# CHAPTER13 : FLUID MECHANICS¶

## Example E1 : Pg 260¶

In [1]:
import math
#example 13.1
#calculation of the force exerted by the water on the bottom

#given data
h=20.*10.**-2.#height(in m) of the flask
r=10.*10.**-2.#radius(in m) of the bottom of the flask
P0=1.01*10.**5.#atmospheric pressure(in Pa)
rho=1000.#density of water(in kg/m**3)
g=10.#gravitational acceleration(in m/s**2) of the earth

#calculation
P=P0+(h*rho*g)#pressure at the bottom
A=math.pi*r**2.#area of the bottom
F=P*A#force on the bottom

print '%s %.2f %s' %('the force exerted by the water on the bottom is',F,'N\n')

the force exerted by the water on the bottom is 3235.84 N



## Example E2 : Pg 262¶

In [2]:
import math
#example 13.2
#calculation of the volume of the cube outside the water

#given data
m=700.*10.**-3.#mass(in kg) of the cube
l=10.*10.**-2.#length(in m) of the cube
rho=1000.#density of water(in kg/m**3)

#calculation
V=m/rho#weight of displaced water = V*rho*g
Vtotal=l**3.#total volume of the cube
Vout=Vtotal-V#volume of the cube outside the water

print '%s %.2f %s' %('the volume of the cube outside the water is',Vout*10**6,'cm**3')

the volume of the cube outside the water is 300.00 cm**3


## Example E3 : Pg 264¶

In [3]:
#example 13.3
#calculation of the speed of the outgoing liquid

#given data
A1=1.*10.**-4.#area(in m**2) of the inlet of the tube
A2=20.*10.**-6.#area(in m**2) of the outlet of the tube
v1=2.#speed(in cm/s) of the ingoing liquid

#calculation
v2=A1*v1/A2#equation of continuity

print '%s %.2f %s' %('the speed of the outgoing liquid is',v2,'cm/s\n')

the speed of the outgoing liquid is 10.00 cm/s



## Example E4 : Pg 265¶

In [4]:
#example 13.4
#calculation of the difference in the pressures at A and B point

#given data
A1=1.*10.**-4.#area(in m**2) at point A of the tube
A2=20.*10.**-6.#area(in m**2) at point B of the tube
v1=10.*10.**-2.#speed(in m/s) of the ingoing liquid
rho=1200.#density of the liquid(in kg/m**3)

#calculation
v2=A1*v1/A2#equation of continuity
#by Bernoulli equtation.....P1 + (rho*g*h1) + (rho*v1**2/2) = P2 + (rho*g*h2) + (rho*v2**2/2)
deltaP=(1./2.)*rho*(v2**2.-v1**2.)

print '%s %.2f %s' %('the difference in the pressures at A and B point is',deltaP,'Pa\n')

the difference in the pressures at A and B point is 144.00 Pa



## Example E5 : Pg 266¶

In [15]:
import math
#example 13.5
#calculation of the speed of the water coming out of the tap

#given data
h=6.#depth(in m) of the tap
g=9.8#gravitational acceleration(in m/s**2) of the earth

#calculation
v=math.sqrt(2.*g*h)#torricellis theorem

print '%s %.2f %s' %('the speed of the water coming out of the tap is',round(v),'m/s\n')

the speed of the water coming out of the tap is 11.00 m/s



# WORKED EXAMPLES¶

## Example E1w : Pg 267¶

In [6]:
import math
#example 13.1w
#calculation of the force exerted by the mercury on the bottom of the beaker

#given data
h=10.*10.**-2.#height(in m) of the mercury
r=4.*10.**-2.#radius(in m) of the beaker
g=10.#gravitational acceleration(in m/s**2) of the earth
P0=1.*10.**5.#atmospheric pressure(in Pa)
rho=13600.#density of mercury(in kg/m**3)

#calculation
P=P0+(h*rho*g)#pressure at the bottom
A=math.pi*r**2.#area of the bottom
F=P*A#force on the bottom

print '%s %.2f %s' %('the force exerted by the mercury on the bottom of the beaker is',F,'N\n')

the force exerted by the mercury on the bottom of the beaker is 571.02 N



## Example E2w : Pg 267¶

In [7]:
import math
#example 13.2w
#calculation of the height of the atmosphere to exert the same pressure as at the surface of the earth

#given data
P0=1.*10.**5.#atmospheric pressure(in Pa)
rho=1.3#density of air(in kg/m**3)
g=9.8#gravitational acceleration(in m/s**2) of the earth

#calculation
h=P0/(g*rho)

print '%s %.2f %s' %("the height of the atmosphere to exert the same pressure as at the surface of the earth is",round(h),"m\n")

the height of the atmosphere to exert the same pressure as at the surface of the earth is 7849.00 m



## Example E3w : Pg 268¶

In [8]:
#example 13.3w
#calculation of the height of the water coloumn

#given data
h1=2.*10.**-2.#difference in the height(in m)
s=13.6#specific gravity of mercury

#calculation
#P = P0 + (h*rho*g)........using this equation
h=h1*s#height of the water coloumn

print '%s %.2f %s' %('the height of the water coloumn is',h*10**2,'cm\n',)

the height of the water coloumn is 27.20 cm



## Example E5w : Pg 268¶

In [9]:
#example 13.5w
#calculation of the force applied on the water in the thicker arm

