Solved Problems on EFV

1. Solved Problems on EFV Dr. Sahar El-Marsafy Professor – Chemical Engineering

2. 3-Equilibrium Flash Vaporization (EFV) calculation (cont.) Example 2-4: A well fluid having the following composition in mole fraction: C1 = 0.4, C2 = 0.2, C3 = 0.1, C4 = 0.1, C5 = 0.1, C6 = 0.05, C7+ = 0.05, is to undergo a gas-oil separation process using a 3-stage GOSP. The last stage is operated at atmospheric pressure. The first stage is fixed at a pressure of 500 psia. Calculate (P2)o using two different empirical correlations: Solution: Using equations [2-22] and [2-23] 1st calculate the value of Mav as given in the following table:

3. 3-Equilibrium Flash Vaporization (EFV) calculation (cont.)

4. 3-Equilibrium Flash Vaporization (EFV) calculation (cont.) Determine M = Mav / M air = 39.92 / 28.96 = 1.38 Where : Mair = 28.96 2nd Find the value of A from figure (2-16) using Z = 0.4 + 0.20 + 0.1 = 0.7 and M = 1.38. A is found to be = 0.421. .

5. 3rd Since M > 1.0, equation [2-22] applies. By substituting the values of A and P1 in equation [2-22]; (P2)o, the most favorable pressure, is calculated and found to be (P2)o = 50 psia. • Using equation [2-24] Direct substitution for the values of P1 = 500 psia and P3 = 14.7 psia, in this equation yields (P2)o = 85.5 psia

6. Illustrative Example • A feed mixture consisting of 30 mole percent ethane (c2) and 70 mole percent n-butane (n-c4) is subjected to flash distillation in a separator operating at 130oF and 200 Pisa. *- Calculate the B.P. and D.P. temperatures of the feed mixture at the flashing pressure *- Check on the flashing temperature such that TB.P < Tf < TD.P *- Find the value of (V/F) and the concentration of c2 in both the vapor and liquid streams. Solution • Part (a) : To determine T.B.P:- i- As a first assumption, take T.B.P. = 80o F. Hence, the equilibrium constants as determined from the corresponding equilibrium charts are: Kc2 =2.4 and Kn-c4=0.24, ii- Hence Yc2=Kc2 Xc2= (2.4) (0.3) =0.72

7. Example (cont.) iii- Yn-c4=(0.24)(0.7)=0.17 iv- then = 0.89 <1.00 *- Now, we have to assume another temperature greater than 80 F, say 100 F. v- Repeating as above, we find: vi- Yc2 = (2.9) (0.3) = 0.87 Vii- Yn-c4= (0.31) (0.7) = 0.217 Viii- then • For rough estimation, it is considered good enough to select 100oF as the Bubble point temperature.

8. Example (cont.) • To determine T.D.P. i- As a first assumption, take T.D.P= 160 oF . Hence, the equilibrium constants as determined from Charts are: Kc2 =4.2 and Kn-c4=0.66 ii- Hence Xc2 = Yc2 / Kc2 = XFc2 / Kc2 = 0.3 / 4.2 = 0.071 Iii- and Xn-c4 = 0.7 / 0.66 = 1.060 iv- Then = 1.131 >1.00 *-Now , we have to assume another temperature greater than 160oF, say 170oF. V- Repeating as above, we find: Vi- Xc2 = 0.3/4.4 =0.07 , Xn-c4=0.7/0.72=0.97 Vii- then =1.04 • For rough estimation, it is considered good enough to select 170 F as the Dew point temperature. • Thus the following inequality holds: 100<130<170

9. Example (cont.) • Part (b)Flashing calculations: i-Assume V/F=0.5 for the 1st trial, using Flashing Equation get: • Since is greater than 1, Assume a lower value for V/F, say 0.3 get :