# Ch-6, Unconventional Machining Process¶

## Problem 1 on page no. 332¶

In [1]:
from math import pi, sqrt
from __future__ import division
# Given that
a = 5 # Side of the square hole in mm
t = 4 # Thickness of tungsten plate in mm
d = 0.01 # Diameter of abraisive grains in mm
F = 3.5 # Force for feeding in N
A =25e-3 # Amplitude of tool oscillation in mm
f = 25e3 # Frequency in Hz
Hw = 6900 # Fracture hardness of WC in N/mm**2

Z = (1/2)*(4*a**2)/(pi*d**2)
lamda = 5
d1 = (d**2)
h_w = (sqrt((8*F*A)/(pi*Z*d1*Hw*(1+lamda))))
Q = (2/3)*((d1*h_w)**(3/2))*Z*f*pi
t = (a**2)*t/(Q*60)
print "The approximate time required = %0.2f min"%(t)
# Answer in the book is given as 13.66 min

The approximate time required = 14.26 min


## Problem 2 on page no. 334¶

In [2]:
# Given that
r = 1/3 # Ratio of hardness values of copper and steel
R_Q = (r)**(3/4)
R_t = 1/R_Q
P_R = (1-(1/R_t))*100
print "Percentage change in cutting time when tool is changed from coppper to steel = %d percent(reduction)"%(P_R)

Percentage change in cutting time when tool is changed from coppper to steel = 56 percent(reduction)


## Problem 3 on page no. 345¶

In [3]:
# Given that
m = 5 # Romoval rate in cm**3/min
A = 56 # Atomic gram weight in gm
Z = 2 # Valence at which dissolation takes place
D = 7.8 # Density of iron in gm/cm**3
I = (m/60)*(D*Z*96500)/(A)
print "Current required = %d amp"%(I)

Current required = 2240 amp


## Problem 4 on page no. 345¶

In [4]:
# Given that
I = 1000 # Current in amp
p1 = 72.5 # Percentage(by weight) of Ni in Nimonic 75 alloy
p2 = 19.5 # Percentage(by weight) of Cr in Nimonic 75 alloy
p3 = 5 # Percentage(by weight) of Fe in Nimonic 75 alloy
p4 = 0.4 # Percentage(by weight) of Ti in Nimonic 75 alloy
p5 = 1 # Percentage(by weight) of Si in Nimonic 75 alloy
p6 = 1 # Percentage(by weight) of Mn in Nimonic 75 alloy
p7 = 6 # Percentage(by weight) of Cu in Nimonic 75 alloy
# From the table 6.3 given in the book
D1 = 8.9 # Density of Ni in g/cm**3
D2 = 7.19 # Density of Cr in g/cm**3
D3 = 7.86 # Density of Fe in g/cm**3
D4 = 4.51 # Density of Ti in g/cm**3
D5 = 2.33 # Density of Si in g/cm**3
D6 = 7.43 # Density of Mn in g/cm**3
D7 = 8.96 # Density of Cu in g/cm**3
A1 = 58.71 # Gram atomic weight of Ni in gm
A2 = 51.99 # Gram atomic weight of Cr in gm
A3 = 55.85 # Gram atomic weight of Fe in gm
A4 = 47.9 # Gram atomic weight of Ti in gm
A5 = 28.09 # Gram atomic weight of Si in gm
A6 = 54.94 # Gram atomic weight of Mn in gm
A7 = 63.57 # Gram atomic weight of Cu in gm
Z1 = 2 # Valence of dessolation for Ni
Z2 = 2 # Valence of dessolation for Cr
Z3 = 2 # Valence of dessolation for Fe
Z4 = 3 # Valence of dessolation for Ti
Z5 = 4 # Valence of dessolation for Si
Z6 = 2 # Valence of dessolation for Mn
Z7 = 1 # Valence of dessolation for Cu
# Above values are given in table 6.3 in the book
D = 100/((p1/D1)+(p2/D2)+(p3/D3)+(p4/D4)+(p5/D5)+(p6/D6)+(p7/D7))
Q = ((0.1035*(10**-2))/D)*(1/((p1*Z1/A1)+(p2*Z2/A2)+(p3*Z3/A3)+(p4*Z4/A4)+(p5*Z5/A5)+(p6*Z6/A6)+(p7*Z7/A7)))
R = Q*I*60
print "Removal rate = %0.1f cm**3/min"%(R)

