import math
#variable initialisation
R=0.06 #Diamter in m
r=0.02 #radical distance
u=0.6 #velocity in m/s
#Calculation
#For a laminar flow,
#(-ga/4*mu*()dh/ds
K=(u)/(R**2-r**2)
u_m=K*R**2 #Maximum velocity
print "u_m=",u_m,"m/s"
M=u_m/2 #Mean velocity
print "M=",M,"m/s"
Q=math.pi/4 *(R**2)*(M) #Discharge
print "Q=",round(Q*1000,3),"L/s"
#variable initialisation
mu=0.08 #viscosity in kg/ms
D=0.1 #Diameter in m
u_m=1.4 #Maximum velocity
L=1.0 #Length in m
R=0.05 #Radius in m
#Calculation
#u=1.4(1-r/R**2)
tau_0=2*mu*1.4*R/(R**2) #shear stress
print "tau_0=",tau_0,"N/m^2"
#Maximum velocity um occurs at r=0
V=u_m/2 #mean velocity
delp1=(32*mu*V*L)/D**2 #Pressure drop per unit length of the pipe
print "(-delP)1=",delp1,"N/m^2"
#Alternate method for pressure drop
Pd=4*tau_0*L/(D)
print "(-delP)1=",Pd,"N/m^2"
#variable initialisation
ga=9.81 #gravity constant
h_f=20 #Head loss
L=3000 #length of pipe in m
R=0.30 #Radius in m
r=0.10 #radial distance in m
mu=0.15 #absolute viscosity in Pa.s
D=0.30 #Diameter in m
rho=0.85*998 #relative density in kg/m^3
#Calculation
tau_0=round((R/2)*(ga*rho*h_f/L),3) #Wall shear stress
tau=(tau_0/(R/2))*r #shear stress
print "tau=",tau,"Pa"
V=round((h_f/L)*(rho*ga*(D**2))/(32*mu),2)#Volume
Re=round(V*D*rho/mu,1) #Reynold's number
#Re is less than 2000.hence laminar flow is correct.
f=64/Re #friction factor
print "f=",round(f,5)
import math
#variable initialisation
mu=1.50 #absolute viscosity
rho=1260 #relative density in kg/m^3
V=5.0 #Velocity in m/s
D=0.10 #Diameter in m
ga=9.81*rho #density
L=12 #Length in m
#Calculation
Re=rho*V*D/mu #Reynold's number
#As,Re value less than 2000,the flow is conduit
tau_0=8*mu*V/D
print "tau_0=",int(tau_0),"Pa"
h_f=32*mu*V*L/(ga*(D**2)) #In laminar flow the head loss
print "h_f=",round(h_f,1),"m"
#P=ga*Q*h_f
A=(math.pi*(D**2)/4) #Area
Q=A*V #Discharge
P=ga*Q*h_f #Power
print "P=",round(P/1000,2),"kW"
#variable initialisation
R=2000 #maximum Reynolds number
rho=950 #density in kg/m^3
mu=8E-2 #Viscosity in Pa.s
L=200 #length in metre
D=50 #diameter in cm
#solution
D=15/100 #diameter in m
#Re=(V*D)/v
V=R*(mu/rho)/D
#by,Hagen-Poiseuille equation,
h_f=(32*mu*V*L)/((rho*9.81)*(D**2)) #headloss
print "Maximum difference in oil surface elevations h_f=",round(h_f,3),"m"
import math
#variable initialisation
D=0.075 #Diamter in m
mu=0.1 #visocity of oil in N.s/m^2
P=5.4 #power in kW
rho=0.90*998 #relative density
E=60 #efficiency in %
L=1000 #length in m
#solution
P=P*(E/100) #power spent in fluid friction
Q=round(math.sqrt(((P*1000)*(math.pi)*(D**4))/(128*mu*L)),5) #quantity
print "Quantity of oil=",round(Q*1000*60,3),"L/min"
V=round((Q)/((math.pi/4)*(D**2)),3) #Velocity in m/s
Re=(V*D*rho)/mu #Reynolds number
print "Reynolds number=",int (Re)
import math
#variable initialisation
mu=1.5 #absolute viscosity
rho=0.9 #relative density in kg/m^3
ga=9.81 #gravity
Pa=200 #Pressure of guage A in kPa
Za=20 #Pressure guages in m
Pb=500 #Pressure of guage B in kPa
Zb=0 #Pressure guages in m
L=30 #Length in m
D=0.03 #Diameter in m
#Calculation
mu=mu/10 #in Pa.s
rho=rho*998 #density in kg/m^3
ga=(ga*rho)/1000 #in kN/m^3
ha=(Pa/ga)+Za #piezometric head at A
hb=(Pb/ga)+Zb #piezometric head at B
print "Since hb>ha,the flow is from B towards A.i.e.Upwards"
h_f=hb-ha
V=(h_f*ga*(D**2))/(32*mu*L)*1000
Re=round(V*D*rho/mu,1) #Reynold's number
#As Re<2000,the flow is laminar and our initial assumption is correct.
