'''Finding the density'''
#let the total mass of mud be 100lbm
#variables
m_total=100.0; #lbm
#70% by wt of mud is sand(SiO2)and remaining is water
m_sand=0.7*m_total; #lbm
m_water=0.3*m_total; #lbm
rho_sand=165.0; #lbm/ft^3
rho_water=62.3; #lbm/ft^3
#calculation
#rho=mass/volume
rho_mud=m_total/((m_sand/rho_sand)+(m_water/rho_water));
#result
print "The density of mud=" ,rho_mud, "lbm/ft^3"
'''
Calculate the shear stress at the surface of the inner cylinder
'''
import math
# variables
D1=25.15 #mm
D2=27.62 #mm
dr=0.5*(D2-D1) #mm
f=10. #rpm
# calculations
Vo=math.pi*D1*f/60. #mm/s
#Let D denote d/dr
DV=Vo/dr #s^-1
tow=0.005 #Nm
L=92.37 #mm
s=2*tow/D1**2/(math.pi)/L*(10**6) #N/m^2
# result
print "The stress at the surface of the inner cylinder is %f N/m^2"%s
#problem on surface tension
# variablees
l=0.10; #m (length of sliding part)
f=0.00589; #N (pull due to 0.6 gm of mass)
#calculation
f_onefilm=f/2; #N
#surface tension=(force for one film)/(length)
sigma=f_onefilm/l;
# result
print "The surface tension of fluid is",sigma,"N/m"
#Convert 327 miles/hr into ft/s
# variables
V=327. #miles/hr
#1 mile = 5280 ft
#1 hour = 3600 sec
# calculation
V1=V*5280/3600.0#ft/s
# result
print "327 miles/hr = %f ft/s"%V1
#Convert 2.6 hours into seconds
# variables
t=2.6 #hr
#1 hr = 3600 s
# calculations
t1=2.6*3600 #s
# result
print "2.6 hours = %f seconds"%t1
#Calculate the acceleration in ft/min^2
# variables
m=10. #lbm
F=3.5 #lbf
#1 lbf.s^2 = 32.2 lbm.ft
#1 min = 60 sec
# calculations
a=F*32.2*60**2/m #ft/min^2
# result
print "The acceleration provided is %f ft/min^2" % a
#Calculate the wt of metallic aluminium deposited in an electrolytic cell
# variables
I=50000. #Ampere or Coulumbs/sec
#1 hr = 3600 sec
I1=50000*3600. #C/hr
#calculation
#96500 C = 1 gm.eq
#1 mole of aluminium = 3 gm.eq
#1 mole of aluminium = 27 gm
m=I1*(1.0/96500)*(27/3.0)/1000.0 #Kg/hr
#result
print "the wt of metallic aluminium deposited in an electrolytic cell is %f Kg/hr"%m