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HW_6_Solution

# HW_6_Solution - Problem 3-2 Pressure(P = 14.7psia...

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Problem 3-2 Pressure (P) = 14.7psia Density (ρ) = 0.103 lb/ft3 Therefore, the molecular weight of the compound is: M = 12n + 2n = 14n (1) Applying the ideal gas law, Therefore, Applying the values of ρ, R, T and P, from equation (1) M = 14n = 42.102 Therefore, n = 3.0 Temperature(T) = 100 o F = 560 o R R = 10.73 psia-ft 3 /lbmol- o R The general formula for this compound is C n H 2n From the general formula for this compound ,C n H 2n , The compound is C 3 H 6 à Propene RT M m nRT PV * = = P RT RT PV m M ρ = = * 42.102 7 . 14 560 * 73 . 10 * 103 . 0 = = M

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Problem 3-21 Component Mole Fractions Critical Pressure Critical Temperature Molecular weight Methane 0.8700 666.4 579.8 -116.7 343.3 298.7 16.04 13.96 Ethane 0.0490 706.5 34.6 89.9 549.9 26.9 30.07 1.47 Propane 0.0360 616.0 22.2 206.1 666.1 24.0 44.10 1.59 i-Butane 0.0250 527.9 13.2 274.5 734.5 18.4 58.12 1.45 n-Butane 0.0200 550.6 11.0 305.6 765.6 15.3 58.12 1.16 1.0000 660.8 383.3 19.63 1.67 7.04 From Fig 3-7, Z 0.96 For P= 4650 Psia and T=(180+460) o R y i P ci , psia y i * P ci T c , o F T c , o R y i * T ci P pc T pc T pr = T / T pc P pr = P / P pc i M i i M y ZRT PM V m density ZRT M m PV nZRT PV = = = = ft lb/cu 84 . 13 640 * 10.73 * 0.96 19.63 * 4650 density = =
Problem 3-23 0.932 Z 580 * 10.73 * 0.0862 0.5 * 1000 nRT PV Z is pressure at this factor - Z the Therefore moles - lbm 0.0862 0.0923 - 0.1785 moles of

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HW_6_Solution - Problem 3-2 Pressure(P = 14.7psia...

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