# Chapter 2 : Fluid Statics¶

## Example 2.1 Page no 44¶

In [4]:
# Force applied on the piston

from math import *

# Given

d = 10                        # diameter of hydraulic press in meters

d1 = 1                       # diameter of piston in meters

W = 1000                        # weight in Newtons

Ap = math.pi*d1**2/4         # Area of piston in m**2

Ar = math.pi*d**2/4        # Area of rram in m**2

# Solution

p = W/Ar                       # pressure to be supplied by the oil in N/cm**2

F = p*Ap                       # Force applied on the piston

print "Using the pascal's law a weight of 1000N can be lifted by applying a force only of ",round(F,1),"N"


Using the pascal's law a weight of 1000N can be lifted by applying a force only of  10.0 N


### Example 2.2 Page no 53¶

In [5]:
# Pressure in kN/m**2

# Given

h = 1                           # ocean depth below the surface in km

gma = 10070                     # Specific weight of sea water

# Solution

P =gma*h                        # Pressure in kN/m**2

print "Pressure = ",round(P),"kN/m**2"

Pressure =  10070.0 kN/m**2


### Example 2.3 Page no 53¶

In [15]:
# Pressure at the bottom of the tank

# Given

p1 = 150*10**3                  # Pressure at point 1 in kN/m**2

Sg = 0.85                       # Specific gravity of oil

h = 0.8                         # height of oil 2 i tank in meters

g = 9810                        # specific gravity

h1 = 2.0                        # height of oil 3 in tank

# Solution

p2 = (p1 + Sg*h*g)

p3 = (p2 + g*h1)/1000

print "Pressure at the bottom of the tank is",round(p3,1),"kN/m**3"

Pressure at the bottom of the tank is 176.3 kN/m**3

### Example 2.4 Page no 54¶

In [6]:
# Height of the mountain

# Given
from math import *

from __future__ import division

Po = 570                         #  mercury barometer reading in mm

T = 273 -5                       # temperature in K

p = 750                          # mercury barometer reading in mm

n = 1.2345                       # for polytropic atmosphere

R  =287                          # univerasl gas constant in J/Kg-K

g = 9.81

r = p/Po

# Solution

y = -(R*T/(g*0.19))*(1 - (r)**((n-1)/n))

print "Height of the mountain is",round(y,0),"m"

Height of the mountain is 2208.0 m


### Example 2.5 Page no 57¶

In [7]:
# Pressure in the pipe

# Given

h1 = 500           # height in mm

h2 = 950           # height in mm

S1 = 1             # specific gravity fo water

S2 = 1.5           # specific gravity of liquid 2

w = 9810           # specific weight of water

# Solution

ha = ((h2*S2)-(h1*S1))/1000

Pa = w*ha/1000     # Pressure in kPa

print "Pressure in the pipe = ",round(Pa,3),"kPa"

Pressure in the pipe =  9.074 kPa


### Example 2.6 Page no 66¶

In [18]:
# Force required to open the gate

# Given

from __future__ import division

from math import *

d = 1                  # diameter of the gate in m

w = 9810               # specific weight in N/m**3

A = pi*d**2/4          # area of the gate

Ig = pi*d**4/64        # mamm moment of inertia

theta = pi/2

y1= 5+0.5

# Solution

F = w*round(A,3)*(y1) # the answer will come out to be different as they have used the value of Area as 0.78

# depth of the COP

h1 = y1 + (Ig*math.sin(theta)*math.sin(theta)/(A*y1))

F1 = (F*(h1 - 5))/d

print "Magnitude of the force required to open the gate = ",round(F1,0),"N"
#The area calculated in the book is 0.785 and that calculated from code is 0.78.
#This difference of 0.005 is causing the answer to change from the original

Magnitude of the force required to open the gate =  21659.0 N


### Example 2.7 Page no 67¶

In [3]:
# total force ; position of the center of pressure

from math import *

# Given

l =2             # length of the plate in m

b =1             # width of the plate

theta = pi/3

h = 0.75         # depth of the plate

w = 9810         # specific weight of water

# Solution

A = 1*2

y1 = h + 1*sin(theta)

F = w*A*y1/1000

print "(a) Total force = ",round(F,2),"kN"

Ig = (b*l**3)/12

h1 = y1 + (Ig*sin(theta)*sin(theta)/(2*y1))

print "(b) Position of center of pressure = ",round(h1,3),"m"

