from __future__ import division
print "Example 19.1 page no 246\n\n"
#we have to find pressure at different point in a oil tank
#apply manometer equation between point 1 and 2
#since rho1=rho2,z1=z2
#it gives P1=P2
#applying manometer equation between points 2 and 3
rho_oil=0.8*1000#density of oil
#since rho3=rho_oil=rho2
rho3=rho_oil
z_32=.4#height difference between point 2 and 3
g=9.807#grav. acc.
P7=0#pressure at point 7,on gauge basis
z_76=0.8#height difference between point 6 and 7
rho_hg=13600#density of mercury
P6=P7 + rho_hg*g*z_76#pressure at point 6
P5=P6#pressure at point 5
rho_air=1.2#density of air
z_54=1#height difference between point 5 and 4
P4=P5 + rho_air*g*z_54#pressure at point 4
P3=P4#pressure at point 3
P2=P3 + rho_oil*g*z_32#pressure at point 2
P1=P2#air pressure in the oil tank
print "\n pressure P1=%0.2e Pag"%(P1)#
from __future__ import division
from math import sqrt
print "Example 19.2 page no 250\n\n"
#pitot tube is located at the center line of a horizontal pipe transporting air
rho=0.075#density of gas ,lb/ft**2
h=0.0166667#height difference,ft
g=32.2#gravitational acc. lb/ft**2
rho_m=62.4#density of medium which is air
v=sqrt(2*g*h*(rho_m-rho)/rho)#velocity
print "\n velocity v=%0.2f ft/s"%(v)#
v_max=v#because at that point where the reading was taken is the centerline
print "\n maximum veocity v_max=%0.2f ft/s"%(v_max)#
#since the flowing fluid is air at a high velocity the flow has a high probability of being turbilent .from chapter 14,assume
#v_av/v_max=0.815
v_av=v_max*0.815
print "\n average velocity v_av=%0.2f ft/s"%(v_av)#
from __future__ import division
print "Example 19.3 page no 251\n\n"
#refer to example 19.3
S=0.785#cross sectional area,ft**2
v_av=24.4#average velocity,ft/s
q=v_av*S*60#flow rate,factor 60 for minute
print "\n flow rate q=%0.2f ft**3 min"%(q)#
rho=0.075#density
m_dot=q*rho*60#mass flow rate
print "\n m_dot mass flow rate=%0.2f lb/hr"%(m_dot)#
from __future__ import division
from math import pi,sqrt
print "Example 19.4 page no\n\n"
#water flow ina circular pipe,a pitot tube is used to measure the water velocity
h=0.07#manometer height,m
rho=1000#density of water,kg/m**3
rho_m=13600#density of mercury,kg/m**3
g=9.807
v=sqrt(2*g*h*(rho_m-rho)/rho)
print "\n water velocity v=%0.2f m/s "%(v)#
D=0.0779#pipe inside diameter,by using table A.5 in the appendix for a 3 inch schedule 40 pipe
S=(pi/4)*D**2
print "\n cross sectional area S=%0.2f m**2"%(S)#
q=v*S#flow rate
print "\n flow rate q=%0.2f m**3/s"%(q)#
meu=0.001#viscosity of water,kg/m.s
R_e=rho*v*D/meu#reynolds number
print "\n reynolds no R_e=%0.2f "%(R_e)#
from __future__ import division
from math import pi,sqrt
print "Example 19.5 page no 254\n\n"
#a venturi meter has gasoline flowing through it.
h=0.035#height of venturi meter
D1=0.06#upsteeam diameter,m
D2=0.02#throat diameter,m
rho_m=13600#density of mercury
rho=680#density of gasoline
g=9.807
v2=sqrt((2*g*h*(rho_m-rho)/rho)/1-D2**4/D1**4)#velocity of gasoline at the the throat
print "\n velocity at throat v2=%0.2f m/s"%(v2)#
q=(pi/4)*D2**2*v2#flow rate
print "\n flow rate q =%0.2f m**3/s"%(q)#
P1=101325#upstream pressure,Pa
P2=P1-g*h*(rho_m-rho)#pressure at throat P2
print "\n pressure P2=%0.2f Pa"%(P2)#
P_d=P1-P2#pressure difference
P_l=.1*P_d#pressure loss is 10 %
print "\n pressure loss P_l=%0.2f Pa"%(P_l)#
W_l=q*P_l#power loss
print "\n power loss W_l=%0.2f W"%(W_l)#
from __future__ import division
from math import pi,sqrt
print "\n Example 19.6 page no. 255\n\n"
#refer to example 19.5
#if gasoline has vapor pressure of 50000Pa ,we have to calculate flow rate at whhich cavitation to occur
P1=101325#upstream pressure,Pa
P2=50000#given vapor pressure,Pa
D1=0.06#upstream diameter,m
D2=0.02#throat diameter,m
rho=680#density of gasoline
v2=sqrt((2*(P1-P2))/rho*(1-D2**4/D1**4))#velocity
print "\n velocity v2=%0.2f m/s"%(v2)#
q=(pi/4)*D2**2*v2#flow rate
print "\n flow rate q=%0.2e m**3/s"%(q)#
from __future__ import division
from math import pi,sqrt
print "Example 19.7 page no 258\n\n"
#an orifice meter is equipped with flange top is installed to measure the flow rate of air in a circular duct
D1=0.25#diameter of circular duct,m
D2=0.19#orifice diamter,m
v2=4/(pi*D2**2)#velocity through orifice
print "\n velocity through orifice v2=%0.2f m/s"%(v2)#
C_o=1# assuming orifice discharge coefficient
rho=1.23#density of air,kg/m**3
P=rho*v2**2*(1-(D2**4/D1**4))/2#pressure
print "\n pressure P=%0.2f Pa"%(P)#
meu=1.8e-5# absolute viscosity
R_e=rho*v2*D2/(meu)#reynolds no.
print "\n Reynolds no. R_e=%0.2f "%(R_e)#
C_ac=0.62#actual discharge cefficient,from fig.19.8
P_ac=P/(C_ac)**2#actual pressure drop
P_rec=14*(D2/D1) + 80*((D2/D1)**2)#equation for percentage pressure recovery
P_loss=100-P_rec#precentage pressure loss
P_l=round((P_loss/100)*P_ac)#actual pressure drop after recovery
print "\n actual pressure drop P_l=%0.2f Pa"%(P_l)#
from __future__ import division
print "\n Example 19.9 page no 259\n\n"
#air at ambient condition is flowing in a pipe
rho=0.075#density of air ,lb/ft**3
m_dot=0.5#mass flow rate ,lb/s
q=m_dot/rho#volumatric flow rate
print "\n volumatric flow rate q=%0.2f ft**3/s"%(q)#