Chapter 1 : Basic Concepts

Example 1.1 Page No : 5

In [1]:
import math 
			
#Initialization of variables
weight = 9800. 			#Kg
g = 9.81      			#m/s**2
a = 2.   		    	#m/s**2
			
#calculations
m = weight/g
Wm = m*a
			
#results
print "Density on earth  = %.2f Kg/m**3"%(m)
print " Weight on moon  =  %.2f N"%(Wm)
print " Density on moon remains unchanged and is equal to %.2f Kg/m**3"%(m)

Density on earth  = 998.98 Kg/m**3
 Weight on moon  =  1997.96 N
 Density on moon remains unchanged and is equal to 998.98 Kg/m**3

Example 1.2 Page No : 14

In [3]:
import math 
			
#Initialization of variables
w = 150. 			#N
theta = 30. 			#degrees
l = 0.8 			#m
b = 0.8 			#m
dy = 0.12 			#cm
v = 20. 			#cm/s
			
#calculations
Tau = round(w*math.sin(math.radians(theta)) /(l*b),2)  #shear stress
rd = v/dy                                              #rate of deformation
vis = Tau/rd                                           #viscosity

#results
print "Viscosity of the fluid  =  %.2f N s/m**2"%(vis)

# incorrect solution for 'rate of deformation' in textbook

Viscosity of the fluid  =  0.70 N s/m**2

Example 1.3 Page No : 14

In [11]:
import math 
			
#Initialization of variables
vis = 2.5/10 			#N s/m**2
D = 15. 			#cm
N = 180.
dy = 0.0001 			#m
l = 0.15 			#length - m
b = 0.25 			#breadth - m
r = 0.152 			#radius - m
			
#calculations
dv = math.pi*D*N/60/100
Tau = vis*dv/dy
Tor = round(Tau*math.pi*l*b*r/2,1)
P = Tor*2*math.pi*N/60
print 			
#results
print "Power required  =  %d W"%(P)

# Note :  The answer is different due to rounding off error in textbook.

Power required  =  595 W

Example 1.4 Page No : 15

In [16]:
import math 
			
#Initialization of variables
w = 1 			#rad/s
T = 0.4 			#N/m**2
			
#calculations
mu = T/math.tan(w)
			
#results
print "Viscosity  =  %.2f N s/m**2"%(mu)

Viscosity  =  0.26 N s/m**2

Example 1.6 Page No : 19

In [7]:
import math 
			
#Initialization of variables
d = 0.05*10**-3 	#diameter - m
T = 72.*10**-3 		#surface tension of water - N/m
P = 101. 			#pressure - kN/m**2
			
#calculations
Pi = P*1000 + 2*T/(d/2)
			
#results
print "Pressure  =  %.2f kN/m**2"%(Pi/1000)

Pressure  =  106.76 kN/m**2

Example 1.7 Page No : 19

In [17]:
import math 
			
#Initialization of variables
rho = 981.  			#dyn/cm**2
sigma = 72. 			#dyn/cm
theta = 0. 	    		#degrees
d = 0.5 		    	#cm
depth = 90. 			#cm
			
#calculations
h = 4*sigma*math.cos(math.radians(theta)) /(rho*d)
Td = depth-h
			
#results
print "True depth  =  %.3f cm"%(Td)

True depth  =  89.413 cm