Removal rate = 2.2 cm**3/min


## Problem 5 on page no. 352¶

In [5]:
# Given that
V = 10 # DC supply voltage in Volt
k = 0.2 # Conductivity of electrolyte in ohm**-1-cm**-1
f = 0.1 # Feed rate in m/min
Vo = 1.5 # Total overvoltage in Volt
F = 96500 # Faraday constant in coulombs per mole

A = 55.85 # Atomic gram weight of iron in gm
Z = 2 # Valency of dissolation of iron
rho = 7.86 # Density of iron in gm/cm**3
Yc = k*A*(V-Vo)/(rho*Z*F*(f/60))
print "Equilibrium gap = %0.2f cm"%(Yc)

Equilibrium gap = 0.04 cm


## Problem 6 on page no. 353¶

In [6]:
# Given that
S_I1 = 5 # Surface irregulation in micro meter
S_I2 = 8 # Surface irregulation in micro meter
V = 12 # DC supply voltage in Volt
k = 0.2 # Conductivity of electrolyte in ohm**-1-cm**-1
Vo = 1.5 # Total overvoltage in Volt
F = 96500 # Faraday constant in coulombs per mole
Y_min = (S_I1+S_I2)*(10**(-4))
A = 55.85 # Atomic gram weight of iron in gm
Z = 2 # Valency of dissolation of iron
D = 7.86 # Density of iron in gm/cm**3
f_max = (k*A*(V-Vo)/(Z*D*F*Y_min))*60
print "Largest passible feed rate = %0.1f mm/min"%(f_max*10)

Largest passible feed rate = 35.7 mm/min


## Problem 7 on page no. 355¶

In [7]:
# Given that
f  = 0.2 # Feed rate in cm/min
l = 2.54 # Length of tool face in cm
w = 2.54 # Width of tool face in cm
T_b = 95 # Boiling temperature of electrolyte in °C
Nita = 0.876e-3 # Viscosity of electrolyte in kg/m-sec
D_e = 1.088 # Density of electrolyte in g/cm**3
c = .997 # Specific heat of electrolyte
V = 10 # DC supply voltage in Volt
k = 0.2 # Conductivity of electrolyte in ohm**-1-cm**-1
T = 35 # Ambient temperature in °C
Vo = 1.5 # Total overvoltage in Volt
F = 96500 # Faraday constant in coulombs per mole
A = 55.85 # Atomic gram weight of iron in gm
Z = 2 # Valency of dissolation of iron
D = 7.86 # Density of iron in gm/cm**3
Ye = k*A*(V-Vo)*60/(D*Z*F*f)
J = k*(V-Vo)/(Ye)
D_T = T_b -T
v = (J**2)*(l)/(k*D_T*D_e*c)
Re = ((D_e*v*2*Ye)/Nita)*(0.1)
p = 0.3164*D_e*(v**2)*l/(4*Ye*(Re**0.25))*(10**-4)
A = l*w
F = p*A*(10**-1)*(1/2)
print "Total force acting on the tool = %d N"%(F)
# Answer in the book is given as 79 N

Total force acting on the tool = 103 N


## Problem 9 on page no. 378¶

In [8]:
from math import log
# Given that
a = 10 # Side length of a square hole in mm
t = 5 # Thickness of low carbon steel plate in mm
R = 50 # Resistance in relaxation circuit in ohm
C = 10 # Capacitance in relaxation circuit in micro F
V = 200 # Supply voltage in Volt
V_ = 150 # Minimum required voltage for discharge in Volt

E = (1/2)*C*(10**-6)*(V_**2)
tc = R*C*(10**-6)*log(V/(V-V_))
W = (E/tc)*(10**-3)
v = t*a**2
Q = 27.4*(W**(1.54))
T = v/Q
print "The time required to complete the drilling operation = %d min"%(T)
# Answer in the book is given as 306 min

The time required to complete the drilling operation = 300 min


## Problem 10 on page no. 382¶

In [9]:
# Given that
R = 50 # Resistance in relaxation circuit in ohm
C = 10 # Capacitance in relaxation circuit in micro F
V = 200 # Supply voltage in Volt
V_ = 150 # Minimum required voltage for discharge in Volt
E = (1/2)*C*(10**-6)*(V_**2)
tc = R*C*(10**-6)*log(V/(V-V_))
W = (E/tc)*(10**-3)
Q = 27.4*(W**(1.54))
Hrms = 1.11*(Q**0.384)
print "Surface roughness = %0.3f micro meter"%(Hrms)
# Answer in the book is given as 5.16 micro meter which is wrong

