Energy Engineering - Chemical Processes and Technologies

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1 Chemical Processes and Technologies (Ing. Giorgia De Guido) Exam June 22 nd, 2022 - Exercise: Solution Exercise no. 1 (of 2) A mixture of i-butane and n-butane is separated in the so-called “C 4 splitter” (a standard trayed distillation column), which is operated at 9 bar. The mixture is available at its bubble-point temperature at the operating pressure of the column and is characterized by a content [mol%] of i-butane as reported in the Data section below. The aim is to obtain a top product that contains 99 mol% i-butane and a bottom product that contains 98 mol% n-butane. Assuming an ideal thermodynamic behaviour for the system and a total molar flow rate of the feed stream as reported in the Data section below, please determine: 1. the total molar flow rate [kmol/h] of the distillate and bottom products and how much i-butane is recovered, respectively, in these two products; 2. the value of the average relative volatility, explaining how it was calculated; 3. whether it is possible to operate the distillation column with a reflux ratio (R) equal to 12. If it is possible to operate the column with the value of R given at the previous point, determine: 4. the number of theoretical stages needed to achieve the desired separation, respectively, for the upper (enriching) section and for the lower (stripping) section of the column; 5. the thermal duty (in kW) required at the top condenser. If it is not possible to operate the column with the value of R given at point 3, assume a reflux ratio equal to 1.2·R min and solve points 4 to 5. Data: Mole percentage of i-butane in the feed stream = 5 30.5 M+ =[mol%]= Molar flow rate of the feed stream== 6 (6 27) 100 3 M ⋅− + =[kmol/h]= where:= M 5== fifth=digit of student number from the left=(Student number/matricola = XXXXXX) M 6 = sixth digit of student number from the left (Student number/matricola = XXXXXX) The vapour pressure is calculated with the Antoine equation: ln( [ ]) [] ev B P mmHg A TK C = − + . Component A B C cPV [kJ/kmol/K] = ΔH ev=(@T NBP ) [kJ/mol] = TNBP = [K] = TC== [K] = PC= [atm] = i-butane = 15.5381 2032.73 -33.15 88.39 21.30 261.3 408.10 36.00 n-butane 15.6782 2154.90 -34.42 92.07 22.40 272.7 425.20 37.47 2 Results: The results are reported in the following assuming M 5 = M 6 = 0. Therefore: • mole percentage of i-butane in the feed stream, z F = 30.5 [mol%], • molar flow rate of the feed stream, F = 91 [kmol/h]. Point no. 1 Solving the overall and component material balance for the more volatile component (i.e., i-butane) F DB F DB Fz Dx Bx = +   = +  == considering the given two purity specifications (x D = 0.99, (1-x B) = 0.98), it is possible to obtain: D = 26.7371 [kmol/h]; B = 64.2629 [kmol/h]. The amount of i-butane recovered in the two products is, respectively, given by: recovery in the distillate = 100 10 26.4697 27.7 0 550 D F Dx Fz ⋅ ⋅=⋅ = ⋅ 95.3693 %; recovery in the bottoms = 100 1001.2853 27.7550 B F Bx Fz⋅ ⋅=⋅ = ⋅ 4.6307 %. Point no. 2 To calculate the relative volatility of i-butane with respect to n-butane at the thermal conditions of the top of the column, of the bottom of the column and of the feed stream (since the feed stream is at its bubble point), the following three temperatures are determined: • dew point temperature of the distillate product: T d,D = 335.7606 [K]; • bubble point temperature of the bottom product: T b,B = 348.2188 [K]; • bubble point temperature of the feed stream: T b,F = 344.2465 [K]. The corresponding relative volatilities are: • αtop = 1.3413 ; • αbottom = 1.3170 ; • αfeed = 1.3245 . Thus, the average relative volatility is 3 top bottom feed αα α α α ==⋅⋅ == 1.3275 (this average is used in this case, thus also including α feed , because the feed is at its bubble point, i.e. q = 1). 