- = = We are absorbing n-pentane from a light gas into a heavy oil at 300 kPa and 21°C. The flow rate of the inlet gas i

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- = = We are absorbing n-pentane from a light gas into a heavy oil at 300 kPa and 21°C. The flow rate of the inlet gas i

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We Are Absorbing N Pentane From A Light Gas Into A Heavy Oil At 300 Kpa And 21 C The Flow Rate Of The Inlet Gas I 1
We Are Absorbing N Pentane From A Light Gas Into A Heavy Oil At 300 Kpa And 21 C The Flow Rate Of The Inlet Gas I 1 (96.64 KiB) Viewed 16 times
- = = We are absorbing n-pentane from a light gas into a heavy oil at 300 kPa and 21°C. The flow rate of the inlet gas is Vn+1 = 150 kmol/h and the mole fraction n-pentane in the inlet gas is Yn+1 = 0.003. The inlet solvent flows at Lo = 75 kmol/h and contains no n-pentane, Xo = 0. We want an exit vapor with y1 = 0.0004 mole fraction n-pentane. Use the DePriester chart for equilibrium data. Assume the light gas is insoluble and the heavy oil is non-volatile. a) Find the mole fraction of n-pentane in the outlet liquid, Xn. b) Find the number of equilibrium stages that is sufficient for this separation using McCabe-Thiele method. c) Use a suitable form of Kremser equations to calculate the number of stages required. d) Find the number of equilibrium stages required using McCabe-Thiele method if a Murphree liquid efficiency of 30 % is given.
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