density of the gas entering the tower is ρv = 1.095 (kg/m3), while the density of the alkali is ρL = 0.996×103 (kg/m3). The values wv = 133.57 (kg/h), L = 25.182 (kmol/h), and wL = 546 (kg/h) are also given. Then (wL/wv)(ρv/ρl)0.5 = 546÷133.57×(1.095÷996)0.5 = 0.14. According to Echter’s general correlation diagram, by looking up the value at the horizontal axis of 0.15, the corresponding value on the vertical axis is 0.08; that is: uf2φψρvμL0.2/(gρL)=0.08. For a 25 mm rectangular saddle, the packing factor φ = 300, the liquid-phase correction coefficient ψ = 1,000/996 = 1.004, and the viscosity of the base is μL = 0.8 mPa•s; therefore, uF = 1.57 (m/s). Take the empty tower gas velocity to be 60% of the flooding velocity, i.e., u = 0.60uF = 0.60 × 1.57 = 0.942 (m/s). Then the tower diameter D = (4×0.0278)/(π×0.942) = 0.194 (m); therefore, D is taken as 0.2 m. The gas velocity in the empty tower is then calculated as: u=4VS/(πD2)=4×0.0278/(3.14×0.22)= 0.885 (m/s). The safety factor u/uF = 0.885/1.57 = 56%. Check whether the spray density under operating conditions is greater than the minimum spray density; since the packing size is less than 75 mm, (Lw)min = 0.08 (m3•m-1•h-1) is adopted. Given that σ=200 m2/m3, then Umin=(Lw)minσ =0.08×200=16 (m3•m-2•h-1). The spray density under operating conditions is: U=(546÷996)÷(π/4×0.22)=17.5 (m3•m-2•h-1)>Umin. The pressure drop across the packing layer is calculated as u2φψρvμ0.2L/(gρL)=0.562×0.08=0.025, and the abscissa is (wL/wv)(ρv/ρl)0.5=0.14. The operating point of the tower is determined in the Thueckert universal correlation diagram based on the values of the horizontal and vertical coordinates; this point lies at Δp/Z=360 (Pa/m), so the total pressure drop across the packing layer is 360×2.6=936 Pa