Energy Engineering - Chemical Processes and Technologies

Full exam

1 Chemical Processes and Technologies (Ing. Giorgia De Guido) Exam July 10 th, 2020 - Exercise: Solution Exercise no. 1 Consider a liquid mixture composed of propane, n-butane, n-pentane and n-hexane, which is at a temperature of 340 K and at a pressure of 25 bar. The mass flow rate of each component in it is reported in the data section. This mixture is depressurized to 15 bar and fed to a flash drum in order to recover 95% of propane in the vapour phase. 1. At which temperature [°C] does the flash drum have to be operated? 2. Which are the molar flow rates [kmol/h] of the vapour and liquid outlet streams? 3. Which thermal power [kW] has to be supplied to/removed from the flash drum? Data Component Mass flow rate Molecular weight A B C im[kg/h] MW i [kg/kmol] propane 1808 + β 44.10 15.7260 1872.46 -25.16 n-butane 1163 58.12 15.6782 2154.90 -34.42 n-pentane 722 72.15 15.8333 2477.07 -39.94 n-hexane 2500 86.17 15.8366 2697.55 -48.78 β = 5 30 ( 4.5) M ⋅− [kg/h] M 5 = fifth digit of student number from the left (Student number/matricola = XXXXXX) The vapour pressure is calculated with the Antoine equation: ln( ) ev B PA TC = − + , with T in [K] and P ev in [mmHg] Component TNBP ΔH ev (@T NBP ) Tc [K] [kcal/kmol] [kcal/(kmol K)] [kcal/(kmol K)] [K] propane 231.1 4491 15.06 28.670 369.9 n-butane 272.7 5351 21.99 34.140 425.2 n-pentane 309.2 6160 31.02 37.614 469.6 n-hexane 341.9 6896 38.86 51.700 507.4 2 Results The results are reported in the following assuming M 5 = 0. Therefore: • mass flow rate of propane in the liquid feed mixture = 1673 [kg/h]. Point no. 1 The molar flow rate of each component in the liquid mixture, , () Fi Fz⋅ , can be computed as:= , () i Fi im Fz MW ⋅=  = and:= , , , 1 () () Fi FiNC Fi i Fz z Fz = ⋅ = ⋅ ∑ = Component zF,i= [-]= propane = 0.391234 n-butane 0.206364 n-pentane 0.103200 n-hexane 0.299201 To calculate T, since P and the recovery of propane (i.e., component r) in the vapour phase are given, it is possible to use the following equation: ( ) ( ) , 1 1 ()0 NC Fj j r j jr r r zK fT Vy KK K Fz = − == −+ ∑ = T FLASH = 132.2547 [°C]. Point no. 2 From the recovery of propane in the vapour stream (RRV r), it is possible to calculate the split factor for it (SF r). Since the T has been determined at the previous point, the relative volatility of each component with respect to propane, α i,r, can be determined and, thus, the split factor of each ith component, SF i. Then, the recovery of each ith component in both the vapour and liquid phases (respectively, RRV i and RRL i) can be calculated, so that their molar flow rates in the two phases (respectively, V·y i and L·x i) are computed. Therefore: ( ) 1 NC i i Vy = = ⋅∑ V = 78.8538 [kmol/h], ( ) 1 NC i i Lx = = ⋅∑ L = 18.1124 [kmol/h]. 3 Point no. 3 From an energy (enthalpy) balance on the flash drum (considering a control volume that includes the feed stream before expansion as inlet material stream and the vapour and liquid products as outlet material streams), it is possible to write: (@ )(@ ) (@ )LVL FEED FLASHFLASHFLASH FHT Q VHT LHT⋅ + = ⋅ +⋅    == If we assume liquid @T FEED as reference state for the enthalpies (so that ()L FEED HT = =0=[kcal/kmol]),= with the given data we can write:= ,,, , ,, 1 , 1 6037.5920 [ (@ ) ( ) ( ) (@ /] 2956.0 )( ) (@ ) ( ) () 909 [ / ] NC VLLV FLASHi i FEED P i NBP i FEED ev i NBP i P i FLASH NBP i i NC LLL FLASHi i FEED P i FLASH FEED i kcal kmol kcal kmol H T yhT cT T H T cT T H T xhT c T T = =  =+ − +∆+ − =  =  = + −=  ∑ ∑       = Therefore:= (@ ) (@ )VL FLASHFLASHFLASH Q V H T LH T= ⋅ +⋅   = 615.8407 [kW] (it is supplied to the flash drum). 