# 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)
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)
#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)
#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 (Hô€€€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)
#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)
# 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)
#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)
#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)
#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)
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)
#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)
#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))
#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)
#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)