# Chapter 5:Flow of Multiphase Mixtures¶

## Example no:5.1,Page no:190¶

In :
#Variable declaration
id=75e-3  # internal diameter of pipe
f_r_s=0.05  # Flow rate of steam in (kg/s)
f_r_w=1.5  # Flow rate of water in (kg/s)
T=330  # Mean Temperature
P=120  # Mean Pressure drop

#Calculation
import math
area=math.pi*id**2/4.0  # Cross-sectional area for flow
f_r_w_m3s=f_r_w/1000.0  # Flow of water
wtr_vel=f_r_w_m3s/area  #Water velocity
rho_steam=18*273*120.0/(22.4*330.0*101.3)  # density of steam at 330 K and 120 kN/m**2
f_r_s_m3s=f_r_s/rho_steam  #Flow of Steam
steam_vel=f_r_s_m3s/area  #Steam velocity
meu_steam=0.0113e-3
meu_water=0.52e-3
Rel=id*wtr_vel*1000/meu_water
Reg=id*steam_vel*rho_steam/meu_steam
DPl=4*0.0025*(1000*wtr_vel**2)/id
DPg=4*0.0022*(rho_steam*steam_vel**2)/id
X=(DPl/DPg)**0.5
phi_l=4.35
phi_g=3.95
DP_tpf=phi_g**2*DPg/1000

#Result
print"Pressure drop per unit length of pipe =",round(DP_tpf,3),"kN/m**2"

Pressure drop per unit length of pipe = 0.298 kN/m**2


## Example no:5.2,Page no:212¶

In :
#Variable declaration
M_p_d=0.2e-3  # Mean particle diameter
f_r_w=0.5  #Flow rate of water
id=25e-3  #Diameter of pipe
l=100  #length of pipe
t_vel=0.0239  #Terminal velocity of falling sand particles

#Calculation
import math
import sympy
Um=f_r_w/(1000*math.pi*id**2/4)
Re=id*Um*1000/0.001
#Assuming e/d = 0.008, then, from Figure 3.7:
phi=0.0046
f=0.0092
#From, equation 3.20, the head loss is:
hf=4*phi*l*Um**2/(9.81*id)
iw=hf/l
i=300*1000/(1000*9.81*100)
# Substituting in equation 5.20:
C=(iw/(i-iw)*(1100*9.81*id*(2.6-1)*t_vel)/(Um**2*Um))**-1
#If G kg/s is the mass flow of sand, then:
G=sympy.Symbol('G')
p=sympy.solve(2600**-1*G-0.30*(2600**-1*G+.0005))

#Result
print"\n Mass flow of sand =",round(p,2),"kg/s=",round(p*3600.0/1000.0),"tonne/h"

 Mass flow of sand = 0.56 kg/s= 2.0 tonne/h


## Example no:5.3,Page no:225¶

In :
#Variable declaration
p_s=1.25e-3  # Particle size of sand
rho_sand=2600.0  #Density of sand
flow_sand=1.0  #flow rate of sand in air
l=200.0  #length of pipe
d=100e-3  # taking nearest standard size of pipe
Uo=4.7
Ug=30.0
Meu_air=1.7e-5  # viscosity of air
rho_air=1  # Density of air

#Calculation
import math
flow_air=flow_sand/5.0
vol_flow_air=1*flow_air
area=flow_air/Ug
pipe_dia=math.sqrt(4*area/math.pi)
Ug=flow_air/(math.pi*(0.1**2)/4.0)
Ug=round(Ug,1)
Us=Ug-(Uo/(0.468+(7.25*(Uo/rho_sand)**0.5)))

Re=(d*Ug*rho_air/Meu_air)
phi=0.004
DP_air=(4*phi*l/d)*rho_air*Ug**2/2.0
DP_x=2805.0*DP_air/(Uo*Us**2)
DP=DP_air+DP_x

#Result
print"Volumetric flow rate of air =",vol_flow_air,"m**3/s"
print"The cross-sectional area of a 100 mm ID. pipe =",round(pipe_dia*1000),"mm\nThe nearest stand size is 100 mm"
print"Pressure drop due to air =",round(DP_air/1000,1),"kN/m**2"
print"Pressure drop due to sand particles =",round(DP_x/1000,1),"kN/m**2"
print"The total pressure drop =",round(DP/1000,1),"kN/m**2"

Volumetric flow rate of air = 0.2 m**3/s
The cross-sectional area of a 100 mm ID. pipe = 92.0 mm
The nearest stand size is 100 mm
Pressure drop due to air = 10.4 kN/m**2
Pressure drop due to sand particles = 16.4 kN/m**2
The total pressure drop = 26.8 kN/m**2