4 Exercise no. 2 The depropanizer is a distillation column that is used in the natural gas industry to isolate propane from a mixture containing butane and other heavy hydrocarbons. Consider a mixture composed of propane, n-butane, n-pentane and n-hexane, which enters the plate distillation column as a saturated liquid at a flow rate of 100 kmol/h. The column has a total condenser and a partial reboiler, and is operated at 17 bar. The composition of the feed stream (in terms of molar fractions, z Fi) is reported in the data section. The aim is to recover 0.7% of n-butane in the distillate product and a percentage of propane in the bottom product as reported in the data section. 4. Specify which are the key components, determine their molar fraction in the top and bottom product streams and the total molar flow rate, D and B [kmol/h], of these streams. 5. Calculate the minimum number of equilibrium stages (N min ) and the actual number of equilibrium stages (N), assuming that: i) the average relative volatility of the light-key component with respect to the heavy-key component is 2.9, ii) the minimum reflux ratio, R min , is 0.8, and iii) the actual reflux ratio (R) is 1.3∙R min . Data Recovery of propane in the bottom product = 4 2 ( 4.5) 1.5 9 M⋅− + [%]= M 4== fourth digit of student number from the left (Student number/matricola== XXuXXX) The vapour pressure is calculated with the Antoine equation: ln( ) ev B PA TC = − + , with=T in [K] and P ev in [mmHg] Component zc,i= A=B=C= TNBP = x-]= [K] = propane = 0.41 15.7260 1872.46 -25.16 231.1 n-butane 0.20 15.6782 2154.90 -34.42 272.7 n-pentane 0.10 15.8333 2477.07 -39.94 309.2 n-hexane 0.29 15.8366 2697.55 -48.78 341.9 Results Results are reported in the following assuming M 4 = 0, therefore: • recovery of propane in the bottom product = 0.5 [%]. 5 Point no. 4 Considering the given recovery specifications for propane and n-butane, with propane more volatile than n-butane, it is possible to infer that propane is the light-key component and n-butane is the heavy-key one. From the material balances on each component, it is possible to obtain their molar flow rates in the top and bottom product streams and, then, the total molar flow rates D and B and their composition (in particular, the molar fractions of the two key components in such streams). Component D∙x D,i [kmol/h] B∙x B,i [kmol/h] xD,i x-] xB,i x-] propane 40.7950 0.2050 0.996580 0.003471 n-butane 0.1400 19.8600 0.003420 0.336240 n-pentane 0.0000 10.0000 0.000000 0.169305 n-hexane 0.0000 29.0000 0.000000 0.490985 D = 40.9350 [kmol/h], B = 59.0650 [kmol/h]. Point no. 5 From Fenske’s equation, it is possible to calculate the minimum number of equilibrium stages ( lkα =2.9): ,, ,, min ln ln lkD hkB lkB hkD lk Dx BxBx Dx N α ⋅    = = 9.6253. The actual number of equilibrium stages can be calculated by using the Gilliland graphical correlation, or alternatively: • the Eduljee equation: φ(N) = 0.5270  N = 21.46  N = 22; • the Molokanov equation: φ(N) = 0.5360  N = 21.90  N = 22. Thus, in both cases (namely, either using the Eduljee equation or the Molokanov equation) the actual number of equilibrium stages required to perform the desired separation is 22.