from __future__ import division
print "Example 14.1 page no 148\n\n" # a liquid flow through a tube
meu=0.78e-2#viscosity of liquid,g/cm*s
rho=1.50#density,g/cm**3
D=2.54#diameter,cm
v=20#flow velocity
R_e=D*v*rho/meu#reynolds no
print "\n Reynolds no R_e = %.2f "%(R_e)#
from __future__ import division
print "Example 14.2 page no 148\n\n" # a fluid is moving through a cylinder in laminar flow
meu=6.9216e-4#viscosity of fluid,lb/ft*s
rho=62.4#density,lb/ft**3
D=1/12#diameter,ft
R_e=2100#reynolds no
v=R_e*meu/(D*rho)#minimum velocity at which turbulance will appear
print "\n velocity v = %.2f ft/s"%(v)#
from __future__ import division
from math import log10
print "Example 14.3 page no 152\n\n" # calculate the friction factor by using different equation's
R_e=14080#reynolds no
K_r=0.004#relative roughness (a) by PAT proposed equation
f_a=0.0015+(8*(R_e)**0.30)**-1
print "\n fanning friction factor f_a=%0.2f "%(f_a)# equation for 5000<R_e>50000
f_b1=0.0786/(R_e)**0.25
print "\n friction factor f_b1=%0.2f "%(f_b1)# equation for 30000<R_e>1000000
f_b2=0.046/(R_e)**0.20
print "\n friction factor f_b2=%0.2f "%(f_b2)# equation for the completely turbulent region
f_c=1/(4*(1.14-2*log10(K_r))**2)
print "\n friction factor f_c=%0.2f "%(f_c)# equation given by jain
f_d=1/(2.28-4*log10(K_r+21.25/(R_e**.9)))**2
print "\n friction factor f_d=%0.2f "%(f_d)#
f_e=0.0085 #from figur 14.2
print "\n friction factor f_e=%0.2f"%(f_e)#
f_av=(f_a+f_b1+f_b2+f_c+f_d+f_e)/6
print "\n average friction f_av=%0.2f "%(f_av)#
from __future__ import division
from math import pi
print "Example 14.4 page no 154\n\n" # for turbulent fluid flow in across section (a) for a rectangle
w=2#width of a rectangle,in
h=10#height of rectangle,in
S_a=h*w#cross sectional area
P_a=2*h+2*w#perimeter of rectangle
D_eq_a=4*S_a/P_a#equivalent diameter
print "\n equivalent diameter D_eq_a=%0.2f in"%(D_eq_a)# (b) for an annulus
d_o=10#outer diameter of annulus
d_i=8#inner diameter
S_b=pi*(d_o**2-d_i**2)/4#cross sectional area
P_b=pi*(d_o-d_i)#perimeter
D_eq_b=(4*S_b)/(P_b)#eq. diameter
print "\n equivalent diameter D_eq_b=%0.2f cm"%(D_eq_b)# (c) for an half- full circle
d_c=10#diameter of circle
S_c=pi*d_c**2/8# cross sectional area
P_c=pi*d_c/2#perimeter
D_eq_c=4*S_c/P_c#eq. diameter
print "\n equivalent diameter D_eq_c=%0.2f cm"%(D_eq_c)#
from __future__ import division
from math import pi
print "Example 14.5 page no 157\n\n" # air is transported through a circular conduit
MW=28.9#molecular weight of air
R=10.73#gas constant
T=500#temperature
P=14.75#pressure,psia applying ideal gas law for density
rho=P*MW/(R*T)#density
rho=0.08#after round off
meu=3.54e-7#viscosity of air at 40 degF assume flow is laminar
q=8.33#flow rate ,ft**3/s
L=800#length of pipe,ft
P_1=.1#pressure at starting point
P_2=.01#pressure at delivery point
D=((128*meu*L*q)/(pi*(P_1-P_2)*144))**(1/4)#diameter
print "\n pipe diameter D=%0.2f ft"%(D)# check the flow type
meu=1.14e-5
R_e1=4*q*rho/(pi*D*meu)#reynolds no print "\n reynolds no R_e=%0.2f "%(R_e)# from R_e we can conclude that laminar flow is not valid
P_drop=12.96#pressure drop P_1-P2 in psf
f=0.005#fanning friction factor
g_c=32.174
D=(32*rho*f*L*q**2/(g_c*pi**2*P_drop))**(0.2)#diamter from new assumption strat the second iteration with the newly calculated D
k=0.00006/12#roughness factor
K_r=k/D#relative roughness
C_f=1.321224
R_e_n=4*q*rho/(pi*D*meu)#new reynolds no print "\n new reynolds no R_e=%0.2f "%(R_e)#
f_n=0.0045#new fanning friction factor
D=(((8*rho*f_n*L*q**2)/(g_c*pi**2*P_drop))**(0.2))*C_f#final calculated diameter because last diameter is same with this
print "\nD=%0.2f "%(D)# iteration may now be terminated
S=pi*(D**2)/4#cross sectional area of pipe
v=q/S#flow velocity
