# Chapter 16 : Two Phase Flow¶

## Example 16.2 Page no 183¶

In :
from __future__ import division
print " Example 16.2 page no 183\n\n"
#cal. pressure drop if the flow for both phases is turbulent
#a. since the flow is tt and 1<X<10 ,apply equatuion 16.16b to obtain Y_g
X=1.66
Y_g=5.80+6.7143*X+6.9643*X**2-0.75*X**3
print "\n Y_g=%0.2f "%(Y_g)#
#the value of Y_g is an excellent agreement with the values provided by lockhart and Martinelli
#then pressure drop is
P_drop_g=2.71
P_drop_t=Y_g*P_drop_g
print "\n P_drop_t=%0.2f psf/100 ft"%(P_drop_t)#
#b. applying eq. 16.17b to generate Y_l
Y_l=18.219*X**-.8192
print "\n Y_l =%0.2f "%(Y_l)#
#pressure drop from eq. 16.2
P_drop_l=7.50
P_drop=Y_l*P_drop_l
print "\n P_drop=%0.2f psf/100ft"%(P_drop)#

 Example 16.2 page no 183

Y_g=32.71

P_drop_t=88.63 psf/100 ft

Y_l =12.03

P_drop=90.21 psf/100ft


## Example 16.3 Page no 185¶

In :
from __future__ import division
print " Example 16.3 page no 185\n\n"
#if the flow for the gas phase is turbulent and the liquid phase is viscous
#cal. pressure drop total
X=1.66#from ex. 16.1
Y_G_tv=20-21.81*X+16.357*X**2-1.8333*X**3
print "\n Y_G_tv=%0.2f "%(Y_G_tv)#
#pressure drop from eq 16.1
P_drop_g=2.71
P_drop_a=Y_G_tv*P_drop_g
print "\n pressure drop P_drop_a=%0.2f psf/100 ft"%(P_drop_a)#
#b. applying eq 16.20b to generate Y_l
Y_l_tv=11.702*X**-0.7334
print "\n Y_l_tv=%0.2f "%(Y_l_tv)#
#pressure drop from equation 16.2
P_drop_l=7.50
P_drop_b=Y_l_tv*P_drop_l
print "\n P_drop_b=%0.2f psf/100 f"%(P_drop_b)#

 Example 16.3 page no 185

Y_G_tv=20.48

pressure drop P_drop_a=55.51 psf/100 ft

Y_l_tv=8.07

P_drop_b=60.52 psf/100 f


## Example 16.4 Page no 187¶

In :
from __future__ import division
print "Example 16.4 page no 187\n\n"
#if flow for both phases is laminar then cal pressure drop total
#a. apply eq. 16.22b to obtain Y_G
X=1.66
Y_G=10-10.405*X+8.6786*X**2-0.9167*X**3
print "\n Y_G=%0.2f "%(Y_G)#
#pressure drop from eq 16.1
P_drop_g=2.71
P_drop=Y_G*P_drop_g
print "\n pressure drop P_drop=%0.2f psf/100 ft"%(P_drop)#
#b. apply eq 16.23b to generate Y_l
Y_l=6.4699*X**-0.556
print "\n Y_l =%0.2f "%(Y_l)#
#pressure drop from eq. 16.2
P_drop_l=7.50
P_drop_b=Y_l*P_drop_l
print "\n pressure drop P_drop_b=%0.2f psf/100 ft"%(P_drop_b)#

Example 16.4 page no 187

Y_G=12.45

pressure drop P_drop=33.74 psf/100 ft

Y_l =4.88

pressure drop P_drop_b=36.61 psf/100 ft


## Example 16.6 Page no 191¶

In :
from __future__ import division
from math import pi
print "\n Example 16.6 page no 191\n\n"
#a mixture of air(a) and kerosene(k) are flowing in a horizontal pipe
rho_a=0.075#density of airlb/ft**3
meu_a=1.24e-5#viscosity of air ,lb/ft.s
q_a=5.3125#flow rate ft**3/s
rho_k=52.1#density of kerosene,lb/ft**3
meu_k=0.00168#viscosity lof kerosene,lb/ft.s
q_k=1.790#flow rate ft**3/s
D=.19167#diameter of pipe ,ft
S=(pi/4)*D**2#cross sectional area,ft**2
print "\n S=%0.2f "%(S)#
#superficial velocity of each phase can be obtained by applying either eq, 16.7 and 16.8
v_a=q_a/(S*60)#for air
v_k=q_k/(S*60)#for kerosene
print "\n velocity v_a =%0.2f ft/s\n velocity v_k=%0.2f ft/s"%(v_a,v_k)#
R_e_a=D*rho_a*v_a/meu_a#reynolds no. of Air
R_e_k=D*rho_k*v_k/meu_k#reynolds no. of kerosene
print "\n R_e_a=%0.2f\nR_e_k=%0.2f "%(R_e_a,R_e_k)#

 Example 16.6 page no 191

S=0.03

velocity v_a =3.07 ft/s
velocity v_k=1.03 ft/s

R_e_a=3557.49
R_e_k=6145.91