HW #7 Solution!

# HW #7 Solution! - Problem 4.1 Consider a high-pressure...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 4.1 Consider a high-pressure wellbore with a casing and tubing of dC = 0.23m and dT = 0.051m ID, respectively, as shown schematically in Fig. 4.28. The production from the perforation initially flows through the casing until it is diverted by a packer into the tubing. At this location, the densities of the oil and gas are 0.1m/s 1) Determine the existing flow pattern in the casing and in the tubing. 2) Run FLOPATN computer code (see Sec. 7.6) to generate the flow-pattern maps for the casing and the tubing, and show the operational points on the maps dT = 800kg/m3 and = 100kg/m3; the corresponding viscosities are µL = 5 x10-3kg/ms and µG = 2 x 10-5 kg/ms; and the surface tension is σ = 0.03N/m. the superficial velocities in the casing are vSL(c) = 0.1m/s and vSG(c) = dC Solution Casing: Checking Bubble-Slug transition - E . 1.53 0.205 0.1 0 1.53 0.1 / 9.81 / 0.1 / 1 0.289 700 / 800 / 0.205 0.25 , 0.03 / . 0.205 / 0.1 0.1 0.1 1 0.205 0.405 0.205 Checking Dispersed Bubble-Slug transition – F-G 0.2 / 1 2 2 800 1 0.005 . 0.289 597.7 0.23 . 100 0.2 0.289 1.039 597.7 / / 0.725 4.15√ 0.2 597.7 1.039 Thus 0.725 0.000348 4.15√0.289 0.03 800 . 0.000348 . 0.158 2 0.4 . 2 0.4 0.03 9.81 800 100 . 0.0026 0.158 , 0.0026 Checking Annular Flow transition J Solve for vG in the transition boundary . 3.1 . 3.1 9.81 / 700 / 100 / 1.17 0.03 / . . 1.17 / 0.1 / Flow is not annular, thus in the Casing Flow Pattern is Slug Flow Tubing: 0.23 4 0.051 2.03 / 4 Similarly; 4 Checking Annular Flow transition J Solve for vG in the transition boundary . 0.23 4 0.051 2.03 / 3.1 . 3.1 9.81 / 700 / 100 / 1.17 0.03 / . . 1.17 / 2.03 This implies that bubbles have enough velocity to maintain Annular Flow in the Tubing Flow pattern map for Casing showing operating points Flow pattern map for Tubing showing operating points Problem 4.2 Air and Water flow upward in a vertical 5-cm-ID pipe. When the flow rates are vSL = 0.02m/s and vSG = 0.3m/s, the observed flow pattern is slug flow. The fluid physical properties are as follows: 1000kg/ m3 µL = 0.7 x 10-3kg/ms σ = 0.075N/m Using the simplified Sylvester (1987) model, Calculate: 1. 2. vTB and vGTB: the Taylor-bubble-rise velocity and the gas-phase velocity in the Taylor-bubble, respectively. vLLS, vGLS, and vLTB : the liquid and gas velocities in the slug body and the liquid-film velocity around the Taylor-bubble respectively 3. 4. 5. Solution 0.3 / 1000 0.02 / 0.32 / 22,857.14 2.48 1 10 1.53 1.2 0.32 0.35√9.81 0.7065 . . = 1.2kg/m3 µG = 1.9 x 10-5 kg/ms LS, LF, LU, δL : the slug-length distribution and the film thickness around the Taylor-bubble. HLSU : the average liquid holdup in the slug unit –dp/dL : the total pressure gradient across the slug unit. 0.32 0.05 0.0007 0.35 10 1.2 0.05 0.629 / 0.466 0.54 1.0 Ã 1 7.85 0.0567 1 0.466 0.629 9.916 0.466 0.32 1 0.089 1 1 . 0.466 1.53 9.81 1000 1.2 0.075 1000 1 Ã 0 . 0.466 . 0.442 By Iteration: 9.916 1 1 . 9.916 9.81 0.05 1 √1 0.089 . 1.482 / 0.629 1.2 0.629 1.53 1.482 0.089 0.466 . 0.226 / . 1.2 0.32 9.81 1000 1.2 0.075 1.53 1000 . 0.466 . 0.556 / 1 0.629 1 0.629 0.556 1 1 1 0.466 0.089 0.586 / 1 1 0.226 0.466 0.02 1.482 0.089 0.226 0.466 0.36 / 20 1 20 1 0.05 1 0.36 1.562 1.0 1.562 1 0.562 √1 0.089 1.14 10 2 1 1 0.05 1 2 1 0.629 0.466 0.556 1 0.629 0.466 0.3 0.461 To accelerate the computation of the pressure drop, we will neglect the pressure drop of the Taylor-bubble/liquid film and use the first two terms in the Taitel and Barnea model Thus ∆ 1000 4 0.00196 0.32 / 0.466 1.2 1 0.466 466.641 / (a) Computing the value of the friction factor 1000 0.32 0.05 0.0007 . 22,857.14 0.00618 0.2, 0.046 . 0.046 22,857.14 2 Thus ∆ Hence, ∆ 0.00618 466.641 0.32 2 0.148 466.64 9.81 1 1 0.148 0.05 0.00196 4589.57 4589.57 1.562 2938.27 / (b) Using a constant Friction factor f =0.005 466.641 0.32 2 2 Thus ∆ Hence, ∆ 0.005 0.1195 466.64 9.81 1 1 0.1195 0.05 0.00196 4587.32 4587.32 1.562 2936.82 / Difference is not very significant ~ about 0.05% for both friction factors Problem 4.3 The following data have been obtained from a gas well in the Alagoas basin, Brazil. The well is vertical, with an ID of d = 0.062m and roughness of ε = 0.00018m. The production rates are qG = 188000 STD m3/D with a specific gravity of γG=0.65, and q0 = 8.5 STD m3/D, expected at standard conditions (101.3kPa and 25OC). At one location in the wellbore, the following fluid properties are measured: 800kg/ m3 = 120kg/m3 µG = 1.5 x 10-5 kg/ms µ0 = 0.3 x 10-3kg/ms σ = 0.025N/m Using the dimensionless graphs developed by Alves et al. (1991), determine the total pressure gradient at the given location. Assume incompressible liquid-phase and no mass transfer between the phases. Solution 4 , 0.00301 1 1 1 1 188000 2.175 24 3600 24 3600 1 1 1 1 8.5 0.0000984 24 3600 24 3600 101300 29 0.65 0.77071 / 8314 298 , 2.175 0.0139 0.00301 0.0000983 0.00301 4.618 / 0.0326 / 0.77071 120 0.0139 / 10 Using Fig 4.20 0.69 Hence, . 10 4.618 0.000015 120 800 0.025 . 10.731 4.618 4.618 0.0326 0.9929 1 1 4.618 800 0.0071 120 0.9929 124.83 0.9929 / 0.000017 / 0.0003 0.0326 0.0071 0.69 0.000015 4.64 / 124.83 4.64 0.062 0.000017 800 0.0326 0.062 0.0003 2 10 10 / 2112417.32 5389.866 , 0.2, 0.2, 2 0.046 0.046 0.001375 1 0.00671 Thus, 4 2 4 2 Hence, 0.001375 1 10 0.00018 0.062 10 2112417.32 / . 2 800 4.64 0.062 0.0326 0.046 0.062 5389.866 . 0.226 2 124.83 0.00671 581.72 0.226 581.72 √0.0003885 1 Also, 0.0003885 0.0197 0.0003885 1 0.69 . 0.00687 0.01 9.81 800 124.83 581.72 from Fig. 4.23 1.75 1.75 11.39 10 From Fig. 4.22 0.003 581.72 124.83 9.81 2242.59 / Problem 4.9 Water is circulated through a vertical pipe loop injecting air at the bottom of one leg, as shown in Fig.4.31. All the gas is vented at the top of the leg. Determine the circulated liquid-flow rate, vSL, for an injected gas rate of vSG = 0.04m/s. the following additional data are given: d = 0.051m 1000kg/ m3 h = 5m = 1.2kg/m3 µG = 2 x 10-5 kg/ms µL= 1 x 10-3kg/ms Assume: 1. 2. Solution Bubble flow in the aerated leg with the bubble-swarm-rise velocity of 0.24m/s. Short horizontal top and bottom sections with negligible pressure drop. The Pressure drop in the first section is equal to the pressure gain in the second section. Bubble flow exist in the first section, hence, 1.2 0.04 1.2 1 1 1000 1 0.046 0.046 0.24 1.2 0.04 0.033 1 1 1.2 0.24 1.2 1000 1 0.001 1 1.2 1000 0.051 0.001 9.81 . 1.2 0.00002 1 0.0053 . 0.0053 0.033 0.0053 0.033 1 . 0.2 . 1000 0.051 0.001 . 0.0053 . 1 0.24 2 2 0.0052 0.033 1 . 1 0.2 . 1000 1 0.051 1.2 1 0.033 1 0.2 2 0.0052 0.033 1 0.2 . 1000 1 0.051 1.2 0.033 1 0.2 0.204 0.033 2 0.2 2 1 1000 1 1 0.2 1.2 . 0.033 1000 1 0.051 0.2 1.2 1 0.033 1 0.24 0.0052 0.033 . 0.204 0.033 0.2 1000 1 1.2 0.033 0.24 Considering the gravitational pressure gradient: The Pressure balance in the system will be The Pressure gradient balance in the system will be Thus 0 This gives 1000 1 1.2 9.81 0.204 0.033 1 . 1 . 0.2 1000 1 1.2 1.2 0.033 1 1 0.033 0.24 1 0.2 1000 9.81 0.204 0.033 1 0.2 . 1000 1 0 1 . 9798.23 0.204 0.033 0.2 1.2 1000 1 0.033 1 1.2 0.24 0 0.033 0.2 0.204 0.033 0.2 1000 1 Solving for Hence, 0.0306 Thus, 0.033 (gives 2 Values; 0.0306 or -0.0428) 1 0.24 0.033 1 0.0306 0.24 0.84 / ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online