Energy Engineering - Chemical Processes and Technologies

Full exam

1 Chemical Processes and Technologies (Ing. Giorgia De Guido) Exam June 23 rd, 2021 – Exercise: Solution Exercise no. 1 (of 3) A saturated liquid mixture, having the flow rate and the composition reported in the Data section below, is available at 1 atm and has to be separated in a flash drum operated at atmospheric pressure and 280 K. Please, determine: 1. the vaporization ratio, α; 2. the composition (mole fractions) of the vapour and liquid products; 3. the thermal duty [kW] to be supplied to the flash drum. Data: Molar flow rate of the feed stream = 6 (8 36) 100 2 M⋅− + [kmol/h] M S = 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 z F [mol/mol] A B C T NBP [K] T C [K] ΔH ev (@T NBP ) [kcal/kmolz [kcal/(kmol∙K)] [kcal/(kmol∙K)] propane 0.1500 15.7260 1872.46 -25.16 231.1 369.90 4491 15.06 28.67 0 n-butane 0.5000 15.6782 2154.90 -34.42 272.7 425.20 5351 21.99 34.14 0 n-pentane 0.3500 15.8333 2477.07 -39.94 309.2 469.60 6160 31.02 37.614 Results The results are reported in the following assuming M 6 = 0. Therefore: • molar flow rate of the feed stream = 82 [kmol/h]. Point no. 1 Using the Rachford-Rice equation: α = 0.485365. Dipartimento di Chimica, Materiali e Ingegneria Chimica “Giulio Natta” POLITECNICO di MILANO 2 Point no. 2 Then, the following composition (mole fractions) of the vapour and liquid products is obtained. Component x y [-] [-] propane 0.045537 0.260763 n-butane 0.434579 0.569366 n-pentane 0.519884 0.169871 Point no. 3 By writing the energy (enthalpy) balance, it is possible to write the following expression for the thermal duty to be supplied to the flash drum: ( ) ( ) ( ) ( ) ( ) ( ) @@ @ @ (1 ) @@ VL L flashflashfeed VLL flashflashfeed QVHT LHT FHT FH TLH T FH T αα = ⋅ +⋅ − ⋅ = ⋅ ⋅ + − ⋅⋅ − ⋅      =====(1)= where T feed is the bubble point temperature at the feed pressure (1 atm) of the feed stream, equal to 265.8314 [K]. If Eq. (1) is solved assuming as reference state for the enthalpies the one of a liquid at T feed , Eq. (1) reduces to: ( ) ( ) @ (1 ) @ VL flashflash Q F H TLH T αα= ⋅ ⋅ + − ⋅⋅  = where: = ( ) ,,, , , , 1 @(@ ) ( ) (@ ) ( ) NC VLLV flashi i feed P i NBP i feed ev i NBP i P i flash NBP i i H T y h T cT T H T cT T =  = ⋅ + − +∆+ −  ∑    = = ( ) , 1 @(@ ) ( ) NC LLL flashi i feed P i flash feed i H T xh T cT T =  =⋅ +−  ∑    = where (@ )Li feedhT == 0 [kcal/kmol] due to the chosen reference state.= As a result:= Q = 281.4183 [kW] (using the conversion factor 1 kcal = 4.186 kJ). 3 Exercise no. 2 (of 3) A wastewater stream containing ammonia (NH 3) is treated in a stripping trayed tower using air. The column is operated at 25 °C and atmospheric pressure. The NH 3 content in the wastewater stream is reported in the Data section below. The air stream, which does not contain any ammonia, is fed to the tower bottom with a flow rate of 20 [Nm 3AIR /kmol WATER ]. Please, determine: 4. whether the system can be treated as a dilute system or not, and explain why; 5. which number of theoretical stages is needed to lower the content of ammonia in the wastewater stream to 5 ppm (on a molar basis). Data: NH 3 content in the inlet wastewater stream = 5 1000 (30 135)M +⋅− =[ppm=on=a molar basis]= M 5== fifth digit of student number from the left=(Student number/matricola = XXXXXX) KNH3 = 1.41 at 25°C, where K = y/x Results The results are reported in the following assuming M 5 = 9. Therefore: • NH 3 content in the inlet wastewater stream = 1135 [ppm on a molar basis]. Point no. 4 The mole fractions and ratios are computed for the transferred component (i.e., ammonia), respectively in the inlet and outlet wastewater streams and in the inlet air stream. Stream Mole fraction Mole ratio Absolute Difference Inlet wastewater (x N+1 , X N+1 ) 1.13500E-03 1.1363E-03 1.29E-06 Outlet wastewater (x 1, X 1) 5.00E-06 5.00E-06 2.5E-11 Inlet air (y 0, Y 0) 