Chapter1: Fluid Statics

Example 1.1,Page 3

In [2]:
# Variable Decleration
rho=924; #density
g=9.81; #grivity
H=2; #height
d=2; # depth 

#Calculation
p=rho*g*H;
a=d*H;
F=p*a/2;

#result
print" Total force exerted over the wall in(N) =",round(F,3)
 Total force exerted over the wall in(N) = 36257.76

Example 1.2,Page 5

In [5]:
import math

#Variable declaration
p_v =50*1000; #pressure
r =1; #m
p_atm =101.3*1000; #atmospheric pressure
rho =1000; #density
H =2.5; #m
g =9.81; #m/s^2


#Calculation
F= p_v*math.pi*r*r;
p= p_atm + p_v + rho *g*H;
Fd =( p_v+ rho *g*H)*math.pi*r*r+rho*g*2*math.pi*r*r/3;

#result
print " Total vertical force tending to lift the dome (N)", round(F,3)
print " Absolute pressure at the bottom o f the vessel (Pa)",round(p,3)
print "Downward force imposed by the gas and liquid (N)", round(Fd,3)
 Total vertical force tending to lift the dome (N) 157079.633
 Absolute pressure at the bottom o f the vessel (Pa) 175825.0
Downward force imposed by the gas and liquid (N) 254673.208

Example 1.3,Page 7

In [3]:
#Variable Decleration
a1 =0.3; #m^2
m =1000; # kg
a2 =0.003; #m^2
rho_oil =750; #kg /m^3
H =2; #m
g =9.81; #m/ s ^2

# Calcualtion
F1=m*g;
F2=a2 *( F1/a1 - rho_oil *g*H);

#result
print "The force on the plunger(N)",round(F2,3)
The force on the plunger(N) 53.955

Example 1.4,Page 8

In [2]:
#variable Decleration
rho_0 =800; # kg /m^3
rho_aq =1100; # density of aqueous solution

#calculation
H =0.5* rho_aq /( rho_aq - rho_0 );
# For a fixed length of chamber o f 3 m, the
#interface between the two phases is determined
#from the p r e s s u r e i n the chamber and d i s c h a r g e
#p o i n t .
#r h o 0 gH1+rho a q gH2=rho a q g (H􀀀0.5) ;
#H=H1+H2
rho_0 =600; #kg /m^3
H1 =0.5* rho_aq /( rho_aq - rho_0 );

#result
print "H(m)",round(H,3)
print "The lowest possible position of the interface in the chamber below the overflow (m)",round(H1,2)
H(m) 1.833
The lowest possible position of the interface in the chamber below the overflow (m) 1.1

Example 1.5, Page11

In [2]:
#variable decleration
rho_o =900; #kg /m^3
rho_n =1070; #kg /m^3
H =1; #m
g =9.81; #m/ s ^2
dp =10*10**3; #change in pressure

#H=H1+H2
#calculation
H1 =(dp - rho_n *g*H)/( rho_o - rho_n )/g;

#result
print "The position of the interface between the legs (m)", round(H1,2)
The position of the interface between the legs (m) 0.3

Example 1.6,Page 13

In [12]:
# Variable defining
dp =22*10**3; #N/m^2
g =9.81;#m/ s ^2
H =1.5;#m
rho =1495.00;#kg /m^3
rho_s =1270;#kg /m^3
rho_c =2698;#kg /m^3

# Calculation
p=dp/g/H;
#rho=f 1  r h o s+f 2  r h o c
#f 1+f 2=1
f2 =( rho - rho_s )/( rho_c - rho_s );

#result
print " the density of the solution with crystals (kg/m^3)",round(p,3)
print "The fraction of crystals ", round(f2,3)
 the density of the solution with crystals (kg/m^3) 1495.073
The fraction of crystals  0.158

Example 1.7, Page 15

In [3]:
#Variable defining
p_atm =101.3*10**3; #N/m^2
rho =1000; #kg /m^3
g =9.81; #m/ s ^2
H1 =3; #m
a =0.073; #N/m
r1 =5*10**( -4) ; #m
p1= p_atm + rho *g*H1 +2* a/r1;

#calculation
# p2=p atm+rho gH2+2a / r2 ;
# p1 4/3%pi r1 ^3=p2 4/3%pi r2 ^3
# Solving above two equations we get
r2 =0.053; #//mm

#result
print"Radius of the bubble(mm)",round(r2,3)
Radius of the bubble(mm) 0.053

Example 1.8,Page 17

In [9]:
#Variable defining
H =0.2; #m
rho =1000; #kg /m^3
rho_Hg =13600; #kg /m^3
g =9.81; #m/ s ^2

#calculation
dp =( rho_Hg -rho)*g*H;

#result
print " Differential pressure(Pa)",round(dp,2)
 Differential pressure(Pa) 24721.2

Example 1.9,Page 19

In [1]:
#variable defining
rho =1000;
g =9.81; # m/ s ^2
H =0.4; #m

#calculation
dp=rho*g*H;

#result
print" Pressure drop in the pipe(Pa)",round(dp)
 Pressure drop in the pipe(Pa) 3924.0

Example 1.10,Page 21

In [11]:
import math

#variable defining
dp =20*10**3; #N/m^2
rho_Hg =13600; # kg /m^3
rho =700; #kg /m^3
g =9.81; #m/ s ^2
d =0.02; #m


#calculation
H=dp*(rho_Hg - rho)**-1/g;
V= math.pi /4*d*d*H;

#result
print"Quantity of mercury to be removed (m^3)",round(V,9)
Quantity of mercury to be removed (m^3) 4.965e-05

Example 1.11,Page 23

In [1]:
#Variable defining
import math
from math import sin
rho =800; #kg /m^3
g =9.81; #m/ s ^2
L =0.12;


#calculations
theta = math.pi /180*20; #radians
dp=rho*g*L*(sin(theta));

#result
print "The gauge pressure across the filter(Pa)",round(dp,2)
The gauge pressure across the filter(Pa) 322.100890178

Example 1.12,Page 25

In [13]:
#Variable defining
mc =100; #kg
g =9.81; #m/ s ^2
rho =1000; #kg /m^3
rho_c =7930; #kg /m^3

#calculation
m=mc*rho/ rho_c ;
F=mc*g-m*g; #tension

#result
print("The tension in the cable(N)",round(F,2))
('The tension in the cable(N)', 863.28)

Example 1.13,Page 27

In [5]:
#Variale defining
rho =1000.0;
x =0.06;
rho_0 =800.0;
x_0 =0.04;
rho_L =900.0;

#calculation
L=( rho*x- rho_0 *x_0)/( rho - rho_0 );
x_L =L-rho/ rho_L *(L-x);

#result
print " Length of the stem above the liquid of SG 0.9 is (m)",round(x_L,3)
 Length of the stem above the liquid of SG 0.9 is (m) 0.05

Example 1.14,Page 29

In [1]:
#variable defining
m_s =5*10**6; #kg
T2 =4.5; #m
T1 =3; #m
rho_hc =950; #kg /m^3
Q =125; #m^3/h

#calculation
m_hc =m_s *( T2/T1 -1);
t= m_hc / rho_hc /Q;

#result
print"Quantity delivered(kg)",round(m_hc,2)
print "Time taken(hours)",round(t,3)
Quantity delivered(kg) 2500000.0
Time taken(hours) 21.053
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