# Chapter 1 : Introduction¶

### Example 1.1 page no : 6¶

In :
'''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"

The density of mud= 110.401675438 lbm/ft^3


### Example 1.2 page no : 8¶

In :
'''
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

The stress at the surface of the inner cylinder is 0.054481 N/m^2


### Example 1.3 page no : 15¶

In :
#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"

The surface tension of fluid is 0.02945 N/m


### Example 1.4 page no : 20¶

In :
#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

327 miles/hr = 479.600000 ft/s


### Example 1.5 page no : 21¶

In :
#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

2.6 hours = 9360.000000 seconds


### Example 1.6 page no : 24¶

In :
#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

The acceleration provided is 40572.000000 ft/min^2


### Example 1.7 page no : 24¶

In :
#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

the wt of metallic aluminium deposited in an electrolytic cell is 16.787565 Kg/hr