Question 1. Problem statement. Problem solution

Transcription

1 TP III Fluid Mechanics Problem Sheet 5: Two-phase Flows in Pipes 1 Prof. Omar K. Matar, Department of Chemical Engineering, Imperial College London, ACEX 517, o.matar@imperial.ac.uk. Question 1 Problem statement Gas and oil flow at volumetric flow rates G 10 ft 3 /s and L 0.5 ft 3 /s in a horizontal pipe of internal diameter D 0.5 ft. Estimate the fraction of gas, the pressure gradient, and the mean velocities of each phase. The physical properties are: ρ g 0.15 lb m /ft 3. ρ l 50 lb m /ft 3. µ g lb m /ft hr. µ l 20 lb m /ft hr. friction factor, f Problem solution The following conversion factors are useful: 1 ft 12 in. 1 psi 1 lb f.in 2. 1 lb f 32.2 lb m.ft.s 2. The cross-sectional area of the pipe is A πd2 π ft 2. (1) The superficial velocities: j g G A ft/s, j l L A The Reynolds numbers Re g ρ gj g D µ g Re l ρ lj l D µ l 2.55 ft/s. (2) , (3) 1 Taken from J. O. Wilkes, Fluid Mechanics for Chemical Engineers, Prentice-Hall, New Jersey,

2 So, the flow is turbulent in the gas and liquid. Hence, n.12 from the Table in lectures on two-phase flows. To determine X 2, we note that the diameters and friction factors are the same for all phases so that X 2 φ2 g φ 2 l ρ lj 2 l ρ g j 2 g X () Now, we calculate φ g : φ g (1 + X 2/n ) n/2 ( /.12 ).12/ (5) The void fraction, ε, is now given by ε 1 ( X 0.58 ) (6) 2.82 ( ) 2.82 The gas phase pressure gradient is ( ) 2ρ gjgf 2 D psi/ft. (7) The two-phase pressure gradient can now be calculated: ( ) ( ) φ 2 g ( ) psi/ft. (8) We can finally calculate the mean speeds of the gas and liquid phases v g G εa ft/s, v l L (1 ε)a 0.5 ( ) (9) 12.3 ft/s. 2

3 Question 2 Problem statement A gas-lift pump is shown in Fig. 1. Here, the buoyant action of a gas flowing upwards with a volumetric flow rate rate G lifts a liquid with a volumetric flow rate L from a height H 0 in a liquid reservoir to a height H in a vertical pipe of pipe D and cross-sectional area A. The flow regime in the pipe is slug flow. You may neglect friction in both phases. (a) Derive an expression for the void fraction ε in the pipe in terms of H and H 0. (b) If L 0, that is, if the liquid just manages to get to the top of the pipe, then prove that H/H 0 is given by where u b 0.35 gd. (c) If G/A u b then what is H/H 0? H H 0 1.2G/A + u b 0.2G/A + u b, (10) (d) If D 0.1 m, ε 0.5, G/A 2L/A, then what is the value of G/A in m/s? Liquid delivery H Liquid reservoir H 0 Gas inlet Figure 1: Problem sketch for Question 2: gas-lift pump. 3

4 Problem solution (a) The hydrostatic pressure at the bottom of the pipe is approximately equal to ρ l gh 0 ; this is obtained by considering the pressure at the bottom of the reservoir and neglecting the small pressure drop in the pipe connecting the reservoir and the vertical pipe. The pressure at the bottom of the pipe is also given by ρgh where the density ρ is given by ρ l (1 ε)+ρ g ε ρ l (1 ε) since ρ g ρ l. Equating the two expressions: ρ l gh 0 ρ l gh(1 ε) gives ε 1 H 0 H. (11) (b) From the notes in lecture 5, we have the following expression for the speed at the nose of the slug u s 1.2 G + L A + u b. (12) Here G εu s A whence u s G/εA so G εa 1.2G + L A + u b. (13) But, from part (a) ε 1 H 0 /H and L 0 so, after some algebra we get (c) If G/A u b then H/H 0 6, so H 6H 0. H H 0 1.2G/A + u b 0.2G/A + u b. (1) (d) Substitution of the data provided into Eq. (13) gives 2 G ( G A 1.2 A + G ) , (15) 2A whence G/A m/s.

5 Question 3 Problem statement Gas and oil flow upwards at volumetric flow rates G 10 ft 3 /s and L 0.1 ft 3 /s in a vertical pipe of internal diameter D 0.25 ft. Estimate the fraction of gas, the pressure gradient, and the mean speeds of each phase. The physical properties are: ρ g 0.15 lb m /ft 3. ρ l 50 lb m /ft 3. µ g lb m /ft hr. µ l 20 lb m /ft hr. friction factor, f The gravitational acceleration, g, is 32.2 ft/s 2. Problem solution The following conversion factors are useful: 1 ft 12 in. 1 psi 1 lb f.in 2. 1 lb f 32.2 lb m.ft.s 2. The cross-sectional area of the pipe A πd2 The superficial velocities are π ft 2. (16) j g G A ft/s, j l L A ft/s. (17) The dimensionless superficial velocities are jg ρ g j g gd(ρ l ρ g ) ( ) 3.9, jl ρ l j l gd(ρ l ρ g ) ( ) (18) 5

6 Make a decision about the flow regime: slug to annular flow transition if jg jl, at least. So, ( ) j g j l < 3.9 annular flow. (19) The pressure gradients of the individual phases are ( ) 2ρ gjgf 2 D l0 2ρ lj 2 l f D The Reynolds numbers Re g ρ gj g D µ g Re l ρ lj l D µ l psi/ft, psi/ft , (20) 590. (21) So, the flow is turbulent in the gas and liquid. Hence, n.12 from the Table in lectures on two-phase flows. The pressure gradient of the two-phase flow is given by the following expressions from lectures: ( ) ( ) φ 2 g ρ g g ( ) 0.085φ g φ2 g psi/ft, ( ) φ 2 l [ερ g + (1 ε)ρ l ] g l φ [0.15ε + 50(1 ε)] l φ 2 l ε psi/ft. (22) Recall that X 2 φ 2 g/φ 2 l so φ 2 l φ 2 g/x 2, so use this to eliminate φ l, and equating the two above expressions for (/), we get the following expression φ 2 g( ) 21(1 ε). (23) X2 We also have φ 2 g (1 + X 2/.12 ).12/2 and ε 1/( X 0.58 ) 2.82 which could be used to eliminate φ g and ε, respectively, from Eq. (23): (1 + X 2/.12 ).12/2 ( X 2 ) 21(1 1/( X0.58 ) 2.82 ). (2) 6

7 Solution of the resultant equation gives X 0.195, φ g 2.115, ε (25) We can now calculate the pressure gradient for the two-phase flow: ( ) ( ) φ g ρ g g psi/ft (26) Finally, we can determine the mean speeds for each phase: v g G εa ft/s, v l L (1 ε)a ft/s. (1 0.92)0.091 (27) 7