3 Point no. 3 The minimum reflux ratio is R min = 9.8518 (it can be determined using the formula available for q = 1, since, in this problem, the feed stream is at its bubble point). Since the given R (R = 12) > R min and R/R min = 1.22 (which falls in the typical range of values for the multiplying constant based on economic considerations), then it is possible to operate the distillation with the given reflux ratio. Point no. 4 The problem is solved using the analytical McCabe-Thiele method. Upper section of the column (including the values of the coefficients in the Riccati equation): a = 3.053115, b = -4.308375, c = 0.251882, N upper = 31.5699. x F- is , then, determined. x F- = 0.300715. Lower section of the column (including the values of the coefficients in the Riccati equation): a’ = 3.053115, b’ = -3.423802, c’ = -0.009528, N lower = 26.0767. Therefore, 32 and 27 stages are needed, respectively for the upper and lower sections of the column. Point no. 5 The thermal duty required at the top condenser ( cQ ) can be determined writing the energy (enthalpy) balance across the condenser. ,, ( ) ( )( )VL d D cbD VH T Q D L H T=++  ,, ( 1)[ ( ) ( )] VL cd DbD Q DR H T H T=+−   where T b,D = 335.7219 K. If the reference state is chosen to be vapour at T d,D , then: , ,, 1 () () NC VV dD iD i dD i H T x hT = = ⋅∑   4 , ,, 1 () () NC LL bDiD i bD i H T x hT = = ⋅∑   where , , ,, , ,, () ( ) ( ) ()LVVi bD i d D Pi bD d D evi bDhT hT c T T H T = + − −∆    = , ( )0Vi dDhT =  =kJ/kmol (due to the chosen reference state)= and ,,() ev i b DHT∆  =is computed using Giacalone�s formula.= As a result:= cQ= 1574.1897 [kW]. Exercise no. 2 (of 2) An air stream containing 0.6 mol% SO 2 has to be treated in an absorption trayed tower by contacting it with water. The aim is to lower the content of SO 2 in the air stream exiting the absorber to 10 ppm (on a molar basis). Assume that: • the equilibrium constant for SO 2 is K = 38; • the water stream fed at the top of the tower is free of SO 2; • the liquid-to-gas ratio (L/G) is [50 + (M 4 - 4.5)], where M 4 is the fourth digit of the Student number from the left (Student number/matricola = XXXXXX). Please, determine: 1. whether the given value of L/G is feasible or not; 2. the number of theoretical stages needed to perform the desired separation and the composition (on a molar basis) of the outlet streams if the given L/G is feasible. Otherwise, determine the number of theoretical stages needed to perform the desired separation and the composition (on a molar basis) of the outlet streams assuming (L/G) = 1.5∙(L/G) min . Results: The results are reported in the following assuming M 4 = 9. Therefore: (L/G) = 54.5. Point no. 1 First, mole ratios are calculated for the three streams for which the mole fraction of SO 2 (the transferred component) is known: 0 0 0 0.00603 1 6 y Y y = = − =, 1 0.000010 N N N y Y y = = − , 1 1 1 0 1 N N N x X x + + + = = − == 5 Since x ≈ X and y ≈ Y, the assumption of dilute solutions applies (linear cascade). Therefore, both the equilibrium relation and the material balance equation are represented by straight lines in the X/Y diagram. In particular, the equilibrium relation can be written as Y=K∙X, where K= 38 (given). For the following calculations, we will set Y 0 = y 0 = 0.006, Y N = y N = 0.00001, X N+1 = x N+1 = 0. (L/G) min corresponds to the slope of the straight operating line that intersects the equilibrium straight line at Y 0. Thus: 0 min01 N eqN YY LY G XX X + − ∆  = = =  ∆−  37.9367 L G kmol kmol   , where 0 0 1.58 04 eq Y X K e = = − . Since (L/G) > (L/G) min and their ratio is 1.44, it is possible to operate the absorption using the given value of (L/G). Point no. 2 Using the Kremser equation for design problems: ( ) ( ) ( ) ( ) 10 0 100 ln ln eqNN eqNN YYYY Y Y EY Y N E + + −− − −− − = , where 11 0 eqNNY KX ++ =⋅= , and 0.6972 G EK L = ⋅= , N = 14.44  N = 15. The composition (on a molar basis) of the outlet streams is as follows: 8.15 6 NNY e y ≈= − , using the Kremser equation for simulation problems (with N = 15), 11 1.10 4 X e x≈= − , from the material balance around the entire column: ( ) 10 1 NN G X YY X L + = −+ .