(a) Total force =  31.71 kN
(b) Position of center of pressure =  1.616 m


### Example 2.8 Page no 68¶

In [9]:
# Hydrostatic force  and point of location

from math import *

# Given

d = 6                   # diameter of the gate in ft

A =pi*d**2/4            # area of the gate

p1 = 600                # pressure on top in psia

y1 = 10 +2 + 3*sin(pi/6)

F = 62.4*A*y1

F1 = p1*A

# Solution

Tf = F+F1

print "Total hydrostatic force =",round(Tf,0),"lbs"

Ig = pi*d**4/64

h1 = y1 + ((Ig*sin(pi/6)*sin(pi/6))/(A*y1))

H = ((F*h1)+(F1*y1))/Tf

print "point of location on the center plate = ",round(H,2),"ft"

# method 2

Hf = p1/62.4            # equivalent fluid height

y2 = Hf+y1

Tf1 = 62.4*A*y2

h2 = y2 + ((Ig*sin(pi/6)*sin(pi/6))/(A*y2))

H1 = y2-Hf

print " OR point of location on the center plate from method 2 = ",round(H1,2),"ft"

Total hydrostatic force = 40783.0 lbs
point of location on the center plate =  13.52 ft
OR point of location on the center plate from method 2 =  13.5 ft


### Example 2.10 Page no 74¶

In [10]:
# Horizontal and Vertical components

import math

# Given

R = 4                 # radius of the gate in ft

w = 1                # width of the gate in ft

gma =62.4            # specific weight of water

y1 = 4               # distance of center of the gate

xv1 = 2              # distance in ft

xv2 = 1.7           # distance in ft

# Solution

Fh = R*y1*gma

Ig = w*R**3/12

yh = y1 + (Ig/(R*y1))

Fv1 = R*2*gma

Fv2 = pi*R**2*gma/4

Fv = Fv1 + Fv2

Xv = (Fv1*xv1+Fv2*xv2)/(Fv)

print "Horizontal component acting on the plate = ",round(Fh,2),"lbs"

print "Vertical component acting on the plate = ",round(Fv,2),"lbs"

print "location of Xv =",round(Xv,2),"ft"

Horizontal component acting on the plate =  998.4 lbs
Vertical component acting on the plate =  1283.34 lbs
location of Xv = 1.82 ft


### Example 2.11 Page no 77¶

In [19]:
# Vertical and Horizontal components

from math import *

from __future__ import division

# Given

p = 50                         # pressure in psia

gma = 62.4                     # specific weight in ft

h1 = p*144/gma                 # equivalent height of water surface in ft

R = 4                          # radius of gate in ft

w = 1                          # width of the gate in ft

A = R*w

y1 = h1 + 2.5 + 2              # center of pressure

xv1 = 2                        # center of pressure1 for x direction force

xv2 = 1.7                      # center of pressure2 for x direction force

# Solution

Fh = gma*A*y1                 # hiorizontal force

Ig = 5.33                      # moment of inertia

yh = y1 + (Ig/(A*y1))          # location of horizontal component

y2 = h1+2.5

Fv1 = gma*(R*y2)               # vertical force component 1

Fv2 = gma*(pi*R**2/4)               # vrtical force component 2

Fv = Fv1 + Fv2                # vertical force component

Xv = (Fv1*xv1+Fv2*xv2)/(Fv)

print "(a) Horizontal component acting on the plate = ",round(Fh,0),"lbs" # The answer for horizontal force in the book is wrong

print "(b) Vertical component acting on the plate = ",round(Fv,2),"lbs"

print "location of vertical component Xv =",round(Xv,2),"ft from the left wall"

(a) Horizontal component acting on the plate =  29923.0 lbs
(b) Vertical component acting on the plate =  30208.14 lbs
location of vertical component Xv = 1.99 ft from the left wall


### Example 2.13 Page no 84¶

In [12]:
# Depth to which water would rise

from math import *

from __future__ import division

# Given

l = 3                    # length in m

b = 4                    # breadth in m

h = 15                    # height in m

S = 0.9                 # specific gravity of barge

Sw  = 1.09              # specific gravity of water

Wd = 150*10**3          # addditional weight in kN

V = l*b*h               # volume in m**3

Wb = gma*S*V          # weight of barge

Tw = Wb + Wd            # total weight in kN

gma = 9810              # specific density

# Solution

Fb = Tw                 # since barge is floating

V1 = Fb/((gma/1000)*Sw)    #  volume in m**3

d = (V1/(h*b))/1000

print "Depth to which water would rise = ",round(d,2),"m"

