Chapter 20 : Performance of Materials in Service ( Fracture, Fatigue, and Corrosion and its Control)

Example 20.1 page no : 537

In [1]:
import math 

# Variables
l = 1.5*10**-6;			#crack length in m
e = 70*10**9;			#Young's modulous in N/m**2
y_e = 1.05;			#specific surface energy in j/m**2

# Calculations
a_f = math.sqrt((2*y_e*e)/(3.14*l));
a_f1 = a_f*10**-6;			#in MPa
r = a_f/e;			#ratio

# Results
print "Fracture strength (in MPa)  =  %.2f"%a_f1
print "Ratio of fracture strength to Youngs modulous  =  %.2e"%r

Fracture strength (in MPa)  =  176.66
Ratio of fracture strength to Youngs modulous  =  2.52e-03

Example 20.2 pageno : 547

In [2]:
import math 

#Variables
m_m = 58.71;			#molecular weight of ni
m_c = 74.71;			#molecular weight of nio
p_m = 8900;			#density of ni in kg/m**3
p_c = 7080;			#desity of nio in kg/m**3

#Calculations & Reults
x = m_m/p_m;			#molar volume of ni in m**3/mol
print "Mc/Pc (in m**3/mol) %.2e m**3/mol"%x
y = m_c/p_c;			#molar volume of nio in m**3/mol
print "Mm/Pm (in m**3/mol) %.2e m**3/mol"%y
print "           Mc/Pc > Mm/Pm Hence protective layer of NiO will form over Ni        ";

Mc/Pc (in m**3/mol) 6.60e-03 m**3/mol
Mm/Pm (in m**3/mol) 1.06e-02 m**3/mol
           Mc/Pc > Mm/Pm Hence protective layer of NiO will form over Ni

Example 20.3 pageno : 548

In [3]:
import math 

# Variables
x1 = 0.1;			#in mm
t1 = 25.;			#in hours
t2 = 300.;			#in hours

# Calculations
x2 = x1*math.sqrt(t2/t1);			#in mm

# Results
print "Oxidation loss in 300 hours (in mm)  =  %.3f mm"%x2

Oxidation loss in 300 hours (in mm)  =  0.346 mm

Example 20.4 pageno : 551

In [4]:
# Variables
p_mg = 1.74;			#density of magnesium in gm/cm**3
p_mgo = 3.65;			#density of magnesium oxide in gm/cm**3
m_mg = 24.;			#mol wt ogf mg
m_mgo = 40.3;			#mol wt of mgo

# Calculations
PBR = (m_mgo/p_mgo)/(m_mg/p_mg);

# Results
print "PBR  =  %.2f"%PBR
print "Since PBR < 1. So porous film will form which will be non protective";

PBR  =  0.80
Since PBR < 1. So porous film will form which will be non protective

Example 20.5 pageno : 562

In [5]:
# Variables
m = 0.0243;			#one mole of magnesium in kg

# Calculations
c = 2.*96490;			#in C
j = 20.*10**-3;			#in A/m**2
t = 15.*365*24*3600;			#in sec
x = j*t;        			#in A s
w_mg = m*x/c;	    		#in kg/sqm

# Results
print "Amount of Magnesium needed (in Kg/m**2)  =  %.2f kg/m**2"%w_mg

Amount of Magnesium needed (in Kg/m**2)  =  1.19 kg/m**2