A=(math.pi*(D**2)/4) #Area
Q=A*V #Discharge
print "Q=",round(Q*1000,4),"L/s"
#variable initialisation
mu=2.5 #absolute viscosity
rho=0.9 #relative density in kg/m^3
ga=9.81 #gravity
L=500 #Length in m
D=100 #Diameter in m
Q=2 #Discharge
P_b=0 #atmospheric pressure in Pa
Z2=f=20 #elevation of pump2,friction factor
e=0.65 #efficiency of pump
#Calculation
mu=mu/10 #in Pa.s
rho=rho*998 #density in kg/m^3
D=D/1000
Q=Q/1000
V=Q/((math.pi/4)*(D**2))
Re=round(V*D*rho/mu,3) #Reynold's number
print "Re=",Re
#As Re<2000,the flow is laminar.
h_f=round((32*mu*V*L)/(ga*rho*(D**2)),2) #In laminar flow the head loss
#If A is the pump end and B is the outlet,
P_A=int((Z2+h_f)*(ga*rho))
print "P_A=",int(P_A/1000),"kPa"
#H=static head+friction head
H=h_f+f #Overall Head
P=int(ga*rho*Q*H) #Power required for pumping
Pi=round(P/e,1) #power input required
print "Pi=",Pi,"W"
import math
#variable initialisation
v=1.5 #kinematic viscosity in stokes
rho=900 #relative density in kg/m^3
L=300 #length in m
D=8 #Diameter in cm
e=60 #Overall efficiency
Pi=5000 #input power
#Calculation
v=v/10000 #viscosity in m^2/s
P=e/100*Pi #power delivered to the fluid
D=D/100 #diameter in m
mu=rho*v #Dynamic viscosity in Pa.s
Q= math.sqrt((P*math.pi*(D**4))/(128*mu*L))
print "Q=",round(Q*1000,2),"L/s" #Rate of flow
V=round(Q/((math.pi/4)*(D**2)),3) #Velocity in m/s
Re=V*D/v #Reynolds number
print "Re=",round(Re,1)
#As Re<2000,the flow is laminar
import math
#variable initialisation
Q=850 #Discharge in L/min
D=0.15 #Diameter in cm
del_P=95000 #change in pressure
ga=9.81 #gravity
L=800 #Length in m
rho=917 #density
#Calculation
Q=Q/(1000*60) #Converting into m^3/s
V=Q/((math.pi/4)*(D**2)) #velocity
Hf=del_P/(ga*rho) #Head loss
mu=(Hf*ga*rho*(D**2))/(32*V*L)
print "mu=",round(mu,5),"Pa.s."
Re=(V*D*rho)/mu #Reynolds number
print "Re=",int(round(Re,))
print"As this value of Re is less than 2000,the flow is laminar as assumed initially"
import math
#variable initialisation
rho=0.92 #relative density
mu=0.085 #dynamic viscosity
Q=0.04 #discharge
ga=9.81 #gravity constant
#Calculation
rho=rho*998#relative density in kg/m^3
#Assuming laminar flow,h_f/L=0.0-4
h_f=0.04
D=((((128*mu*Q))/(math.pi*ga*rho))/h_f)**(1/4) #diameter
V=Q/((math.pi/4)*(D**2)) #Velocity
Re=(V*D*rho)/(mu) #Reynold's number
#As this value is >2000,the flow is not laminar.Hence the required diameter is controlled by critical Reynolds number
Re_crit=2000
D=round(((rho/mu)*(Q/(math.pi/4)))/Re_crit,3) #Diameter
h_f=(128*mu*Q)/(math.pi*rho*(D**4)) #Headloss/L
#As this is less than permissible gradient of 0.04 the diameter is acceptable.
print "Hece the least diameter =",D*100,"cm"
#variable initialisation
Re=1200 #Reynolds number
v=1.92E-3 #kinematic viscosity in m^2/s
g=9.81 #gravity constant
#Calculation
VD=Re*v #equation 1
#Z1-Z2=L
#Se=(Z1-Z2)/(Z1-Z2)=1.0
Se=Sf=1.0 #energy gradient
#Sf=32*v*V/g*D**2
VD2=(g*Sf)/(32*v) #Equation 2
#From VD/VD2,
D=(1/(VD2/VD))**(1/3) #solving to find D
print "D=",round(D,3),"m"
#variable initialisation
B=0.12 #thickness in mm
L=50 #length in cm
h_f=5.0 #headloss in m
ga=9790 #Specific weight
mu=(998)*(0.01E-4) #viscosity
#solution
V=((h_f)*(ga)*((B/1000)**2))/(12*mu*(L/100)) #calculating V in m/s
q=(1*V*(B/1000))*60*1000 #Discharge per metre width
print round(q,3),"L/min per metre width of crack"
#variable initialisation
V=1.40 #mean velocity in m/s
B=1.2 #distance between plates in cm
mu=1.05 #dynamic viscosity
y=0.002 #distance in m
ga=0.92*998*9.81 #relative density
L=25 #length in m
#Calculation
B=B/100 #Converting into m
mu=mu/10 #Converting into Pa.s
#For two dimensional laminar flow between parallel plates
um=(3/2)*V #Maximum velocity
print "Um=",round(um,1),"m/s"
#Let us take (-dp/dx) as dp
#V=(-dp/dx)*(B**2/12*mu)
dp=12*mu*V/(B**2)
tau_0=(dp)*(B/2) #Boundary shear stress
print "tau_0=",tau_0,"Pa"
#let y=0.002m
tau=(dp)*(B/2-y) #Shear stress
print "tau=",int(tau),"Pa"
v=(((1/(2*mu))*dp))*(B*y-y**2) #Velocity
print "v=",round(v,3),"m/s"
h_f=12*mu*V*L/(ga*(B**2)) #Head loss
print "h_f=",round(h_f,1),"m"