Depth to which water would rise =  0.25 m


### Example 2.14 Page no 85¶

In [13]:
# Level at which cylinder will float

from math import *

from __future__ import division

import numpy as np

# Given

W = 0.4 * 9.81              # weight of the solid cylinder in N

# Solution

A = np.array([(1,-0.96),(1,1)])

b = np.array([0,6.37])

x = np.linalg.solve(A,b)

X = x[0]

Y = x[1]

print "X = ",round(X,2),"cm"

print "Y = ",round(Y,2),"cm"

b = 8 -x[0]

print "The bottom of the solid cylinder will be ",round(b,2),"cm above the bottom"

X =  3.12 cm
Y =  3.25 cm
The bottom of the solid cylinder will be  4.88 cm above the bottom


### Example 2.15 Page no 86¶

In [14]:
# Weight of the pan and the magnitude of righting moment

from math import *

# Given

l =100                # length of the pan in cm

w = 20                # width of the pan in cm

d = 4                 # depth of the pan in cm

L = 1.5               # load in N/m

gma = 9810              # sepcific weight

# Solution

Fb = gma*(d*w*l/(2*l**3)) # weight on the pan

W = Fb-L                # weight of the pan

X1 = w/3

X2 = w/2

theta = math.atan(d/w)*180/pi

x = ((X2-X1)*cos(theta*pi/180))

# momentum equation

M = W*x

print "Weight of the pan =",round(W,1),"N"
print "Magnitude of right momentum = ",round(M,1),"N.cm"

Weight of the pan = 37.7 N
Magnitude of right momentum =  123.4 N.cm


### Example no 2.16 Page no 90¶

In [15]:
# Metacentric height and rightning moment

from math import *

from __future__ import division

# Continued from example 2.15

# Given

Io = 15*4**3/12                     # moment of inertia in m**4

V = 15*4*2.71                       # Volume in m**3

Gb = ((3/2)-(2.71/2))

W = 1739.2                          # weight of the barge from the previous example in kN
# Solution

Mg = (Io/V)-Gb                     # metacentric height in m

print "(a) Metacentric height = ", round(Mg,3),"m"

M = W*Mg*sin(pi*6/180)

print "(b) Righting moment =",round(M,1),"kN.m"

(a) Metacentric height =  0.347 m
(b) Righting moment = 63.1 kN.m


### Example no 2.17 Page no 92¶

In [16]:
# Maximum pressure in the tank

from math import *

from __future__ import division

# Given

l=6                  # length of the tank

w =2                 # width of the tank

d = 3                # depth of the tank

a = 3                # acceleration in m/s**2

theta = pi/6

W = 9810            # specific weight

X = 0

po=0               # pressure at the origin

# Solution

A = np.array([(1,-1),(1,1)])

b = np.array([-1.38,3.0])

x = np.linalg.solve(A,b)

Y1 = x[0]

Y2 = x[1]

# AMximum pressure at the bottom of the tank

P = po - W*(2.61*X/9.81) - W*(1+(1.5/9.81))*(-Y2)

print"Amximum pressure occurs at the bottom of the tank =",round(P,3),"N"

Amximum pressure occurs at the bottom of the tank = 24768.9 N


### Example 2.18 Page no 98¶

In [17]:
# Height of the paraboloid revolution; maxm=imum pressure and location ; pressure at the point 0.2 from the center

from math import *

from __future__ import division

# Given

d = 1                          # diamter of the jar in ft

h =2                           # height of the jar in ft

H = 1                          # height of water in the jar in ft

N = 150                        # RPM

g = 32.2                       # acceleration due to gravity in ft/s**2

# Solution

w = 2*pi*N/60

ho = H+((w**2*(d/2)**2)/(4*g))
print "(a) Height of paraboliod revolution of base = ", round(ho,2),"ft"

Pmax = 62.4*ho

print "(b) Maximum pressure corresponding to maximum height = ",round(Pmax,1),"lbs/ft**2"

z = H - ((w**2*(d/2)**2)/(4*g))

r = 0.2                # distance from center

y = -(0.52-0.25)

P = po + (62.4*w**2*r**2/(2*g))-(62.4*y)

print "(c) Pressure =",round(P,1),"lbs/ft**2"

(a) Height of paraboliod revolution of base =  1.48 ft
(b) Maximum pressure corresponding to maximum height =  92.3 lbs/ft**2
(c) Pressure = 26.4 lbs/ft**2

In [ ]: