Beggs and Brill method

1 Beggs and Brill method The Beggs and Brill method works for horizontal or vertical flow and everything in between. It also takes into account the different horizontal flow regimes. This method uses the general mechanical energy balance and the average in-situ density to calculate the pressure gradient. The following parameters are used in the calculations. gDuNmFR2= (2-38) mlluu= (2-39, 40) = =lL (2-41, 42) =lL =lL Determining flow regimes Segregated if l < .01 and NFR < L1 or l >= .01 and NFR < L2 Transition if l >= .01 and L2 < NFR <= L3 Intermittent if.

2 01 <= l <.4 and L3 < NFR <= L1 or l >= .4 and L3 < NFR <= L4 Distributed if l < .4 and NFR >= L1 or l >= .4 and NFR > L4 For segregated, intermittent and distributed flow regimes use the following: 0llyy= cFRbllNay =0 (2-43, 44) with the constraint of that yl0 >= l. ()()[] +=C ()()gFRfvlellNNdC ln1 = (2-45,46) Where a, b, c, d, e, f and g depend on flow regimes and are given in the following table For transition flow, the liquid holdup is calculated using both the segregated & intermittent equations and interpolating using the following: ()()ntIntermitteBySegregatedAyylll+= (2-47) 233 LLNLAFR = AB =1 (2-48,49) ggllyy +=_144sin_ cPEggdldp= (2-50,51) The frictional pressure gradient is calculated using.

3 DgufdldpcmmtpF22 = (2-52) ggllm += ntpntpffff= (2-53,54) The no slip friction factor fn is based on smooth pipe ( /D =0) and the Reynolds number, mmmmDuN 1488Re= where ggllm += (2-55,56) ftp the two phase friction factor is (2-57) Sntpeff=where [][]()42)ln( )ln( )ln( )ln(xxxxS+ + = (2-58) and 2llyx = (2-59) Since S is unbounded in the interval 1 < x < , for this interval ) ( =xS (2-60) Using Beggs & Brill qo = 2000 bpd qg = 1 mmcfpd Temp = 175 oF o = 2 cp g = .0131 Pipe = o = lb/ft3 g = lb/ft3 Pressure = 800 psi First find the flow regime, calculate NFR, l, L1, L2, L3, and L4.

4 NFR = , l = .35, L1=230, L2=.0124, L3= .456, L4= 590. So .01 < l < .4 and L3 < NFR < L1 so flow is intermittent. Using the table to get a, b and c: * Find C and , d, e, f and g from table: ()()()() * *35.* = = gFRfvlellNNdC [][] ) (sin333.) ( ) (sin333.) (133= += += C Find yl * llyy The in-situ average density is 3_ *) ( * +=+= Potential gradient is * For friction gradient First find the mixture density and viscosity 3 * * +=+= * *2=+=+= The Reynolds Number *203.* * From Moody plot fn is .0045, solve for S [][]()42)ln( )ln( )ln( )ln(xxxxS+ + = [] []()379.) ( ) ( ) ( ) (42=+ + =S Solve for ftp Find the friction gradient * * *0066.

5 *22322==== 1)Using the Beggs and Brill method find the length of pipe between the points at 1000psi and 500 psi with the following data. Both vertical and horizontal cases. d = g = .65 oil 22o API qo = 400 stb/day qw = 600 bpd g = .013 cp o = 30 dynes/cm w = 70 dynes/cm GLR = 500 scf/stb @ average conditions = Rs = 92 scf/stb o = 17 cp w = .63 z = .91 Pipe Fittings in Horizontal flow To find the pressure drop through pipe fitting such as elbows, tees and valves an equivalent length is add to the flow line. This will account for the additional turbulence and secondary flows which cause the additional pressure drop. These equivalent lengths have been determined experimentally for the most of the fittings.

6 These are found in the following tables. They are given in pipe diameters, which are in feet. So to find the equivalent length for a 45o elbow in 2 inch pipe, find the equivalent length for the elbow in the table, 16, and multiply it by .166 feet, which gives feet. This is added to the length of the flow line, the pressure drop for the system is then calculated using one of the methods for horizontal flow.