Surface roughness = 1.350 micro meter


## Problem 11 on page no. 391¶

In [10]:
# Given that
w = 150 # Width of slot in micro meter
t = 1 # Thickness of tungsten sheet in mm
P = 5 # Power of electron beam in KW
C = 12 # Specific power consumption for tugsten in W/(mm**3/min) from the table 6.7 given in the book
v = (P*(1000)/C)*(1000/(w*t))*(1/600)
print "Speed of cutting = %0.1f cm/sec"%(v)

Speed of cutting = 4.6 cm/sec


## Problem 12 on page no. 392¶

In [11]:
from math import ceil
# Given that
V = 150e3 # Acceleration voltage in V
D = 76e-7 # Density of steel in kg/mm**3
Delta = 2.6*(10**-17)*((V**2)/D)
print "Electron range = %d micro meter"%(ceil(Delta*(10**3)))

Electron range = 77 micro meter


## Problem 13 on page no. 395¶

In [12]:
# Given that
w = 0.015 # Width of slot in cm
t = 1 # Thickness of tungsten sheet in mm
P = 5e3 # Power of electron beam in W
rho_c = 2.71 # Value of volume specific heat for tugsten in J/cm**3
k = 2.15 # Thermal conductivity of tungsten in W/cm-°C
T_m = 3400 # Melting temperture in °C
Z = t/10 # In cm
v = (0.1**2)*(P**2)/((T_m**2)*(Z**2)*(k*w*rho_c))
print "Speed of cutting = %0.1f cm/sec"%(v)

Speed of cutting = 24.7 cm/sec


## Problem 14 on page no. 399¶

In [13]:
# Given that
I = 1e5 # Power intensity of laser beam in W/mm**2
T_m = 3400 # Melting temperture of tungsten in °C
rho_c = 2.71 # Value of volume specific heat for tugsten in J/cm**3
k = 2.15 # Thermal conductivity of tungsten in W/cm-°C
p_a = 10 # Percentage of beam absorbed
alpha = k/rho_c
H = (p_a/100)*(I)*(100)
tm = (pi/alpha)*((T_m*k)/(2*H))**(2)
print "Time required for the surface to reach the melting point = %f sec"%(tm)

Time required for the surface to reach the melting point = 0.000053 sec


## Problem 15 on page no. 400¶

In [14]:
# Given that
I = 1e5 # Power intensity of laser beam in W/mm**2
d = 200 # Focused diameter of incident beam in micro meter
T_m = 3400 # Melting temperture of tungsten in °C
rho_c = 2.71 # Value of volume specific heat for tugsten in J/cm**3
k = 2.15 # Thermal conductivity of tungsten in W/cm-°C
p_a = 10 # Percentage of beam absorbed
H = (p_a/100)*(I)*(100)
alpha = k/rho_c
zeta = 0.5 # Fr0m the standard table
# By solving the equation T_m = ((2*H)*(sqrt(alpha*tm))/k)*((1/sqrt(pi))-ierfc(d/(4*sqrt(alpha*tm))))
tm = 1/((200**2)*(zeta**2)*(alpha))
print "Time required for the centre of the circular spot to reach the melting point = %f sec"%(tm)
# Answer in the book is given as 0.00013 sec

Time required for the centre of the circular spot to reach the melting point = 0.000126 sec


## Problem 16 on page no. 401¶

In [15]:
# Given that
d = 200 # Diameter of focussed laser beam in micro meter
T_m = 3400 # Melting temperture of tungsten in °C
k = 2.15 # Thermal conductivity of tungsten in W/cm-°C
p_a = 10 # Percentage of beam absorbed
H = 2*k*T_m/(d*10**-4)
I = H/(p_a/100)
print "Minimum value of beam power intensity to achieve the melting = %0.2e W/cm**2"%(I)

Minimum value of beam power intensity to achieve the melting = 7.31e+06 W/cm**2


## Problem 17 on page no. 403¶

In [16]:
# Given that
I = 1e5 # Power intensity of laser beam in W/mm**2
t = 0.5 # Thickness of tungsten sheet in mm
d = 200 # Drill diameter in micro meter
P = 3e4 # Energy required per unit volume to vapourize tungsten in J/cm**3
p_e = 10 # Percentage efficiency
T_m = 3400 # Melting temperture of tungsten in °C
k = 2.15 # Thermal conductivity of tungsten in W/cm-°C
H = (p_e/100)*(I)*(100)
v = H/P
T = t*(0.1)/(v)
print "The time required to drill a through hole = %0.4f sec"%(T)

The time required to drill a through hole = 0.0015 sec