Chapter 5: Surface Tension

Exa 5.1

In [1]:
#If the surface tension of a soap bubble is 0.4 N/m, what work is required to
#increase the diameter from 5 to 15 cm?
import math
#initialisation of variables
St= 0.04 															#N/m
d1= 5. 																#cm
d2= 15. 															#cm
#CALCULATIONS
W= St*1000.*2*4*math.pi*(math.pow((d2/2),2)-math.pow((d1/2),2))		#Work
#RESULTS
print '%s %.2e' % ('Work (dyn cm or erg) = ',W)
raw_input('press enter key to exit')
Work (dyn cm or erg) =  5.03e+04
press enter key to exit
Out[1]:
''

Exa 5.2

In [2]:
#By much does the pressure inside a 0.017 in droplet of water at 20C
#exceed the outside pressure.
import math
#initialisation of variables
R= 0.017 												#in
sigma= 72.8 											#m N/m
#CALCULATIONS
P= (2*sigma*0.005*0.017)/(72.8*R*7.08*math.pow(10,-4))	#Presuure difference
#RESULTS
print '%s %.2f' % ('Presuure difference (lbf/ft^2) = ',P)
raw_input('press enter key to exit')
Presuure difference (lbf/ft^2) =  14.12
press enter key to exit
Out[2]:
''

Exa 5.3

In [3]:
#A 0.4 mm diameter glass tube stands vertically in a dish of mercury at 20C.
#Determine the difference between the level of mercury in the dish and in
#the tube. The specific gravity for mercury is 13.6 and the contact angle 
#with glass is 130C
import math
#initialisation of variables
d= 13.6 							#gm/cm^3
g= 980 								#cm/s^2
D= 0.4 								#mm
angle= 130*math.pi/180. 			#radians
s= 514. 							#dyn/cm
#CALCULATIONS
h= (4*s*10*math.cos(angle))/(d*g*D)	#Depression in mercury level
#RESULTS
print '%s %.2f' % (' Difference in mercury level (cm (depression)) = ',h)
raw_input('press enter key to exit')
 Difference in mercury level (cm (depression)) =  -2.48
press enter key to exit
Out[3]:
''