Chapter 10 : Phase Diagram and Equilibrium Diagram

Example 10.1 pageno : 251

In [1]:
# Variables
c = 2.;     			#for 2 component system
p = 4.;

# Calculations
d = c-p+2;			#degree of freedom
t = p*(c-1)+2;			#no. of total variables

# Results
print "D = C-P+2"
print "Total no. of variables  =  P*(C-1)+2"
print "Degree of freedom  =  ",d
print "when p  =  ",p
print "Two component system cannot have more than 4 phases in an equilibrium"

D = C-P+2
Total no. of variables  =  P*(C-1)+2
Degree of freedom  =   0.0
when p  =   4.0
Two component system cannot have more than 4 phases in an equilibrium

Example 10.2 page no : 257

In [3]:
# Variables
p_pb = 11364.1; 	    		#density of lead in kg/m**3
p_sn = 7220.14;	    	    	#density of tin in kg/m**3

# Calculations
p_e = 100./((38./p_pb)+(62./p_sn));			#density of eutectic composition at point D
w = .88*p_e;	    		#in kgf
w_pb = .38*w;		    	#of lead in kgf
w_sn = .62*w;			    #of tin in kgf
p_b = 7300.;        		#density in beta phase in kg/m**3
w1 = .12*p_b;		    	#in kgf
w1_pb = .03*w1;			    #of lead in kgf
w1_sn = .97*w1;			    #of tin in kgf
w2_pb = w_pb+w1_pb;			#Total weight of lead in kgf
w2_sn = w_sn+w1_sn;			#Total weight of tin in kgf
sn = (w2_sn/(w2_sn+w2_pb))*100;

# Results
print "Density of eutectic composition at point D (in kg/m3)  =  %.1f kg/m**3"%p_e
print "Total weight of lead (in kgf)  = %.2f kgf"%w2_pb
print "Total weight of tin (in kgf)  =  %.1f kgf"%w2_sn
print "%% of Sn  =  %.2f"%sn

Density of eutectic composition at point D (in kg/m3)  =  8381.6 kg/m**3
Total weight of lead (in kgf)  = 2829.07 kgf
Total weight of tin (in kgf)  =  5422.7 kgf
% of Sn  =  65.72

Example 10.4 pageno : 266

In [4]:
# Variables
c_be = 100;
c_e = 1.65;
c_o = 10;

# Calculations
w = (c_be-c_o)/(c_be-c_e);

# Results
print "weight fractions  =  %.3f"%w

weight fractions  =  0.915

Example 10.5 pageno : 267

In [6]:
# Variables
#m be amount of tin
w_sn = 900.;	    		#weight of tin
w_pb = 1000.;		    	#weight of lead

# Calculations
m = ((w_pb*0.97)-w_sn)/(1-.97);			#in grams
m1 = m/1000.;			                #maximum mass of tin in kgm

# Results
print "Maximum weight of tin that can be added without changing systems temperature (in kgm)  =  %.2f kg"%m1

Maximum weight of tin that can be added without changing systems temperature (in kgm)  =  2.33 kg

Example 10.6 page no : 270

In [7]:
# Variables
c = 0.83;			#carbon
f = 0.; 			#ferrite
ce = 6.67;			#cementite

# Calculations
w_a = (ce-c)/(ce-f);
w_b = (c-f)/(ce-f);

# Results
print "Weight fraction of errite  =  %.3f"%w_a
print "Weight Fraction of Cementite  =  %.3f"%w_b

Weight fraction of errite  =  0.876
Weight Fraction of Cementite  =  0.124