Example 8 calculate the pressure drop per mile in an

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Example 8 Calculate the pressure drop per mile in an NPS 16 pipeline, 0.250 inch wall thickness, flowing crude oil (Sg = 0.895 and viscosity = 15.0 cSt) at 100,000 bbl/day. Assume pipe absolute roughness = 0.002 inch. Compare results using Colebrook-White, Miller and MIT equations. 8
Solution Pipe inside diameter D = 16.0 – 2 x 0.250 = 15.50 in. Next, calculate the Reynolds number R = 92.24 x 100000/(15.0 x 15.50) = 39,673 Since the flow is turbulent we use the Colebrook-White equation in the first case to determine the friction factor. The relative roughness = (e/D) = 0.002/15.5 = 0.000129 1/ f = -2 Log10[(0.000129/3.7) + 2.51/(39673 Solving for f by trial and error we get f = 0.0224 The pressure drop due to friction using Colebrook-White equation is Pm= 0.0605 x 0.0224 x (100000)2(0.895/(15.5)5) =13.56 psi/mi For the Miller equation we assume Pm= 14 psi/mi as the first approximation. Then calculating the parameter M M = Log10((15.5)3 0.895 x 14 /(15 x 0.895)2) + 4.35 where viscosity in cP = 15 x 0.895 = 13.425 Therefore, M = 6.7631 Next calculate the value of Pmfrom Equation 2.27 100000 = 4.06 x 6.7631 x [(15.5)5Pm/0.895]0.5Solving for Pm, we get Pm= 13.27 psi/mi, compared to the assumed value of 14.0 Recalculating M from this latest value of Pm, we get M = Log10((15.5)3 0.895 x 13.27 /(15 x 0.895)2) + 4.35 Therefore, M = 6.74 Next calculate the value of Pmfrom Equation 2.27
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100000 = 4.06 x 6.74 x [(15.5)5Pm/0.895]0.5Solving for Pm, we get Pm= 13.36 psi/mi Repeating the iteration once more, we get the final value of the pressure drop with Miller Equation as Pm= 13.35 psi/mi Next we calculate the pressure drop using MIT equation. The modified Reynolds number is Rm = R/7742 = 39673/7742 = 5.1244 Since R>4,000, the flow is turbulent and we calculate the MIT friction factor from Equation 2.34 as f = 0.0018 + 0.00662(1/5.1244)0.355= 0.0055 The pressure drop due to friction is calculated from Equation 2.35 Pm= 0.241 x 0.0055 x 0.895 x (100000)2/(15.5)5 = 13.32 psi/mi Therefore, in summary, the pressure drop per mile using the three equations are as follows: Pm= 13.56 psi/mi using Colebrook-White equation Pm= 13.35 psi/mi using Miller equation Pm= 13.32 psi/mi using MIT equation It is seen that in this case all three pressure drop equations yield approximately the same value for psi/mi, the Colebrook-White equation being the most conservative (highest pressure drop for given flow rate).
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Using the previously discussed equations such as Colebrook-White, Hazen-Williams, etc. we can easily calculate the frictional pressure drop in a straight piece of pipe. Many appurtenances such as valves and fittings installed in pipelines also contribute to pressure loss. Compared to several thousand feet (or miles) of pipe, the Pressure Losses throughfittings and valvesare small. Therefore, such pressure drops through valves, fitting and other appurtenances are referred to as minor losses. These pressure drops may be calculated in a couple of different ways. Using the equivalent lengthconcept, the valve or fitting is said to have the same frictional pressure drop as that of a certain length of straight pipe. Once the equivalent length of the device is known, the pressure drop in that straight length of pipe can be calculated. For example, a gate valve is said to have an equivalent length to the diameter ratio of 8. This

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