print "\n flow velocity v=%0.2f ft/s"%(v)##printing mistake in book in the value of meu in the formula of D is first time that's why this deviation in answer
from __future__ import division
print "Example 14.6 page no. 159\n\n" # ethyl alcohol is pumped through a horizontal tube
rho=789#density .kg/m**3
meu=1.1e-3#viscosity ,kg/m-s
k=1.5e-6#roughness,m
L=60#length of tube,m
q=2.778e-3#flow rate
g=9.807
h_f=30#friction loss
A=(L*q**2)/(g*h_f)
A=1.574e-7
D=0.66*(((k**1.25)*(A**4.75)+meu*(A**5.2)/(q*rho))**.04)
D=0.0377 # calculate velocity of alcohol in the tube
S=3.14*(D)**2/4#surface area
v=q/S#velocity
v=3.93#velocity
neu=1.395e-6#dynamic viscosity
R_e=D*v/neu#reynolds no
print "\n R_e=%0.2f "%(R_e)##printing mistake in book
print "\n since R_e is more than 4000 flow is turbulent" #
from __future__ import division
print "Example 14.7 page no 160\n\n" # kerosene flow ina lng ,smooth ,horizontal pipe
rho=820#density,kg/m**3
D=0.0493#iside diameter of pipe by appendix A.5,m
R_e=60000
meu=0.0016#viscosity,kg/m.s
v=(R_e*meu)/(D*rho)# flow average velocity
print "\n average velocity v=%0.2f m/s"%(v)#
S=(pi/4)*D**2#cross sectional area
print "\n S=%0.2f "%(S)#
q=v/S#flow rate
print "\n flow rate q=%0.2f m**3/s"%(q)##printing mistake in book
m_dot=rho*q#mass flow rate
print "\n mass flow rate m_dot=%0.2f kg/s"%(m_dot)##printing mistake in book in the value of v
n=7#seventh power apply
v_max=v/(2*n**2/((n+1)*(2*n+1)))#maximum velocity
print "\n v_max=%0.2f m/s"%(v_max)# check the assumptioon of fully developed flow
R_e=60000#reynolds no
L_c=4.4*R_e**(1/6)*D#critical length
print "\n length L_c=%0.2f m"%(L_c)# since L_c <L th eassumption is valid
from __future__ import division
print "\n Example 14.8 page no 161\n\n" # refer to example no 14.7
rho=860#density
R_e=60000#reynolds no
f=.046/R_e**.2#fanning friction factor
print "\n fanning friction factor f=%0.2f "%(f)#
L=9#length of tube
v=2.38#velocity
D=.0493#diameter of tube
g=9.807
h_f=4*f*(L*v**2)/(D*2*g)#friction loss
print "\n h_f friction loss=%0.2f m "%(h_f)# applying bernoulli equation
P_drop=rho*g*h_f#pressure drop in pa
P_drop_a=P_drop/10**5#pressure drop in atm
print "\n P_drop_a =%0.2f atm"%(P_drop_a)#
from __future__ import division
print " Example 14.9 page no 161\n\n" # refer to example 14.7
D=0.0493#diameter of tuube
S=pi*D**2/4#cross sectional area\
P=8685#pressure
F=P*S#force required to hold the pipe,direction is opposite the flow
print "\n Force required to hold pipe F=%0.2f N"%(F)#
from __future__ import division
from math import sqrt
print "Example 14.10 page no 163\n\n" # a fluid is moving in the turbulent flw through a pipe a hot wire anemometer is inserted to measure the local velocity at a given point P in the system following readings were recorded at equal time interval instantaneous velocities at subsequent time interval
vz=[43.4,42.1,42,40.8,38.5,37,37.5,38,39,41.7]
vz_bar=0#
n=10#
i = 0#
sums=0#
for i in range(0,10):
sums=sums+vz[i]#
vz_bar=sums/n#
print "\n vz_bar=%0.2f"%(vz_bar)#
sigma=0#
for i in range(0,10):
sigma=sigma+(vz[i]-vz_bar)**2#
vz_sqr=sigma/10#
print "\n vz_sqr=%0.2f"%(vz_sqr)
I = sqrt(vz_sqr)/vz_bar#intensity of turbulance
print "\n intensity of turbulance I=%0.2f "%(I)#
from __future__ import division
from math import pi
print "Example 14.11 page no 164\n\n" # a fluid is flowing through a pipe
D=2#inside diameter of pipe,in
v_max=30#maximum velocity,ft/min
A=(pi/4)*(D/12)**2#cross sectional area (a) for laminar flow
v_a=(1/2)*v_max#average velocity
q_a=v_a*A#volumatric flow rate
print "\n flow rate q_a=%0.2f ft**3/min"%(q_a)# (b) for plug flow
v_b=v_max#average velocity
q_b=v_b*A#volumatric flow rate
print " \n flow rate q_b=%0.2f ft**3/min"%(q_b)# (c)for turbulent flow
v_c=(49/60)*v_max#average velocity
q_c=v_c*A#volumatric flow rate
print "\n flow rate q_c=%0.2f ft**3/min"%(q_c)#