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### ChE200 Quiz#1 F07

Course: CHE 200, Fall 2009
School: W. Alabama
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Word Count: 444

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F07 Quiz ChE200 #1 ID#____________________ NAME:____________________________ P# 1: Consider the liquid-liquid extraction shown below, equivalent to one ideal stage. Water is the extraction solvent used to remove the alcohol from the gasohol stream (feed solution of 20 % by volume ethanol and 80 % by volume gasoline). If we wish to recover 90% of the amount of ethanol initially present in the gasohol stream fed...

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F07 Quiz ChE200 #1 ID#____________________ NAME:____________________________ P# 1: Consider the liquid-liquid extraction shown below, equivalent to one ideal stage. Water is the extraction solvent used to remove the alcohol from the gasohol stream (feed solution of 20 % by volume ethanol and 80 % by volume gasoline). If we wish to recover 90% of the amount of ethanol initially present in the gasohol stream fed to the mixer/settler equilibrium stage, how much should be the ratio of liters of water per liter of gasoline? Assume that the extraction solvent (Water) and the diluent (Gasoline) are totally immiscible. The equilibrium curve is given by: Y*E =3XR, where Y*= Liters of Ethanol per 1L of H2O and X= L of ethanol per 1 L of gasoline. Water (MSA) Gasoline Phase Raffinate phase (gasoline and ethanol) R, XR 100 L/min Gasohol 20 L Ethanol 80 L Gasoline MIXER Solution: Water Phase Settler Extract phase ( water and ethanol) E , YE Basis: 100 L/min Gasohol feed, F=80L gasoline We assume that gasoline and water are not soluble in one-another and only ethanol is transferred between phases in the mixer-settler arrangement. Let XF = L of Ethanol per 1L of gasoline =20/80 =0.25, and Y= L of Ethanol per 1L of H2O Ethanol Balance: FXF +W Y = R..XR +E.YE ............(1) Substituting F = R = 80L = W E, and (R..XR / FXF )= 0.1 we have: XR =0.1XF Therefore, XR =0.025 L Ethanol per 1 L of gasoline Based on Equilibrium conditions, Y*= YE =3XR .... .(2) Therefore, YE =3XR =3 x 0.025 = 0.075 L of Ethanol per 1L of H2O Now, (FXF - R..XR ) = ( W YE ) - W Yin . from Equ.(1) and (W/F) = (XF XR)/( YE Y in) = (0.25-0.025)/(0.075 0.0 ) = 3 L of Water per 1 L of Gasol. Using the T-x,y diagram for the binary mixture shown above, answer the following questions: Q1: what is the K value of normal heptane(n-C7) at T = 135 0C and Pt = 101.3 kPa? KA (at T = 135 0C and Pt = 101.3 kPa) = (y*A/xA) = (0.48 /0.18) = 2.667 Q2: What is the relative volatility value AB at T =135 0C? AB at T =135 0C = KA/KB=(y*A/xA) /[(1 - y*A)/( 1 - xA) ]= 4.2 Q3: what is the dew point temperature of a va...

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