homework6-solutions(1) - Homework 6 Solutions Problem...

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Unformatted text preview: Homework 6 Solutions Problem 3 ­3 Component Moles MW Mass (g) Weight Frac Vol Frac Methane 0.870 16.040 13.955 0.711 0.870 Ethane 0.049 30.070 1.473 0.075 0.049 Propane 0.036 44.100 1.588 0.081 0.036 i ­Butane 0.025 58.120 1.453 0.074 0.025 n ­Butane 0.020 58.120 1.162 0.059 0.020 Problem 3 ­7 The key to solving this problem is to realize that the second container is initially evacuated, and thus the number of moles of gas remains constant throughout the experiment. !! = !! Substitute for the ideal gas law: !! !! 5 − 1.3 + 1.3! = 5 + 5 − 1.3 + 1.3! !" !" Cancel RT, 760!!!" ∙ 5 − 1.3 + 1.3! = 334.7!!!" ∙ 5 + 5 − 1.3 + 1.3! Solving for porosity: ! = !. !"#$$% Problem 3 ­9 This problem is a simple application of the ideal gas law. Our strategy will be to calculate the initial moles of gas, then add the additional moles of gas when the dry ice sublimes. !" 14.7!"# ∙ (40!! ! − 0.157!! ! ) !! = = = 0.101!"#$!"# !! ! ∙ !"# !" 10.73159 ∙ 80 + 460 !"#$! ∙ °! The moles contained in the 15 lb block of dry ice is computed as follows: 15!" !! !! = = 0.341!"#$! !! 44.01!" !"#$! Homework 6 Solutions Thus, the final number of moles is calculated: !! = !! + !! !! = 0.101!"#$!"# + 0.341!"#$! !! = 0.442!"#$ Compute the final temperature using the ideal gas law: !!! ∙ !"# !. !!"!"#$ ∙ !". !"#$% !"#$! ∙ °! ∙ !" + !"# !"# != = = !". !"!"# ! !"!!! Problem 3 ­11 Our strategy for this problem will be to find the initial moles of gas contained in the container (the gas above the fluid), then we will determine the change in volume required to reduce the pressure inside the tank by the threshold crush pressure. Using the flow rate given in the problem, it will then be possible to determine the time until the tank is crushed. Assume that the gas is ideal, and that the pressure inside and outside the tank are 30 in Hg initially. !" 14.7329334!"# ∙ ! ∙ 40! ∙ 10 740558 !! = = = !" !" !" Now, we require a pressure differential of 1 oz/sq in. to crush the tank. This is the same as 0.0625 psi. So the pressure inside the tank will have to be reduced to the crush pressure: !!"#!! = 14.7329334 − 0.0625 = 14.6704334!"# Let’s calculate the new volume: 740558 !! !" !" ∙ !" = 50479.627!! ! !!"#$! = = !!"#$! 14.6704334!"# Let’s find out the change in volume: Δ! = !! − !! = 50265.4825 − 50479.627!! ! = 214.14!! ! Convert flow rate: !!" !! ! 15000 = 58.49 !"# !"# Homework 6 Solutions Determine the time required to remove the necessary volume: !"#. !"!!! !!"#$% = = !. !!!"#$%&' !!! !". !" !"# The force applied to the roof is the calculated as follows: ! ! = ! ! = ! ∙ ! = !. !"#$!"# ∙ !"#$"". !"#!!! = !"#$%. !"!!! Note that the pressure used is the pressure difference between the inside of the tank, and the outside (crush pressure). Also note the considerable amount of force on the roof of the tank for such a small delta ­P. If more oil was in the tank before collapse, it would serve to reduce the time required to crush the tank. The more air that is in the tank, the more volume change is required to drop the pressure. Problem 3 ­19 Component Partial Pressure Mole Frac MW Mole Frac*MW Methane 17.8 0.89 16.040 14.2756 Ethane 1 0.05 30.070 1.5035 Propane 0.4 0.02 44.100 0.882 i ­Butane 0.2 0.01 58.120 0.5812 n ­Butane 0.6 0.03 58.120 1.7436 SUM 20 18.9859 The specific gravity of the gas is computed with air as a reference: !!""