Chapter 03:Pressure and Fluid Statics

Example 3.3-1, Page No:75

In [1]:
import math

#Variable Decleration
Pvac=40 #Vaccum Gauge Reading in kPa
Patm=100 #Atmospheric Pressure in kPa


#Calculations
Pabs=Patm-Pvac #Absolute Pressure reading in kPa


#Result
print"The absolute Pressure is",Pabs,"kPa"

The absolute Pressure is 60 kPa

Example 3.3-2, Page No:80

In [3]:
import math

#Variable Decleration
h=0.74 #Barmoetric Pressure in m
g=9.805 #Acceleration due to gravity in m/s^2
rho=13570 #Density of mercury in kg/m^3
c=1000 #Conversion factor in N/m^2


#Calculations
Patm=(rho*g*h)/c #Atmospheric Pressure in kPa


#Result
print"The Atmospheric Pressure is",round(Patm,1),"kPa"

The Atmospheric Pressure is 98.5 kPa

Example 3.3-3, Page No:80

In [8]:
import math

#Variable Decleration
h_arm_bottle=1.2 #Height of bottle above armlevel in m
rho=1020 #Density of IV fluid in kg/m^3
g=9.81 #Acceleration due to gravity in m/s^2
P_gauge=20 #Gauge pressure in part b calculations in kPa
c=1000 #Conversion factor in kg.m/s^2


#Calculations

#Part(a)
Pgaugearm=rho*g*h_arm_bottle/c #Gauge Pressure in kPa

#Part(b)
h_arm_bottle=(P_gauge*c)/(rho*g) # Height of the surface of the IV Fluid in m

#Result
print"The gaguge pressure  at 1.2m is",round(Pgaugearm),"kPa"
print"The height of the surface of the IV Fluid at 20kPa is",round(h_arm_bottle),"m"

The gaguge pressure  at 1.2m is 12.0 kPa
The height of the surface of the IV Fluid at 20kPa is 2.0 m

Example 3.3-4, Page No:81

In [36]:
import math

#Variable Decleration
H=4 #Thickness of gradient zone in m
rho_o=1040 #Density at the water surface in kg/m^3
s=4 #Depth in m
g=9.81 #Acceleration due to gravity in m/s^2
h1=0.8 #Surface zone thickness in m
c=1000 #Conversion Factor in kg.m/s^2
pi=3.14


#Calculations
P1=(rho_o*g*h1)/c # Gauge Pressure at bottom of surface zone in kPa


#After carrying out the integral


#The next three steps are for mere mathametical simplicity
theta=(pi*s)/(4*H) #Angle Conversion into degrees
x=round(tan(theta)) #Tangent calculation
y=arcsinh(x) #Sine Inverse Hyperbolic calculation
P2=P1+((y*rho_o*g*4*H)/(pi*c)) #Gauge Pressure at the bottom in kPa 



#Result
print"The gauge pressure at the bottom is",round(P2),"kPa"

The gauge pressure at the bottom is 54.0 kPa

Example 3.3-5, Page No:83

In [39]:
import math

#Variable Decleration
SG=0.85 #Specific Gravity of the Fluid 
h=0.55 #Manometer column height in m
rho_H2O=1000 #Density of water in kg/m^3
Patm=96 #Local Atmospheric Pressure in kPa
g=9.81 #Acceleration due to gravity in m/s^2
c=1000 #Conversion Factor in N/m^2


#Calculations
rho=rho_H2O*SG #Density of the fluid in kg/m^3
P=Patm+((rho*g*h)/c) #Absolute Pressure in the tank in kPa



#Result
print"The absolute pressure in the tank is",round(P,1),"kPa"

The absolute pressure in the tank is 100.6 kPa

Example 3.3-6, Page No:84

In [53]:
import math

#Variable Decleration
H=1400 #Altitude of tank in m
Patm=85.6 #Atmospheric Pressure at H altitude in kPa
h1=0.1 #m
h2=0.2 #m
h3=0.35 #m
rho_water=1000 #Density of water in kg/m^3
rho_oil=850 #Density of oil in kg/m^3
rho_mercury=13600 #Density of mercury in kg/m^3
g=9.81 #Acceleration due to gravity in m/s^2
c=1000 #Conversion Factor in N/m^2



#Calculations

#For Simplicity we do the following
X=g*(rho_mercury*h3-rho_water*h1-rho_oil*h2)
P1=Patm+((X/c)) #Pressure at point 1 in kPa

#Result
print"The pressure at point1 is",round(P1),"kPa"

The pressure at point1 is 130.0 kPa

Example 3.3-8,Page No:92

In [56]:
import math

#Variable Decleration
s=8 #Distance of top of the edge of the door in m
b=1.2 #Height of the door in m
w=1 #width of the door
rho=1000 #Density of water in kg/m^3
g=9.81 #Acceleration due to gravity in m/s^2
c=1000 #Conversion Factor in kg.m/s^2

#Calculations
Pavg=(rho*g*(s+(b/2)))/c #Average Pressure in kN/m^2

#Hydrostatic Force
Fr=Pavg*(b*w) #Hydrostatic Force in kN
yp=s+(b/2)+(b**2/(12*(s+(b/2)))) #Center of pressure in m

