Homework+Solutions+Ch.+12

Homework+Solutions+Ch.+12 - PETE 2032 Homework Solutions...

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Unformatted text preview: PETE 2032 Homework Solutions Chapter 12 12-9 Six pound moles of ethane, three lb-moles of propane and one lb-mole of n- butane are mixed in a closed container and the temperature is adjusted to 75 o F. What is the bubble-point pressure? What is the dew-point pressure? Calculate the compositions of the gas and liquid at equilibrium at 75 o F and 300 psia. You may assume that the mixture behaves like an ideal solution. Component Composition mole fraction z j Vapor pressure at 75 o F p vj z j p vj z j /p vj C 2 0.6 610 366 0.00098 C 3 0.3 135 40.5 0.00222 n-C 4 0.1 35 3.5 0.00286 1.00 P b =410 psia 0.00606 p d = 1/0.00606 = 165 psia Equation 12-8. This is a trial-and-error solution; only the calculation with the final trial value of n g = 0.532 will be shown. Component Composition mole fraction z j Vapor pressure at 75 o F p vj Eq 12-8 x j Eq. 12-2 y j = x j p vj /p C 2 0.6 610 0.387 0.787 C 3 0.3 135 0.424 0.191 n-C 4 0.1 35 0.189 0.022 1.00 1.000 1.000 Since x j = 1.000, the trial value of n= 1....
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Homework+Solutions+Ch.+12 - PETE 2032 Homework Solutions...

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