Chapter 4: viscosity

Exa 4.1

In [2]:
#At 68F the experimental value of absolute viscosity for water is 1 cp.
#What is the equivalent value in (a)SI units (b) BTU and (c)reyns?
import math
#initialisation of variables
V= 1. 								#cp
#CALCULATIONS
#An extra 100 is multiplied to convert poise to lbf s 
SI= V/100./10. 						#Pa.s
BE= (V*32.2*100/100)/(4.788*100.) 	#lbf s/ft^2
RE= V*100/100./(4.788*100*144.) 	#reyn
#RESULTS
print '%s %.2e' % ('SI units (Pa s) = ',SI)
print '%s %.2e' % (' \n BE units (lbm/ft s) = ',BE)
print '%s %.2e' % (' \n Reyns units (reyn) = ',RE)
raw_input('press enter key to exit')

SI units (Pa s) =  1.00e-03
 
 BE units (lbm/ft s) =  6.73e-02
 
 Reyns units (reyn) =  1.45e-05
press enter key to exit
Out[2]:
''

Exa 4.2

In [4]:
#Determine the kinematic viscosity of water at 68F in (a)SI units (b) BTU
import math
#initialisation of variables
T= 68 							#F
d= 1.0					 		#gm/cm^3
mu= math.pow(10,-2) 			#gm/cm s
SIm= math.pow(10,-4) 			#m^2/s
m= 10.76 						#ft
#CALCULATIONS
SI= mu*SIm 						#kinematic viscosity in SI
BU= SI*m 						#kinematic viscosity in BTU
#RESULTS
print '%s %.2e' % ('SI Units (m^2/s) = ',SI)
print '%s %.2e' % (' \n British Units (ft/s) = ',BU)
raw_input('press enter key to exit')

SI Units (m^2/s) =  1.00e-06
 
 British Units (ft/s) =  1.08e-05
press enter key to exit
Out[4]:
''

Exa 4.3

In [5]:
#Determine the kinematic viscosity of 40 SUS oil.
import math
#initialisation of variables
Ku= 40. 						#SUS
#CALCULATIONS
SU= 0.0022*Ku-(1.8/Ku) 			#stoke units
#RESULTS
print '%s %.4f' % ('Stoke Units (stoke) = ',SU)
raw_input('press enter key to exit')

Stoke Units (stoke) =  0.0430
press enter key to exit
Out[5]:
''

Exa 4.4

In [6]:
#Air at 70F flows through a duct at 50 fps. Calculate the Reynolds number
#if the duct is (a) 10 in in diameter (b) 10 in square
import math
#initialisation of variables
v= 50 									#fps
mu= 1.6*math.pow(10,-4) 				#ft^2/s
d1= 10. 								#in
d2= 10. 								#in square
#CALACULATIONS
D= (math.pi*4*d1*d1/4)/(math.pi*d2*12) 	#Modified diameter
Re= (v*D)/mu 							#Reynolds number
D1= (d1*d1/(4*d2*3)) 					#Modified diameter
Re1= (v*D1)/mu 							#Reynolds number
#RESULTS
print '%s %.3e' % ('Re= ',Re)
print '%s %.3e' % (' \n Re=  ',Re1)
raw_input('press enter key to exit')

Re=  2.604e+05
 
 Re=   2.604e+05
press enter key to exit
Out[6]:
''

Exa 4.5

In [7]:
#A fluid with a viscosity of 1.75x 10^-3 pa.s. flows through a tube of 0.5 mm
#diameter. if the pressure drop across a 1m length of the tube is 1 MPa, and 
#the flow is fully developed and laminar, what is the flow rate?
import math
#initialisation of variables
v= 1.75*math.pow(10,-3) 									#pa s
l= 1 														#m
P= 1 														#Mpa
d= 0.5 														#mm
#CALCULATIONS
Q= (math.pi*P*1000000.*math.pow(((d/2)/1000.),4))/(l*8*v)	#Flow rate
#RESULTS 
print '%s %.2e' % ('Q (m^3/s) = ',Q)
raw_input('press enter key to exit')

Q (m^3/s) =  8.77e-07
press enter key to exit
Out[7]:
''