!#$%& !". !"#! !" = = = !. !"" !" !" Homework 6 Solutions Problem 3 ­22 The following table is constructed using the real gas law: !" = !"#$ Volume for each container was arbitrarily assumed to be 50 cu. Ft. (the value you choose is irrelevant as long as each tank has the same volume). Z ­values are determined using charts beginning on page 106. Component Moles yi Tc (°!) YiTc Pc (psia) yiPc Methane 8.26 0.63 343.30 215.24 666.40 417.82 Ethane 4.91 0.37 549.90 205.12 706.50 263.54 SUM 13.18 420.37 681.36 The final pressure is then determined as follows: !! ! ∙ !"# !! !! !" !! 8.26 + 4.91!"#$% ∙ 10.73159 !"#$! ∙ °! ∙ (140 + 460) !! = = !! 100!! ! = !! ∙ 848.37 We do not know zf as this is a mixture and not a pure component. We know that the pseudo ­reduced temperature is calculated as follows: ! 140 + 460 !!" = = = 1.43 !"# 420.37 + 460 Pseudo ­reduced pressure is then calculated: !! ∙ 848.37 ! !!" = = = 1.24511 ∙ !! !!" 681.36 At this Tpr, only one Ppr will match a certain zf according to the above equation. Thus z is found by trial and error on Figure 3 ­7 on page 112. !! = 0.865 Finally, the final pressure is calculated: !! = !!" ∙ ! = !. !"# ∙ !"!. !" ∙ !!". !" = !"". !"!"# Partial Pressures can then be calculated, shown below: Homework 6 Solutions !! = !!! Component Partial Pressure (psia) Methane 460.11 Ethane 273.73 Problem 3 ­30 Using Fig. 3 ­11 on page 119, determine the pseudo ­critical properties of this gas: !!" = 664!"#$ !!" = 387°! Calculate Pseudo ­reduced properties based on the given reservoir conditions: 445!"#$ !!" = = 0.685 664!"#$ 90 + 460°! !!" = = 1.42 387°! Using Fig. 3 ­7 on page 112, determine the z factor: ! = 0.91 Use the real gas law to compute the molar volume: !!! ∙ !"# !. !" ∙ !". !"#$% !"#$! ∙ °! ∙ !" + !"#°! ! !"# !!! = = = !!. !" ! ! !""!"#$ !"# Homework 6 Solutions Problem 3 ­32 Create the table below by looking up the critical temperature and pressure values for each of the pure components. Component Mole Percent Mole Frac (yi) Tc Pc yiTc yiPc Hydrogen Sulfide 8.67 0.09 672.40 1300.00 58.30 112.71 Carbon Dioxide 1.40 0.01 547.90 1071.00 7.67 14.99 Nitrogen 0.45 0.00 227.50 493.10 1.02 2.22 Methane 79.81 0.80 343.30 666.40 273.99 531.85 Ethane 5.28 0.05 549.90 706.50 29.03 37.30 Propane 1.71 0.02 666.10 616.00 11.39 10.53 i ­Butane 0.55 0.01 734.50 527.00 4.04 2.90 n ­Butane 0.53 0.01 765.60 550.00 4.06 2.92 i ­Pentane 0.24 0.00 829.10 490.00 1.99 1.18 n ­Pentane 0.19 0.00 645.80 488.60 1.23 0.93 Hexanes 0.24 0.00 913.60 436.90 2.19 1.05 Heptanes+ 0.93 0.01 1090.00 365.00 10.14 3.39 SUM 100.00 1.00 405.05 721.97 The critical values for the Heptanes+ are found in Fig. 3 ­10 on page 116. Now, we must adjust our pseudo ­critical properties for the presence of non ­ hydrocarbon gasses (CO2 and H2S). Use Fig 3 ­12 on page 122: ! !!" = 405.05 − ! = 405.5 − 17 = 388.05°! ! !!" !!" 721.97!"#$ ∙ 388.05°! ! !!" = = = 689.381 !!" + !!! ! 1 − !!! ! ! 405.05°! + 0.0867 1 − 0.0867 ∙ 17 To find the z ­factor, simply compute the pseudo ­reduced properties: ! 5709!"#$ !!" = ! = = 8.28 !!" 689.381!"#$ 293 + 460°! !!" = = 1.94 388.05°! From Fig 3 ­7 on page 112: ! = !. !"# ...
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