Production Engineering Test 7

Concept:

\(BLU = \frac{{Lead\;screw\;pitch}}{{steps\;per\;revolution}}\)

Calculation:

Given:

Stepper motor = 300 steps per revolution, Lead screw pitch = 5 mm

Now,

\(BLU = \frac{5}{{300}}\)

∴ BLU = 0.0167 

Concept:

The power density is given by \(J = η \frac{{VI}}{A}\)

Calculation:

Given:

Volatge (V) = 45 kV, Current (I) = 60 mA, diameter (d) = 0.25 mm, heat transfer factor (η) = 0.87

\(J = 0.87 \times \frac{{45 \times 1000 \times 60 \times {{10}^{ - 3}} \times 4}}{{\pi \times {{0.25}^2}}} = 47853.435\;W/m{m^2} \)

∴ J = 47.853 kW/mm²

Explanation:

Important G codes

• G 00 – Rapid Transverse
• G 01 – Linear Interpolation
• G 02 – CW Circular Interpolation
• G 03 – CCW Circular Interpolation
• G 04 – Dwell
• G 97 – Spindle Speed

Important M codes

• M 00 – Program Stop
• M 01 – Optional Stop
• M 02 - End of Program
• M 03 – Spindle Start (CW)
• M 04 – Spindle Start (CCW)
• M 05 – Spindle Stop
• M 06 – Tool Change 
• M 08 – Coolant on
• M 09 – Coolant off
• M 10 – Clamp-on
• M 11 – Clamp off

Concept:

\(Metal\;removal\;rate,\;Q = \frac{{AI}}{{\rho ZF}} = c{m^3}/sec\)

A = Gram atomic weight of the metallic iron, I = Current (amp), ρ = Density of anode (g/cm³)

Z = Valence of the cation, F = Faraday = 96500 coulombs

Calculation:

Given:

\(Q = 1\frac{{c{m^3}}}{{min}} = \frac{1}{{60}}c{m^3}/sec\),

A = 56 gm, Z = 2, ρ = 7.8 gm/cm³, F = 1609 amp-min = 1609 × 60 amp-s (coulombs)

Now,

\(Q = \frac{{AI}}{{\rho ZF}} \)

\(∴ \frac{1}{{60}} = \frac{{56 \times I}}{{7.8 \times 2 \times 1609 \times 60}}\)

Concept:

800 pulses = 1 revolution of motor = 1 revolution of lead screw = 2 × pitch of lead screw

Calculation:

800 pulses = 2 × pitch of lead screw

∴ 800 pulses = 2 × 4 mm

∴ 800 pulses = 8 mm

Now,

Position accuracy of motor = distance travelled in 1 pulses

∴ positional accuracy \(= \frac{8}{{800}}\)

Concept:

Energy released per spark

\(E = \frac{1}{2}CV_d^2\)

For maximum Power ⇒   V_d = 0.72 V_s

\({\rm{Cycle\;time\;}}\left( {{{\rm{t}}_{\rm{C}}}} \right){\rm{\;}} = RC\ln \left( {\frac{{Vs}}{{Vs - Vd}}} \right)\)

\({\rm{Average\;power\;input\;}}\left( W \right) = \frac{E}{{{t_C}}}\)

Calculation:

V_d = 0.72 V_s

V_d = 0.72 × 250

∴ V_d = 180 V

Now,

\(E = \frac{1}{2}CV_d^2\)

\(E = \frac{1}{2} \times 10 \times {10^{ - 6}} \times {180^2} = 0.162\;J\)

Now,

\({t_C} = 45 \times 10 \times {10^{ - 6}}\ln \left( {\frac{{250}}{{250 - 180}}} \right)\)

∴ t_c = 5.728 × 10^-4 sec

Now,

\(W = \frac{{0.162}}{{5.728\; \times \;{{10}^{ - 4}}}} = 282.80\;W\)

MRR = θ = 27 (0.2828)^1.70

∴ MRR = 3.15 mm³/min

Now,

Material to be removed = 12 × 8 × 6 mm³

\({\rm{Time\;required\;}} = \frac{{12\; \times \;8\; \times \;6}}{{3.15}}min\)

∴ Time required = 3.043 hour

Concept:

Material removal rate (MRR) \(= \frac{{AI}}{{\rho ZF}}\)

A = atomic weight, I = current, ρ = density, F = faraday constant, Z = valency

\(I = \frac{{{\rm{\Delta }}V}}{R} = \frac{{Voltage}}{{Resistance}}\)

R = ρ_sL/ A

Electrode feed rate = s × S

Electrode feed rate = specific MRR × current density

\(s = \frac{e}{{F\rho }},\;s = \frac{{{\rm{\Delta }}V}}{{{\rho _s}L}},\;e = \frac{A}{Z}\)

ρ_s = specific resistance

Calculation:

A = 56, Z = 3, ρ = 7700 kg/m³ = 7.70 gm/cc, ρ_s = 4 Ω cm, ΔV = 14 V, A_r = 3 × 3 cm², L = 0.3 mm = 0.03 cm

e = A/Z = 56/3

\(I = \frac{{{\rm{\Delta }}V}}{R} = \frac{{14}}{{4 \times \frac{{0.03}}{{3\; \times \;3}}}} = 1050\;A\)

\({\rm{MRR\;}} = \frac{{56\; \times \;1050}}{{96500\; \times \;3\; \times\; 7.70}}\)

∴ MRR = 0.02638 cm³/s

Now,

s = e/FP

\(s = \frac{{\frac{{56}}{3}}}{{96500\; \times \;7.70}}\)

Now,

\(S = \frac{{\Delta V}}{{{P_s}L}}\)

\(s = \frac{{14}}{{4\; \times \;0.03}}\)

Now,

\({\rm{Feed\;rate\;}} = \left( {\frac{{\frac{{56}}{3}}}{{96500\; \times \;7.70}}} \right) \times \frac{{14}}{{4 \times 0.03}}\)

∴ Feed rate = 0.00293 cm/s