0 0 0 Since x ≈ X and y ≈ Y, the system can be considered a dilute system ( linear cascade). From now on, it will be assumed that: X N+1 = x N+1 = 1.13500E-03; X 1 = x 1 = 5.00E-06; Y 0 = y 0 = 0. Point no. 5 Assuming as basis of design a unitary value for L (solute-free liquid molar flow rate), the volumetric flow rate of the solute-free gas ( GV ) is 20 Nm 3 of air/unit of time. The corresponding molar flow rate (G) can be obtained as: • first approach: normal G normal PV G RT ⋅ = ⋅  ; 4 • second approach: GV G v =   , where 22.414 [ / ] v l mol=  = is the=molar volume of an ideal gas=at normal conditions (i.e., at P normal = 101325 Pa and T normal = 273.15 K). So, the absorption factor, A = L/(G∙K NH3 ), is equal to 0.7948 and the number of theoretical stages can be determined using the Kremser equation for design problems (where X 0eq = x 0eq = y 0 / K NH3 = 0)  N = 17 (by rounding the fractional number N = 16.8). 5 Exercise no. 3 (of 3) The debutanizer fractionator, typically referred to just as a “debutanizer”, is a distillation column used to separate butane during the refining process. Consider a debutanizer equipped with a total condenser and a partial reboiler, and operated at 13 bar. The feed stream is saturated liquid at the operating pressure of the column, and enters the debutanizer at a flowrate of 100 kmol/h. Its composition (mol%) is reported in the Data section below. The following specifications have to be met: - recovery of n-butane in the distillate of 99.9%; - recovery of n-pentane in the bottoms of 99%. Determine: 6. the temperature [K] of the feed stream; 7. the mole fraction of each component in the two products, namely the distillate and the bottoms; 8. the minimum number of stages for performing the desired separation, specifying the value used for the average relative volatility of the light key component. Data: The vapour pressure is calculated with the Antoine equation: ln( [ ]) [] ev B P mmHg A TK C = − + . Component zF= [mol %]=A =B=C= TNBP = [K]= TC= [K]= n-butane= 4 29 35 2 M⋅− + 15.6782 2154.90 -34.42 272.7 425.20 n-pentane 4 29 35 2 M⋅− − 15.8333 2477.07 -39.94 309.2 469.6 n-hexane 30 15.8366 2697.55 -48.78 341.9 507.4 M 4 = fourth digit of student number from the left (Student number/matricola = XXXXXX) Results The results are reported in the following assuming M 4 = 5. Therefore, the feed stream has the following composition: Component zF= [mol %]= n-butane= 35.5 n-pentane 34.5 n-hexane 30.0 Point no. 6 The temperature of the feed stream is its bubble point temperature at 13 bar (i.e., the operating pressure of the distillation column)  T feed = T b,F = 398.4909 [K]. 6 Point no. 7 Material balances can be solved using the two recovery specifications for n-butane (lk component) and n-pentane (hk component) assuming that n-hexane (heavy nonkey) is recovered in the bottoms only. The mole fraction of each component in the two products, namely the distillate and the bottoms, is reported in the following table. Component Mole fraction in the distillate= Mole fraction in the bottoms= n-butane= 0.990366= 0.000553= n-pentane= 0.009634= 0.532088= n-hexane= 0.000000= 0.467359= Point no. 8 The Fenske’s equation is used to determine N min . To use it, it is necessary to calculate 3 ,, , lk lk top lk bottom lk feedα αα α =⋅⋅ , with the three relative (relative to the=hk component) volatilities calculated at the following three temperatures (also the thermal conditions of the feed stream are considered since it is at its bubble point): • dew point temperature of the distillate product: T d,D = 367.0587 [K]; • bubble point temperature of the bottom product: T b,B = 428.6155 [K]; • bubble point temperature of the feed stream: T b,F = 398.4909 [K] (as also reported above). Thus, the average relative volatility of the lk component 3 ,, , lk lk top lk bottom lk feedα αα α =⋅⋅ == 2.3203== and = = = Nmin = 13.67  N min = 14.