# Chapter8-Internal Incompressible Viscous Flow¶

## Ex1-pg317¶

In [1]:
import math
##Operation pressure of hydraulic system(in kPa):
p1=20000.;
##Operation temperature of hydraulic system(in C):
T=55.;
##Piston diameter(in mm):
D=25.;
##Viscosity of SAE 10W at 55C(in kg/(m-s):
u=0.018;
##Mean radial clearance of a cylinder(in mm):
a=0.005;
##Gauge pressure on lower pressure side of piston(in kPa):
p2=1000.;
##Lenth of piston(in mm):
L=15.;
##Denity of water(in kg/m^3):
dw=1000.;##Leakage flow rate##

##Leakage flow rate (in mm^3/sec):
Q=math.pi/12.*D*a**3.*(p1-p2)*10**3/u/L
##Velocity of flow(in m/sec):
V=Q/math.pi/D/a/1000.
##Specific gravity of SAE 10W oil:
SG=0.92;
##Reynolds Number:
Re=SG*dw*V*a/u/1000.
##As Re<1400, flow is laminar.
print("RESULTS")
print'%s %.2f %s'%("Leakage flow rate: ",Q," mm^3/sec")

RESULTS
Leakage flow rate:  57.57  mm^3/sec


## Ex2-pg320¶

In [2]:
import math
##temperature fo operation(in F):
T=210.;
##Diameter of te bearing(in inches):
D=3.;
##Diametral clearance(in inches):
a=0.0025;
##Length of shaft(in inhes):
L=1.25;
##Speed of rotation of the shaft(in rpm):
N=3600.;
##Viscosity of the oil(in lbf-s/ft^2):
u=2.01*10**-4.;
##Specific gravity of SAE 10W:
SG=0.92;
##Density of water (in slug/ft^3)
p=1.94;##Torque and power##xec(filename)
##Shear stres .in lbf/ft^2):
Tyx=u*N*2.*math.pi/60.*D/2./(a/2.)
##Torqe(in inches-lbf):
T=math.pi/2.*Tyx*D**2.*L/144.
##Power dissipated in the bearing(in hp):
P=T*N/60.*2.*math.pi/12./550.
##Reynolds number:
Re=SG*p*N*2.*math.pi/60.*1.5*a/2./u/144.
print("RESULTS")
print'%s %.2f %s'%("Torque: ",T," inches-lbf")
print'%s %.2f %s'%("Power dissipated in the bearing: ",P," hp")

RESULTS
Torque:  11.16  inches-lbf
Power dissipated in the bearing:  0.64  hp


## Ex4-pg328¶

In [3]:
import math
##Flow rate through capilarry viscometer(in mm^3/sec):
Q=880.;
##Tube length(in m):
L=1.;
##Tube diameter(in mm):
D=0.5;
##Pressure drop(in kPa):
p=1000.;
##Density of oil(in kg/m^3):
d=999.;
##Viscosity of fluid##
#Viscosity of the liquid(in N-s/m^2):
u=math.pi/128.*p*1000.*D**4./Q/L/1000.
##Velocity(in m/sec)
V=Q/(math.pi/4.*D**2.)/1000.
##Reynolds number:
Re=d*V*D/u/1000.
print("RESULTS")
print'%s %.4f %s'%("Viscosity of fluid ",u," N-s/m^2")

RESULTS
Viscosity of fluid  0.0017  N-s/m^2


## Ex5-pg352¶

In [5]:
import math
##Volme flow rate of water(in m**3/sec):
Q=0.0084;
##Length of horizontal pipe(in m):
L=100.;
##Diameter of pipe(in m):
D=0.075;
##Density of water(in kg/m**3):
d=999.;
##Friction factor:
f=0.017;
##Minor lossses coefficient:
K=0.5;
##Viscosity(in kg/m-s):
u=10**-3.;
##Acceleration due to gravity(in /sec**2):
g=9.8;
##required##