#given data
A1=1.*10.**-4.#area(in m**2) of arm 1
A2=10.*10.**-4.#area(in m**2) of arm 2
f=5.#force(in N) applied on the water in the thinner arm

#calculation
#P = P0 + (h*rho*g)........using this equation
F=f*A2/A1#force applied on the water in the thicker arm

print '%s %.2f %s' %('the force applied on the water in the thicker arm is',F,'N\n')

the force applied on the water in the thicker arm is 50.00 N



## Example E6w : Pg 268¶

In [10]:
#example 13.6w
#calculation of the elongation of the spring

#given data
m=10.*10.**-3.#mass(in kg) of the copper piece
l=1.*10.**-2.#elongation(in m) in the spring
g=10.#gravitational acceleration(in m/s**2) of the earth
rho=9000.#density of copper(in kg/m**3)
rho0=1000.#density of water(in kg/m**3)

#calculation
k=m*g/l#spring constant
V=m/rho#volume of copper
Fb=V*rho0*g#force of buoyancy
x=((k*l)-Fb)/k#elongation of the spring

print '%s %.2f %s' %('the elongation of the spring is',x*10**2,'cm\n')

the elongation of the spring is 0.89 cm



## Example E7w : Pg 268¶

In [11]:
#example 13.7w
#calculation of the maximum weight that can be put on the block without wetting it

#given data
l=3.*10.**-2.#length(in m) of the edge of the cubical block
rho=800.#density of wood(in kg/m**3)
k=50.#spring constant(in N/m)
g=10.#gravitational acceleration(in m/s**2) of the earth
rho0=1000.#density of water(in kg/m**3)

#calculation
s=rho/rho0#specific gravity
hin=l*s#height inside water
hout=l-hin#height outside water
V=l**3.#volume of the block
Fb=V*rho0*g#force of buoyancy
Fs=k*hout#force exerted by the spring
Wdash=V*rho*g#weight of the block
W=Fb+Fs-Wdash#maximum weight

print '%s %.2f %s' %('the maximum weight that can be put on the block without wetting it is',W,'N\n')

the maximum weight that can be put on the block without wetting it is 0.35 N



## Example E8w : Pg 269¶

In [12]:
import math
from math import acos,sqrt
#example 13.8w
#calculation of the angle that the plank makes with the vertical in equilibrium

#given data
l=1.#length(in m) of the planck
h=0.5#height(in m) of the water level in the tank
s=0.5#specific gravity of the planck

#calculation
#A = OC/2 = l/(2*cosd(theta)
# mg = 2*l*rho*g
#buoyant force   Fb=(2*l*rho*g)/cosd(theta)
#m*g*(OB)*sind(theta) = F(OA)*sind(theta)
theta=acos(math.sqrt(1./2.))*57.3

print '%s %.2f %s' %('the angle that the plank makes with the vertical in equilibrium is',theta,'degree\n')

the angle that the plank makes with the vertical in equilibrium is 45.00 degree



## Example E10w : Pg 269¶

In [13]:
import math
#example 13.10w
#calculation of the rate of water flow through the tube

#given data
A1=30.#area(in cm**2) of the tube at point A
A2=15.#area(in cm**2) of the tube at point B
deltaP=600.#change in pressure(in N/m**2)
rho0=1000.#density of the water(in kg/m**3)

#calculation
r=A1/A2#ratio of area
#from equation of continuity  vB/vA = A1/A2 = r = 2
#by Bernoulli equtation.....P1 + (rho*g*h1) + (rho*v1**2/2) = P2 + (rho*g*h2) + (rho*v2**2/2)
#take vB = vA*2
vA=math.sqrt(deltaP*(r/(r+1.))*(1./rho0))
Rflow=vA*A1#rate of water flow

print '%s %.2f %s' %('the rate of water flow through the tube is',Rflow*10**2,'cm**3/s\n')

the rate of water flow through the tube is 1897.37 cm**3/s



## Example E11w : Pg 270¶

In [14]:
import math
#example 13.11w
#calculation of the velocity of the water coming out of the opening

#given data
AA=.5#area(in m**2) of the tank
AB=1.*10.**-4.#area(in m**2) of the cross section at the bottom
m=20.#mass(in kg) of the load
h=50.*10.**-2.#height(in m)of the water level
g=10.#gravitational acceleration(in m/s**2) of the earth
rho=1000.#density of the water(in kg/m**3)

#calculation
#from the equation............P = P0 + (h*rho*g)#pressure at the bottom
r=m*g/AA#in above equation it is the value of (h*rho*g)
#on solving,we get............PA = P0 + (400 N/m**2)
#from Bernoulli equtation.....P1 + (rho*g*h1) + (rho*v1**2/2) = P2 + (rho*g*h2) + (rho*v2**2/2)
#we get
vB=math.sqrt((2.*(r+(rho*g*h)))/rho)

print '%s %.2f %s' %('the velocity of the water coming out of the opening is',vB,'m/s\n')

the velocity of the water coming out of the opening is 3.29 m/s