#Result
print"The Average Pressure is",round(Pavg,1),"kN/m^2"
print"The resultant Hydrostatic Force is",round(Fr,1),"kN"
print"The center of pressure is at",round(yp,2),"m"

The Average Pressure is 84.4 kN/m^2
The resultant Hydrostatic Force is 101.2 kN
The center of pressure is at 8.61 m

Example 3.3-9,Page No:95

In [68]:
import math

#Variable Decleration
R=0.8 #Radius in m
s=4.2 #m
g=9.81 #Acceleration due to gravity in m/s^2
rho=1000 #Density of water in kg/m^3
b=1 #m
h_bottom=5 #Depth in m
c=1000 #Conversion Factor in kg m/s^2

#Calculations

#Part(a)

#Horizontal Force 
Fh=(rho*g*(s+(R/2))*(R*b))/c #Horizontal Force in kN

#Verticla Force
Fy=(rho*g*h_bottom*R*b)/c #Vertical Force in kN

#Weight of fluid block
W=(rho*g*(R**2-((pi*R**2)/4))*b)/c #Weight of fluid block in kN
Fv=Fy-W #Net upward Force in kN
Fr=(Fh**2+Fv**2)**0.5 #Resultant Force in kN
thet=arctan(Fv/Fh) #Theta in radians
theta=(180/pi)*thet #Thetain degrees

#Part(b)

#Weight of the cylinder per m length
Wcyl=Fr*sin(thet) #Weight of the cylinder in kN

#Result
print"The magnitude of hydrostatic force is",round(Fr,1),"kN"
print"The direction of the hydrostatic force is",round(theta,1),"degrees"
print"The weight of the cylinder per m length is",round(Wcyl,1),"kN"

The magnitude of hydrostatic force is 52.3 kN
The direction of the hydrostatic force is 46.4 degrees
The weight of the cylinder per m length is 37.9 kN

Example 3.3-10,Page No:98

In [70]:
import math

#Variable Decleration
rho_w=1000 #Density of water in kg/m^3
h_sub=0.1 #m
R=0.005 #Radius in m

#Calculations

#Part(a) is theoretical hence not coded here

#Part(b)
V=pi*R**2*h_sub #Volume in m^3
m=rho_w*V # Mass in kg

#Result
print"The mass is",m,"kg"

The mass is 0.00785 kg

Example 3.3-11,Page No:99

In [72]:
import math

#Variable Decleration
rho_f=1025 #Density of sea-water in kg/m^3
rho_concrete=2300 #Density of concrete in kg/m^3
g=9.81 #Acceleration due to gravity in m/s^2
l=0.4 #length of block in m
b=0.4 #breadth of block in m
h=3 #height of block in m
c=1000 #Conversion factor in kg.m/s^2

#Calculations

#Part(a)
V=l*b*h #Volume in m^3
Ft_air=(rho_concrete*g*V)/c #Tension in the rope in kN
W=Ft_air #kN

#Part(b)
Fb=(rho_f*g*V)/c #Force of Buoyancy in kN

Ft_Water=W-Fb #Tension in the rope under water in kN

#Result
print"The tension in the string under water is",round(Ft_Water),"kN"

The tension in the string under water is 6.0 kN

Example 3.3-12, Page No:106

In [78]:
import math

#Variable decleration
vo=0 #Initial Velocity in km/h
vf=90 #Final Velocity in km/h
t=10 #Acceleration time in s
h=0.8 #Height of the tank in m
b1=2 #Breadth of the tank in m
b2=0.6 #Width of the tank in m
g=9.81 #Acceleration due to gravity in m/s^2
c=3.6 #Conversion factor in km/h
az=0 #Acceleration in the z direction in m/s^2
d=100 #Conversion from m to cm

#Calculations
ax=((vf-vo)/t)/c #Acceleration in x direction in m/s^2
thet=arctan(ax/(g+az)) #Theta in radians

#CASE 1
delta_zs1=d*(b1/2)*tan(thet) #cm

#CASE 2
delta_zs2=d*(b2/2)*tan(thet) #cm

#Result
print"Assuming the tipping is not a problem,The tank should be definitely be placed in such a"
print"way that the short side is parallel to the direction of motion"
print"As",round(delta_zs2,1),"cm <",round(delta_zs1,1),"cm"

Assuming the tipping is not a problem,The tank should be definitely be placed in such a
way that the short side is parallel to the direction of motion
As 7.6 cm < 25.5 cm

Example 3.3-13,Page No:109

In [81]:
import math
g=9.81 #Acceleration due to gravity in m/s^2
H=0.6 #Height of vertical cylinder container in m
ho=0.5 #Height of partially filled container in m
rho=850 #Density of liquid in kg/m^3
R=0.1 #Radius in m
c=60 #Conversion factor in s

#Calculations
w=((4*g*(H-ho))/R**2)**0.5 #Max rotational Speer in rad/s
n_dot=(w/(2*pi))*c #RPM

#Result
print"The Rotational speed at which the liquid just starts to spill is",round(n_dot),"RPM"

The Rotational speed at which the liquid just starts to spill is 189.0 RPM