##Reservoir depth required to maintain flow(in m):
D1=8.*Q**2./(math.pi)**2./D**4./g*(f*L/D+K+1.)
##Reynolds number:
Re=4.*d*Q/((math.pi)*u*D)
print("RESULTS")
print'%s %.2f %s'%("Reservoir depth required to maintain flow: ",D1+1.6," m")

RESULTS
Reservoir depth required to maintain flow:  6.06  m


## Ex6-pg353¶

In [6]:
import math
##Flow rate of crude oil(in bbl):
Q=1.6*10**6.;
##Inside diamete of pipe(i inches):
D=48.;
##Maximum allowable pressure(in psi):
p2=1200.;
##Minimum pressure required to keep gases dissolves(in psi):
p1=50.;
##Specific gravity of crde oil:
SG=0.93;
##Viscosity at 140 F(in lbf-s/ft**2):
u=3.5*10**-4.;
##Efficincy of pump:
Effp=0.85;
##Density(in slug/ft**3):
d=1.94;
##Viscosity (in lbf-sec):
u=3.5*10**-4.;
##Friction factor:
f=0.017;
##Maximum and power##

##Velocity of flow(in ft/sec):
V=Q/24./3600./(math.pi/4.*(D/12.)**2.)*42./7.48
##Maximum spacing(in ft):
L=2./f*D/12.*(p2-p1)/(SG*d)/V**2.*144.
##Power needed at each pump(in hp):
Win=1./Effp*V*math.pi/4.*(D/12.)**2.*(p2-p1)/550.*144.
print("RESULTS")
print'%s %.2f %s'%("Maximum spacing: ",L," feet")
print'%s %.2f %s'%("Power needed at each pump: ",Win," hp")

RESULTS
Maximum spacing:  630853.59  feet
Power needed at each pump:  36832.62  hp


## Ex7-pg355¶

In [7]:
import math
##Height of standpipe(in ft):
l=80.;
##Length of longest pipe(in ft):
L=600.;
##Diameter of pipe(in inches):
D=4.;
##Friction factor:
f=0.031;
##Acceleration due to gravityin ft/sec**2):
g=32.2;
##Volume low##

##Velocity(in ft/sec):
V2=math.sqrt(2.*g*l/(f*((L+l)/D*12.+8.)+1.))
##Volume flow rate(in gpm):
Q=V2*math.pi*(D/12.)**2./4.*7.48*60.
print("RESULTS")
print'%s %.2f %s'%("Volume low rate: ",Q,"")

RESULTS
Volume low rate:  350.07


## Ex8-pg357¶

In [8]:
import math
##Length of copper wire(in ft):
L=10.;
##Inner diameer of pipe(in inches):
D=1.5;
##Dischare(in ft**3/sec):
Q=0.566;
##Level of reservoir above pipe centreline(inn feet):
h=85.1;
##Kinematic viscosity at 70 F(in ft**2/s):
v=1.05*10**-5.;
##Acceleration due to gravity(in ft/sec**2):
g=32.2;
##Loss Coefficient##
##Average velocity (in ft/s):
V2=4./math.pi*Q/D**2.*144.
##Reynolds number:
Re=V2*D/v/12.
##For this value,
f=0.013;
##Power law exponent:
n=-1.7+1.8*math.log10(Re)
##Value of V/U:
v_u=2.*n**2./(n+1.)/(2.*n+1.)
##Value of alpha:
alpha=(1./v_u)**3.*2.*n**2./(3.+n)/(3.+2.*n)
##Loss Coefficient for a square edged entrance:
K=2.*g*h/V2**2.-f*L/D*12.-alpha;
print("RESULTS")
print'%s %.2f %s'%("Loss Coefficient for a square edged entrance: ",K," ")

RESULTS
Loss Coefficient for a square edged entrance:  0.50


## Ex9-pg359¶

In [9]:
import math
##Length of copper wire(in ft):
L=10.;
##Inner diameer of pipe(in inches):
D=1.5;
##Dischare(in ft**3/sec):
Q=0.566;
##Level of reservoir above pipe centreline(inn feet):
h=85.1;
##Kinematic viscosity at 70 F(in ft**2/s):
v=1.05*10**-5;
##Acceleration due to gravity(in ft/sec**2):
g=32.2;
##Loss Coefficient##
#Average velocity (in ft/s):
V2=4./math.pi*Q/D**2*144.
##Reynolds number:
Re=V2*D/v/12.
##For this value,
f=0.013;
##Power law exponent:
n=-1.7+1.8*math.log10(Re)
##Value of V/U:
v_u=2.*n**2./(n+1.)/(2.*n+1)
##Value of alpha:
alpha=(1./v_u)**3.*2.*n**2./(3.+n)/(3.+2.*n)
##Loss Coefficient for a square edged entrance:
K=2.*g*h/V2**2.-f*L/D*12.-alpha;
print("RESULTS")
print'%s %.2f %s'%("Loss Coefficient for a square edged entrance: ",K," ")

RESULTS
Loss Coefficient for a square edged entrance:  0.50


## Ex10-pg362¶

In [2]:
import math
##Nozzle exit diameter(in mm):
D=25.;
##N/R1 value value:
N_R=3.;
##AR value:
A_R=2.;
##Static head available from the main(in m):
z0=1.5;
##Acceleration due to gravity(in m/sec**2):
g=9.8;
##Value of Cp:
Cp=0.45;
##Volume and increase##

##Velocity V1(in m/s):
V1=math.sqrt(2.*g*z0/1.04)
##Volume flow rate(in m**3/sec):
Q=V1*math.pi*D**2./4.
Kdiff=1.-1./A_R**2.-Cp
##For 2nd case:
##Velocity(in m/s):
V1=math.sqrt(2.*g*z0/0.59)
##Volume flow rate(in m**3/s):
Qd=V1*math.pi*D**2./4.
##Increase in discharge after addition of diffuser is:
dQ=(Qd-Q)/Q*100.
print("RESULTS")
print'%s %.2f %s'%("Volume flow rate in case1: ",Q," m^3/sec")
print'%s %.2f %s'%("Volume flow rate in case 2: ",Qd," m^3/sec")
print'%s %.2f %s'%("Increase in discharge after addition of diffuser is: ",dQ," percent")

RESULTS
Volume flow rate in case1:  2609.92  m^3/sec
Volume flow rate in case 2:  3465.11  m^3/sec
Increase in discharge after addition of diffuser is:  32.77  percent


## Ex11-pg366¶

In [10]:
import math
##Volume flw rate of ai(in m**3/sec):
Q=1.;
##Diameter of pipe(in m):
D=0.25;
##Density of air (in kg/m**3):
d1=1.23;
##Acceleration due to gravity(in m/s**2):
g=9.8;
##Density of water(in kg/m**3):
d2=999.;
##Maxmum range of manometer(in m):
h=0.3;
##Kinematic viscosity(in m**2/s):
v=1.46*10**-5;

##Value of K*B**2:
K_B=Q/(math.pi/4.*D**2.)*math.sqrt(0.5*d1/g/d2/h)
##Reynods number:
ReD1=4./math.pi*Q/D/v
##By trial and error method, the value of beta is fixed at:
betta=0.66;
##K is then:
K=K_B/betta**2
##Diameter of orifice plate(in m):
Dt=betta*D
##Value of p3-p2(in N/m**2):
P1=d1*Q**2./(math.pi/4.*D**2.)**2.*(1./0.65/betta**2.-1.)
##Value of p1-p2(in N/m**2):
P2=d2*g*h
##Head loss between sections 1 and 3(in N-m/kg):
hLT=(P2-P1)/d1
##Expressing the permanent pressure as a fractio of the meter differential:
C=(P2-P1)/P2
print("RESULTS")
print'%s %.2f %s'%("Diameter of the orifice: ",Dt," m")
print'%s %.2f %s'%("Head loss between secions 1 and 3: ",hLT," N-m/kg")

RESULTS
Diameter of the orifice:  0.17  m
Head loss between secions 1 and 3:  1337.12  N-m/kg